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n (a) Distinguish and longitudinal vibrations. (b) Define: complete vibration, ampli- transverse tude, frequency, period. (a) What is necessary for the transmission of sound? (b) Compard^the transmission of sound through the three states of matter and suggest a theoretical explanation for any differences. 4. (a) Distinguish between transverse and longitudinal waves. (b) Define: amplitude, wave-length, period, frequency. (c) By what fype of wave-motion is sound transmitted? What are the components of each complete sound wave? 5. 6 . (d) What is the fundamental characteristic of wave-motion? Explain fully. (a) Establish the wave formula. (b) Calculate the velocity of sound in air if its frequency is 250 v.p.s. and its wave-length is 4.4 ft. (a) What is meant by the terms "in phase” and "out phase” as applied to any wave-motion? (b) How are standing waves produced? of (c) Define: node, loop. (d) Explain why the distance between two successive nodes is one-half a wave-length. 7. 8. (a) State the laws of reflection of sound waves. (b) Distinguish echoes and reverbera- tions. (a) In what three ways may the velocity of sound be determined? (b) What is the effect of a change of 61 Chap. 6 SOUND 7 . At what temperature will the velocity of sound be (a) 1,1 19 ft. per sec. (b) 336 metres per sec.? 8. A thunder-clap is heard 5 seconds after the lightning-flash was seen. How far away was the flash if the temperature of the air were 1 5°C.? 9 . When the temperature of the air is 15°C., calculate the wave-length in metric units of the sound from a tuning-fork having a frequency of 256 v.p.s. TO. When the temperature of the air is 25’/2°C., and the wave-lengfh of a sound is 4.40 ft., calculate the frequency of the sound. TT. A signal of 128 v.p.s. has a wavelength of 279. cm. (a) Find the velocity of sound in air. (b) What would be the temperature of the air to the nearest degree centigrade? Express your answer in metric units. T2. The human ear is incapable of disindividual sounds unless they tinguishing are separated by a time interval of at least ]/(q sec. (a) Calculate the length of the shortest auditorium that would give a distinct echo, (b) What would be the effect if the auditorium were shorter? (Assume that the temperature of the air is 20°C.] T3. A 220 yd. dash over a straight course was timed at 23.2 sec. What would the time have been had the timer started the watch on hearing the sound instead of seeing the flash? (Temperature of air -20°C.) T4. Calculate the minimum speed of an aircraft in miles per hour which has broken sound the air = 5.5°C.) barrier. (Temperature of temperature on the velocity of sound In air? 9 . The sound of a gun was heard 1 0 sec. after the flash was seen. If the distance to the gun was 1 1,500 ft., calculate the probable velocity of sound in air. Why is this merely a probable velocity? TO. Two and one-half seconds elapse between shouting across a river 1,375 ft. wide and hearing the opposite bank, (a) Find the velocity of sound, (b) Compare the accuracy of this velocity with that of question 5. from echo the B 1. The horizontal distance between the end points in the swing of a pendulum Is 7.5 cm. What is (a) the amplitude, (b) the distance covered by the bob in one com- plete vibration? 2 . The pendulum in question 1 makes 45 complete vibrations in 30 seconds. Calculate the period of vibration. What would amplitude were be the doubled? period the If 3 . Calculate the velocity of sound in air if the frequency of sound and its wave1 80 v.p.s. and 6.50 ft., length are (a) (b) 360 v.p.s. and 80.0 cm. respectively. if the 4 . Calculate the frequency of a sound and are (a) 1,120 ft. per sec. and 3.50 ft., (b) 340 metres per sec. and 51.0 cm. wave-length velocity 5 . What are the wave-lengths of the notes when (a) the frequency is 900 v.p.s. and velocity of sound 1 ,350 ft. per sec. and (b) the frequency is 625 v.p.s. and velocity of sound is 350 metres per sec.? 6 . Calculate the velocity of sound in air in (a) feet per second and (b) metres per second at: 5°C., - 17°C., 23°C. 62 CHAPTER 7 CHARACTERISTICS OF MUSICAL SOUNDS 11:10 INTRODUCTION (a) How a Musical Sound Differs from a Noise As we have seen, people are quick to classify the sounds they hear, as either musical sounds or noises. Almost everyone finds noises, like the slamming of a door or the rumble of machinery unpleasant, and musical sounds like those or pleasant. of We can establish a more objective difference than this between noise and music. When an oscilloscope is used to compare the sound of machinery and that of a tuning-fork, a trace similar tuning-fork violin a Oscilloscope Tracing of Fig. 7:1 A— Noise B—Musical Sound It is obvious to that in Fig. 7 : 1 results. that the musical sound is caused by rapid regular (periodic) vibrations, while the noise is the result of irregular (non- periodic) vibrations. (b) How Musical Sounds Differ from Each Other If one tuning-fork is struck lightly, and another more vigorously, the second will Sounds that differ emit a louder sound. in loudness differ in intensity. If the sounds from two vibrating tuning-forks with differing frequencies are compared, the fork with the greater frequency emits the “higher” sound. Sounds that differ Further, in “highness” differ in pitch. if the sounds of a vibrating tuning-fork and a vibrating string of the same frequency are compared, there is no difficulty in identifying the origin of each Sounds from different sources sound. may be distinguished because they differ in quality. Thus musical sounds differ from each other in intensity, pitch and quality. It is the purpose of this chapter to study these three characteristics fur- ther. 11:11 THE INTENSITY OF SOUND If a bell that is rung cannot be heard at a distance, we know that we may be able to make it heard by striking it harder. This is so because striking with greater force transfers more energy to the vibrating body, increasing what we call the sounds. This increases the amplitude (energy of vibration) and gives If the sound is it a greater intensity. still inaudible, the other obvious thing that can be done to make it heard is 63 Chap. 7 SOUND the shorten distance between the to It might seem source and the receiver. from this that there are just two factors affecting the intensity of sound, namely amplitude and distance, but the experience of undersea workers in caissons and diving-bells where the air is under great pressure, is that quite ordinary sounds are unexpectedly loud there. This greater intensity of the sound transmitted results from the fact that increased pressure on a gas crowds the molecules closer and increases the density of the medium. For the same reason, as is well known, sounds transmitted by solids and liquids are louder than when transmitted by air. Thus, a third factor affecting intensity of sound is density of medium. fles, while neighbouring buildings can be protected by double windows. The importance of such efforts to reduce Threshold of Painful Noise Airplane Engine Riveting Machine Heavy Traffic Motor Truck Ordinary Conversation Vacuum Cleaner Average Office Quiet Home, Quiet Conversation Rustle of Leaves Quiet Whisper —;130 —h20 -^110 — 100 -^90 — 80.^ -70\|.( —;60 Q — 50 — 40 -30 — 20 — 10 “0 The particular intensity of sound are: laws governing the Threshold of Hearing 1. The intensity of sound varies directly as the square of the energy of vibration (amplitude) of the source. 2. The intensity of sound varies in- versely as the square of the distance of the receiver from the source. 3. The intensity of sound increases with an increase in the density of the transmitting medium. in vary intensity. (deci — 1/10), The intensity of sound is measured in bels and decibels the decibel being the faintest sound that can be perceived by a normal human ear. Fig. 7:2 shows the amount of noise or “noise level” in decibels in a few common locations, for, like musical sounds, Naturally, noises noises are inevitable wherever there is machinery, but unnecessary discomfort can be avoided by the noise level being measured and steps being taken to eliminate all unnecessary vibration. Noise levels around machinery can be reduced by the use of rubber mountings, mufflers and the like; walls and ceilings can be covered with sound-absorbent wallboard, and air-ducts with sound-absorbent baf- Fig. 7:2 Noise Levels. noise levels is indicated by the fact that temporary or permanent deafness and many other illnesses can result from long proximity to noisy machinery. 11:12 THE PITCH OF SOUND That musical sounds vary in highness or pitch was stated in a previous section. In chap. 10, exp. 6, Savart’s toothed Fig. 7:3 Savart's Toothed Wheel. wheel ( Fig. 7:3) was used to show what determines pitch. The card held against the teeth of the rapidly rotating wheel received a sequence of taps, and a note, whose pitch increased with the speed of 64 CHARACTERISTICS OF MUSICAL SOUNDS Sec. 11:12 rotation, was heard. If the rate of rotation was doubled, the number of taps per second (the frequency) was doubled. The second note, whose frequency is twice as great, is said to be an octave higher than the first. Thus for a note of high pitch to be sounded, the body must be vibrating rapidly, whereas a slow rate of vibration produces a note of low pitch. In short, the pitch of sound depends on the frequency. The frequency of the sound produced may be determined by applying the formula: frequency of note produced ( v.p.s.) = number of teeth on Savart’s wheel X speed of rotation revolutions per second. in The limits of frequency for notes of are from about 20 to pitch audible 20,000 v.p.s. for the average ear (Fig. 7:4). The range of frequencies used in music is from about 30 to 5,000 v.p.s., the keyboard of a piano extending from 27 to 3,500 v.p.s. A man’s speaking voice embraces frequencies ranging between 100 and 150 v.p.s. approximately. Extreme limits of audibility (for very sensitive ears) may be taken as extendi
ng from 20 to 35,000 v.p.s. The lowest notes of a large organ are in the neighbourhood of this lower audible limit, while the squeak of a bat or the noise of a cricket are examples of frequencies in the region of the upper audible limit. Frequencies above this point, referred to as ultrasonic frequencies, are becoming of increasing importance (Sec. 11:33). sounds heard by the human ear wind instruments string instruments frequency limits of human hearing ^ Fig. 7:4 Frequency Limits of the Human Ear. Bell Telephone Company of Canada. 65 Chap. 7 SOUND 11:13 THE DOPPLER EFFECT When a car with its horn sounding approaches a pedestrian at high speed, the pitch of the sound appears to be higher than the true pitch which the driver hears (Fig. 7:5). After the car has passed, the pitch appears to be lower than the true pitch. Similar changes in pitch occur when the origin is stationary and the observer moves past it. To determine the cause, let us take the case of the origin approaching the In any second, a uniform observer. number of wave-fronts are sent out and they will be a uniform distance apart. However, because the origin is approaching the observer, there will be more than the usual number packed in a given space, i.e,, they will be closer together than before. As a result, more than the normal number will be received by the observer in one second. This will cause an apparent rise in the pitch of the sound. This phenomenon is called the doppler effect. Fig. 7:5 The Doppler Effect. 11:14 THE SONOMETER In stringed instruments, the stretched strings of steel, gut, or silk, are set in a state of transverse vibration by being struck, bowed, or picked as in the piano, violin and guitar respectively. Examination of a piano and violin will reveal that sounds of different pitches are obtained by the use of strings of different lengths, tensions, diameters and densities. To ensure that the sound is loud enough to be distinctly heard, the string is attached to a sounding-box or board. The natural frequency of this is not that of the string but the string will set it in vibration with forced vibrations will increase the volume of the sound. that The laboratory device embodying all these features is the sonometer shown in Fig. 7:6. It consists of a hollow wooden box (A) on which one or more strings (B) are stretched. Permanent (C) and movable (D) bridges and a means of varying the tension (E) are provided. CD AD B C 66 CHARACTERISTICS OF MUSICAL SOUNDS Sec. 11:15 \Ve shall now proceed to use it to study the laws of vibrating strings. 11:15 THE LAWS OF VIBRATING STRINGS (a) The Relationship between Frequency and Letigth After doing experiment 7, chapter 10, results similar to those in the following table are found. Examination of trials 1, 4 and 5 will show that while the frequency of 256 is emitted by 34.4 cm. of string, v.p.s. twice that frequency is produced by onehalf that length and four times the frequency by one-quarter that length. Other possible results would be that three times the frequency is produced by one-third the length, five times the frequency by one-fifth, and so on. Thus we see, as in column 3 below, that the product of frequency times the length of vibrating string is constant within the limits of experimental error. Therefore, the frequency of the note produced by a vibrating string varies inversely as its length. This is the law of lengths. Results for Law of Lengths Constant Tension — 1000 gm. Length of Wire Producing Unison Frequency of Fork Frequency X Length “ 1. 2. 256 v.p.s. 320 384 512 5. 1024 4. 3. “ “ “ 8806 8832 8832 8806 8806 34.4 cm. 27.6 cm. 23.0 cm. 17.2 cm. 8.6 cm. Example If 30 cm. of wire at a certain tension produces a note with a frequency of 256 v.p.s., what would be the frequency when the length is 40 cm.? Solution 1 The ratio of the new length to the old = — 40 30 the frequency varies inversely as the length. the new frequency is — of the old. 40 . the new frequency = 256 X — = 192 v.p.s. 30 40 . Solution II the frequency varies inversely as the length, frequency X length is constant. 67 Chap. 7 SOUND the new frequency X the new length old length. the old frequency X the Let the new frequency be x a; X 40 = 30 X 256 ;c = 40 = 192 the new frequency =192 v.p.s. (b) Relationship between Frequency and Tension Adjust the length and tension of a string on a sonometer until it is in unison with a tuning-fork of frequency 256 v.p.s. Keeping a constant length of 25 cm., adjust the weights that stretch the string to get unison with a second, and then again to get unison with a third fork. Record the results as follows, and from them determine the relationship. it After comparing the results of trials 1 and 2, is noted that twice the frequency is caused by four times the tension. A comparison of trials 1 and 3 shows that 4 times the frequency is caused by 16 times the tension. Since the multiplier for the frequency is the square root of the multiplier for the it follows that the frequency tension, of the note emitted by a vibrating string varies directly as the square root of the tension. This is the law of tensions. Frequency of Fork 1. 128 v.p.s. 2. 256 v.p.s. 3. 512 v.p.s. Tension 385 gm. 1540 gm. 6160 gm. Example A string with tension of 2000 gm. produces a note with a frequency of 300 v.p.s. What would be the frequency of the note if the tension were 4500 gm.? Ratio of the new tension to the old = 4500 9 , = 4 2000 ’ . the frequency varies directly as the square root of the tension. the new frequency will be the new frequency = 300 X 4500 of the old 2000 ^ = 300 X - = 450 v.p.s. 4 2 (c) Relationship between Frequency and Diameter (d) Relationship between Frequency and Density By an experiment somewhat analagous to (a) we may determine the law of diameters, i.e., the frequency of the note emitted by a vibrating string varies inversely as the diameter. frequency of The law of densities states that the the note emitted by a vibrating string varies inversely as the square root of the density of the material. All four laws may be illustrated by 68 CHARACTERISTICS OF MUSICAL SOUNDS Sec. II; 16 the tuned, examination of a violin. To increase the frequency of a note, the violinist shortens the string with his finger. When a violin tension to is increase the frequency and decreased to It will also be decrease the frequency. noted that those strings with greatest densities and diameters the lowest notes. increased produce is 11:16 HOW A STRETCHED STRING VIBRATES We have just considered the simplest mode of vibration of a stretched string wherein the string vibrates as a whole. There is a loop in the centre with a node at each end (Sec. 11:6). When vibrating thus, the string is emitting the note of lowest frequency, the fundamental. However, the in other ways to produce notes of higher frequency (Chap. 10, Exp. 8). string may vibrate When the string vibrates in halves, the note produced has twice the frequency of the fundamental and one-half the wave-length. This is the first over- wave-lengths tone or .second harmonic. When vibratthirds, quarters and fifths, the ing in frequencies produced are three, four and five times that of the fundamental, and onethe are quarter and one-fifth of the fundamental. The notes are called the second overtone or third harmonic and so on. Fig. 7 : 7 shows some modes of vibration of a stretched string. one-third, fundamental when that We see that a string may vibrate in parts, and as a whole as well. When it is vibrating in parts, the frequency of the note is a multiple of that of the fundamental and the notes are called the harmonics or overtones of the string. Frequently these overtones accompany the is sounded, and give quality to the sound Very produced (Chap. 10, Exp. 9). few sources, on the other hand, produce the fundamental free of overtones. The tuning-fork is one that does, but even in it the overtones are present at the bevibration, vanishing as ginning of its It is this absence of overtime goes on. note FREQUENCY WAVE-LENGTH n I Fundamenfal l\ & First Overtone A\ N A\ N N Second Overtone N A\ A\ A\ N N N N Third Overtone N N 2N 3N 4N Fig. 7:7 Nodes and Loops in a Vibrating String. 'A I 'A I 'A I 69 Chap. 7 SOUND 8 Fig. 7:8 Oscilloscope Tracings of Tuning Forks Sounded (a) Singly A—Tuning-fork B—Organ Pipe (b) Pairs tones that makes the tuning-fork valuable in the study of sound, though at the same time it makes the note dull and uninteresting, for it is the overtones that make a note rich and interesting to the listener. 11:17 QUALITY OF SOUND If respectively. Let us analyse the vibration of a fundamental and its harmonics by means of a cathode-ray oscilloscope. the fundamental has a frequency of 128, the first two overtones have frequencies of 256 and 384, shows the results of sounding tuningforks of these frequencies singly and in groups. When sounded singly, we see differences in frequency and wave-length of the notes. When sounded in groups, we see complicated wave forms which represent the blending of the fundamental and one or more overtones. Fig. 7 : Next, with the aid of an oscilloscope, let us analyse notes from a tuning-fork and several other different sources having the same fundamental frequency. Fig. 7:9 shows several traces made in such a way. The regular trace of the is emitting tuning-fork indicates that it but a single tone. The complicated wave forms of the others indicate that: D—Violin Fig. 7:9 Oscilloscope Tracings. 70 CHARACTERISTICS OF MUSICAL SOUNDS Sec. II : 18 1. They contain tones (overtones) in addition to the fundamental. 2. That some have more of these over- tones than others. 3. That in some, the overtones are more prominent than in others, We may conclude therefore that the quality of a musical note is dependent on the number and relative prominence of the overtones that occur along with the fundamental. is the quality of It the sound that enables us to distinguish notes of the same pitch and intensity from different sou
rces. 7. QUESTIONS II : 18 1. A between (a) Distinguish sound and a noise. (b) What are the three distinguishing characteristics of musical sounds? musical a 2. (a) Define intensity of sound. (b) State three factors that affect the intensity of sound. Illustrate each with a suitable example. 3. (a) Define pitch. (b) Describe an experiment to illustrate upon what the pitch of sound depends. (c) Make up a set of observations and show how the frequency of a given note may be calculated. 4. (a) Describe the sound of a train whistle as the train moves rapidly away from you. (b) Explain this phenomenon fully. 5. Describe a sonometer and state the purpose of each of its parts. 6 . (a) State four factors that affect the frequency of a vibrating string. the Law of Lengths. A (b) State stretched string 50 cm. long vibrates with a frequency of 1 50 v.p.s. What will be its frequency when the length Is (i) 10 cm., (ii) 75 cm.? (c) State the Law of Tensions. A stretched length vibrates with a frequency of 1 25 v.p.s. when its tension is 900 gm. What will be its frequency when the fixed string of tension is (i) increased to 2500 gm. (ii) decreased to 144 gm.? (a) Describe an experiment to show the modes of vibration of vibrating strings. (b) Define; fundamental, overtone. (c) What governs the quality of a note? B 1. A certain note has a frequency of the frequencies 480 v.p.s. (a) Determine of notes that are one, two, and four octaves above the given note. (b) Find the frequencies of notes that are one, three and five octaves below the note. 2. A toothed wheel with 40 teeth Is rotated at the rate of 360 revolutions per minute while a card is in contact with the teeth. Calculate the frequency of the note heard. 3. A toothed wheel having 66 teeth is rotated while in contact with a card. What will be the speed of rotation in revolutions per minute when the frequency of the note produced Is 352 v.p.s.? 4. How many teeth Savart’s wheel have if the speed of rotation is 540 revolutions per minute and the frequency of the note produced Is 1 350 v.p.s.? will a 5. Savart’s toothed wheels generally are arranged in sets of four on a common shaft with 12, 15, 18, 24 teeth respectively. 71 Chap. 7 SOUND . frequency of 540 v.p.s. What would be the weight the became 16000 gm., 1000 gm.? frequency stretching If 10 A piano string is 60.0 in. long. It vibrates at 260 v.p.s. A piano tuner changes the tension from 25.0 lb. to 36.0 lb. What will be the new frequency? n. A string 40.0 cm. long and having a tension of 1 600 gm. emits a note of frequency 1 28 v.p.s. Determine the tension of this string when it vibrates with a frequency of: 64 v.p.s., 1 60 v.p.s. 12 . A string 36.0 in. long under a tension of 1 6.0 lb. vibrates with a frequency of 256 v.p.s. What is the vibration frequency if the length is increased to 54.0 in. and the tension is increased to 81.0 lb.? long 13 . A string 100 cm. under a tension of 4900 gm. has a frequency of 280 v.p.s. What is the frequency if the 1 25 cm. and the length is increased to tension reduced to 2500 gm.? When rotating at the average rate of 21 Vz revolutions per second, they produce frequencies corresponding to C major chord at middle C on the piano, i.e., CEGC'. Determine the frequency of each note in the chord. 6. A stretched string 45.0 cm. long emits a note with a frequency of 300 v.p.s. What would be the frequency if length became 15.0 cm., 60.0 cm., 20.0 cm.? the 7. The A string of a violin vibrates at 440 v.p.s. The string is 40.0 cm. long from If the violinist moves his bridge to nut. finger so that only 30.0 cm. of the string vibrates, what will be the frequency of vibration? 8. A certain vibrating string 50.0 cm. long emits a note with a frequency of 320 v.p.s. Whaf length of string would vibrate with the following frequencies: 640 v.p.s., 200 v.p.s., 457 v.p.s.? 9 . A string 30.0 cm. long stretched by a weight of 4000 gm. emits a note having a 72 CHAPTER 8 RESONANCE AND INTERFERENCE PHENOMENA I I , I ! 11:19 THE MEANING OF RESONANCE As resonance is a new idea, we shall find out what it means using the apparatus shown in Fig. 8:1. This consists of several pendulums attached to a cord tied between two supports. When one is vibrated transversely, the motion will be transmitted the supporting cord. Any other pendulum through the rest to Fig. 8:1 Mechanical Illustration of Resonance. will vibrate erratically, starting, stopping, but never accomplishing the persistence of vibration referred to above. Now, when impulses from one body affect another having the same period of vibration, the second will begin to vibrate with increasing amplitude. If it is already in motion, the amplitude will become greatThis effect is known as resonance er. and will be very valuable in explaining the phenomena that follow. 11:20 RESONANCE IN AIR COLUMNS That air columns can be set in vibration and made to produce sounds of a definite pitch is well illustrated by such simple experiments as blowing across empty test-tubes of various lengths. Such air columns have a natural period of vibration depending on their length. If the fluctuations of pressure at the end of the column (caused by blowing) have the same period as that of the air column, resonance will occur. The column will be in a state of violent sympathetic vibration, and a strong note will be heard. In our study of air columns, we shall use tubes of uniform cross-section. If the tube is closed at one end, it is called a closed tube, while if it is open at both ends it is designated an open tube. of the same length and period of vibration will take up the vibratory motion and move with increasing amplitude. The others that have different periods (a) The Closed Tube After performing the experiment to demonstrate resonance in a closed tube (Chap. 10, Exp. 10), let us consider 73 Chap. 8 SOUND what occurred. When the sound that proceeds down the tube is reflected at the closed end the wave returns without change of phase. Thus a condensa- Condensation goes down and is reflected up. li Wave-length Fig. 8:2 Closed Tube in Resonance with Tuning-fork. phase occurs tion is reflected as a condensation and a rarefaction as a rarefaction. As the sound arrives at the open end a change as follows : when a in rarefaction reaches the open end some air is taken into the tube, and a condensation goes down the tube; when a condensation reaches the open end some air spills out of the tube and a rarefaction goes down the tube. Thus a condensation returns as a rarefaction, and a rarefaction as a condensation from the open end. half vibration, diagram (b), a rarefaction will go into the tube which tends to reinforce the rarefaction already proceeding downward. This process continues until the air in the tube is vibrating with such wide amplitude that it becomes the major source of the sound heard. It should be apparent that the sound travelled twice the length of the closed tube during one-half vibration, and therefore the length of the closed tube is equal to one-quarter of the wave-length. (b) The Open Tube Open tubes will also vibrate in resonance with sources of sound such as It has been found that tuning-forks. during one vibration the sound travels twice the length of the tube, and therefore the length of the open tube is equal to one-half of the wave-length. It will be evident, therefore, that the open tube that vibrates in resonance with a tuningfork of a certain frequency is twice the length of the closed tube. Closed Tubes Open Tubes in resonance with it. Fig. 8:2 shows a vibrating tuningfork held over a closed air column which For greater is simplicity we shall confine our attention to the movements of the lower prong, since movements of the upper one do not alter the final result. As the prong of the tuning-fork traces one-half a vibration, diagram (a), a condensation is sent down the tube and is reflected at the closed end as a condensation. When it reaches the open end it will be reflected down the tube as a rarefaction while the air that spills out forms a condensation the condensation pro- that duced above the prong of the fork. the length of the tube is such that the fork is about to execute the next one- reinforces If 74 Fig. 8:3 Modes of Vibration in Air Columns. 11:21 MODES OF VIBRATION IN CLOSED AND OPEN TUBES In vibrating air columns (Fig. 8:3), there will be nodes and loops ( Sec. 11:6). In the closed tube there will be RESONANCE AND INTERFERENCE PHENOMENA Sec. 11:22 a node at the closed end and a loop at the open end. The length of closed tube will be one-quarter of a wave-length when it is responding to its fundamental For the overtones, the distance tone. from node to loop must also be onequarter of a wave-length. After examination of the diagram, it will be evident that the closed tube can be in overtones whose resonance the frequencies are odd-number multiples of that of the fundamental. with In the open tube there will be a loop at the open ends and a node will occur in the middle when in resonance with the fundamental. The length of the tube will be one-half a wave-length. For overtones, the tube must be capable of containing several half wave-lengths. Figure 8:3 shows how this can be done and makes it clear that the open tube can be in resonance with all the over- tones. I I j [ V I II : 22 DETERMINING THE VELOCITY OF SOUND IN AIR BY RESONANCE IN AIR COLUMNS (a) Closed Tubes As explained in section II: 20(a), the condensation travelled twice the length of the tube during one-half a vibration of the prong of the tuning-fork. Hence, sound must travel four times the length of the tube during one vibration of the prong. Since the distance energy travels during one vibration is one wave-length, the wave-length of sound must be four times the length of the closed tube which is in resonance with the tuning-fork. In the true wave-length actual practice, must be augmented by .3 times the diameter of the tube (see note) but for purposes we may disregard our Therefor
e, wave-length (/) of sound = 4 X length of closed tube (L) giving resonance. it. Example A tuning-fork whose vibration frequency is 256 v.p.s. produces resonance with a closed tube 13.0 inches long. Calculate the velocity of sound in air. ? n = 256 v.p.s. l:=4L = 4x222. 12 ft. (Sec. 11:5) T = 256 X =1109 4 X 13.0 12 .'. Velocity =1109 feet per second Since the measurements are accurate to three significant digits, the proper answer is 1 1 1 X 10 feet per second ( Sec. 1:3). What would be the approximate temperature of the air in the above example? (b) Open Tubes In section II: 20(b) ; it was stated that sound travels twice the length of the open tube during one vibration. There- fore, the wave-length (/) of sound = 2 X the length of the open tube (L) (see note), A tuning-fork whose vibration frequency is 1024 v.p.s. produces resonance with a tube 17.2 cm. long. Calculate the velocity of sound in air. Example 75 Chap. 8 SOUND V= ? n = 1024 v.p.s. l = 2L = 2 X 17.2 cm. \'V^nl V = 1024 X 2 X 17.2 100 = 352.3 Velocity = 352.3 metres per second or 352 metres per second. What would be the approximate temperature of the air in the above example? Note In very accurate work a correction must be made for the change in pressure influencing the sound waves a short distance from the end of the tube. In a first course in physics, however, this factor need not be considered. For closed tubes, the end correction of .3 X the diameter of the tube must be added. For open tubes, you make the same correction for each end. The phenomenon is caused by the inertia of the air molecules. 11:23 SYMPATHETIC VIBRATIONS Two tuning-forks frequencies are mounted on hollow wooden boxes each open at one end (Fig. 8:4). identical of Fig. 8:4 Identical Tuning-forks to Illustrate Sympathetic Vibrations. The size of the air column is such that its natural period of vibration is the same as that of the fork. The boxes are placed a short distance apart with their open ends facing each other. When one fork is vibrated and then silenced shortly afterwards, a sound of the same pitch is still heard (Chap. 10, Exp. 11.) It is found to originate from the other fork. This response of one body to the sound 76 waves caused by the vibrations of another is called sympathetic vibrations. When a piece of plasticine is attached to the prongs of one fork its frequency is alIf the above experiment were tered. repeated this fork would not set up vibrations in the other. This shows that the two forks must have identical periods for sympathetic vibrations to occur. Vibrations from the fork cause forced vibrations in the box and in turn the air in the box vibrates in resonance with them. The vibrations pass through the air to the other box, through the same sequence of events as above, but in reverse order, causing the second fork to vibrate. 11:24 INTERFERENCE OF SOUND WAVES (a) Silent Points around a Tuning-Fork When a vibrating tuning-fork is held vertically and rotated near the ear alternate loud and faint sounds will be It will be heard (Chap. 10, Exp. 12). found that the faint sounds are obtained when the tuning-fork is held cornerwise to the ear. To understand this phenomenon, we should recall that in transverse RESONANCE AND INTERFERENCE PHENOMENA Sec. 11:24 vibration the prongs of the fork move together for half a vibration and apart for the next half. When they approach a compression each other (Fig. 8:5) R. tube (Chap. 10, Exp. 13). It consists of two U-shaped tubes that telescope in and out of each other (Fig. 8:6). One side has a speaking-tube, the other an listen. When a opening at which to sound is sent in and the length of the two paths is adjusted, faint sounds are heard in some positions and loud ones at If the two paths that the sound others. follows differ by one-half a wave-length, or an odd number of half wave-lengths, the two parts of the sound will arrive out of phase at the observer and a faint If the two paths sound will be heard. are equal, or differ by a whole number of wave-lengths, the two parts of the I j rarefaction (Ri) at either side. These r two waves spread out in all directions and, because they are in opposite phase, interfere with each other, producing silence when they meet at the corners (S). It is often noticed that the intensity of sound varies in different parts of an auditorium without any obvious cause. One possible reason for this is the interference of direct waves with reflected If these are out of phase they will produce a faint i sound as happened at the corners of the 1 waves from walls and ceiling. I I tuning-fork above. Another cause is discussed in Sec. 11:30. (b) The Herschel Divided Tube Fig. 8:6 The Herschel Divided Tube. I be can produced Interference by dividing a wave disturbance into two parts, conducting each along a separate path, and then blending the two. This ' can be accomplished by the Herschel sound will arrive in phase at the observer and a loud sound will be heard. This phenomenon is used to find the wavelength of sound. If the frequency of the 77 Chap. 8 SOUND note is known, the velocity of sound can be calculated. (c) Beats If the prongs of one of two middle C tuning-forks are loaded with plasticine, the frequency of its vibration will be slightly lower than that of the other. If these forks are sounded together, a sound will be heard which periodically increases and decreases in intensity. The alterations in the loudness of the sound are called beats. Consider each of the waves sent out by the forks as transverse waves. By representing one with a solid line and the other by a dotted line, as in Fig. 8:7 (a), we see that they become progressively more out of step until one cancels the other. If the lines are continued they will eventually arrive in step, although one will be a wave-length in front of the other. Now, when waves are out of phase they interfere with each other with consequent reduction in amplitude of vibration or loudness of the sound. When completely out of phase will be no movement and no there sound. When completely in phase, there will be a greater amplitude and a louder sound. cases there will be a gradual increase or de- Between these extreme 78 RESONANCE AND INTERFERENCE PHENOMENA Sec. 11:25 crease in both amplitude and loudness. 8: 7(b) shows the result of such Fig. : interference. i I i I j j Fig. easily. 8; 7(c) Flaving studied beats with reference to transverse waves, we may understand their production in sound waves more shows two sound , waves of slightly different frequencies being produced simultaneously. Assuming that they begin in phase, two condensations or rarefactions will occur together, producing a loud sound or a beat. As 1 the two waves get out of phase, the sound will become fainter and the more out of phase they are, the fainter the be. When completely out of phase, as when a condensation from i one and a rarefaction from the other ' occur together, silence results. As they ‘ become progressively more in phase, the a ! maximum, at which point a loud sound ' sound will increases intensity reaches until it I ' or beat will occur as before. A repetition of this sequence of events gives us noticed when two tuningforks of slightly different frequencies are effect the vibrated together. If the forks being used had a difference in frequency of one v.p.s., one beat per second would result. Similarly, a difference of two v.p.s. would produce two beats per second and so on. In general, the number of beats per second equals the difference between the frequencies of the two notes. This provides a convenient means of determining the frequency of a sound. (How could Moreover, musical inthis be done?) struments are tuned by listening for beats. The fewer the beats, the more nearly alike are the two frequencies. When unison is achieved, no beats may be discerned II : 25 QUESTIONS (a) What is meant by resonance? fully. 5. (a) How are beats produced? Explain (b) Explain resonance in closed tubes. 2. (a) A closed tube 1 2 in. long is in resonance with a tuning-fork whose is 300 v.p.s. vibration frequency Calculate (i) the wave-length of the sound (ii) the velocity of the sound in air. (b) What would be the length of a tube open at both ends that would be in resonance with the tuning-fork used in part (a)? (b) What determines the number of beats per second? 6 . A person holds down the "loud pedal” of a piano and sings a note. Account for the humming sound heard. Why does more than one string respond? 7. Account for the sound produced by blowing across the top of an empty testtube. What would be the wave-length of such a note? 3. Explain and give examples of sym- pathetic vibrations. 8. Account for the rise in the pitch of sound heard as a cylinder is gradually 4. (a) What is the cause of interference in sound? (b) Explain (i) silent points around a tuning-fork, (ii) variations in the loud- ness of sound as Herschel’s divided tube is elongated. filled with water. 9. Explain why a wavy sound is frequently heard when a tuning-fork mounted on a sounding-box, the open end of which faces a wall, is moved towards and away from the wall. Try it. 79 Chap. 8 SOUND B 1. (a) Calculate the wave-length of a note that gives resonance with (i) a closed tube 15 in. long, (ii) an open tube 6 In. long. (Disregard the cor- rection for diameter.) 5. A closed tube 4.0 ft. long responds to a frequency of 70 v.p.s. Find the temperature of the air. 6 . A closed tube 40.0 cm. long responds the to a frequency of 220 v.p.s. temperature of the air. Find 7. A resonance box is to be made for (n = 440 v.p.s.). When the a tuning-fork velocity of sound in air is 330 metres per second, what would be the shortest length of box, closed at one end, that would resonate with it? 8. Two forks, having frequencies of 384 and 380 v.p.s. respectively, are sounded together. How many beats per second will be produced? 9. When a tuning-fork (n — 4S0 v.p.s.) is sounded with another of slightly different pitch and there ar
e 6 beats per second, what are the possible frequencies of the second fork? How would you determine whether its frequency would be higher or lower than the other? (b) Determine the velocity of sound (n = 220 results shown in tuning-fork gives v.p.s.) the air a in if question 1 (a). 2 . Compare the frequencies to which a closed tube 1 2 in. long and an open tube of the same length will respond, the temperature of the air being 1 91/2° C. 3. Find the frequency of a note that resonates with a closed tube 1 0.5 in. long, the temperature of the air being '\5Vi°Q. 4 . Find the length of closed tube that will respond to a frequency of 288 v.p.s., the temperature of the air being 25V2°C. Express the answer in both British and metric units. 80 CHAPTER 9 APPLICATIONS OF SOUND ing between these membranes sets them in vibration in a way similar to blowing between the strands of a stretched elastic band. The faster the air moves, the the intensity of the sound greater is produced. The particular quality of the sound depends on resonance in the cavities of the mouth (m) and (n). Variaare caused by muscles altering tions the size and shape of these cavities. In the mouth, the tongue measure, makes the major changes. large (t), in 11:27 THE EAR The ear is the most wonderful sound It con- receiver that can be imagined. the outer ear, the sists of three parts: middle ear, and the inner ear. Sounds are collected by the pinna (Fig. 9:2) and directed into the ear canal to the eardrum. The ear-drum consists of a thin (3/1000 inch thick) tightly stretched membrane that is set in vibration by the sound waves and serves as the gateway to the middle ear. the ear there Within middle are three bones named because of their shape, the hammer, the anvil and the stirrup. The hammer is in contact with the eardrum, the stirrup with the oval window leading to the inner ear, and the anvil connects the two so that the vibrations of the ear-drum are transmitted to the inner ear. The middle ear is joined to the throat by the eustachian tube, the purpose of which is to equalize the air pressure on either side of the ear-drum. ^ ! j I! : 26 THE VOICE Of all sources of sound the voice is the most wonderful. The vocal cords in the larynx (Fig. 9:1) are two (c) elastic membranes whose thickness, length and tension affect the pitch in response to the will of the person and in keeping with the maturity and sex of the individual. Air (a) from die lungs (1) pass- Chap. 9 SOUND This adjustment of pressure can be felt when motoring in hilly country. The inner ear contains a spirallyshaped organ, the cochlea, containing a fluid which is agitated when the oval window vibrates. Movement in this fluid will cause hair-like projections to vibrate, transmitting small nerve impulses through the auditory nerve to the brain. In addition to the cochlea, the inner ear contains another organ known as the semicircular canals, which is associated with posture and balance. 11:28 MUSICAL SCALES The story of the evolution of the existing musical scale is a long and interesting one. The scale which gives maximum pleasure to us is one in which the frequencies of the notes are in the simple ratios shown (page 83). This scale is known as the diatonic scale, which on the tonic sol-fa corresponds to the notes doh, ray, me, fah, soh, lah, te, doh, or more familiarly perhaps, C, D, E, F, G, A, B, C'. The first note on the scale is called the tonic, and the last note, of twice the frequency of the first, the octave. The number of the note (counting from the tonic) defines a musical interval on the scale; thus the interval from C to D is a second, that from C to E a third, and so on. These intervals correspond respectively to frequency ratios of 9/D _ 288\ 8\C ~ 256/’ 4\C ~ 256 / 5/E _ The last row of figures gives the ratios 82 6 APPLICATIONS OF SOUND Sec. 11:28 The Diatonic Scale No. of note Notation Absolute frequencies (scientific pitch)* Frequency ratios 2 D 1 3 4 C 256 288 320 341.3 384 426.6 480 512 C 15 8 Interval ratios 9 8 10 9 16 15 9 8 10 9 9 8 16 15 *In science C represents a frequency of 256 v.p.s. but for concert work however, C is usually tuned to 261 v.p.s. I of the frequencies between successive These are known notes on the scale. as the interval ratios, of which it will be seen that there are three. These are 9 / 0 8VC ^ and 288 \ = — ), — ( — 10/E _320 9\D "~288 256/ 16/F 15\E ~ 341.3\ 320 } 1 etc. A note is sharpened when raised by an interval of 25 - e.g., Clt = — X 256 = 266.7 v.p.s. 25 24 24 A note is flattened when lowered by an interval of 24 — , e.g., Db = — X 288 = 276.5 v.p.s. 24 25 25 9 10 The ratios - and — are called major 9 and minor tones respectively, those of — being called semitones. The intervals o 1 15 between major and minor tones are too large for musical requirements and accordingly intermediate notes are inserted, known as sharps and flats. These are the black notes on the piano, the one between C and D, for example, being C sharp (CJj:) or D flat (D[)). not exactly equal in Thus on the diatonic scale Cjf and Db frequency. are Further practical difficulties arise with the diatonic scale when changing from key to key—a process known as modulation. Thus in the key of D (the scale obtained using D as the tonic) it will be seen that intervals obtained using the white notes of the piano occur in a different sequence (minor tone, semitone, major tone, . .) from those obtained when using the key of C (major tone. . Fig. 9:3 The Piano Keyboard Showing Two Octaves of the Chromatic Scale. 83 Chap. 9 SOUND The Scale of Equal Temperament or the Chromatic Scale Number of Note Notation Absolute frequencies Interval ratios 1 G 3 2 C# D Db 4 D# Eb 5 E 6 F 7 FJt Gb 8 9 10 G G# A Ab 11 A# Bb 12 B 13 C' 256 271 287 304 323 342 362 384 406 431 456 483 512 1.059 1.059 1.059 etc. minor tone, semitone, . . .). In order to correct this in all the various keys the number of “black” notes required would be very large indeed. This would make keyed instruments such as the piano and to play, although organ very difficult with the violin and the human voice, where a continuous range of frequencies is possible, there is no difficulty in producing the notes required. fixed The difficulty is overcome with instrupitch by adopting a ments of scale of equal temperament or the chroin which the matic scale differences in the intervals of the diatonic the octave being scale divided into twelve equal semitone intervals each of which equals are abolished, (Fig. 9:3) 1.059, e.g., — = C 256 =r 1.059. There is no difference between 0 % and Db, etc., on this scale and difficulties of modulation are overcome. It should be noted that a small amount of discord is inevitably present in instruments of fixed pitch, such as the piano and organ, which are tuned according to this scale. Thus reference to the tables will show that a chord of the three notes C, E, G, which is known as a major triad, does not have exactly the desired frequency ratio 4: 5: 6 as on the diatonic scale. However, in spite of this imperfection, this scale meets all the requirements admir- ably. 84 Photo by Everett Roseborou^h Ltd, APPLICATIONS OF SOUND Sec. 11:29 11:29 MUSICAL INSTRUMENTS (a) Stringed Instruments (Fig. 9:4). extensive list of these The reader will be able to suggest an All ! hav'e a sounding-box or board over which one or more strings are stretched. This is made to vibrate at the same frequency as the vibrating strings to give greater ' intensity of sound. The frequencies of the notes emitted by the strings are de, termined by their lengths, tensions, diameters, and densities. Some instruments, : e.g., the banjo-like group, have a fretboard to which the string is pressed, thus pre-determining the length required for a certain note. Those of the violin group have no frets and the performer : must rely on his ear to obtain the desired I note when he presses and vibrates the string. In the piano, there is a string of tension and All stringed in- !' density for each note. diameter, certain length, I j I j j struments produce notes with one or more overtones, their number depending on the manner of vibrating the string and the place where it is bowed, picked, or struck. (b) Wind Instruments These include the pipe and reed organs, the wood-winds and the brass instruments (Fig. 9:5). They involve a means of vibrating a resonant air column either of the open or closed variety. In some the air column is of fixed length and in others it may be varied by means of valves or a sliding telescoping device. The closed pipe or flute type of pipe of an organ is pictured in Fig. 9: 6(a). Compressed air (A) enters the space (C), is forced through the slit (S) and, on striking the lip (L), causes periodic variations in pressure. The length of the pipe is adjusted so that it resonates with these and gives out a musical tone, including overtones. Being a closed pipe. Cello Violin Guitar Harp Turner Musical Instruments Lyon Healy, Chicago. Fig. 9:4 Stringed Instruments. Chap. 9 SOUND Fig. 9:5 Wind Instruments. Greene Music Co. Gtd it has a node at the closed end and a loop at the open end and the sound contains the overtones whose frequencies are oddnumbered multiples of the fundamental. Owing to the greater number of higher overtones pipes, organs generally contain that kind, but they have the disadvantage of being twice as long for a given note. obtained open with An organ pipe of the reed-type is pictured in Fig. 9: 6(b). Here air (A) a metal-covered chamber (C) enters containing the stem of the pipe (S) with the reed (R). The note is determined by the reed and the air column serves resonance and improve the give to quality of the note. Instruments such as the flute and piccolo are like the flute-type of organ pipe. Different notes are made by opening holes along the length of the air column. The saxophone, bassoon, clarinet, and oboe use a reed to set the air column in vibration, and the pitch is varied by opening and closin
g holes to vary the length of this column. In brass instru- 86 ments the lips of the performer act as a double reed. Differences in pitch are produced by changing the length of air column with “valves” as in the cornet and similar instruments. complished by telescopic sliding U-shaped part of the tube in or out as in the trombone, or by over-blowing to produce the overtones as in the bugle. This is the ac- All musical instruments require frequent tuning due to mechanical defects or changes in temperature. Temperature affects not only the lengths of the strings in stringed instruments but the frequency of a resonant air column. The latter may be understood when we recall that the velocity of sound in air changes with the temperature. Since V — nl and I is constant, then changes in V cause a corresponding change in n. (c) Drums There are several types of drums used in bands and orchestras (Fig. 9:7). The bass and snare drums and the tympani They may have are some examples. APPLICATIONS OF SOUND bEC. 11:30 either one or two “heads” made of calfskin or some similar material stretched over a metal or wooden shell, and are sounded by being struck with some form of drumstick. The first two types are used to give a “beat” to the music and certain other effects. The latter is tuned during a musical number to give a sound of low pitch in harmony with the orIn each, the vibrations of the chestra. head are amplified through forced vibrations of the whole instrument. (d) The Electric Organ y 1 Electric organs are without pipes or air columns. The sounds are produced electrically by various means. In the Hammond Organ (Fig. 9:8), an iron “tone wheel” revolving near a magnet in a coil of wire generates alternating currents in the coil which, after they are an amplified, rise amplifier. Many of these are required to give the whole range of frequencies. to sound in give Reed Type Fig. 9:6 Organ Pipes. II: 30 ACOUSTICS OF BUILDINGS As was stated earlier, one of the difficulties encountered in the construction Tympani Bass Fig. 9:7 Types of Drums. Greene Music Co. Ltd. Snare 87 Chap. 9 SOUND Fig. 9:8 The Hammond Electric Organ. (a) Organ. Hammond Organ Western Export Corp. of auditoria is reverberation (Sec. 11:7). As performers and speakers are entertaining bigger audiences in larger and larger halls, the study of reverberation has engaged the attention of acoustic Halls with too much reverberation are useless, those with too little are dead and uninteresting. Hence the achieve a happy medium aim is (Fig. 9:9). engineers. to means Reverberation may be reduced by including upholstered various seats, the use of coffers or hanging drapes from arched roofs, tapestries and special acoustic wall finishes. Since the size of the audience would materially affect the acoustic qualities of an auditorium, upholstered seats with the absorbing power of one person, are emacoustics become ployed. Thus ideal curtains, APPLICATIONS OF SOUND Sec. 11:30 Fig. 9:9 Acoustics in Auditoria independent of the number of people present. Focusing effects from curved surfaces often cause sounds of undesirable inten. sides at various places. Such effects may i be reduced by acoustic finishes on walls or hangings of sound-absorbing material. Outside noises frequently present ! I acoustic problems. Street and hall noises may be reduced by double doors, and Sound abdouble or triple windows. sorbent baffles in ventilating shafts will minimize source. building may be Machinery mounted on rubber to absorb the vibraBuildings so protected should be sounds from that the in tions. free from outside noise. A Fig. 9:10 A. Magnified Section Showing Grooves on a Recording. B. A Sound Reproducer. B 89 Chap. 9 SOUND Fig. 9:11 Tape Recorder. Dominion Sound Co. Ltd. 11:31 THE REPRODUCTION OF SOUND The Phonograph was invented by Edison in 1877. He spoke into a tube which activated a sharp needle whose vibrations cut a straight groove of varying depth in a sheet of tin-foil. On the “play-back’' another needle followed the groove and these up-and-down movements were transmitted to a diaphragm which set the air behind it in vibration. These early records were cylindrical. Later improvements suggested by Alexander Graham Bell, and others, inof a wax recording cluded material, a disc- type record and a groove of constant depth which wavered from side to side, instead of a straight groove of varying depth. the use A simple diagram of the grooves on a modern record is shown in Fig. 9:10 (a), and of a modern reproducer in Fig. 9: 10(b). Recordings are cut by a steel needle (stylus) in a polished slab of wax. This is dusted with a powder such as graphite (an electrical conductor) then immersed in an electrolytic bath of bluestone, becoming copper plated on the passage of 90 electricity through the bath. The slab emerges as a negative which is now reinforced by iron and is used to impress duplicates of the original on plastic discs. The electrical reproducer transforms the vibrations of the needle as it follows the groove into electric currents which are amplified until they are able to actiSuch reproducers vate a loudspeaker. make a sound more nearly like the original than did the former mechanical type. is of the other locations Another form of recording of great use in music, in the office and in a tape host recorder (Fig. 9:11). The principle first used was to transform sounds into electric currents and use them, when amplified, to vary the magnetism in a wire. Later models used paper tape, coated with a thin layer of magnetic material. This has the advantage of being compact and can be readily edited. In addition, any recorded matter may be erased, permitting the same tape to be used many times. Another type of sound recording is the sound track used on movie film @ © 0 0 (3 APPLICATIONS OF SOUND Sec. 11:32 Ob\|ective Lens Light-valve ribbon of duralumin which regulate the width of the slit. Recording Lomp 90 ft. /minutes. Pholc C fir'll^ lens _(3 1 / Fig. 9:12 A. Recording Sound. B. Variable Density Sound Track. C. Reproducing Sound from the Track. Film Strip: Rank Film Distributors of Canada Ltd. stations Si, S 2 and S 3 (Fig. 9:13) are located in carefully surveyed positions and an enemy gun is located at G. The times taken for each station to hear the I (Fig. 9:12). Sound is transformed into electric currents which are amplified so that the strength of two magnets can be varied. These magnets vai'y the width of a slit through which light passes from Such a film is a source to the film. 'said to be of the variable-density type. Reproduction from the sound track is accomplished by shining through it light ijfrom a source of constant strength to a As a photo-electric cell result, the cell will produce a current of i varying strength. This can then be amplified so that iit is strong enough to operate a loudspeaker. (Sec. V:82). I 11:32 LOCATING UNSEEN OBJECTS (a) Sound Ranging j ' Let us imagine that three receiving- 91 Chap. 9 SOUND report are recorded and the differences are determined. Knowing the velocity of sound and the extra time required for the sound to get to S 2 and S 3 , the respective distances from the gun (G) may be calculated (distance =.V y. t). Construct a circle with centre S 2 and radius the calculated extra distance. Repeat for S 3 . Construct a circle through Si and tangent to these other two. Locate its centre and that is the position of the gun. (b) Sounding the Sea process, In this a ship produces a sound (S) under water (Fig. 9:14) and receives it by a hydrophone (R) also For the source (S), it has been found to use ultrasonic vibrations, best i.e., having frequencies above the audible These can be beamed and on range. being detected are of such a nature as not to be confused with other vibrations in the water. 11:33 THE FUTURE OF SOUND the No one can foresee future of sound. The properties of sounds in the audible range, that is, 20-20,000 v.p.s., are well understood. However, there is much to be learned about the ultrasonic vibrations whose frequencies are from 20,000 to 500,000,000 v.p.s. The dog whistle (20,000 v.p.s.) and the squeak of a bat (30,000 v.p.s.) are at the lower limit of this group of vibrations. sea was used at Some present uses for ultrasonic vibrations follow. When used at an intensity of 160 decibels or more they have been used to remove the dust and soot from chimney gases. During World War II Sonar (Sound, Navigation, and Rangfor sounding, ing) locating submarines or other ships, and for underwater communication. Peacetime underwater uses include locating schools of fish and sunken ships. Another use is to cause molten metals to set more thus giving them finer grain quickly, structure and, thereby, greater strength. Conversely, it is used by large organizastructural tions materials such as concrete. Still another is in homogenizing milk. The physiological these high-frequency sound waves are only Research beginning to be understood. IS proceeding on measuring the body’s a view to tolerance deriving possible curative values. Truly, the future of sound may be amazing! discover them, effects flaws with of to to in Sea Bed Fig. 9:14 Determining the Depth of the Sea. By means of a timingunder water. device the interval may be determined. This must be halved when finding the depth of the sea. (Why?) Knowing the velocity of sound in water and the time, the depth can be determined. The hydrophone is a special receiver for underwater work designed to respond to vibrations from one direction only In time of war, it serves a useful purpose in locating enemy submarines. 92 APPLICATIONS OF SOUND Sec. 11:34 QUESTIONS II : 34 1. 2. A (a) Describe the larynx. (b) How do we produce sound in the larynx? (c) Give varying reasons the for quality of different voices. (a) What are the three main divisions of the ear? Name the parts and purpose of each. (b) Describe how we hear. 3. (a) Define: tonic, octave, major tone, minor tone, semitone, major triad. (
b) Distinguish between diatonic scale, and scale of equal temperament or chromatic scale. orchestra under the headings: (a) stringed (b) wood-wind (c) brass Give at least three examples of each. percussion. (d) 5. Describe how the acoustics of lecture halls may be improved. 6. Describe several ways of recording sounds. 7. In sounding a lake, the time lapse between producing a sound and hearing the echo is 0.75 sec. The velocity of sound in water is 4750 ft. per sec. Calculate the depth of the lake at that point. 8. (a) Distinguish ultrasonic and supersonic. between the terms (b) Give several uses for ultrasonic 4. Classify the instruments of a school vibrations. 93 CHAPTER 10 EXPERIMENTS ON SOUND EXPERIMENT 1 To study transverse vibrations. (Ref. Sec. II;3)j Apparatus A simple pendulum and support, stop-watch Fig. 10:1 Method 1. Attach the pendulum to the support. Draw the bob aside and let it swing freely. 2. With the bob at rest, mark its position by a chalk mark on the table. Draw the bob aside and measure the distance it travels to either side of the rest position. 3. While the bob is swinging measure the time required for 30 complete vibrations. Calculate (a) the number of vibrations per second and (b) the time required for one vibration. 4. Repeat part 3 using a greater and a smaller amplitude. 5. Repeat pai't 3 with a longer and a shorter pendulum. Observations 1. (a) In the simple pendulum, what is the direction of motion relative to the length of the vibrating object? (b) What is one complete vibration? 94 2. 3. EXPERIMENTS ON SOUND How do the distances that the bob swings to either side of the rest position compare in magnitude? Time for 30 VIBRATIONS The number OF VIBRATIONS PER SEC. The time for 1 VIBRATION The given pendulum The same with greater amplitude The same with smaller amplitude The pendulum made shorter The pendulum made longer Conclusions 1. (a) What type of vibratory motion is illustrated by the pendulum? (b) Define complete vibration. 2. Define amplitude of vibration. 3. Define (a) frequency of vibration, (b) period of vibration. 4. What efTect has changing the amplitude on the frequency and period of vibration? 5. What effect has changing the length of the pendulum on its fre- quency and period? Questions 1. Why do we call these vibrations transverse? 2. What changes take place in amplitude as the body is allowed to vibrate for a long time? What effect has this on the period or frequency of vibration? 3. Why is the period not dependent on the amplitude? 4. Why is the pendulum a suitable device for controlling a clock? EXPERIMENT 2 To study longitudinal vibrations, (Ref. Sec. 11:3) Apparatus A coil spring, weight, support, stop-watch. Method 1. Suspend the weight from the support by the coil spring. Draw the bob down and release it. Fig. 10:2 95 Chap. 10 SOUND 2 . When the weight is at rest, mark its position by a chalk mark on some vertical object such as a ruler. Draw the weight down and note the distance that it travels above and below the rest position. 3. While the weight is moving measure the time required for 30 complete vibrations. Calculate (a) the number of vibrations per second and (b) the time required for one vibration. Observations 1. (a) In the coil spring what is the direction of motion relative to the length of the vibrating object? (b) What is one complete vibration? 2. How do the distances that the weight moves to either side of the rest position compare in magnitude? 3. (a) What is the number of vibrations in 30 seconds? (b) What is the number of vibrations in one second? (c) What is the length of time for one vibration? Conclusion 1. What type of vibratory motion is illustrated by the coil spring? 2. Give the meaning of the terms complete vibration, amplitude, fre- quency and period. Questions 1. Why do we describe these vibrations as longitudinal? 2. What would be the mode of vibration of a tuning-fork at a, b, c, d? Test your answers by exploring the fork with a pith ball. a b c d EXPERIMENT 3 To determine whether or not sound requires a material medium for its transmission, (Ref. Sec. II: 4) Apparatus Bell-in-vacuo (Fig. 6:5), exhaust pump, wax or vaseline, electric wires, two dry cells, switch. Method 1. Seal the bell-in-vacuo onto the pump plate with the wax and connect with the exhaust pump. Connect the bell, cells and switch. Close the switch. 2. Start the pump and gradually evacuate the jar. 3. Stop the pump and let the air slowly return. 96 . EXPERIMENTS ON SOUND Observations What changes in loudness are observed? Conclusion What would you be led to conclude about the ability of sound to be transmitted in the absence of a material medium? Questions 1. Why was sound not entirely eliminated? 2. What changes in the apparatus would improve this experiment? 3. What effect does changing the density of the medium have on the transmission of sound? 4. Verify your answer to question 3 experimentally, using different media, e.g., wood, water and air, between your ear and a sounding object, e.g., a waterproof watch. 5. Does light require a material medium for its transmission? EXPERIMENT 4 To illustrate the different kinds and fundamental characteristics of wave motion. (Ref. Sec. II; 5) Apparatus A length of rubber tubing, a long coil spring, two rigid supports. (a) Method 1. Tie one end of the rubber tubing to one of the supports. Tie a piece of string to the tube, leaving one end dangling. Vibrate the free end of the tube up and down with the hand. Note the effect of the vibration on the tube. 2. Attach the coil spring between the two supports. Tie a piece of string Squeeze several coils together and release them. to it as in part ( 1 ) Note the movements of the string. 97 Chap. 10 SOUND Observations 1. (a) What is observed when the end of the tube is vibrated? (b) What is the direction of the motion of the particles of the tubing relative to the length? 2. (a) What is observed when the coils of the spring are released? (b) What is the direction of motion of the coils of the spring relative to the length? Conclusions 1. What are the two parts of transverse waves? Define and give examples of transverse wave-motions. 2. What are the two parts of a longitudinal wave? Define and give examples of longitudinal wave-motion. 3. Define; crest, trough, condensation, rarefaction, wave-train. 4. How is the energy from the source of disturbance transmitted through a medium? 5. What is the fundamental characteristic of wave-motion? Questions 1. What is the meaning of wave-length, the amplitude of the wave, the frequency of the wave, the period of the wave? 2. Establish the relationship between velocity, frequency and wave- length. EXPERIMENT 5 To show standing waves in a stretched cord. (Ref. Sec. 11:6) Apparatus Electric bell with gong removed, rigid stand and clamp, pulley and support, pan, with weights, length of light silk cord, batteries, connecting wire. Method Tie one end of the cord to the clapper of the bell and the other to the pan of weights. Assemble the apparatus as in diagram and close the circuit. Adjust the tension of the cord by changing the weights on the pan until the cord takes up a steady appearance. Observations 1. What effect has the clapper on the cord? 98 EXPERIMENTS ON SOUND 2. Wliat happens to this disturbance when it reaches the support at the distant end of the cordi’ 3. What is the appearance of tlie cord? Conclusions 1. What causes standing waves in a stretched cord? 2. Define; node, loop. Explain the cause of each. EXPERIMENT 6 To show what determines the pitch of sound. (Ref. Sec. 11:12) Apparatus Savart’s toothed wheel, rotator, cardboard card. Method 1. Assemble the apparatus as in Fig. 7:3 and rotate the disc while holding the card against it. Note the pitch of the sound produced. 2. Repeat part 1 while rotating the disc at a slower rate. 3. Repeat part 1 while rotating the disc at a faster rate. Observations What is observed in each of the above parts? Conclusion What determines the pitch of sound? Questions 1. How would you determine the frequency of the note produced in the above experiment? 2. What is the frequency of the note an octave higher than another? 3. What kind of sound would have been produced by a disc with irregularly spaced teeth? Explain. /Vote Experiment 6 may be done using a perforated disc and a jet of compressed air. EXPERIMENT 7 To determine the law of lengths for vibrating strings. (Ref. Sec. 11:15) Apparatus Sonometer (Fig. 7:6), one steel string, movable bridge, tuning-forks with different frequencies such as 256, 320, 384, 512, 1024 v.p.s. Method 1. Using a string 100 cm. long, adjust its tension until the pitch of sound that it produces is the same as that of the fork whose frequency is 256 v.p.s. Record the frequency of this note and the length of string that produces it in the table below. 2. Without changing the tension, adjust the length of the string by 99 Chap. lO SOUND inserting the movable bridge beneath it. Determine the lengths of string that will produce notes of the same pitch as the other forks provided. Tabulate these frequencies and lengths. Observations 1. What length of string produces a sound of low frequency? 2. What change in frequency of sound occurs as the string is shortened? 3. Table of Results. Frequency OF Note Length of String Ratio of Frequencies Ratio of Lengths t Product of Frequency and Length of String 256 v.p.s. 100 320 v.p.s. 384 v.p.s. 512 v.p.s. 1024 v.p.s. Explanation 1. What is true of the ratio of the frequencies compared to the ratio of the lengths? 2. What is true of the product of the frequency of the note times the length of the string producing it? Conclusion State the law of lengths for vibrating strings.- Questions 1. In the above experiment, what length of string would have a frequency of 768 v.p.s., 128 v.p.s.? 2. In the experiment, what frequency would be produced by a string of length 150 cm., 20 cm.? 3. In part 1, what effect does changing the tension have on the
fre- quency of the note produced? 4. List two factors that affect the frequency of a note produced by a vibrating string. EXPERIMENT 8 To illustrate the modes of vibration of vibrating strings, (Ref. Sec. II; 16) Apparatus Sonometer, one steel string, bow, several V-shaped paper riders. 100 EXPERIMENTS ON SOUND Method 1. Place three paper riders at equal intervals along the string. Bow the string at its centre and note the effect on the riders and the pitch of the note produced. Record your results in the table. 2. Place three riders on the string as before. Touch the string lightly at its centre and bow the string in the middle of one of the halves. Make the same observ^ations as before and tabulate them. 3. Repeat part 2 using five riders, touching the string one-third of its length from one end and bowing in the middle of this third. 4. Repeat using seven riders, touching the string one-quarter of its length from one end and bowing in the middle of this quarter. 5. Repeat using nine riders, touching the string one-fifth the length of the string from one end and bowing in the middle of this fifth. Observations Position of Damping Effect on Riders Diagram to Si-iow Mode OF Vibration of String Frequency of Note Produced 1. None 2. 1/2 3. 1/3 4. 1/4 5. 1/5 1 1 Conclusions 1. What is the manner of vibration of a vibrating string when producing its fundamental? Its various overtones? 2. What are the frequencies of the various overtones compared to that of the fundamental? 3. Define: fundamental, overtone. Question How do you account for differences in quality of the same note from various sources? EXPERIMENT 9 To study the effect of the superposition of waves on the quality of sound produced, (Ref. Sec. II; 17) Apparatus Four tuning-forks (n = 256, 320, 384, 512 v.p.s.), rubber mallet, hard mallet, bow, tuning-fork on resonance box. Method 1. (a) Vibrate each tuning-fork separately by striking it with the rubber mallet and note the quality and pitch of the sound produced by each. 101 ' Chap. 10 SOUND (b) Vibrate the forks, n = 256 v.p.s. and 320 v.p.s., simultaneously and note the quality of the note. (c) Repeat part 1 one, two or three of the others. using the fork, n = 256 v.p.s., and with (b) 2. (a) Bow the -tuning-fork on the resonance box and observe the quality of sound produced. (b) Vibrate the same fork by striking it with the hard mallet and again note the quality of sound produced. (c) Vibrate the same fork by bowing it and striking it with the hard mallet simultaneously. Note the quality of this sound. Observations What is observed in the above parts of the experiment? Conclusions 1. What determines the quality of sound? 2. Explain in terms of superposition of waves. EXPERIMENTIO To demonstrate resonance in a closed air column and to find the velocity of sound in air, (Ref. Sec. 11:20) > Apparatus Retort stand, clamp, cylinder, water, open tube about 15 inches long, several tuning-forks (256, 384, 512 v.p.s.). Method 1. (a) Fill a tall glass jar with water at room temperature to the three-quarter mark and place a smaller glass tube, open at both This tube should be clamped to a ends, in the water as shown. retort stand so that it may be raised, lowered or secured at will. Sound a tuning-fork {n = 256 v.p.s.) and hold it close to the open 102 EXPERIMENTS ON SOUND end of the tube. Raise or lower tlie tube until a position is found where the air column resounds most loudly. Measure the length of the air column. (b) Repeat (a) with the other tuning-forks. Observations 1. What was the room temperature? 2. What is observed when the tuning-fork is brought to the top of the tube before and after adjustment of the length? 3. Table of Results. Frequency of Fork Length of Closed Tube in Resonance 256 384 512 Explanation What is the cause of resonance? Calculations 1. What is the wave-length of the note produced by each tuning-fork? 2. Calculate the velocity of sound in air from the result obtained for each fork and average your answers. Conclusions 1. What is resonance? 2. What is the average velocity of sound in air? Questions 1. From the temperature recorded during the experiment, calculate the velocity of sound in air and compare it with the experimental value. 2. Determine the percentage error in your experimental result. 3. What is the relationship between the length of vibrating air column and the frequency of fork that gives resonance? EXPERIMENT 11 To study the production of sympathetic vibrations. 11:23) (Ref. Sec. Apparatus Two matched tuning-forks mounted on resonance boxes (Fig. 8:4), rubber mallet, wax or plasticine. Method 1. Place the tuning-forks close together with the open ends of the Strike one fork and silence it after a short boxes facing each other. time. Note what happens. 2. Strike the other fork and repeat part 1. 103 Chap. 10 SOUND 3. Load the prongs of one of the forks with wax or plasticine. Sound the forks separately. What is observed? 1 and 2 using the forks as they are now. What 4. Repeat parts happens? Observation State your observations for each part above. Explanation Account for your observations. Conclusions 1. What are sympathetic vibrations? 2. Explain their cause. Questions 1. What would occur in this experiment if one fork were an octave higher than the other? Try it, and explain. 2. Suggest other examples of sympathetic vibrations. EXPERIMENT 12 To illustrate interference of sound waves by means of a study of silent points round a vibrating tuning-fork. (Ref. Sec. II: 24(a) Apparatus A tuning-fork, rubber mallet, closed air column as in experiment 10. Method 1. Vibrate the tuning-fork and hold it over the closed air column. Adjust the length of the column until resonance results. 2. Now slowly rotate the vibrating tuning-fork while holding it above the air column and observe. Observations Note the changes in the intensity of the sound heard and the position of the tuning-fork for each change. Fig. 10:7 Explanation With the aid of a diagram account for these changes. Conclusion What is meant by interference of sound waves? 104 EXPERIMENTS ON SOUND EXPERIMENT 13 To illustrate the interference of sound waves by the use of Herschel's divided tube, (Ref. Sec. II; 24(b) Apparatus Tuning-fork, rubber mallet, Herschel’s divided tube (Fig. 8:6), rubber hose and ear-trumpet. Method 1. Adjust Herschel’s tube so that the two paths CAD and C B D are equal. Hold a vibrating tuning-fork in front of the opening C and listen at the ear-trumpet joined to D. 2. Gradually draw out A and note the positions of minimum and maximum loudness of sound. Measure the difference in lengths of the two paths for each such position. Tabulate your results. Observations 1. Describe carefully what is heard. 2. Table of Results. Intensity of Sound Difference in Lengths OF Paths (C A D-C B D) ( 1 ) Maximum (2) Minimum (3) Maximum (4) Minimum Explanation 1. Account for the variations in the intensity of sound heard. 2. How could the wave-length of the note be calculated? Calculation Calculate the wave-length of the note. Conclusion How does Herschel’s divided tube illustrate interference of sound waves? Question Knowing the frequency of a sound, how would you determine the velocity of sound in air? 105 Chap. 10 SOUND EXPERIMENT 14 To study the production of beats. (Ref. Sec. II: 24(c) Apparatus Two matched tuning-forks on resonance boxes, rubber mallet, wax or plasticine. Method 1. Vibrate the two forks together and note the result. 2. Change the frequency of one slightly by putting a little wax or plasticine on its prongs (or rearranging the weights if your forks are of this type) and repeat part 1. 3. Make a still greater difference in the frequencies of the forks (how?) and repeat part 1. Observations What do you observe in each part? Conclusions 1. How are beats produced? 2. What governs the number of beats per second? 3. Suggest an explanation for beats. Question How does a musician use beats to tune his musical instrument? 106 UNIT HEAT This? in Role Important an Plays Effect Heat What Jet-Propulsion. by Air the Through Sent is Aircraft This CHAPTER 11 THE NATURE AND SOURCES OF HEAT III : 1 HISTORICAL BACKGROUND Perhaps the greatest single advance made by mankind was the discovery of The presence of how to make fire. ashes and charcoal in caves inhabited in very ancient, times, shows that the discovery may have taken place in the Stone Age or earlier. Man’s mastery of fire has enabled him to protect himself from exposure to cold, to kill diseaseproducing organisms and to frighten off It has allowed him to smelt wild beasts. copper and iron from which new weafashioned. pons Without this discovery man might not have risen from barbarism. could and tools be friction between two dry History reveals the various fire-producing devices used. One was the igniting of dry moss or leaves by heat from the sticks. Another was the production of sparks when a piece of pyrite (fools’ gold, a sulphide of iron) was struck a glancing blow with a sharp piece of flint (a hard, compact mass of silica, the mineral of which sand is composed). Both of these are a far cry from the use of such modern devices igniters etc., although ihe principle of the flint gas and still employed in as matches, electrical igniter is cigarette lighters. The many developments in the production of heat that have occurred since early times give ample evidence of man’s creativeness. Equally interesting are the forward strides made in the understanding of the nature of heat and the laws that govern its use. Not only has man learned how to produce heat but he can now control it, retain it, measure it, transfer it from place to place and convert it into motive power. This study of heat is sufficiently large and important to warrant a special branch of physics, namely, “thermodynamics”. Ill : 2 THE NATURE OF HEAT (a) Caloric Theory Until well into the nineteenth cen- in most the spite people believed tury, Caloric Theory of Heat in of questionable proofs
to support it. This theory insisted that all empty spaces in matter contained a fluid called caloric and that the warming or cooling of a body was due to the gain or loss in the amount of this fluid. The early Greeks speculated that heat was the rapid vibratory motion of the molecules of a body. Francis Bacon produced some promising experimental evidence to (1561-1626) this effect. In an 1798 American, Benjamin Thompson, who later became Count Rumford, made further investigations. While directing the boring of cannon, he became interested in the amount of 109 Chap. 11 HEAT heat produced in the process and decided to investigate the problem of the nature of heat. The adherents of the caloric theory argued that caloric came out when iron shavings were formed and that the shavings had more of it than the iron. To test this theory, he applied a blunt drill (to produce few shavings) to the cannon with the whole assembly immersed in a box filled with water. In a short time the assembly became warm and finally the water boiled. Because heat was long the mechanism continued to turn, Rumford concluded that anything which could be as heat had produced without limit, been, could not be a material substance (caloric). He reasoned that heat must be caused by a vibratory motion in the produced as as material. In 1799 Sir Humphrey Davy of England dealt the caloric theory the coup de grace by rubbing together two pieces of ice in a vacuum at a temperature The below the melting-point of caloric theory held that as ice contained no caloric it could not melt under these conditions. But melt it did and in doing so afforded yet more proof that heat must be a product of motion. ice. (h) The Kinetic Theory of Matter According to theory, matter is this composed of numerous, tiny, moving called molecules, each being particles separated from its neighbour by empty space. A molecule is defined as the smallest particle of a substance which can exist alone and possess the properties of that substance. Scientists are forced to accept such a theory because they know that gases, and to a lesser extent liquids and solids, diffuse, and that gases do not settle but maintain a uniform pressure on the walls of the container. All the evidence indicates that molecules of gases are separated by comparatively vast distances, e.g., water vapour molecules are no more than ten times as far apart as water molecules, and possess great freedom of movement. Molecules of liquids have freedom of movement but must be fairly close together, since liquids resist compression. In solids the molecules exhibit great cohesion as illustrated by their rigidity. Their closeness is shown by their resistance ,to compression. Their movement is said to consist of vibration about certain positions in a prearranged pattern. (c) Heat—A Form of Energy Energy is the ability to do work. All moving bodies can do work and therefore have energy. Heat, which is caused by the motion of the molecules of a body, is capable of doing work and therefore must be a form of energy. THE NATURE AND SOURCES OF HEAT Sec. 111:3 that This may be illustrated with the aid of the apparatus shown in Fig. 11:1. When the tube is heated an effect is imitates somewhat the observed motion of the molecules. The mercury vaporizes and the molecules of mercury vapour drive the bits of glass upward. The pieces move about erratically, colliding with mercury molecules and each Heat is causing the motion of other. the particles (molecules and glass) and hence heat must be a form of energy. Under other circumstances heat energy may be transformed into other forms of energy such as electricity, light, etc. if from the Law of Conservation of Energy. This fundamental law of nature states that energy can neither be created nor destroyed although it may be transformed into any of its many forms. In a moving automobile is practice, brought to a stop by its brakes, the disappears enei‘gy changes to heat energy in the brakes. From this and numerous other examples, it is evident that heat is produced at the expense of some other form of energy. A brief discussion of several sources of heat (Fig. 11:2) will follow. motion which of (a) Mechanical Action Ill : 3 SOURCES OF HEAT The origin of heat may be inferred In every mechanical operation, the output of work is always somewhat less (a) MECHANICAL (a) MECHANICAL (a) MECHANICAL Compression (c) ELECTRICAL Combustion Fig. 11:2 Sources of Heat. Ill Chap. 11 HEAT than the input of energy, the loss being equal to the amount of energy converted into heat. This is stated thus: input = output -\|- heat. All such operations involve friction, because no surface is perfectly smooth. When one body slides or rolls over another, as when the hands are rubbed together, some of the energy devoted to the purpose is converted to heat. Frequently, percussion, which is the sudden stopping of a moving object when it collides with one at rest, e.g., hitting a nail, is employed in mechanienergy of cal motion is converted into molecular motion within both bodies which is mani- Here the operations. fested as heat. Some operations involve compression of gases. We know from operating a bicycle pump that some of the energy applied changes into heat. In the Diesel engine (Sec. 111:31), the heat produced by compression of the air in the cylinder is sufficient to ignite the fuel. (h) Chemical Change When a fuel is burned in air, new are formed and energy is materials released. Such a happening is called a chemical change. Every substance has its share of stored chemical energy, a form of potential energy, and when it burns the products of combustion generally possess less of it than do the original materials. The difference in the amount two represents changed into heat. For example: energies the (c) The Electric Current it This Whenever a conductor carries elecbecomes warmer. tricity, is because electrical energy encounters resistance to its flow in much the same way as water encounters resistance (friction) while flowing through a pipe. Just as the moving water loses energy as it overcomes resistance, so too does an electric current. The electrical energy lost becomes transformed into heat. Heatingelements clearly demonstrate this. (d) The Atom All matter in the universe is made up from about one hundred different kinds of elements. Elements are simple substances that have not been decomposed by ordinary chemical means. They are composed of atoms which are the building-blocks for the molecules of all sub- like those Some atoms, this energy. of stances. uranium or radium, are very large and complex and change into new atoms As they do so a small spontaneously. amount of mass changes into energy some of which becomes heat. There are more details of this process in chap. 32. Huge structures called atomic piles conContrary to popular trol belief, the energy from this source is not amazingly limitless large. For example, it has been estimated that one pound of uranium can produce three million times as much energy as one pound of coal or one pint of oil. Imagine how little uranium would be required to heat your home for one year! nevertheless, but, fuel + oxygen carbon dioxide + water + heat. The total energy of the fuel and of the oxygen equals the total energy of the carbon dioxide and of the water plus the energy which was converted into heat. 112 (e) The Sun Few of us realize the importance of the sun as a source of energy. We accept its daily warmth, and take for granted its energy stored in plant and animal products, in the water vapour of the air, in water and air currents, and in the THE NATURE AND SOURCES OF HEAT Sec. Ill: 3 Fig. 11:3 Trapping Solar Energy. fossil fuels—oil, coal and gas. If the sun were to be suddenly extinguished and all the sources of energy at our disposal were tapped at one time, our accustomed temperature would be maintained for After that we would only three days. quickly freeze to death! In spite of the fact that the earth is only a tiny dot in space, it receives a million-trillion kilowatt hours (Sec. V:75) of energy per year, of which all but .05 per cent slips from our grasp. Green plants trap the major part of this percentage, as follows: carbon dioxide + water + light energy (in the presence of chlorophyll) carbohydrates + oxygen This process is known as photosynthesis. The sun’s internal temperature, estimated at about 20 million degrees centi- grade, is maintained by a complicated process which is essentially the union of 4 atoms of hydrogen to form 1 atom of helium gas. Dr. Hans Bethe at The Bell Telephone Company of Canada. Fig. 11:4 The Solar Battery. 113 Chap. 11 HEAT Cornell University in 1938 showed that there is a decrease in the mass during the process and that this is converted to heat. For some time man has dreaded a world scarcity of fuel, knowing that at our present rate of consumption we shall be at that critical point in two or three centuries. Accordingly, research workers are constantly seeking ways and means of using solar energy. Some pin their hopes in part on utihzing an improved photosynthetic process, while others are investigating the use of light-sensitive 11:3, 11:4). As the chemicals (Figs. sun will yield its fabulous supply of energy at the present rate for an estimated 10 billion years, our future is assured, provided inexpensive ways and means of sunlight can be found. collecting that III : 4 QUESTIONS A 1. (a) Present an argument to show that heat is a form of energy. (b) Describe the changes in size and state that occur on intensely heating a piece of iron. Explain each by means of the kinetic theory of matter. 2. Explain each of the following: (a) A bullet is found to be warmer after hitting a target. when concentrated sulphuric acid is added to water. (d) A fuse burns out in an overloaded electrical circuit. (e) The origin of heat from the splitting of the atom. (f) The origin of energy from the sun. 3. State all the energy transformatio
ns that are involved in the sequence: sunlight, water-power, electricity, heat from a toaster. (b) Bearings frequently "burn out" when they run short of oil. (c) Considerable heat is produced 4. If the heat from 6 tons of coal will heat a home for one year, what mass of uranium (U235) will do the same thing? 114 CHAPTER 12 EXPANSION CAUSED BY HEAT (Fig. 12:2) consists of equal lengths of iron and brass welded together. On being heated, the bar bends with the III : 5 EXPANSION OF SOLIDS Almost all bodies expand on being heated and contract on being cooled. The ball and ring experiment (Chap. 15, Exp. 1), demonstrates the expansion and contraction of metals (Fig. 12:1). That different metals expand and contract by different amounts when heated or cooled through a given change of Fig. 12:2 Unequal Expansion of Solids. Since the brass on the outside. distance round the outside of a curved path is longer than round the inside, the brass has expanded more than the iron for a given change in temperature. On being cooled in a freezing mixture, it bends with the brass on the inside. For this reason, we say that brass contracts more than iron for a given temperature change. The expansion of metals, though small, exerts great force and is of particular concern to the engineer. For example, rivets are heated before being put into 115 temperature is shown by the “compound bar” experiment (Chap. 15, Exp. 2). The compound bar or bimetallic strip Chap. 12 HEAT Fig. 12:3 Compensating tor Expansion, fa) Bridge fb) Balance Wheel of a Watch. is it the necessary provided To determine the clearances and allowances required, to know exactly the small changes in length, area and volume resulting from temperature changes. These can be calcuappropriate lated that coefficient of expansion is known. Coefficients of expansion are of two kinds, namely, linear and volume. The linear coefficient of expansion of a substance is a number representing the increase in length of a unit length of the substance when its temperature is raised one centigrade degree. The volume coefficient may be defined in a similar way. The Note (Circled) the effect o f atmos- pheric heat on the bolts in the struc- ture of this bridge. place so that they will contract and hold the parts securely in place when cool. Steel bridges are subject to considerable contraction and expansion and for this reason are provided with a small gap at the end while the end may rest on a steel roller supported by the abutment (Fig. 12:3a). For similar reasons, there are expansion bends in metal piping, and gaps between the of railway tracks; hydro and telephone wires are left slack and a slight clearance is provided between the piston and the cylinder walls of internal combustion engines to avoid “seizing”. sections 116 EXPANSION CAUSED BY HEAT Sec. III:6 experimental determination of these co- is beyond the scope of this efficients course in physics but an explanation of their meaning makes the various applications very much more easily understood. Coefficients of Linear Expansion OF Solids Substance Aluminum Copper Iron Nickel Platinum Silver Zinc Brass Invar Steel Glass (flint) (soda) Silica (fused) Pyrex* Coefficient 0.000025 0.000017 0.000012 0.000013 0.0000089 0.000019 0.000026 0.000019 0.0000009 0.000011 0.0000088 0.0000085 0.0000004 0.0000036 Pyrex consists of 80 per cent silica and 20 per cent various oxides of metals, chiefly of boron. 111:6 APPLICATIONS OF EXPANSION OF METALS Some applications of expansion have been mentioned already and it is clear that the expansion of metals, though small, must always be taken into consideration. The errors in using metal surveying tapes have been largely overcome by the use of invar steel, a nickel-steel alloy containing 36 per cent of nickel and having a coefficient of linear expansion which is almost negligible. The same material is used for the pendulums of clocks to ensure almost constant length and accurate time-keeping. Watches are controlled by a metal balance-wheel (Fig. 12:3b) and hairoscillation of the spring, wheel being determined by its diameter. A rise in temperature would cause the diameter of the wheel to increase and. the time of consequently, the watch would lose time. To compensate for this defect the rim of the wheel is made in segments, each being a bimetallic strip of brass and steel with the more expansible metal on the outside. When the temperature rises, the segments curl inward, reducing the “efthe wheel and fective of compensating for the troublesome increase in diameter that would otherwise occur. diameter” strip, with The principle of the bimetallic strip finds other applications. One is the dial thermometer (Fig. 12:4a). The essential part of this instrument is a coiled bithe more again metallic expansible metal on the outside. One end of the coil is firmly attached to the case of the instrument and the other is connected to the pointer. As the coil winds or unwinds with a rise or fall in temperature, the movement of the free end is transmitted to the pointer moving over a scale graduated in degrees. Although not as accurate as other thermometers to be described in Sec. Ill: 8, is a robust instrument and has the it advantage of containing no liquid to vaporize or solidify. Such a device is also an essential part of the thermograph or Fig. 12:4 (a) The Dial Thermometer. 117 Chap, 12 HEAT fb) The Thermo- graph. Compound Bar continuous recording thermometer (Fig. 12:4b). Thermostats for automatically regulating temperature contain a bimetallic Such as the controlling feature. strip devices are pictured in Fig. 12:5. Fig. 12:6 shows a simple circuit controlled Thermostats control the temby one. perature of ovens, refrigerators, rooms, homes and hot-air furnaces. The par- Thermostats fa) Household Fig. 12:5 Type, (b) Type with Mercury Switch. Fig, 12:6 A Simple Thermostat. 118 EXPANSION CAUSED BY HEAT Sec. Ill: 8 ticular type used in hot-water furnaces is called an aquastat. Thermostats may be also used to activate fire-alarms and for a variety of other purposes, where the bending of the metal closes or opens a switch which controls of electricity to the device. In every case the supply involved. j I heat is Ill : 7 EXPANSION OF LIQUIDS I 1 I When liquids are heated, they too expand and, when cooled they contract (Fig. 12:7). The actual expansion or contraction is always somewhat greater i than that observed because of the ex- pansion and contraction of the container. With suitable apparatus, the expansion in volume per centigrade degree, i.e., the coefficient of expansion in volume, may alcoefficient than ' be determined. Some liquids, T; cohol, show a larger like is It at others. different coefficient Most, except mercury, show a temdifferent peratures. The table shows a few of these coefficients including some coefficients of expansion for water at various temperathis property of a liquid tures. which often determines its suitability for certain therAgain allowance must be mometers. made for expansion wherever liquids are being heated in conhned spaces, such as in hot-water heating systems (Sec. 111:14). purposes, such in as Coefficients of Cubical Expansion OF Liquids Liquid Alcohol Glycerine Mercury Water 5-10°C. 10-20 20-40 40-60 60-80 80-100 Coefficient 0.00110 0.00053 0.00018 0.00005 0.00015 0.00030 0.00046 0.00059 0.00070 III : 8 THERMOMETRY (a) Temperatiire If a scale is attached to the long glass tube (Fig. 12:7) while performing the experiment described in Sec. Ill: 7, a form of thermometer registering an increase or decrease in the temperature will be constructed. The temperature of a body may be defined as that condition which determines the direction of heat flow between it and its surroundings. Thus, a body at a high temperature will give heat to cooler objects while a body at a low temperature will take in heat from warmer objects. This will proceed until all objects are at the same final temperature. (b) Thermometers Originally, man relied solely on his 119 Chap. 12 HEAT sense of touch to measure temperature. Obviously, judgments obtained in this way are not very precise. For example, a door-knob feels colder to the touch than the wooden door. Again, if one hand is placed in a beaker of hot water, and the other in a beaker of cold water, and then both hands are placed simultaneously in lukewarm water, the first hand will get the impression of coolness and the second that of warmth. Clearly, i' B W Fig. 12:8 Filling a Mercury Ther- mometer. scale therefore, some means of measuring temperature that is more sensitive and more reliable than that provided by human sensations is needed for scientific purposes. We must have a precise, conof temperature and an sistent instrument for measuring it accurately. The evolution the modern thermometer is an interesting story. Students are advised to consult a good encyclopedia for the contributions of such men as Galileo, Viviani, Rey, Boulliau and others who have shared in its perfection. of 120 to register rapidly. The modern thermometer is constructed from a length of capillary tubing of uniform bore, sealed at one end by heating it in a flame. By gently blowing down the tube when it is hot, a small bulb is produced at B (Fig. 12:8). This should be very thin if the instrument ‘is After the tube has cooled, a small funnel is attached to the open end A, and clean dry mercury is poured into it. Before the mercury will fill the bore it is necessary to heat and cool the bulb alternately to force the air past the metal. When the tube is full, it is heated to expel any remaining traces of air. The bulb is now placed in a bath of liquid which has a tempera15° higher than the ture 10° of to ther- temperature which the highest mometer will be required to register. Using the fine blow-pipe flame, the tube is sealed at a point just below the free surface of the mercury. On removing the thermometer from the the mercury contracts in the stem, leaving a vacuum in the sp
ace above it. bath, To graduate the thermometer, we choose two fixed temperatures which can be easily obtained, and mark the level of the mercury on the stem when each of these temperatures has been maintained for some little time. The temperatures chosen are the freezing- and boiling-points of pure water at standard atmospheric pressure (760 m.m. of mercury). The former is called the lower fixed point and is marked on the stem by making a groove in the glass with a file at the level of the mercury when it has been standing for some time in melting ice (Fig. 12:9a). To obtain the upper fixed point, the thermometer is placed in the apparatus shown in Fig. 12:9b where the bulb and stem are surrounded by steam. When the mercury level is stationary the upper fixed point is scratched on the stem. If the pressure is not standard, it is necessary EXPANSION CAUSED BY HEAT Sec. Ill: 8 i to apply a correction before making this mark. 100°C. Upper Fixed 212°F. Having determined the positions of the fixed points, we divide the distance n V Point f 100 divs. C. = 180 divs. F. ^ .'. 1 div. C. = — div. F. 9 0°C. Lower Fixed k 32°F. Point CENTIGRADE SCALE FAHRENHEIT SCALE Fig. 12:10 Comparison of Temperature Scales. Daniel Gabriel Fahrenheit (16861736), a German instrument maker at Amsterdam, selected points 212°F. and 32°F. and constructed the scale that bears his name. In choosing fixed the >-55C.° 100°C. 45°C. 30°C. 15°C. °C or Actual Temperatures Changes in Temperatures or Fig. 12:11 The Comparison of °C and C°, 121 j grees. Then we test the thermometer I at various temperatures against a stan- dard instrument for accuracy. II - (c) Temperature Scales Two thermometer scales, the Fahren! heit and Centigrade or Celsius are in common use in English-speaking countries. The former is used in everyday practice while is used in science. In countries that are not English-speaking the centigrade thermometer is used for all purposes. latter the i , 'I I Chap. 12 HEAT those fixed points, he was influenced by the incorrect thought that 0°F. was the lowest temperature that could be reached. The centigrade scale introduced by the Swedish scientist Celsius, in 1742, had the fixed points of 100°C. and 0°C. boilingrepresenting, and freezing-points of pure water. respectively, -the The comparison of these two scales may be seen by reference to Fig. 12:10 and with its help we are able to convert a temperature on one scale to a corresponding temperature on the other. However, before we attempt any con- versions it should be stressed that °C. and °F. refer to actual temperatures whereas C.° and F.° refer -to a change of temperature anywhere on the scale. For example, difference between 15°G. and 15C.° is shown in Fig. 12:11. the (d) Conversion of Temperatures both scales Since temperature are legal, it is important that we be able to convert a centigrade reading into the corresponding Fahrenheit reading, and vice versa. The following examples will show how this is done. Examples 1. Convert 20°C. to a Fahrenheit reading. 20°C. is 20C.° above the freezing-point (0°C.) 100C.° = ISOF.r 100 20C.° = 20 X - = 36F.° 5 5 20°C. is 36F.° above the freezing-point (32°F.) 20°C. = (32 + 36) = 68°F. 2. Convert 14°F. to a centigrade reading. 14°F. is 18F.° below the freezing-point (32°F.) 180F.° = 100C.° lF.o=l^ = ^C." 180 9 18F.° = 18 X - = 10C.° 9 14°F. is 10C.° below the freezing-point (0°C.) 14°F. = (0 — 10) = — 10°C. The above conversions may be accomplished more conveniently by applying the following formula: not to attempt to use the formula until they have mastered the previous solu- tions. °C. =^(°F. -32) However, it is only by a study of the foregoing examples that the reasons for the various operations in the formula will be understood. Students are advised Ill ; 9 EXPANSION OF GASES You will remember from your earlier studies of science that gases expand on heating and contract on cooling (Fig. 12:12). In addition, you will recall that gases expand much more than liquids 122 EXPANSION CAUSED BY HEAT Sec. Ill: 9 and solids for a given change of temperature, i.e., they have a greater coIt may seem expansion. efficient of Fig. 12:12 Expansion ond Contraction of Gases. strange, but is nevertheless true, that all the same gases have This may be coefficient of expansion. exactly almost expressed thus: “At constant pressure, the volume of a given mass of gas Increases by 1/273 of its volume at 0°C. for each centigrade degree rise in temperature”. It should be noted that, because gases are compressible, constant pressure must be relationship prescribed this for to hold. A special use of the coefficient of expansion of gases is in determining “abIf we were provided with solute zero”. a tube containing 273 c.c. of gas at 0°C. it would contain 263 c.c. at — 10°C., 200 c.c. at — 73°C., and theoretically, 0 c.c. at — 273°C. We know, of course, that we cannot destroy matter in this way and the gas would have changed in that state before reaching this temperature. This temperature, — 273°C., is called absolute zero, a temperature at which bodies all molecular motion having ceased. More accurately, absolute zero is — 273.16°C. The lowest temperature so far recorded is .005° above absolute zero. possess no heat whatever, Absolute zero is the lowest point on another temperature scale, the Absolute or Kelvin Temperature Scale, first proposed by Lord Kelvin, a great English scientist (1824-1907). This finds application in the calculation of the volumes of gases and will be used extensively for that purpose in your chemistry course. Fig. 12:13 shows the relationship between centigrade and Kelvin tempera- 273'^C. 546°K. I00°C. -- 373°K. 0°C. -- 273°K. -273‘’C, J- 0°K. Fig. 12:13 Comparison of Centigrade and Kelvin Scales. tures. You will see that to obtain a Kelvin temperture 273 is added to the centigrade reading. 123 Chap. 12 HEAT A practical application of the effect of heat on the volume of a gas is shown in rise the pressure 12:12, the instrument is sealed so that the gas is maintained at constant volume and a rise in temperature causes a proportionate (Fig. This pressure change is read 12:14). directly in degrees. For low temperature work hydrogen or helium is used. Above 500°C. they would diffuse through the bulb and for this reason nitrogen is This device is used to used instead. calibrate thermometers. Ill : 10 THE EFFECT OF EXPANSION ON DENSITY Since changes in temperature cause changes in volume without affecting the mass, densities of substances vary with the temperature. When heat is applied, substances usually expand and a decrease Substances are said in density occurs. to be “lighter” then. The opposite effect occurs when they are cooled. A few exceptions to this rule are known, the most outstanding being water. As was shown in Sec. 1:6, water contracts when its temperature rises from 0°C. to 4°C. and thus its density increases. This is known as the anomalous behaviour of water. Above and below these temperatures water behaves normally. in the gas thermometer. Instead of allowing the gas to escape as in Fig. Ill : 11 QUESTIONS 1. 2. A (a) A threaded metal cover on a glass sealer fits too tightly. How may it be released? Explain your method, (b) What error would be introduced a into surveyor’s tape made of copper? What material is used to avoid such errors? Why? measurement by using a explain and (a) Describe happens when a bimetallic strip heated and then cooled. (b) What is purpose of the what is the 124 balance-wheel of a watch? How does it accomplish its purpose? (a) How is a centigrade thermometer scale calibrated? (b) Under what conditions is mercury preferable to alcohol as the liquid in a thermometer? Give reasons for your answers. (a) When would gases be used in thermometers? Where and when are such thermometers used? (b) What is absolute zero? (a) Define linear and cubical efficients of expansion. co- 3. 4. 5. EXPANSION CAUSED BY HEAT Sec. Ill: 11 (b) Why does increasing the temperature usually cause a decrease . in the density of a substance? B 1. Find the readings on the Fahrenheit thermometer corresponding tol5°C., 200° C, -60°C, -273°C 2 . Find the readings on the centigrade Fahrenheit thermometer scale when the reads: 100°, 350°, -220°, -50°. Fahrenheit and centigrade readings the same? (b) At the Fahrenheit reading double the centigrade reading? temperature what is 4 (a) (i) Express 57°C, — 23°C as Kelvin temperatures. (ii) Convert 298°K., 237°K. to centigrade temperatures. (b) (i) Express 98°F., 0°F. as Kelvin temperatures. (ii) Convert 373°K., 0°K. to 3 . (a) At what temperature are the Fahrenheit temperatures. 125 CHAPTER 13 TRANSFER OF HEAT is transmitted from molecule to molecule along the length of the bar until the far end becomes hot. Metals are generally good conductors of heat, some better than others. The differences in the conductivities of four different metals may be shown by performing experiment 3, chapter 15, using a conductometer similar to that shown in Fig. 13:1. The relative conductivi- ties of some common metals are shown (The figures used in the in the table. table indicate the number of heat units, calories, conducted in one second by a cube 1 cm. to the edge for each centigrade degree.) III; 12 HOW HEAT IS DISTRIBUTED In a previous chapter we studied the sources of heat energy. Here we shall learn how heat is conveyed from the source so that it may be made to go where it is required, or prevented from is not needed. When a going where it saucepan touches a hot stove it becomes warm: heat has travelled by conduction. The current of warm air above a hot radiator is carried upwards by convection. A fire-place sends out heat by radiation. Thus, the three methods of heat transfer are; conduction, convection and radiation. Ill : 13 CONDUCTION (a) Solids If one end of an iron bar is placed in a fire, the other end will soon become warm. The heat energy has been transferred along the bar by the process of conduction. The
rate of vibration of the molecules at the hot end, and therefore their energy, has been greatly increased, and this results in the molecules in sucbar acquiring cessive increased energy by the chain of colliIn this manner heat sions that results. sections the of 126 TRANSFER OF HEAT Conductivities of Some Common Substances Alcohol Petroleum Oil (a) Metals Silver Copper Aluminum Brass Iron German Silver (b) Other Solids Porcelain Glass Cork Sawdust (c) Liquids Mercury Water .97 .92 .50 .26 .16 .10 .0025 .0020 .0001 .0001 .0148 .00139 Canadian Pittsburgh Industries Ltd. One Kind of Window Construction. Fig. 13:2 Methods of Reducing Heat Loss. Sec. 111:13 .000423 .000355 .0000568 ( d ) Gases Air Metals are used for many purposes transfer heat efficiently. they because Boilers^ hot-water tanks, cooking utensils and radiators frequently employ copper because of its superior conductivity and durability. However, aluminum may be is lighter, cheaper and used because it more easily shaped. Iron, though heavier Coarse Brick 1” Air Space_ Pope, 1'll Sheathing 1 ] 'j ' 1 1 ^ Studding I f u „ ' ^ Rock Lath tM : Plaster 127 Chap. 13 HEAT and not as conductive as the others, is used for boilers because it can withstand the forces of high pressure better than the other metals. Certain other solid substances, such as china, plastic material, wood, bone, etc., are employed for their low conduci.e., their inability to pass heat tivity, from molecule to molecule readily. These substances are known as insulators. Further examples are cork used in the walls of refrigerators, rock wool, used in outer walls of homes and the wrapping of water tanks, wool for clothing and bedding, glass and porcelain food containers, sawdust in ice-houses, and many others. Some of these, because of the looseness pockets of which contribute immeasurably to their insulating properties, air being a most contain texture, their air efficient insulator. (h) Liquids When a test-tube full of water or other liquid is held at the bottom while the top is heated by a Bunsen burner is noted that while the (Fig. 13:3), it bottom remains cold the liquid above Fig. 13:3 Poor Conductivity of Liquids. may be boiling. liquids are poor conductors of heat. If the positions of the hand and burner This indicates that 128 are reversed, heat will be transferred quickly from bottom to top, but by convection currents (Sec. 111:14), not by conduction. To avoid confusion, it should be stressed that the rule about the poor conductivity of liquids does not include mercury, which, being a metal, is a good conductor. (c) Gases If the hand is held close to a Bunsenburner flame, the resulting burn is not as intense as when gripping a metal bar at the same distance from the flame. This demonstrates that air (or any gas) is a poor conductor of heat. When we recall that gases are composed of molecules that are very far apart and that heat conductivity depends on the actual contact between molecules, we understand why gases are poor conductors of heat. The above fact concerning gases has many practical applications. In part (a) of this section, reference was made to certain solid insulating materials with loose texture. Many of these, such as fur, wool, sawdust, rock wool, asbestos, snow, etc., depend on the poor conductivity of pockets of air trapped in them for a large insulating properties. Storm-windows, thermopane and the hollow construction of the outer walls of buildings (Fig. 13:2) likewise have insulating value because of the poor heat conductivity of the enclosed their part of air. Ill : 14 CONVECTION (a) Liquids (Chap. If a small crystal of potassium permanganate is dropped into a beaker of cold water heated gently by a Bunsen red burner 15, streaks will be observed as the crystal dissolves (Fig. 13:4). The streaks will rise, move just under the surface of the water for some distance, and fall. Some of the colour may be seen to return to 4A), Exp. TRANSFER OF HEAT Sec. 111:14 If we realize that its point of origin. different parts of the liquid in the beaker have different temperatures, then the streaming of the colour signifies that there are rising and falling currents in the water caused by these differences in temperature. These currents are known as convection currents and are the means by which the heat is circulated through water and liquids. The movement is established because of the expansion and consequent decrease in density of the water immediately above the source of heat. The mass of hot water is being pushed up continually and replaced by the surrounding denser water. Convection, then, is the transfer of heat in a substance by the actual, sometimes observable, motion of its parts. It should Fig. 13:4 Convection Currents in Liquids. Fig. 13:5 Applications of Convection Currents in Liquids. (a) Domestic Hot-water Supply (b) Hot-water Heating 129 Chap. 13 HEAT be clear from the nature of solids, in which the molecules occupy certain fixed positions, that convection is impossible, is possible in liquids and gases but it where the molecules have greater freedom of movement. The applications of convection currents in liquids are numerous but only a few will be mentioned. In nature, ocean drifts, which are produced by differences in the temperature of seawater, are an interesting example. The Gulf Stream is a well-known illustration. Others, such as domestic hot-water supply and hot-water heating are presented diagrammatically 13:5). Students should study these and explain how each operates. above (Fig. (h) Gases If a glass tube of large diameter is lowered over a burning candle (Fig. 13:6a), the flame will burn fitfully and then go out unless an air inlet is proIf some burning vided at the bottom. smoke-paper (blotting-paper soaked in concentrated potassium nitrate solution Hot is It heated, forcing the and dried) is held close to the bottom, the smoke will move into the tube and upward. is evident that the cool, dense air less This movement of dense air upward. the air caused by a difference in the temperatures of its various parts and the resulting difference in densities is called All gases show a convection current. this phenomenon. The apparatus used to illustrate convection currents in gases Fig. 13:6b shows the is quite varied. form used in Exp. 4B, Chap. 15. (air) The applications of convection curare commonplace rents in gases and not too intricate for the student to explain. A partial list includes ventilation, draught in a chimney, circulation of air in a refrigerator, hot-air heating and winds. Although all winds are caused by convection currents, we shall confine our attention to land- and sea-breezes (Fig. 13:7). During the day, the land warms faster than In consequence, land will be hotter than that above the water adjoining water. air above the the the (a) 130 Fig. 13:6 Convection Currents in Gases. TRANSFER OF HEAT Sec. 111:14 Night Fig. 13:7 On-shore and OfF-shore Breezes. (i.e., and a giant convection current, a seaConbreeze on-shore) versely, at night, the land will cool faster by radiation (Sec. 111:15) and the reverse in a land- situation will results. result breeze (i.e., off-shore). Hot-air heating systems (Fig. 13:8) depend upon convection currents for However, both the transfer of heat. bution of heat on cold windy days, when it is hard to heat the windward This situation is side of the building. largely corrected by the use of “forcedair” heating where a motor-driven fan This system accomplishes the transfer. has the further advantage that the air is “conditioned”, that is, dust is filtered out and the humidity is more efficiently Fig. 13:8 Hot-air Heating (a) Pipeless (b) Conduit Type (c) Forced-air. 131 Chap. 13 III : 15 RADIATION (a) Introduction HEAT tions, When you stand before a camp-fire, you are aware of its intense heat. Since the effect may be prevented by holding up a blanket between yourself and the fire, you will conclude that the energy travels in straight lines. Since the same thing happens on all sides of the fire, this energy must radiate in all direci.e., travel along the radii of a sphere with the fire at the centre. This kind of energy is a form of radiant energy and the method of transfer is It is only when this called radiation. energy strikes an object and is absorbed that it changes to heat energy. Transfer of energy by radiation is different from conduction and convection since the latter require a material medium, whereas radiation may proceed through a vacuum. For example, radiant energy from the sun traverses 93 X 10® miles of space, most of which is empty. Or again, energy may radiate from the filament to the glass envelope of an evacuated radio tube. Radiant energy is a wave-motion and has many properties in common with is the radiation light. The major heat effect comes from the infra-red radiations just beyond the red of the visible spectrum (Sec. IV: 38). Subsequent references to radiant energy in this section refer to these infra-red radiations. All bodies whose temperatures are above absolute zero (Sec. III:9) emit this kind of energy at the expense of the energy of motion of their atoms or molecules. The rate of emission and the wave-length of proportional to the temperature: the higher the temperature, the faster the rate and the shorter the wave-length. The waves are believed to be of the transverse variety which, according to one theory, are set up by a minute pulse of energy, called a quantum, from the source. These waves are a part of the great electromagnetic family of waves (Fig. 19:4) that includes visible light. X-rays, ultraviolet rays, radio waves, cosmic rays, all of which have a velocity of etc., 186,000 miles per sec. For their transmission, early physicists invented an imaginary, weightless, all-pervading medium called ether, but the Theory of Relativity proposed by Einstein denies its existence. The nature of the medium still remains a mystery. 132 Light Surfaces. TRANSFER OF HEAT Sec. 111:1
5 (h) The Emission of Radiant Energy (c) The A bsorption of Radiant In the introduction to this section, it was stated that radiant energy is released at the expense of motion of the molecules. It may well be asked whether or not all objects under the same conditions emit this form of energy. To find the answer, experiment 5, chapter 15, should be performed. For the purposes of discussion, let us take two cans (Fig. 13:9), one dark and dull on the outside, the other light and shiny, but identical in other respects. Energy As this objects absorb previously, was suggested heat results when radiant energy is absorbed. A critical thinker will want to know if energy different equally well. To answer this in part, experiment 6, chapter 15, should be performed. Another demonstration (Fig. 13:10) involves two thermometers, one darkened and dulled by the soot from a candle flame, the other left light and shiny, placed at equal distances on either side of a source of heat such as a Bunsen burner. The temperature of the one with the dark, dull surface rises more quickly than that with the light, shiny surface. We know that dark, dull surfaces absorb light without reflecting much of it and a light, shiny one reflects most of the light without absorbing much of it. In Comparing the Ability of Fig. 13:10 Dull Dark and Shiny Light Surfaces to Absorb Radiant Energy. Place a quantity of hot water and a thermometer in each and support them Although on identical insulating-bases. both are at the same temperature initially, the water in the dark, dull can cools more quickly than that in the light, shiny one. No matter how we perform such an experiment we always find that dark, dull surfaces are good emitters of radiant energy, while light, shiny ones are poor in this respect. It is admitted that other factors, such as starting temperature and area of surface also affect the rate but for the purpose of our discussion, these were kept constant. Can Pratt and Whitney Aircraft. Radial Engine of Airplane. Note cooling fins on cylinders. the same manner dark, dull surfaces are good absorbers and poor reflectors of radiant energy, while light, shiny sur- 133 Chap. 13 HEAT faces are poor absorbers and good reflectors. Knowing this we wear lightcoloured clothes in summer and dark ones in winter. (d) The Transmission of Radiant Energy are Certain materials “transparent” or “opaque” toward radiant energy just as some are toward light. As an example, ice does not transmit much radiant energy, while rock salt transmits almost all that falls upon it. Glass, on the other hand, transmits well the shorter wavelengths that originate from a high-temperature source like the sun but does not transmit the longer ones that originate from a low-temperature source such as the earth or a living object. This property of glass makes it greenhouses (part (f) below). useful in (e) Some Detectors of Radiant Energy The simplest device is the darkened air-thermometer, or thermoscope, where radiant energy is converted into molecular motion which manifests itself as a rise of temperature (Fig. 13:11a). The radiometer (Fig. 13:11b) consists of an almost completely evacuated glass bulb in which four light aluminum vanes are mounted so as to turn easily. One side of each vane is blackened while left shiny. When radiant the other is energy falls upon the vanes, the black surfaces become warmer than the others. Accordingly the few air molecules adjacent to the black sides will become heated and will move away from the vanes. The reaction of the vanes causes them to turn about their pivot. The more radiant energy that enters, the faster will the vanes turn. This instrument is very sensitive to small amounts of radiant energy. 134 Fig. 13:11 Some Detectors of Ra- diant Energy, (a) Thermoscope. (b) Radiometer. (f) Applications of Radiant Energy The vacuum or thermos bottle (Fig. 13:12) is a double- walled glass bottle, with a high vacuum between the walls, contained in a suitable protective carrying case. The inner glass walls facing each other are silvered. Liquids, whether hot or cold, will remain at very nearly the same temperature for several hours. The reason is that the bottle is so constructed that it is very difficult for heat to be transferred by any of the three methods described above. We shall consider the storing of a hot liquid here. Similar explanations obtain for a cold liquid. 1. The vacuum prevents the loss of heat by conduction owing to the lack of molecules present. The transfer of heat through the glass and the stopper is slow owing to the poor conducting property of each. 2. Convection from inside the bottle TRANSFER OF HEAT Sec. 111:15 reflected back inside. Thus, these devices act as heat traps for the energy from the sun. A further application, the screening action of the clouds, depends on the inability of water to transmit radiant upwards is prevented by the stopper. Loss by convection in the air space between the glass and the case is prevented by being closed at the top. 3. Heat loss by radiation is prevented by the silvered surfaces of the walls. These reflect back into the bottle any radiant energy that tends to escape. Greenhouses and cold frames (Fig. are heated by radiation. The 13:13) short wave-length radiant-heat energy from the sun is readily transmitted by the glass. This is absorbed by the plants, etc. within; as their temperatures soil, rise, they lose heat by radiation. Since this longer wave-length radiant energy is not transmitted by the glass it is largely Shiny Metal Cap Cork Stopper Double-Walled Glass Bottle Silvered Inside Vacuum Silvered Outside Metal Case Spring Felt Fig. 13:12 The Thermos Bottle, Fig. 13:13 A Greenhouse Acts as a "Fleat Trap". well. The moisture energy present in the atmosphere absorbs much of the sun’s heat by day, thereby preventing the scorching of plant and animal life. At night the clouds provide a blanket which prevents the escape of radiation from the earth’s surface, the temperature of which is largely maintained. On the other hand, in hot, dry, arid regions, the absence of water vapour results in extreme temperature changes, being very hot by day and very cold by night. We have included these few applications for their general appeal since they come within the realm of everyone’s experience. However, many other applications are to be found both in nature and elsewhere. It is to be hoped that with this introduction to the subject, the student will be able to recognize others as he encounters them. 135 Chap. 13 HEAT Installation for radia heating ant in building. Anaconda American Brass Ltd. III : 16 QUESTIONS 1. Name three methods of heat transfer and explain how they are involved in heating water in a kettle over an electric (b) Should the bottom of a kettle be polished for economical heating? Explain. 5. Make a chart comparing conduction, convection and radiation, under the following headings (a) the media in which the transference takes place, (b) direction of the transference, (c) a brief comparison of the theories which explain how the transference occurs. 7. 6. Explain the action of a radiometer. (a) Make a labelled diagram of a thermos bottle. (b) Write a note to show how (i) conduction, (ii) convection, (iii) radiation are reduced to a minimum when a hot liquid is placed in the bottle. heating-coil. it, in fact, 2. (a) On a cold day, why does the metal door handle feel colder than the wooden door? Is colder? (b) Name three good conductors and three good Insulators of heat, and state the use for each. (a) What are convection Explain how they are produced. (b) Explain the production of an on- currents? 3. shore breeze. 4. (a) Why does more rapidly when dirty, than when clean? snow melt 136 CHAPTER 14 MEASUREMENT OF HEAT III; 17 WHY WE MEASURE HEAT We know that heat is a form of energy (Sec. Ill :2c) and that other forms of energy can be changed into heat, but why do we bother to measure it? Were we required to determine the efficiency of an electric heater, the energy yield when a gallon of gasoline, a pound of tablespoonful coal sugar or of is a Fig. 14:1 Distinction between Quan- tity of Heat and Temperature. burned, we should be able to measure the quantities of heat produced. Various fuels and foods are used widely because of their large energy content. Hi: 18 THE COMPARISON OF QUANTITY OF HEAT AND TEMPERATURE When two equal masses of water are heated by the same source for the same length of time, each will show the same If the experiment rise in temperature. is repeated with one mass larger than the other (Fig. 14:1), the smaller mass will show a greater rise in temperature. If two unequal masses of water are heated to the same temperature by the same source, the larger mass will require to be heated for a longer time. is evident that temperature and quantity of heat are entirely different and should never be confused. It Ill : 19 FACTORS THAT INFLUENCE THE QUANTITY OF HEAT We are all familiar with the fact that a basin of hot water may be cooled by the addition of cold water and that the final temperature of the mixture will be lower than that of the hot water and higher than that of the cold. We realize that the hot water becomes cool as ft gives heat to the cold water while the cold water becomes warm because it gains heat from the hot. This is the principle of heat exchange and it applies whenever substances at different temperatures are mixed (Sec. 111:22). 137 Chap. 14 HEAT : at each different equal masses of water, Let us mix two equal masses of water, with temperatures, the same at Since the warmer water temperature. gives rise to the higher final temperature (see example), the mass at the higher temperature obviously contains the greater quantity of the heat. quantity of heat contained in a body varies as its temperature. For example: When 100 gm. of water at 80 °C. are added to 100 gm. of water at 20°C., the final temperature is 50° C. When 100 gm. of water at 40°C. are add
ed to 100 gm. of water at 20°C., the final temperature is 30°C. Therefore, Let us mix two different masses of water, at the same temperature, each with equal masses of water, at the same temperature. As the larger mass gives rise to the higher final temperature (see example), it follows that the larger mass larger quantity of heat. contains the Thus, the quantity of heat contained in a body varies as its mass. For example: When 100 gm. of water at 80° C. are added to 100 gm. of water at 20°C., the final temperature is 50°C. When 200 gm. of water at 80°C. are added to 100 gm. of water at 20°C., the final temperature is 60° C. So far, we have dealt with quantities of water in the above examples, but what would be the effect of using one different substance along with water? us mix two equal For example, let masses, one of water and one of iron filings, at the same temperature, with two equal masses of water, also at the same temperature. The water will give rise to the higher final temperature (see example), because it contains more heat than the iron. Hence, the quantity of heat contained in a body depends upon the nature of the material of which it is composed. For example: When 100 gm. of water at 80°C. are added to 100 gm. of water at 20°C., the final temperature is 50°C. When 100 gm. of iron filings at 80° C. are added to 100 gm. of water at 20°C., the final temperature is 26°C. Ill : 20 THE UNITS FOR MEASURING THE QUANTITY OF HEAT Because water is a common substance and its capacity for heat is so great, it is used as a reference material in defining the units for measuring the quanIn the metric system, the tity of heat. unit of quantity of heat is the calorie. A calorie is the quantity of heat gained or lost when the temperature of one gram of water rises or falls one centigrade degree. How many calories of heat are gained by 100 gms. of water when its Example temperature rises from 20°G. to 80°C.? Change in temperature = 80 — 20 = 60C.° Quantity of heat required to raise the temperature of 1 gm. of water 1C.° = 1 cal. 100 gm. of water 1C.° = 100 X 1 = 100 cal. 100 gm. of water 60C.° = 100 X 1 X 60 = 6000 cal. .. Quantity of heat gained = 6000 calories. 138 MEASUREMENT OF HEAT Sec. 111:21 In the British system, the unit of quantity of heat is the British Thermal Unit. One B.T.U. is the quantity of heat gained or lost when the temperature of one pound of water rises or falls one Fahrenheit degree. Example How many B.T.U. are lost when 100 pounds of water cool from 170°F. to 100°F.? The change in temperature 170 — 100 = 70F.° Quantity of heat lost by: 1 lb. of water cooling 100 lb. of water cooling 100 lb. of water cooling 70F.° = 100 X 1 X 70 = 7000 B.T.U. 1F.° = 1 B.T.U. 1F.° = 100 X 1 = 100 B.T.U. I j 1 1 Quantity of heat lost = 7000 B.T.U. Note 1. The calorie used when measuring the energy content of foods and fuels (sometimes called the kilogram calorie), is equivalent to 1000 of the calories above. 2. 1 B.T.U. is equivalent to 252 calories. Ill : 21 SPECIFIC HEAT To find the quantity of heat gained or lost by a substance other than water, we must take into account the nature of the substance as well as its mass and the change in its temperature. The calculation is done by multiplying the mass by the change in temperature by a quantity, related to the nature of the substance, called the specific heat. The specific heat of a substance is a number representing the quantity of heat gained or lost by a unit mass of substance when its temperature rises or falls one degree. In the metric system, this is the number of calories of heat gained or lost when the temperature of one gram of the substance rises or falls one centigrade degree. is the number In the British system, it of B.T.U. gained or lost when the temperature of one pound of the substance rises or falls one Fahrenheit degree. The Specific Heats of Some Common Substances Substance Specific Heat Substance Specific Heat Water Alcohol Ice Steam Aluminum 1.000 0.548 0.500 0.500 0.214 Iron Copper Silver Mercury Lead 0.110 0.092 0.056 0.033 0.031 139 : : Chap. 14 HEAT Example 1 How much heat is gained by 50 gm. of mercury when its temperature rises from 20°G. to 60°C.? Solution 1 Change in temperature = 60 — 20 = 40C.°. Quantity of heat required to raise the temperature of 1C.° = .033 cal. 1 gm. of mercury 1C.° = .033 X 50 cal. 50 gm. of mercury 50 gm. of mercury 40C.° = .033 X 50 X 40 = 66 cal. Quantity of heat gained = 66 calories. Solution 2 Change in temperature = 60 — 20 = 40C.° Quantity of heat gained = mass X change in temperature X specific heat = 50 X 40 X .033 = 66 cal. Quantity of heat gained = 66 calories. How much heat is lost by a piece of iron weighing 10 lb. when it cools from 150°F. to 70°F.? Example 2 Solution 1 Change in temperature = 150 — 70 = 80F.° Quantity of heat lost by 1 lb. of iron in cooling 1F.° = .110 B.T.U. 1F.° = .110 X 10 B.T.U. 10 lb. of iron in cooling 10 lb. of iron in cooling 80F.° = .1 10 X 10 X 80 = 88 B.T.U. Quantity of heat lost = 88 B.T.U. Solution 2 Change in temperature = 150 — 70 = 80F.° Quantity of heat lost = mass X change in temperature X specific heat = 10 X 80 X .110 = 88 B.T.U. Quantity of heat lost = 88 B.T.U. 111:22 THE PRINCIPLE OF HEAT EXCHANGE IN MIXTURES As was explained in Sec. 111:19, whenever two substances different temperatures are mixed or in contact, heat passes from the warm one to the cool one until both have attained the at same temperature. Because heat is a form of energy and energy can neither be created nor destroyed (Sec. HI: 3), it follows that the quantity of heat lost by the warm body equals the quantity of heat gained by the cool one. This is the principle of heat exchange. Example 1 A piece of lead weighing 200 gm. and at a temperature of 100°C. is placed in water and the final temperature of the mixture is 25 °C. How much heat is transferred to the water! 140 MEASUREMENT OF HEAT Sec. 111:22 M = 200 gm. Lead Change in temperature 100 - 25 = 75C.° 100°G. 25°G. Water Quantity of heat lost by the lead = mass X change in temperature X specific heat 200 X 75 X .031 = 465 cal. Heat lost by the lead = gained by the water. . Quantity of heat gained by the water = 465 cal. ‘ . Example 2 A mass of 200 gm. of mercury at 100°C. is mixed with an unknown mass of water at 20° C. and the final temperature is 25 °C. Find the mass of the water used. Mercury M = 200 gm. S = .033 Change in temperature = 100 — 25 = 75C.° 100°C. 25°G. 20°C. Quantity of heat lost by the mercury = mass X change in temperature X specific heat = 200 X 75 X .033 = 495 cal. Water M = gm. S= 1 Change in temperature 25 - 20 = 5G.° Heat lost by mercury = heat gained by the water 495 = 5x Quantity of heat gained by the water = mass X change in temperature X specific heat = a: X 5 X 1 = 5x cal. Mass of water used 99 gm. 141 Chap. 14 HEAT Stirrer Bakelite Cover Polished Aluminum Inner Vessel Fibre Ring Polished Aluminum Outer Vessel (a, (b) Canadian Laboratory Supplies Ltd. Fig. 14:2 The Calorimeter. Ill : 23 THE CALORIMETER Students will recognize at once that an operation such as is outlined in ex- 142 Fig. possible. ample 2 above cannot be conducted without some heat exchange with the surroundings. An instrument called a calorimeter (literally, heat measurer) is designed to prevent as much of this as Its component parts are seen 14:2. The small metal can, in or inner vessel, of known mass and specific heat is smooth and shiny to avoid radiation of heat from within and absorption from without. a container during experiments and the heat that it absorbs may be calculated It is supported by a fibre ring readily. which is a poor conductor of heat. The outer metal can is smooth and shiny like the inner vessel to reduce radiation and absorption. The dead air space between the cans reduces loss of heat by Loss of heat through conconduction. vection is avoided by the cover. Since is made of wood, a poor conductor of heat, the cover also prevents loss of heat by conduction. The stirrer, of the acts as It it MEASUREMENT OF HEAT Sec. 111:23 same material as is used to permit uniform mixing of the the inner vessel, contents. The materials put into the inner vessel must be so chosen that they do not interact to generate heat. Further, they must not react with the metal. The final temperatures of the materials should not be far above or below room temperature, If the final temperature for best results. is above room temperature, it should be arranged that the initial temperature be the same number of degrees below room temperature and vice versa. An example of a problem involving a a calorimeter follows: Example A mass of 200 gm. of mercury at 113°C. is mixed with an unknown mass of water at 18°C. contained in a calorimeter, the inner can and stirrer of which has a mass of 100 gm. and a specific heat of .22. If the final temperature is 23 °C., find the mass of the water. Mercury Water Calorimeter Quantity of heat lost by the mercury =: mass X change in temperature X specific heat = 200 X 90 X .033 = 594 cal. Quantity of heat gained by the water = mass X change in temperature X specific heat = V X 5 X 1 = 5.V cal. Quantity of heat gained by the calorimeter = mass X change in temperature X specific heat = 100 X 5 X .22 = no cal. Heat lost by the mercury = heat gained by the water + heat gained by the calorimeter. 5v + 110 594 5v = 594 — no = 484 X = 96.8 Mass of water required = 97 gm. 143 Chap. 14 HEAT Another form of calorimeter, the bomb calorimeter (Fig. 14:3) is used in the determination of the energy content of foods and fuels. Some typical results follow: Calorific Values of Some Common Fuels Fuel B.T.U. per lb. Fuel B.T.U. per cu. ft. Gasoline Fuel Oil Alcohol Soft Goal Hard Goal Wood (average) 20,750 18,500 11,600 14,000 11,600 5,000 III : 24 FINDING THE SPECIFIC HEAT OF A METAL The method generally employed is known as the method of mixtures. The substance whose specific heat is to be determined, say a metal, is mixed with a mate
rial which absorbs its heat, say water. From the various observations within the calorimeter the specific heat of the metal can be calculated. A brief summary of the method and a model the Full solution follow. details of Propane Acetylene Natural Gas Goal Gas 2,450 1,450 1,000 300 method are to be found in experiment 7, chapter 15. is A known mass of copper shot heated to a known temperature in a water boiler (Fig. 15:5). The metal is transferred to a certain mass of water at a known temperature contained in the inner vessel of a calorimeter. The mass and specific heat of the calorimeter are known. The mixture is stirred until the highest constant temperature is obtained. A table of data and a model calculation follow. = 100 gm. = 200 gm. Mass of the calorimeter vessel Mass of the vessel and water Mass of the water = 200 — 100 Mass of the copper shot 100 gm. = 200 gm. = 95 °G. Initial temperature of the copper Initial temperature of the water and vessel = 15°C. Specific heat of the water Specific heat of the calorimeter Final temperature of the mixture Let the specific heat of the copper = 1 = .22 = 25.5°C. = x 144 MEASUREMENT OF HEAT Sec. 111:25 Change in temperature zi: 95 - 25.5 = 69.5C.° Quantity of heat lost by the copper = 200 X 69.5 X X = 13900x cal. Copper M = 200 gm. S = a: 95°C. Water Change in temperature = 25.5 - 15 — 10.5C.° Quantity of heat gained = 100 X 10.5 X 1 = 1050 cal. Quantity of heat gained by the calorimeter = 100 X 10.5 X .22 z= 231 cal. Heat lost by the copper = heat gained by the water + heat gained by the calorimeter. 13900x = 1050 + 231 = 1281 1281 X =— = .092 13900 the specific heat of the copper shot = .092 As in all experiments, some error is unavoidable. Some heat, not accounted for in our method, will be absorbed by other parts of the calorimeter, by the thermometer and a small amount will escape by the methods of heat transfer. With care these losses are quite small. Ill : 25 APPLICATIONS OF SPECIFIC HEAT Specific heats affect our lives more than we realize. Water has the highest specific heat of all common substances Substances with a low (Sec. 111:21). specific heat undergo a great rise in temperature when a given quantity of heat is absorbed. When cooled, those same substances undergo a large drop in temperature. On the other hand, water gains or loses a great quantity of heat without much change in temperature. The high specific heat of water makes it useful in the cooling system of an engine and in automobile hot-water In each case it absorbs large heating. quantities of heat and carries it to a 145 Chap. 14 HEAT radiator to be dissipated. Because it has a higher specific heat than land, water does not reach as high a temperature during the summer season or during the day. In the winter season, or at night, water will not cool to as low a temperature as land for the same reason. Thus temperatures over water or near it will always be more moderate than inland. For example the Niagara region has a more moderate climate because of the water round it. The Prairie Provinces, on the other hand, will experience extremes of temperature since there are no moderating influences. The daily differences in temperature referred to above are also responsible for land- and sea-breezes (Sec. 111:14) in coastal reAgriculturalists know well that gions. wet soils do not warm as rapidly in because this high spring specific heat of water. Dry, sandy soils warm up more quickly, produce crops earlier and frequently yield more than one crop in a season. dry, of as Metals, as a rule, have low specific 111:21), and this makes heats them ideal for cooking utensils. (Sec. Ill : 26 HEAT EXCHANGE DURING CHANGES OF STATE gases, states three either physical intermediate in All matter is found in the solid, liquid or gaseous state. These are the of matter. Each state of matter consists of moving molecules separated from each other by spaces that vary with the state, largest liquids and in smallest in solids. The rate of motion of the molecules is faster, and the amount of space between them larger, at higher temperatures. Each state may be converted into one of the others by the addition or removal of heat. In solids and liquids, there is a force of attraction between the molecules known as the force of cohesion which must be overcome by the absorption of heat energy before a liquid or gas state may result. Fig. 14:4 shows these changes in state diagrammatically. The heat exchange during melting (fusion) can be illustrated by stirring some chopped ice or snow with a thermometer while warming it very gently over a low flame. The temperature is 0°C. when we begin and does not rise Heat Added Fig. 14:4 Changes of State. 146 MEASUREMENT OF HEAT Sec. 111:26 a wide range of temperature. We should realize also that freezing occurs at the same temperature as melting and that the same quantity of heat is released during freezing as was absorbed during melting. Moreover, when the temperature of a mass of substance is kept at its freezing-point without any change in the quantity of heat, melting and freezing are both occurring at the same rate, i.e., equilibrium will exist between the ice and the water. It is only when heat is added or removed that one or other process predominates. To illustrate heat exchange during boiling, let us heat a quantity of water from 0°G. to 100°C., stirring constantly with a thermometer. It is found that the temperature does not rise above 100°C., although heat is being absorbed continually. The heat is being used to overcome the force of cohesion rather than the but, since until all the ice has melted. Heat is being temperature absorbed, does not rise, the heat is being used to melt the ice, that is, to overcome the force of cohesion between the molecules. The temperature at which the solid becomes a melting-point. Some substances, ice, sulphur or like salt, have a definite melting-point, while others, like glass, wax and tar, melt over liquid the is Pressure (a) High Pressure Raises Boiling Point. The Pressure Cooker. (b) Low Pressure Lowers Boiling Point. A Commercial Evaporator. (c) Graph Relating Pressure and Boiling Point. Fig. 14:5 The Effect of Changes of Pressure on the Boiling Point of Water. 147 Chap. 14 HEAT and to raise the temperature. The temperaat which the water is changing ture from liquid to vapour throughout the whole mass is called the boiling-point. evaporation Boiling distinguished by the fact that vapour forms throughout the body of the liquid in the former while occurring only at the Because boilingsurface in the latter. points vary with the atmospheric presthey are expressed sure with relation to standard atmospheric 14:5), (Fig. are (760 mm. of mercury). pressure It should be recognized that the boilingpoint is the same as the temperature at which condensation occurs, and the quantity of heat released when a unit mass of vapour condenses is the same as that absorbed during vaporization. Moreover, if the temperature of a mass of liquid is kept at the boiling-point with no change in heat, vaporization and condensation take place simultaneously at i.e., equilibrium exists between liquid and It is only when a change in vapour. the quantity of heat occurs that either process predominates. the same rate, quantity of the Ill : 27 DETERMINING THE HEAT OF FUSION OF ICE In the melting of ice, the quantity of heat required to convert a unit mass to water without a change in of ice temperature is called the heat of fusion of ice. In the metric system, the heat of fusion of ice is the quantity of heat in calories required to change one gram of ice at 0°C. to water at 0°C. To determine it, the method of mixtures is employed again. A brief summary of the method, a set of typical results and a sample calculation are presented below but the method in detail will be found in experiment 8, chapter 15. Small pieces of ice that have been dried with a cloth are allowed to melt with continuous stirring in a known mass of warm water of known temperature, contained in the weighed inner vessel of a calorimeter of known specific heat. When the ice is completely melted, the final temperature is recorded, and, after weighing the vessel and contents, the quantity of ice used may be calculated. Example = 1 00 gm. Mass of calorimeter vessel = 500 gm. Mass of calorimeter and warm water Mass of calorimeter, warm water and melted ice = 592 gm. Mass of warm water = 500 — 100 = 400 gm. Mass of ice used = 592 — 500 = 92 gm. = 32 °C. Initial temperature of water and vessel = 12°C. = .22 = 1 = x Final temperature of water and vessel Let the heat of fusion of ice Specific heat of water * Specific heat of the vessel Before we attempt the calculation, let us examine the heat exchange that occurs. Since the water and the vessel cooled they lost heat. This heat did two things. First it melted the ice at 0°C. to water at 0°C. and then it warmed this water to 12°C. 148 MEASUREMENT OF HEAT Sec. 111:27 Quantity of heat lost by the water = 400 X 20 X 1 = 8000 cal. Quantity of heat lost by the calorimeter = 100 X 20 X .22 = 440 cal. Ice Water M = 92 gm. Heat of F M = 92 gm. S= 1 No change in temperature Quantity of heat gained by the ice melting = 92 X X = 92^: cal. Quantity of heat gained Change in temperature by the ice water warming = 12 - 0 = 12C.° = 92 X 12 X 1 = 1104 cal. Heat lost by the water + heat lost by the calorimeter = heat gained by the ice + heat gained by the ice-water. 8000 + 440 = 92x +1104 92^ = 8000 + 440- 1104 = 7336 92 the heat of fusion of ice = 79.7 cal. There is bound to be a small error in this experiment owing to the use of ice that was not entirely dry and to heat released by the thermometer and other parts of the equipment. With care a fairly accurate result may be obtained. The accepted value for the heat of fusion of ice is 80 calories per gram. In the British system, the heat of fusion of ice is 144 B.T.U. per pound. It is the quantity of heat required to change a pound of i
ce at 32°F. to water at 32°F. 149 Chap. 14 HEAT Some Typical Heats of Fusion and Melting-Points Substance M.-P. rc.) H. of F. ( Cal. per gm.) Ice Aluminum Copper 0 660 1083 80 77 42 Substance Lead Cast Iron Mercury M.-P. (^C.) 327 1230 -39 H. of F. ( Cal. per gm.) 6 5.5 3 III : 28 THE IMPORTANCE OF THE HEAT OF FUSION OF ICE When one gram of ice melts without a change in temperature, 80 calories of heat are absorbed. This is enough heat to raise the temperature of one gram of water at 20°C. to the boiling-point, or of 80 grams of water through one centigrade degree. Since such a large quantity of heat is required to melt ice, it is not difficult to understand why ice is useful in preserving food. There are important consequences of this large heat of fusion in nature. Icebergs float long distances before absorbing enough heat to melt. Large snowfalls melt slowly and disastrous floods are avoided. Since ice on lakes and streams melts slowly in spring, sudden extreme changes in temperature do not occur. On the other hand, the heat released when water freezes is useful. Tubs of water are set near fruits and vegetables in unheated basements when there is a chance of frost damage. As the temperature drops, the water freezes and the heat released prevents the fruits from freezing. The heat released when lakes and rivers freeze has a moderating effect and prevents extremes of temperature. This helps to make the climate of regions like Southern Ontario milder than regions more distant from large bodies of Furthermore, the weather genwater. erally becomes milder before or during a snowstorm because of the heat released when the water vapour changes to solid. 150 These examples and many more may be cited to show the usefulness of the heat of fusion of ice. Ill : 29 DETERMINING THE HEAT OF VAPORIZATION OF WATER the called In boiling water, the quantity of heat required to convert a unit mass of water at its boiling-point to steam without a change in temperature is heat of vaporization of water. In the metric system, the heat of vaporization is the quantity of heat required to change one gram of water at 100°C. to steam at 100°G. (atmospheric pressure being 760 mm. of mercury). Experiment 9, chapter 15, describes the method used to determine it. A brief summary of the method, some typical results and a set of calculations follow. required of heat The quantity to vaporize a unit mass of water at its boiling-point is the same as that released when a unit mass of steam condenses. Since the latter is more easily determined, the method involves using it rather than the former. Hence, steam from a boiler is passed through a steam trap to free it of water (It is now called “dry” or “live” steam). It is then conducted into a quantity of cool water of known mass and temperature contained in the weighed inner vessel of a calorimeter of known specific (Fig. 14:6). The water is continually stirred and, shortly, the final temperature is recorded. The vessel and contents are weighed to find out the weight of steam used. heat MEASUREMENT OF HEAT Sec. 111:29 Mass of inner vessel of calorimeter = 100 gm. = 500 gm. Mass of vessel and cool water Mass of vessel, water and condensed steam = 521.7 gm, Mass of cool water = 500 — 100 = 400 gm. = 521.7 — 500 = 21.7 gm. Mass of steam = 5°C. Initial temperature of water and vessel = 36°C. = .22 = 1 = X Let the heat of vaporization of water Specific heat of the vessel Specific heat of water Final temperature Let us analyse these results before proceeding. The heat absorbed by the water and the vessel originated not only from the condensation of the steam, but also from the cooling of the resulting water from 100°C. to 36°C. We must recognize these two sources of heat in our calculations. Steam Water formed by condensa- tion Cool Water Calorimeter M = 21.7 gm. HofV M = 21.7 gm. S = 1 No change in temperature Quantity of heat lost by the steam condensing = 21.7 X x^2\Jx cal. Change in temperature = 100 — 36 = 64C.° Quantity of heat lost by the resulting water cool- ing = 21.7 X 64 X 1 = 1389 cal. 100°C. 36°G. 5'^C. M = 400 gm. S = 1 M = 100 gm. S = .22 Change in temperature = 36 — 5 = 31C.° Quantity of heat gained by the cool water = 400 X 31 X 1 = 12400 cal. Quantity of heat gained by the calorimeter = 100 X 31 X .22 = 682 cal. 151 Chap. 14 HEAT The heat lost by the steam + heat lost by the resulting water = the heat gained by the cool water + heat gained by the calorimeter. 21.7a: + 1389 = 12400 + 682 21.7a: = 12400 + 682 — 1389 = 11693 11693 ^ =— = 538.8 .’. the calculated heat of vaporization of water is 539 calories. The accepted value is 540 calories. In the British system, the heat of vaporization of water is 972 B.T.U. per pound and is the quantity of heat required to change a pound of water at 212°F. to steam at 212°F. Heats of Vaporization and Boiling-Points of Various Useful Materials Substance Water Ethyl Alcohol Ethyl Ether Chloroform Ammonia Methyl Chloride Sulphur Dioxidfe Freon 12 B.-P. rc.) 100 78 35 61 — 33 — 24 — 10 — 30 H. of V. ( Cal. per gm.) 540 204 84 59 327 102 95 41 III : 30 APPLICATIONS OF HEAT OF VAPORIZATION (a) Water of heat The high vaporization of water is of great practical importance in nature. Evaporation of soil water is slow because of the vast quantity of heat required for this purpose. Thus extremes of drought and torrential that would attend excessive evaporation are avoided. rains Steam-heating, as used in most large ex- is an excellent buildings, public 152 Fig. 14:6 Determining the Heat of Vaporization of Water. ample of heat of vaporization at work. Water is boiled; the steam is conducted to radiators where it condenses and yields its heat of vaporization. The hot water now flows back to the boiler to be used again. This system is cheaper than hotwater heating (Sec. 111:14) as smaller radiators are required and the steam pipe itself often serves as a water return. Cooling is frequently accomplished by evaporation. The faster-moving molecules of the liquid escape first and the Further evaporation temperature falls. . MEASUREMENT OF HEAT Sec. 111:30 occurs only if more heat is absorbed from its surroundings. This is the principle involved in the use of perspiration to regulate body temperature, the sprinkling of city streets with water on hot summer evenings, and the alcohol rub to reduce fever. same principle. The chief requirement is a gas which can be liquefied without difficulty and which has a high heat of vaporization. Some gases of this kind are ammonia, sulphur dioxide, methyl chloride, and—the most widely used—-freon 1 2 ( dichlorodifluorom ethane ) (b) Refrigerators The refrigerator makes use of the a Fig. 14:8. The parts of a refrigerator are shown In it, a small electric in compressor which motor operates forces the gas into the coils of a conHere it condenses because of denser. the pressure and cooling (See Faraday’s experiment below) . A small fan is often provided for cooling purposes, though loss of heat by radiation is frequently sufficient. The liquid proceeds to the cooling-unit where it vaporizes, removing Graph Summarizing Heat Fig. 14:7 Exchanges for the Changes of State of Water. Fig. 14:8 The Operation of an Elec- tric Refrigerator. 153 Chap. 14 HEAT heat from the food compartment. The gas now returns to the compressor to Ice-making machines, be recirculated. silent automatic. The gas refrigerator (Fig. 14:9) involves no moving parts and is, conseThe and quently, sealed tubes contain water, ammonia and hydrogen. Ammonia, which is ordinarily very soluble, is expelled from the water by the heat of the flame. A combination of the pressure created and the simultaneous cooling by the air causes amIt now proceeds to monia to liquefy. absorbs the evaporates, cooling-unit, the water and is heat, ready to repeat the process. Hydrogen is used to prevent the formation of a vacuum which otherwise would be caused by the rapid dissolving of the 'ammonia. This avoids “pounding”. The refrigerator is simple in its principle of operation but its structure is somewhat more complicated than this explanation would lead one to believe. redissolves in Courtesy of Servel (Canada) Ltd. Fig. 14:9 The Operation of a Gas Refrigerator. quick-freeze units etc., all work on the same principle. (c) The Liquefaction of Gases learn more later Michael Faraday, about whom we (Sec. V:57), shall devised an ingenious method of liquefying gases. He filled a thick-walled glass tube of the type shown in Fig. 14:10 with chlorine gas. One end of the tube was surrounded by a freezing mixture of Fig. 14:10 (a) Illustrating Faraday's Method of Liquefying Gases, (b) Producing Cold Artificially. 154 MEASUREMENT OF HEAT Sec. Ill: 30 ice and salt. Heating the gas in the other end caused a rise in pressure while the gas liquefied in the cool end. In this way he was also able to liquefy several other gases. This principle is employed in the gas refrigerator above. methyl chloride and freon 12 are said to be easily liquefied at ordinary tem- peratures. When a compressed gas is allowed As an exto expand, cooling results. ample, the air escaping from an inflated oxygen, hydrogen, Faraday found that certain gases, for example, nitrogen, air and many others, could not be liquefied in this way. It was found that these gases had to be cooled to a certain temperature, called the critical temperabefore any amount of pressure ture, would liquefy them. The pressure required to liquefy the gas at this temperature is known as the critical pressure. It should be noted that when the operating temperature is lower than the critical temperature, the quired for liquefaction pressure reis very much lower than the critical pressure. not difficult to understand, then, why dioxide, ammonia. the sulphur gases It is Some Critical Temperatures AND Pressures Substance C.T. C.P. (°C.) (Atm.-^) 78 132 143 112 157 144 76 66 112 40 Sulphur Dioxide Chlorine Methyl Chloride Ammonia Freon 12 Carbon Dioxide Oxygen Air Nit
rogen Hydrogen Helium *One atmosphere pressure = 760 m.m of mercury. - 119 — 141 — 147 - 240 - 268 73 50 34 37 13 31 2 ABC High Pressure D E F Low Pressure Fig. 14:11 The Production of Liquid Air. 155 Chap. 14 HEAT automobile tire feels cold. In liquefying a gas of low critical temperature, the gas is subjected to the cooling effect of an expanding cold gas around it, in order to reach the low temperature required. This principle is seen in opera- tion in the liquefaction of air (Fig, 14:11). The air is first compressed and then to of the through needle-valve circulated through cooling-coils remove the heat of compression. It passes on the liquefier where it expands, cools, and returns to the compressor by way of a pipe which surrounds the one that carAs these ried it processes are repeated continuously, the gas is soon cooled to its critical temperature and hence it liquefies. to the needle-valve. Liquid air at atmospheric pressure boils at between — 182°G. and — 194°C., the variation depending on the percentIt is so cold tage of nitrogen present. that it will boil when in contact with convert mercury (F.-P. ice and will — 39° C.) into a solid so hard that it can be used as a hammer. It is drawn off into open Dewar flasks of the same construction in which it may be kept for a short time. On standing, the nitrogen will boil off first leaving a pale blue liquid which is largely oxygen. “thermos” bottles as Liquid air is used in industry as a source for the very low temperature needed for causing metals to contract. It is a source of industrial nitrogen used for preparing nitrogen compounds. From it may be obtained oxygen used for welding, hospitals, aviation etc., together with the inert gases which are used in neon signs, argon bulbs and the like. Liquid carbon dioxide is prepared in a similar way. When this liquid is allowed to escape from a valve, its evaporation removes heat from the surroundings and from the escaping liquid converting it to “dry” ice. This solid carbon dioxide 156 a temperature of — 80° C. and has changes to a gas without melting (sublimation), hence the apt use of the word “dry”. As the heat absorbed in sublimation (87 cal. per gm.) is large and only a small amount of solid required for a considerable cooling effect, is used as a refrigerant in shipping. it It saves freight and keeps foods at such a low temperature that is no danger carbon dioxide is also used for fire-extinguishers. spoilage. Liquid there of is 111:31 THE RELATIONSHIP BETWEEN HEAT AND WORK Ancient civilization accomplished work, such as building the Pyramids, by utilizing slave-labour. In recent times various machines have been developed which convert heat into work, so releasing man from much drudgery. be converted Man has known since the time of that Rumford and Davy (Sec. Ill: 2) work could heat. into Later it was shown that heat can be made to do work. When a pound of coal burns, for example, enough heat (14,500 B.T.U.) is generated to raise a ton of material through 5,000 feet. In practice, only a fraction of the energy from fuel is available to do useful work in an engine because of losses through radiation, conduction, convecetc. The small fraction tion, actually available compared to the total possible amount of energy is called the follow efficiency. show the efficiency of various kinds of engines. The examples that friction Efficiencies of Heat Engines Locomotives Steam turbine, condensing steam- 6-8% engine 16-30% 22-28% Gasoline (automobiles, aircraft) Diesel (railway, marine, trucks, buses) 32-38% G.M Corp. Heat engines are of two types (a) the external combustion engine of which MEASUREMENT OF HEAT Sec. 111:31 Eccentric Live Steam from Boiler Slide Valve Steam Chest Crank Bearing II ! i Flywheel Fig. 14:12 The Reciprocating Steam Engine. the steam-engine is an example, and the internal-combustion engine as (b) in automobiles and aircraft. In the former the fuel burns outside of the engine proper while in the burns inside the cylinder. latter it The Steam-Engine In the common steam- or “reciprocating”-engine, the pressure exerted by live steam (dry and much hotter than pushes the piston back and 100°C.) forth in the cylinder. The back-andforth motion is made possible by an automatic slide-valve which first admits the steam at one end of the cylinder, thus driving the piston to the opposite end, and then admits the steam at this end, thus forcing the piston back to its This original motion rotates the fly-wheel which, being heavy, enables the engine to operate 14:12). position (Fig. smoothly. The Steam Turbine It consists of a specially designed paddlewheel against which steam at high pressure is directed through nozzles (Fig. Fig. 14:13 The Steam Turbine. (a) Principle of the Steam Turbine. This type of steam-engine is widely used in power-plants and large ships. 14:13). The force of the vapour rotates the paddle-wheel. are These engines 157 @ CONT«Ot «O0M (?) STATION SERVICE TRANSfORMER © ELECTRIC' GENERATOR © CONDENSER ©TURBINE ©BOOSTER PUMP © FEED PUMP ©STEAM GENERATOR CONTROL ©CONDENSATE PUMP ©STATION SERVICE SWITCHBOARD @ HEATERS ©COAL FEEDER @ TRIPPER © SCALES @ PULVERIZER © STEAM-GENERATOR @ COAL CONVEYOR © STEAM LINE © FORCED DRAFT FAN © AIR INTAKE ® IN OUCED DRAFT FAN ® MECHANICAL OUST COLLECTOR ® ELECTROSTATIC PRECIPITATOR © CRANE Fig. 14:13 (b) How a Steam Generating Station Works. Ontario Hydro Richard L. Hearn Generating Station. Ontario Hydro 158 MEASUREMENT OF HEAT Sec. 111:31 Fig. 14:14 The Four Cycle Internal Combustion Engine. A Sectional View. possible. designed to use the pressure as efficiently They run at very high as speeds and are more efficient and smaller than ordinary steam-engines of the same capacity. The Internal-Combustion Engine If a compact, powerful, mobile source of power is required, the internal-com- bustion engine is the choice. This type of engine can be adapted to the use of any fuel that can be vaporized, such as Coal gasoline, gas is used in some engines, while the Diesel type employs cheap petroleum alcohol, and kerosene. oils. The Gasoline Engine The type generally used is the fourcycle engine, so named because the piston makes four strokes for each explosion of gas in the cylinder. Once started it will run automatically as long as the three necessities of fuel, compression and spark are met. Smoothness of operation is accomplished by the use of a heavy flywheel and of several cylinders which fire Students will have at different times. understanding engine after studying Fig. 14:14. Further treatment than this is beyond the scope of this text. difficulty little this in Chrysler Corporation A Cut-Away Photograph of a Modern Internal Combustion Engine. The Diesel Engine This engine operates like a four-stroke gasoline engine but is without carburetor or electrical ignition system. Air is 159 Chap. 14 HEAT to about one-sixteenth of forced into the cylinder and is compressed its volume and for that reason becomes hot. When oil is forced into this hot gas, it burns, without any need for a spark. The Jet Engine The last few years have seen the very rapid development of a new type of internal combustion engine, the Jet Engine. There are several types of such engines; the commonest, however, is the TurboJet. Air is scooped into the intake at the front of the engine. It is compressed, and consequently heated, by a compressor. This heated air is driven under high pressure into the combustion chamber where fuel is injected in and combustion occurs. The hot expanding gases stream away at a high velocity. A small portion of their energy is used to drive the turbine Most which operates the compressor. of the energy is in the stream of hot gases which is ejected from the rear of the engine. The force exerted by this jet creates an equal of hot gases (action) and opposite force (reaction) that drives the plane forward. Rockets are another type of modern reaction engine that operate very similarly to the jet engine described above. They differ in that they carry their own supply of oxygen to burn the fuel. As a result rockets can travel through outer space where there is no air. Inter-planetary travel, a thing long dreamed about, now seems to be becoming a real pos- sibility. III : 32 QUESTIONS A 1. (a) Distinguish between quantity of heat and temperature. (b) State the factors that govern the quantity of heat possessed by a body. 2. (a) Define: calorie, British Thermal Unit, specific heat. (b) What quantity of heat is needed to: (i) warm 25 gm. of water from 13°C. to 27°C.? (ii) heat 37 lb. of water from 68°F. to 212°F.? 3. of 25 copper 22°C. gm. from (iii) heat (S.H. = .092) 100°C.? (a) What is the exchange in mixtures? (b) How long does heat exchange continue between two substances in principle of heat to contact? (c) What is the purpose of a calorimeter? How does it fulfil its purpose? 160 4. (a) Describe an experiment to find the specific heat of a metal. (b) When a 200 gm. mass of metal at a temperature of 85°C. is immersed in 300 gm. of water at 30°C., the final temperature is 33°C. Calculate the specific heat of the metal. 5. (a) Distinguish between boiling and evaporation. (b) Define melting-point and boiling- point. 6 . (a) Define heat of fusion of ice and state its numerical value in the metric and British systems. 7. (b) Describe how you would determine its value experimentally. (a) How much heat will be released when 50 gm. of water at 0°C. freeze to Ice at 0°C.? (b) How much heat will be absorbed in the melting of 20 gm. of ice at 0°C. to water at 0°C.? (c) How much heat will be required MEASUREMENT OF HEAT Sec. 111:32 to convert 80 gm. of ice at 0°C. to water at 25°C.? 8. (a) Define heat of vaporization of 9. 10. that from water. (b) Why is a burn from steam much more severe hot than water? (a) How much heat will be required to convert 50 gm. of water at 1 00°C. to steam at 1 00°C.? (b) How much heat will be released when 1 5 gm. of steam at 1 00°C. are condensed to
water at 100°C.? (c) How much heat will be released when 35 gm. of steam at 100°C. are condensed to water and the water is cooled to 20°C.? 8. When 25 gm. of water at 100°C. are added to 50 gms. of water at 10°C., what is the final temperature? 9 . When 200 gm. of metal at 100°C. are placed in 200 gm. of water at 1 5.0°C., final temperature becomes 23.0°C. the Calculate the specific heat of the metal. 10 . A brass kilogram weight at a tem- of 90.0°C. is submerged perature in 440 gm. of water at 10.0°C. The final temperature is 24.0°C. Find the specific heat of the brass. 11. 49 gm. of water at 1 3°C. are contained in an aluminum calorimeter weighing 50 gm. If 35 gm. of glass at 87°C. are dropped Into the calorimeter the temperature becomes 21°C. Find the specific heat (a) Explain the principle of operation of the glass. of the electric refrigerator. (b) How is air liquefied? B 1 . How much heat is required to raise the temperature of 2 kg. of water from 25°C. to 75°C.? 2 . How many B.T.U. will be absorbed when 30 gallons of water in a hot-water tank are heated from 70°F. to 200°F.? (1 gallon of water weighs 10 lb.) 3 . How much heat is lost when 1 .3 kg. of water are cooled from 90°C. to 20°C.? 4 . How much heat in will be released when 15 gallons of water cool from 1 65°F. to 1 25°F.? B.T.U. 5 . How many calories of heat must be supplied to heat 200 gm. of cast iron from 20.0°C. to 80.0°C.? 6. How much heat does a silver spoon weighing 30.0 gm. absorb when placed in a cup of coffee that raises its temperature from 20.0°C. to 80.0°C.? 7 . How many grams of water at 85. 0°C. must be added to 100 gm. of water at 10.0°C. to give a final temperature of 37.0°C.? 12 . In an experiment, 500 gm. of lead at 100°C. are placed in 100 gm. of water at 14°C. contained in a copper calorimeter weighing 80 gm. The final temperature is 25°C. Find the specific heat of the lead. 13 . What mass of iron at 90.0°C. when added to 200 gm. of water at 1 5.0°C. contained in a copper calorimeter weighing 100 gm. will give a final temperature of 25.0°C.? 14 . When 400 gm. of silver at 100°C. was placed in water at 1 6.0°C. contained in an aluminum calorimeter weighing 40.0 gm. the final temperature was 24.0°C. What mass of water was used? 15. Calculate the final temperature when 120 gm. of iron at 100°C. are added to 400 gm. of water at 10°C. in a copper calorimeter having a mass of 80 gm. 16 . A copper calorimeter weighing 65 gm. contains 30 gm. of turpentine at 15°C. When 45 gm. of Iron at 98°C. are placed temperature becomes 32°C. Calculate the specific heat of the turpentine. 17. How much ice af 0°C. can be melted the it, in by 1 kg. of water at 100°C.? 18. How much ice at 0°C. will be required 161 Chap. 14 HEAT to cool 1 kg. of drinking-water from 15°C. to 0°C.? ture is 27°C. Find the heat of vaporization of water. 19 . What mass of ice at 0°C. will be required to cool 750 gm. of water from 35°C to 10°C? 20 . When 5,0 gm. of ice at 0°C. are melted in 30 gm. of water at 25°C. the final temperature is 10°C. Find the heat of fusion. 21. A copper calorimeter weighing 55 gm. contains 90 gm. of water at 25°C. When 15 gm. of ice at 0°C. are melted in the water the resulting temperature is 1 1 °C. Find the heat of fusion of ice. 22 . In an experiment 203 gm. of water at 40°C. are contained in a calorimeter vessel weighing 50 gm. having a specific heat of .22. After ice at 0°C. was melted in the water the temperature became 25°C. and the mass 233 gm. Find the heat of fusion of ice. If 15 gm. of ice at — 20°C. are 23 . melted in 50 gm. of water at 40°C. and the resulting temperature is 10°C., cal- culate the specific heat of ice. 24 . How many grams of water can be freezing-point raised boilingpoint by the condensation of 5.0 gm. of steam? from to 25 . To what temperature will 75 gm. of water at 25°C. be heated by the condensation of 3.0 gm. of steam? 26 . When 6.6 gm. of steam at 1 00°C. are passed into 1 80 gm. of water at 6.0°C. contained calorimeter weighing 45 gm., the resulting tempera- aluminum an in 27 . If 15 gm. of steam at 100°C. are added to 150 gm. of water at 20°C. in a calorimeter (S.H. = 0.10) weighing 75 gm. the final temperature is 74°C. Calculate the heat of vaporization of water. 28 . When 160 gm. of water at 7.7°C., contained in an aluminum calorimeter weigh1 26.2 gm., are heated by the condensation of 10.1 gm. of steam, the final temperature is 39.8°C. Find the heat of vaporization of water. ing the Find final 29 . Find the resultant temperature when 8.0 gm. of steam at 100°C. are passed into a vessel of negligible mass containing 40 gm. of ice at 0°C. 30 . temperature when 20 gm. of steam at 1 00°C. are passed into 240 gm. of water at 10°C. contained in an aluminum calorimeter weighing 150 gm. 31. What quantity of heat will convert at — 1 6°C. to steam at 5 gm. of ice 100°C.? 32 . When 45.0 gm. of iron at 95.0°C. are placed in a cavity in a block of ice at 0°C. and the temperature has dropped to 0°C., 6.0 gm. of ice are melted. Knowing the heat of fusion of ice, find the specific heat of iron. 33 . An ice-water mixture weighing 200 gm. is contained in a calorimeter weighing 100 gm. (S.H. = 0.20). When 35 gm. of steam at 100°C. are added the temperature becomes 50°C. Calculate the mass of ice used. 162 CHAPTER 15 EXPERIMENTS IN HEAT INTRODUCTION Before commencing these experiments in heat, students should be familiar with the following. A. The Use of the Bunsen Burner 1. Structure of the Burner Examine a Bunsen burner and identify the gas inlet, the orifice, the air-inlet valve, the mixing tube. Make a labelled diagram of the burner. 2. Lighting the Burner (a) Close the air inlet, turn on the gas and ignite it. Gradually open the air inlet until you have the desired flame. If the flame “strikes back”, i.e., burns at the bottom of the (b) mixing tube, turn off the gas and repeat (a) above. 163 Chap. 15 HEAT 3. Regulating the Size and Temperature (Colour) of the Flame (a) The size of the flame may be changed by increasing or decreasing the supply of gas. (b) When first low for most purposes. When the air inlet is gradually opened, the colour changes from yellow, through blue to nearly colourless, and the temperature increases until it is a maximum at the last stage. lit, the flame is yellow and its temperature is too 4. The Structure of the Flame Fig. 15:2 shows the various regions (cones) of a Bunsen flame. An object to be heated is held just above the turquoise cone. B. The Use of the Thermometer 1. Examine the instrument and find the centigrade scale. (Some ther- mometers have centigrade and Fahrenheit scales.) 2. Wait until the liquid has come to rest before taking a reading. 3. Adjust the thermometer so that you are able to view the top surface of the liquid at right angles in order to avoid the error of parallax. 4. Take readings to a fraction of a degree. EXPERIMENT 1 To study the expansion of solids, (Ref. Sec. III:5) Apparatus Bunsen burner, ball-and-ring apparatus, cold water. Method 1. Try to pass the ball through the ring when both are cold. 164 EXPERIMENTS IN HEAT 2. Heat the ball strongly and try to pass the ball through the ring again. 3. Cool the ball by placing it in cold water and again try to pass it through the ring. Observations What do you observe in the above steps? Conclusion What is the effect of heating and cooling on the volume of a solid? Questions 1. What would be the probable effect of heating the ring and trying to pass the heated ball through it? 2. What would be the probable effect of chilling the ring in a freezing mixture and trying to pass the ball through it? 3. By means of labelled diagrams, show how you would demonstrate the effect of heating and cooling on the volume of (a) a liquid, (b) a gas. EXPERIMENT 2 To compare the expansion of different metals when heated. (Ref. Sec. Ill: 5) Apparatus Bunsen burner, compound bar consisting of strips of copper and iron fastened together, cold water. Method 1. Heat the long straight compound bar in the Bunsen-burner flame and note any change. 2. Cool the bar in cold water and again note the change. Observations 1. Describe the changes that occurred in parts 1 and 2. 2. Which metal was on the outside of the curve? Explanation Account for the changes. Conclusion What is the relative amount of expansion and contraction that occurs when different metals undergo the same change in temperature? Question Make a labelled diagram to show how to use a compound bar as a thermostat to control an oil burner and thus regulate the temperature in a room. 165 Chap. 15 HEAT EXPERIMENT 3 To study the transfer of heat by conduction. (Ref. Sec. 111:13) Apparatus Bunsen burner, metal rod about 12 in. long with a wooden handle, a conductometer (Fig. 13:1), wax. Method 1. Place drops of wax at three-inch intervals along the length of the long metal rod. Hold it by the wooden handle and heat the end of the rod strongly in the flame of the burner. 2. Place drops of wax at equal intervals along the rods of the conductometer and heat the metals simultaneously at the point where the rods meet. Observations 1. (a) What happens to the wax in part 1 ? (b) Is there any change noted in the temperature of the wooden handle? 2. State the differences observed in each metal rod in part 2. Conclusions 1. State the meaning of the term conduction of heat. Explain how it occurs. 2. What have we learned about the heat conductivity of different metals? How does wood compare with metals in this regard? 3. List the metals studied in order of their relative heat conductivities. Questions 1. How do metals compare with other substances in heat conductivity? 2. What use is made of the conductivity of heat through metals? 3. Why are liquids and gases very poor conductors of heat? 4. List some materials that are good insulators and state where they are used for this purpose. EXPERIMENT 4 To study the transfer of heat by convection. (Ref. Sec. 111:14) A. IN LIQUIDS Apparatus A large beaker, reto
rt stand, ring, Bunsen burner, cold water, potassium permanganate. Method Fill the beaker with water and place it on the ring attached to the stand. Make sure that the water is at rest. Drop a crystal of potassium 166 EXPERIMENTS IN HEAT permanganate into the water near the edge. Using the tip of the low Bunsen flame, heat the liquid beneath the crystal. Observations Describe all phenomena. Explanation Account for the changes observed. B. IN GASES Apparatus A candle, smoke-paper, convection apparatus (Fig. 13; 6b). Method Light the candle and place it beneath one of the chimneys. Close the Hold a piece of lighted smoke-paper above the other glass front. chimney. Observations State what occurs. Explanation Account for these results. Conclusion State the meaning of the term convection currents and explain how they occur. Question Show by means of a diagram how heat is transferred from the furnace to an upper room in a home heated by (a) a simple hot- water system (b) a simple hot-air system. EXPERIMENT 5 To compare the abilities of dull/ dark and shiny/ light surfaces to emit radiant energy. (Ref. Sec. Ill; 15(b) Apparatus A differential thermometer with both bulbs blackened with soot from a candle, metal vessel with one side blackened and the other polished, some boiling water. 167 Chap. 15 HEAT Method Mark the levels of the coloured liquid in the arms of the differential thermometer and then place the metal vessel full of boiling water midway between the bulbs. Note any change in the levels of the coloured liquid. Observations What is observed? Explanation Account for your observations. Conclusion What effect has the nature of the surface of an object on its ability to emit radiant energy? Questions 1. Why is this apparatus called a differential thermometer? 2. Why should tea-pots be shiny rather than dull? EXPERIMENT 6 To compare the abilities of dull, dark and shiny, light surfaces to absorb radiant energy. (Ref. Sec. Ill; 15(c) Apparatus A differential thermometer with one bulb shiny and one blackened, dull, dark metal vessel, supply of boiling water. Method Mark the levels of the coloured liquid in the differential thermometer and place the vessel filled with boiling water midway between the two bulbs. Note any changes in the levels of the liquid. Observations What is observed? Explanation Account for the observations. 168 EXPERIMENTS IN HEAT Conclusion What effect has the nature of the surface on its ability to absorb radiant energy? Questions 1. In experiment 5, why were the bulbs of the differential thermometer darkened? 2. In experiment 6, why is a dull, dark vessel used? 3. Why do people wear dark clothing in winter and light-coloured clothes in summer? 4. Examine a radiometer and make a labelled diagram of it. Note and explain what happens when a source of radiant energy is brought near it. EXPERIMENT 7 To determine the specific heat of a metal, (Ref. Sec. 111:24) Apparatus A quantity of copper (or lead) shot, balance and weights, flask, testtube, water, retort stand, ring, gauze, Bunsen burner, calorimeter, two thermometers. Method 1. Fill the test-tube three-quarters full of shot and carefully insert the Place the shot and its container in bulb of a thermometer into it. the flask and boil the water while carrying out parts 2, 3 and 4. Note the temperature of the shot to within a fraction of a degree after the mercury stops rising. 2. Find the mass of the inner vessel of the calorimeter and stirrer. Record the specific heat of the metal of which both are made. 3. Place about 100 ml. of cold tap-water whose temperature is slightly lower than room temperature in the inner vessel. Find the mass of the vessel and water, and determine the mass of the water. 169 Chap. 15 HEAT 4. Place the inner vessel and contents into the outer vessel. Cover with the lid. Stir the water and take its temperature. 5. Open the calorimeter, add the shot to the water, close it, and after stirring the mixture, again take its temperature. 6. Find the mass of the vessel, stirrer, and contents, and determine the mass of the shot. Observations = 1. Temperature of the shot 2. Mass of the inner vessel and stirrer == = 3. Mass of the inner vessel, stirrer and water = 4. Initial temperature of the vessel and water 5. Final temperature of the mixture of water and shot = 6. Mass of the vessel, stirrers and mixture =: = = = Specific heat of the vessel Specific heat of the water Let the specific heat of the shot x Calculations Study the worked example in Sec. 111:24 and calculate the specific heat of the metal from the observations recorded above. Conclusions 1. What is your experimental value? 2. What is the class average? Questions 1. What is the percentage error? 2. What are the sources of error in this experiment? 3. Why is it desirable to have the initial temperature a few degrees lower than room temperature? EXPERIMENT 8 To determine the heat of fusion of ice. (Ref. Sec. 111:27) Apparatus Quantity of ice, paper towels, thermometer, balance and weights, quantity of water at about 10C.° warmer than room temperature, calorimeter. Method 1. Find the mass of the inner vessel and stirrer. 2. Place about 100 ml. of the warm water in the vessel and again find the mass. 3. Place the inner vessel in the outer one. Cover with the lid. Stir the water and take its temperature, estimating it to a fraction of a degree. 170 EXPERIMENTS IN HEAT 4. Wipe dry about 25 gm. of ice with the paper towels and quickly drop it into the warm water. Replace the cover and stir until the ice has melted completely. Record the lowest temperature reached by the water. 5. Find the mass of the vessel, stirrer, and contents, and determine the mass of the ice. Observations = 1. Mass of inner vessel and stirrer = 2. Mass of inner vessel, stirrer and water 3. Initial temperature of the water and vessel = 4. Final temperature of the mixture of original water and melted ice 5. Mass of vessel, stirrer and the mixture Specific heat of the vessel Specific heat of the water Initial temperature of the ice Let the heat of fusion of ice = =: = = = = x Calculations Study the worked example in Sec. 111:27 and calculate the heat of fusion of ice from the observations recorded above. Conclusions 1. What is the experimental value for the heat of fusion of ice? 2. What is the class average? Questions 1. What is the percentage error? 2. What are the sources of error in this experiment? 3. Why is it desirable to have the initial temperature a few degrees higher than room temperature? 4. Why do we dry the ice before placing it in the water? 171 Chap. 15 HEAT EXPERIMENT 9 To determine the heat of vaporization of water, 111:29) (Ref. Sec. Apparatus A calorimeter, quantity of water at about 15C.° below room temperature, thermometer, Bunsen burner, retort stand, ring, gauze, steamboiler, steam-trap, rubber connectors (Fig. 14:6). Method Place the boiler containing water on the ring of the retort stand and heat it. 1 . Find the mass of the inner vessel and stirrer. 2. Put about 100 ml. of the cold water into the inner vessel and find the combined mass. 3. Place the inner vessel in the outer one and put on the lid. Stir the water and take its temperature, estimating it to a fraction of a degree. 4. Connect the steam-trap with the boiler and conduct steam into the water. Stir constantly until its temperature has risen as much above room temperature as it was originally below it. Discontinue passing steam into the water and take the highest temperature reached by the water. 5. Find the mass of the inner vessel and contents. Observations 1 . Mass of inner vessel and stirrer 2. Mass of inner vessel, stirrer and water 3. Initial temperature of the mixture 4. Final temperature of the mixture 5. Mass of vessel, stirrer and mixture Specific heat of the vessel Specific heat of the water Initial temperature of the steam Let the heat of vaporization of water - <2 - X Calculations Study the worked example in Sec. 111:29 and calculate the heat of vaporization of water from the observations recorded above. Conclusions 1. What is the experimental value for the heat of vaporization of water? 2. What is the class average? Questions 1. What is the percentage error? 2. What are the sources of error in this experiment? 3. How may the effect of the sources of error be minimized? 4. Why is it important to use the steam-trap? 172 UNIT IV LIGHT These Searchlights Create an Interesting Construction, Showing both a Converging and a Diverging Pencil of Light Rays. Wheeler Newspaper Syndicate CHAPTER 16 NATURE AND PROPAGATION OF LIGHT IV : 1 NATURE OF LIGHT (sight). Light is that agency which affects the eye and produces the sensation of “seeThat branch of physics ing” that covers all the phenomena pertainlight is called Optics (Greek, ing to ops—eye). Some knowledge of light existed from very early times, though this was limited effects rather than to any fundamental understanding. primarily to As to the nature of light, the early Greeks believed it to consist of streams of minute particles of some sort. There was considerable debate as to whether these particles originated in the eye or Plato (428-348 in the object viewed. B.G.) and Euclid (about 300 b.c.) held to the idea that invisible feelers were emitted from the eye, and that the eye sees a body somewhat as the hand may feel it with a rod. The Pythagoreans, Aristotle (284-322 b.c.) in particular, opposed this view and taught that light consisted of minute particles projected into the eye from the object. Both these conflicting ideas were mere guesses and as such were worthless. However, in the eleventh century Alhazen, an Arabian evidence physicist, provided definite showing that the cause of vision proceeded from the object and not the eye. Even to-day, much mystery still surrounds the nature of light (Sec. IV; 7). In view of the fact that light can be produced from other forms of energy, e.g., heat energy, and that light can be transformed into other forms of ene
rgy, (Sec. V:82), we shall e.g., simply say that light is a form of energy. electricity IV: 2 SOURCES OF LIGHT Few objects give out light and these are termed luminous bodies. Most objects are non-luminous, becoming visible only when they reflect light from some outside source to our eyes (Fig. 16:1). Our main source of light is the sun. When we think how important the sun has always been in human affairs, it not surprising that in prehistoric times it became a prime object of worship. the ancient Egyptian History records Sun-god, Ra, and an ancient Persian god of light, Ahura-mazda. The terms “ray” of light, and “mazda” lamps are derived from these names. is Many objects are rendered luminous by being heated to incandescence. This may be accomplished by mechanical means as shown when sparks are produced by friction between flint and steel in a gas-lighter; by resistance to an electric current in the thin wire used in electric-light bulbs; and by chemical action as in the burning of a fuel. Some objects are luminous at ordinary tem- 175 Chap. 16 LIGHT 16:1 Self-LumFig. inous and Non-Lum- inous Objects. Tigerstedt Studios, Calgary It such as frosted glass or oiled paper, is one that transmits some light, but in doing so distorts or scatters the light so that we cannot see clearly objects on the other side. Opaque substances, like wood, do not transmit light at all and hence we cannot see through them. is common knowledge that light travels in straight lines. Our inability to see around corners, the formation of shadows, and other examples point to this conclusion. This behaviour of light is termed rectilinear propagation. Knowing this, we represent a path of light by a straight line called a ray. (Note: the light that travels along this path is also called a ray.) The direction in which the light is travelling is indicated by arrow-heads placed on the rays. Several parallel rays form a beam of light. Rays of light proceeding towards a point form a converging pencil; when they spread out from a point they form a diverging pencil (Fig. 16:2). IV: 4 PIN-HOLE CAMERA The pin-hole camera is an interesting of electricity of phosphorus, for peratures, a stick example, and fluorescent bodies. Then there is the glow produced by the discharge certain gases, e.g., neon tubes. Another interesting example of “cold light” is that produced by fireflies, and by certain deepsea fish. Probably such light is produced by chemical means. through IV: 3 TRANSMISSION OF LIGHT fact that light Unlike sound, light does not require a material medium for its transmission. Evidence in support of this is supplied by travels through a the vacuum in coming to us from objects in space, and from the glowing filament of an evacuated tube. Further evidence was provided in experiment 4, chapter 10, where we could not hear the bell when the air was evacuated from the jar yet could still see it ringing. Various media diflfer in their ability to transmit light. Transparent objects such as air, glass and water transmit is easy to see light so readily that it through them. A translucent substance. 176 NATURE AND PROPAGATION OF LIGHT Sec. IV: 4 Glass bricks are used in the many construction modern of buildings. What are the advantages? Canadian Pittsburgh Industries Ltd. application of the rectilinear propagaIt consists of an opaque tion of light. box, having a small hole (pin-hole) in the middle of one end, and a translucent screen (piece of ground glass, or oiled paper) at the other. If a lighted candle is placed a little distance in front of the pin-hole, an inverted image of the candle will be seen on the translucent screen. \ / / \ (a) Fig. 16:2 Rectilinear Propagation of Light (a) Ray (b) Beam (c) Converging Pencil. (d) Diverging Pencil. 177 . Chap. 16 LIGHT however, because of the small amount of light admitted through the pin-hole. IV : 5 SHADOWS AND ECLIPSES A shadow is the dark space behind an opaque object, an area from which light has been partially or completely excluded. An opaque object in front of a point source of light will cut off all the light, and a sharply defined shadow is produced. If the light comes from a larger source, the shadow will vary in intensity, the dark central portion of the shadow which receives no light from any part of the source being the umbra 16:4a), the lighter shadow sur(Fig. rounding the umbra which receives some light being the penumbra (Fig. 16:4b). An eclipse of the sun is an interesting shadow phenomenon caused when the moon comes between the sun and earth 16:5). A person located in the (Fig. moon’s umbra will total If in the moon’s eclipse of the sun. observe a This kind of image is formed as a result of very narrow diverging pencils of light from each point of the object passing through the pin-hole, and producing small patches of light, identical in shape to the pin-hole, on the screen. Fig. 16:3 The Pin-Hole Camera. or The resulting image is formed by a large number of these overlapping patches of light, producing an exact replica of the original object. Such an image is a real image. The image so obtained will have a somewhat blurred, out-of-focus appearance due to the circular edges of each patch of light not entirely overlapping. The larger the pin-hole the brighter will be the image on the screen, but the more blurred will be its edges. The image is inverted because the light rays cross at the pin-hole (Fig. 16:3). The size of the image is governed by the size of the object, its distance from the camera, and the distance of the image from the (Chap. 21, Exp. 1 ) pin-hole Size of Image Distance of Image Size of Object Distance of Object Hi _ Di 1h~1^ A consideration of the equiangular (or similar) triangles formed by the rays of should light as shown in 16:3 Fig. establish these relationships fairly readily. If we replace the translucent screen with a light-sensitive paper, or photographic plate, quite acceptable photographs of distant objects can be obtained. A very long exposure is necessary, 178 Fig. 16:4 Shadows (a) Using Point Source. (b) Using Large Source. eclipse of the sun. penumbra, he will be able to see part of the sun. The latter is called a parBecause the tial moon’s orbit around the earth is slightly its distance from the earth Thus it occasionally happens elliptical varies. NATURE AND PROPAGATION OF LIGHT Sec. IV: 6 that the moon comes between the sun and the earth at a time when its umbra does not reach the surface of the earth. A person located on the earth below the tip of the moon’s umbra would see a ring of the sun around the edge of the moon. Such an eclipse is called an annular very eclipse sun and occurs the of rarely. Eclipses of the moon also occur, and at fairly frequent intervals. The moon is a non-luminous body, and is seen only when sunlight is reflected from its surface to the earth. The full moon occurs when the moon is on the opposite side of the earth from the sun. At such a time the moon may pass through the earth’s A partial shadow and be eclipse of the moon is caused when it eclipsed. (a) (b) Fig. 16:5 Eclipses. (a) Total and Partial Eclipse of the Sun. (b) Annular Eclipse of the Sun. (c) Eclipse of the Moon. is partly in the earth’s umbra, and a total eclipse when it is completely in the earth’s umbra. When in the earth’s penumbra, the moon is not eclipsed, but only less bright as it receives, and hence reflects, less light from the sun. IV : 6 VELOCITY OF LIGHT It has long been suspected that light travels with a finite velocity, but early attempts to measure this velocity were Fig. 16:6 Velocity of Light Using the Moons of Jupiter. light too crude to be successful. The first reasonably accurate value was obtained by a young Danish astronomer Olaus Romer in 1676. He found that intervals between the successive eclipses of one of the moons of the planet Jupiter were longer when Earth was receding from Jupiter (going from Ei to E 2 ) and shorter when Earth was approaching (going from Eg to E^, Fig. 16:6). Romer ascribed the discrepancy to the time required for the to diameter of the earth’s orbit. The timelag was found to be about 16.5 minutes or approximately 1000 seconds. Since the diameter of the earth’s orbit is about 186,000,000 miles, the velocity of light is found to be 186,000 miles per second. The first determination of the velocity distances on the of earth was made in 1849 by A. H. Fizeau. His method was to pass a beam of light through one of the gaps in a toothed wheel, and reflect it back on its path from a mirror three or four miles away light over short across travel 179 Chap. 16 LIGHT : (Fig. 16:7). When the wheel was at rest, the return beam passed back through the same gap and was visible on the other side. When the wheel was rotated rapidly, a speed could be found at which Fizeau's ApFig. 16:7 paratus for Measuring Velocity of Scheme of Light. the return way was blocked by the next tooth. The time spent by the wheel in spinning through this small part of a revolution, is also the time required for the light to travel to the distant mirror and back again. Hence, knowing these facts, the velocity of light could be easily calculated. A better method was devised by J. L. Foucault in 1850, who used a rotating mirror instead of a toothed wheel. This method was used in more elaborate form by A. A. Michelson in 1926, with an eight-sided mirror and a considerably increased light path (Fig. 16:8). More recently, Michelson in collaboration with others, set up a mile-long evacuated tube with a mirror arrangement for causing a beam of light to traverse this path back and forth many times before being obAgain using a rotating-mirror served. method, they obtained a quite accurate value for the velocity of light in a vacuum which was found to be slightly higher than its velocity in air. The approximate values for the velo- city of light, C, in air are C = 300,000 kilometres per second or 3 X 10^® centimetres per second or C = 186,000 miles per second. The velo
city of light is a most important physical determination, since it is the speed with which many forms of energy travel through space. It is interesting to note that the vast distances of space are measured in terms of the light-year. This is the distance travelled 180 NATURE AND PROPAGATION OF LIGHT Sec. IV: 7 by light in one year. Some of the more distant stars and nebulae are so remote from the earth that the light by which we now see them set out on its journey to the earth thousands of years ago. IV: 7 THEORIES OF LIGHT The first rational attempt to explain the propagation of light was made by Isaac Newton (1642-1728). His Sir the Emission or Corpuscular theory, Theory, postulated that light energy was conveyed through space by a swiftly moving stream of particles or corpuscles shot out from the luminous body. Most of the properties of light known at Newton’s time such as the rectilinear propagation of light and the effects of reflection (Chap. 17) and of refraction (Chap. 18) were adequately accounted for by this theory. A rival postulate was put forward by Christian Huygens (1629-1695), the son of a Dutch diplomat and poet. He sought to explain the behaviour of light in terms of waves, and hence his theory is called the Wave Theory. Again, reflection and refraction were readily explicable in terms of this wave theory. Difficulties were encountered, however, when seeking to explain the rectilinear propagation of light, and also in the need for postulating the existence of a medium, the ether, completely filling space in which the waves could travel. As a result, the wave theory remained undeveloped and Newton’s corpuscular theory was generally accepted. nineteenth the century, Thomas Young (1773-1829), and A. J. Fresnel (1788-1827), provided valuable Early in experimental support for the wave theory of light. They were able to show that two beams of light could be made to interfere with and to reinforce each other, thereby producing alternate dark and bright lines. This could only be explained by “superposition of waves”. At one position when in opposite phase, a crest with a trough, these waves produce a dark line. At another position when in the same phase, a crest with a crest, or a trough with a trough, they produce a bright line. (Compare with superposition of sound waves. Sec. 11:6.) Another line of experimental support in favour of the wave theory was to show that the velocity of light is smaller in the denser of two media. The corpuscular theory had predicted the exact reverse of this. Recent work seems to favour a combination of the corpuscular and wave theories in the explanation of many of the observed effects. This theory called the Quantum Theory, was first devised by Max Planck in 1901. According to this theory, light is emitted by the atoms of a luminous body in separate packets or bundles of energy called quanta or photons. Probably one or more of the electrons revolving about the nucleus of an atom ( Sec. V : 1 2 ) can be made to jump from one orbit or “energy level” to another. As they do so, one or more quanta of energy, or photons are emitradiates from the ted. luminous body as electromagnetic waves (Sec. IV: 38). The energy content of a photon determines the length and frequency of the wave, and hence the colour of the light observed. energy This 181 Chap. 16 IV : 8 LIGHT QUESTIONS . (b) How many minutes are required for light to travel from the sun to the earth (93X10^ miles)? (c) What is a light-year? 8 What contributions Newton, Huygens and Planck make to a theory of did light? 1. Calculate the distance of an object 12 ft. high whose image is 4 in. high in a pin-hole camera 10 in. long. 2 . Calculate the size of the image of a tree 30 ft. high, 100 yards distant, in a pin-hole camera 8 in. long. 3 . Calculate the height of a building 300 metres distant which produces an image 2.5 cm. high in a pin-hole camera 2.0 in. long. 4 . How long does it take for light to 0'* travel from the moon to the earth (24 X 1 miles)? 5 . Calculate the number of (a) miles, (b) kilometres in 1 light-year. 6 . Sirius, the brightest star in the sky, is 9 light-years away. How far away is this In miles? 7. Our nearest neighbour among the stars, excepting the sun, is Proxima Centauri which is about 25X 1 0’^ miles away. How long does it take for light from this star to reach us? 1. 2 . 3. 4 . 5 . (a) Define optics. (b) What is the nature of light? "luminous” (a) Distinguish between and "illuminated” objects. (b) How may objects be rendered luminous? (a) What evidence have we to prove a that vacuum? through travel can light (b) Distinguish between opaque, and sub- transparent translucent stances. Give examples of each. (a) What Is meant by the "rectilinear propagation of light”? Discuss evidence in support of it. (b) Define: ray, beam, converging pencil, diverging pencil. (a) Describe the image obtained in a pin-hole camera. (b) Explain how it is produced. (c) Name three factors that govern its size. (d) The length of a pin-hole camera is 10 in. An object 6ft. high Is placed at a distance of 30 ft. from the pinhole. Calculate the size of the image produced. 6 . (a) Define: umbra, penumbra. (b) Construct a labelled diagram to show how both total and partial eclipses of the sun are produced. (a) What is the velocity of light in air? 7. 182 CHAPTER 17 REFLECTION OF LIGHTMIRRORS IV: 9 THE LAWS OF REFLECTION In section IV:2, we learned that most are made visible when light objects falling on them is reflected back to our eyes. The rays of light that fall upon a body are called incident rays, while those that are sent back by the body are called reflected rays. A mirror is a smooth, highly polished surface, designed to reflect a maximum amount of light. Mirrors usually consist of pieces of glass silvered on one surface. Some are flat and are called plane mirrors, while others are called Fig. 17:1 The Optical Disc, for Dem- onstrating the Laws of Optics. 183 Chap. 17 LIGHT curved mirrors. However, any smooth surface, such as polished metal, polished wood, or still water, will serve as a mirror. to the reflecting surface and angles meeting it at the point of incidence is called the normal. Rotate the disc and thus cause the incident ray to strike the mirror at different angles. In each case note the direction of the reflected ray and compare the size of angle of incidence (the angle between the incident ray and the normal), with that of the angle of reflection (the angle between the reflected ray and the normal) . In each case these two angles will be found to be equal, and the incident ray, the reflected ray and the normal will all be found to be in the plane of the disc. Hence we can state the two laws of reflection as follows: First Law: The angle of reflection althe angle of ways equals incidence or Z r = Z i Fig. 17:2 Reflection of Light by a Plane Mirror (a) Using Optical Disc. (b) Terms. Second Law: The incident ray, the normal, and the reflected ray all lie in the same plane. To study reflection of light and other is 17:1). optical phenomena an optical disc This consists of a used (Fig. circular flat disc, graduated in degrees, to which various pieces of optical equipment can be fastened by means of thumb-screws. The disc can be turned about a horizontal axis by means of a handle fastened to the back. Surroundis an opaque collar in ing the which is a window containing one or more horizontal slits. disc Mount a plane mirror at the centre of the disc, with the face of the mirror at right angles to the zero line marked on the disc (Fig. 17:2). Allow a ray of light from a lantern to pass through a single slit in the window of the optical falls upon the mirror. disc so that it Adjust so that the point of incidence coincides with the point where the zero line meets the mirror. This line at right 184 are laws quite These simple and straightforward. We apply them daily in games such as handball, tennis and basketball where the bounce of a ball Experiment 2, chapter 21, is is an alternative method for proving these two laws of reflection. utilized. IV : 10 REGULAR AND DIFFUSE REFLECTION a upon rays fall surface, such as When parallel a smooth reflecting plane mirror, they form the same angle of incidence with the surface, and in consequence they will be reflected as a beam of 17:3a). Such reflection is called regular reflection and often produces undesirable glare. For example, one finds it difficult to read from a glazed paper in parallel (Fig. rays sunlight. Diffuse or irregular reflection occurs REFLECTION OF LIGHT—MIRRORS Sec. IV: 11 when light strikes a rough surface (Fig. 17:3b). Such a surface may be conto be composed of a large sidered number of tiny, flat surfaces that face Thus when parallel in all rays of light strike such a surface, the directions. individual rays are scattered or diffused due to their being reflected in different directions. To avoid glare it is often necessary to promote diffusion of light. Unglazed paper is used for newspapers, and for Smooth Surface Rough Surface Regular Reflection Irregular Reflection (a) (b) Fig. 17:3 Regular and Diffuse Reflection. Frequently the woodwork wall-papers. in our homes is left with a dull finish. These, and other devices, all cause dif- bulbs, and lamp-shades. Prism light glass and other types of roughened glass in windows serve the same purpose. IV : 1 1 IMAGES IN PLANE MIRRORS Mirrors have many and varied uses. Large plate-glass mirrors are frequently placed on the walls of our homes and of public rooms to give an impression of spaciousness. Rear-view mirrors are now compulsory in all automobiles. Mirrors are often used in scientific instruments to reflect light onto a scale, in projectors to intensify the light beam, in the view-finder of reflex cameras, in periscopes, and the like. In view of such it should be of real widespread uses, interest to us to study the images produced by plane mirrors (Fig. 17:4). On doing experiment 3, chapter 21, you
will learn the following facts about images in plane mirrors: The Position of the Image. The image is as far behind the mirror as the object is in front, and a line joining the two right the passes through mirror at fuse reflection of light. angles. Diffusion, and hence the elimination of glare, is also obtained by transmission of light through frosted or opalescent The Characteristics of the Image. The image is the same size as the object. it only It is a virtual image, that is, Fig. 17:4 Images in a Plane Mirror (a) The Position and Characteristics of the Image. (b) Lateral Inversion. 185 Chap. 17 LIGHT appears to be there. No light emanates from it. Such an image cannot be proThe image is jected onto a screen. vertically erect, but laterally inverted. When we say “laterally .inverted” we mean that the right and left sides are interchanged (Fig. 17:4b). Note that when the mirror is horizontal, the image is vertically as well as laterally inverted. Recall what you observe on looking into a still body of water to verify this fact. IV: 12 TO LOCATE IMAGES IN PLANE MIRRORS These phenomena observed experimentally can be shown to be true as a geometrical consequence of the laws of reflection, Sec. IV; 9. Let us consider the simplest possible case, that of a point object, O, the image of which, I, is viewed by an eye BD (Fig. 17:5). O is sending out light rays in all directions. Point in a Plane Mirror. lying between OAB and Only those OCD enter the pupil of the eye after being reflected from the mirror. The reflected rays AB and CD appear to be 186 coming from I, their point of inter- section. In order to locate the image of an object in a plane mirror geometrically, the following construction is necessary (Fig. 17:6). From each point of the to object, draw a perpendicular the mirror, and extend it an equal distance behind the mirror. Join the ends of all such lines and you will have an outline of the image of the object. Using the half arrow ( j ) enables you to indicate lateral inversion nicely. To show how the eye sees the image place a diagram of an eye on the same side of the mirror as the object. Draw a cone of rays from the tip of the image to just fill the pupil of the eye. Draw two light rays from the tip of the object to the two points on the mirror where the previous rays met the mirror. Put arrows on the Repeat this real light rays as shown. same procedure for each point on the object. Thus the eye is receiving light that appears to originate from the virtual image behind the mirror, but actually comes from the object. In all diagrams, use faint lines for construction, dotted lines for imaginary rays or virtual image, and solid lines for real rays, real image, object, etc. . REFLECTION OF LIGHT—MIRRORS Sec. IV: 13 Fig. 17:7 Images in Parallel Mirrors. IV: 13 PLANE MIRROR SYSTEMS 1. Parallel Mirrors On looking into a mirror facing a parallel mirror on the opposite wall of a room, such as is used in barber shops, a very large number of images of the room can be seen stretching away almost endlessly. To discover how these are formed study Fig. 17:7 carefully. {ii, 12 , h) A series of images representing the side of O facing mirror Mi are formed (/i, 1 2 , 1 .3 ,) and also a series of images representing the side of O facing mirror These multiple images '^^2 are due to the image formed in one mirror, acting as the object, which in turn forms an image in the second mirror, and so is absorbed at each reflection, each succeeding image is fainter than the one preceding. Since some light on. 2. Mirrors at Right Angles When two mirrors are placed at right angles to each other (Chap. 21, Exp. 4), three images of an object will be 17:8). Ii and h are observed images of O in mirrors Mi and M^. the mirror is the image of Ii (Fig. in Is M 2 produced (Mg produced is really an image of M 2 in Mi) and of I 2 in mirror Mi produced, these two images coinciding. How the these eye sees 1 3. Fig. 17:8 Images in Mirrors at Right Angles. images may be shown by a construction similar to that used previously. Mirrors Inclined at Sixty Degrees When mirrors are inclined, the number of images obtained depends upon the 187 Chap. 17 LIGHT angle between the mirrors. we can say: In general the number of images _ 360 ^ Z Inclination For example, using angles we obtained three images mirrors at If two mirrors are inclined at 60°, five images will be obtained 17:9). (Fig. right IV : 14 CURVED MIRRORS These form multiple images of differently shaped pieces of coloured glass placed between them. Many beautiful and fascinating designs can be produced at every turn of the instrument. Curved mirrors are frequently more suitable than plane mirrors for certain purposes. Curved mirrors are used as rear-vision mirrors, as shaving mirrors and as reflectors for car headlights. One important type of curved mirror is the spherical mirror whose reflecting surface is a portion of the surface of a sphere. If the inner surface is the reflecting surface, it is a concave mirror; if reflection occurs at the outer surface it is a convex mirror. We represent such mirrors in cross-section by the arc of a circle (Fig. 17:10). Silvered Surface Reflecting Surface Silvered Surface Reflecting Surface Fig. 17:9 Images in Mirrors inclined at 60°. Sir David Brewster, of Edinburgh, making use of the images produced by inclined mirrors, in 1819 invented the kaleidoscope. This consists of three mirrors set at angles of 60° to each other. Spherical Mirrors. Fig. 17:10 (a) Concave or Converging. (b) Convex or Diverging. A few terms used in connection with curved mirrors need to be defined (Fig. 17:11): 188 REFLECTION OF LIGHT—MIRRORS Sec. IV: 14 Fig. 17:12 Action of Curved Mirrors. (A) Concave. (B) Convex. Centre of Curvature, C, is the centre of the imaginary sphere from which the mirror was cut. that is, points through which reflected However, light rays are converged. there is only one principal focus as Vertex, F, is the mid point of the mir- defined. ror. Principal Axis, PV, is the line through the centre of curvature and the vertex of the mirror. Secondary Axis, CD, is any other line drawn through the centre of curvature to the mirror. Since all radii of a circle meet the circumference at right angles, and since all axes behave like radii since they pass through the centre of curvature, it follows, that a Normal to the surface of a curved mirror at the point of incidence is simply a secondary axis. to it) pass, close Principal Focus, F, is the point on the principal axis, through which rays, travelling parallel to the principal axis (and fairly after being reflected from the mirror. For a convex mirror, the principal focus is a virtual focus, and is that point from which such rays appear to diverge after reflection from the mirThis point will be found to be ror. midway between the centre of curvature and the vertex of the mirror. There are many other foci possible. Focal Length, FV, is the distance from the principal focus to the vertex of the mirror. The action of curved mirrors (Chap. 21, Exp. 5), may be demonstrated by means of the optical disc. Mount a concave mirror on the optical disc so that the 0°—0° line becomes the prin- cipal axis of the mirror. Insert a metal shield with several parallel slits in the window of the opaque collar and shine light from a projection lantern through these slits onto the mirror (Fig. 17:12). You will note that the reflected rays conIf the disc is rotated so that the verge. incident rays are parallel to the principal axis it will be found that the reflected rays converge through a point. This point is the principal focus of the mirror. Similarly, if we insert a convex mirror in place of the concave one, and repeat the above, it will be found that the reflected rays diverge as though coming from a point behind the mirror. This virtual point locates the principal focus of the convex mirror. 189 Chap. 17 LIGHT IV: 15 IMAGES IN A CONCAVE MIRROR In chapter experiment 21 we 6, studied the characteristics and position of images formed by a concave mirror for difTerent positions of the object. As the object was moved closer to the mirthe image, which was real and ror, Rays Used to Locate the Fig. 17:13 Image, I, of an Object, O, in a Concave Mirror. inverted, gradually grew larger in size and moved farther from the mirror. When the object was located inside the principal focus of the mirror, the image became virtual and erect and was located behind the mirror. All of these results can be verified with fair accuracy by means of simple geometric diagrams. The principles used here are very similar to those used previously with plane mirrors. Each point on the object is sending out rays of light in straight lines in all directions. If we trace the paths of several of these rays from a point on the object, to the mirror, and then their rays back from the mirror, we will find that they intersect. The point of intersection is the point on the image that corresponds to the original point on the object from whence the light rays came. reflected Suitable rays for this purpose, as illustrated in Fig. 17:13, are: 190 1. A ray from the object parallel to the principal axis, which will be reflected through the principal focus. 2. A ray from the object through the centre of curvature which will be reflected back along the same path, since the incident ray strikes the mirror at right angles. 3. A ray from the object through the principal focus, which will be reflected parallel to the principal axis. 4. A ray from the object to the vertex of the mirror, which will be reflected so that the angle of incidence equals the angle of reflection. 5. Any other ray from the object to the mirror will be reflected so that Z f = Z r. In actual practice any two of these rays will suffice. The first two mentioned are the most easily drawn and hence are the most convenient to use. In order to verify the observations obtained in our experiment, it is necessary to make ac
curate scale drawings. To Locate the Image in a Fig. 17:14 Concave Mirror of an Object Placed Beyond F. For example, if the focal length of the concave mirror was 20 cm., the object 60 cm. from the mirror, and 12 cm. high, our construction would be as above (Fig. 17:14), using the scale 10 cm. = 1 cm. REFLECTION OF LIGHT—MIRRORS Sec. IV; 17 Draw principal axis, PV. With centre C, and radius of curvature 4 cm. (twice focal length), draw an arc to represent the mirror. Locate the principal focus F, midway between C and V (2 cm. from mirror) . Locate the object O, 6 cm. from the mirror. Draw the object (OOi) 1.2 cm. high and perpendicular to the axis. From the tip of the object (Oi) draw the two rays and their reflected rays as outlined on previous page. These rays intersect at /i, and hence locate the position of the tip of the (Rays from other points on the image. would corresponding object image points. sufficient.) Draw the image 11 1 , perpendicular to PV. The distance of the image from the mirror and its size can be obtained by measuring accurately. Notice that when the object is beyond the image is between F and C, C, real, smaller than the object, and inverted. See how perfectly this diagram confirms your observations recorded in the table on page 242. produce However, one the is To Locate the Image in a Fig. 17:15 Concave Mirror of an Object Placed Between F and V. of for other positions The student should construct similar diagrams the object, and verify the other observations made. The special case, where a virtual image is obtained, is slightly more difficult. Study the following diagram carefully, and note the method is that identical to that given previously (Fig. 17:15). However, this time the reflected rays diverge, and therefore appear to come from a point behind the mirror, thereby creating the virtual image. IV : 16 IMAGES IN A CONVEX MIRROR One has often observed the images produced by the back of a spoon, the side of a tea-kettle or the shiny fender of a car. Such objects as these are all acting as convex mirrors. In experiment Fig. 17:16 To Locate the Image in a Convex Mirror. 7, chapter 21, we found that the images in a convex mirror were always behind virtual, smaller than the the mirror, object, and erect. The same method is used as with the concave mirror to verify this geometrically (Fig. 17:16). IV: 17 HOW THE EYE SEES THE IMAGE discussed As previously section IV: 12, the eye sees the image by means of rays which actually come from the object, but which appear to come from in Fig. 17:17 How the Eye Sees a Real Image in a Concave Mirror. 191 Chap. 17 LIGHT : the image. Fig. 17:17 and 17:18 show how this is accomplished. In each case the light rays start from the object, are reflected at the mirror, pass through a real image or appear to come from a Fig. 17:18 How the Eye Sees a Virtual Image In a Convex Mirror. virtual image, and enter the eye. In order to construct these diagrams just reverse the previous order. That is, join 7 to the outer edges of the pupil of the eye. This cone is projected back to the mirror or cuts the mirror. From there it is drawn back to the object. !V: 18 THE MIRROR FORMULAE In the previous sections we have been studying the images produced by curved It is possible by means of two mirrors. simple formulae to determine the location and characteristics of the image for various positions of the object. To do a consistent convention regarding so, signs must be followed. Values for distances of objects and real images are always positive; those for virtual images are negative. Similarly, the focal length of a concave mirror (which has a real is positive, that of a principal focus) convex mirror (with a virtual principal is negative. To state this confocus) cisely, the convention of signs is: “real is positive, virtual is negative” The two formulae, with worked examples to show how to use them, follow (a) Magnification Formula Height of Image Distance of Image Height of Object Distance of Object 77o Do Note the similarity of this formula with that obtained for the pin-hole camera (Sec. IV: 4). (b) Distance Formula Distance of Object Distance of Image _ 1 Focal Length 1. An object 2 in. tall is placed 15 in. from a concave mirror whose focal length is 5 in. Find the position and size of the image. Examples Ho= 2 in. Do = 15 in. f = 5 in. Di= ? 1 1 _ 1 — + Do Dr 7 — + 1 1 1 T 1 15 15 1 _ 1 D,“ 5 :.2Di — 15 :.Di =7.5 The image is located 7.5 in. from the mirror. 192 Ho— 2 in. Do = 15 in. Dj — 7.5 in. REFLECTION OF LIGHT—MIRRORS . Hi ^ A ‘ Ho~ Do . Hi _ 7.5 ~ 15 ' 2 ? . ‘ Sec. IV: 19 The image is 1 in. tall. Check these results by making an accurate scale construction to locate the image geometrically. Both position and size of image should agree closely with the above values obtained by calculation. 2. An object 5 cm. high is placed 30 cm. from a convex mirror whose focal length is 20 cm. Find the position, size, and nature of the image. Ho= 5 cm. Do = 30 cm. f = — 20 cm. Di = ? ' Do^ Di f 30 1 _ ~~ ‘ Di . ‘ Di — 20 1 1 20 30 _ -3 -2 ” 60 — 5Di=:60 Di = — 12 cm. Image is 12 cm. from mirror and is virtual (since sign is negative). Ho= 5 cm. Do = 30 cm. 12 cm. Di ? ‘ Ho Do ‘ ’ 5 ~ 30 Hi = 2 cm. Image is 2 cm. high. IV: 19 APPLICATIONS OF MIRRORS Reference has already been made in section IV : 1 1 to the uses of plane mirrors. Many of these uses, however, are more satisfactorily fulfilled by using curved mirrors. For example, a plane mirror is frequently used as a shaving mirror. However, a concave mirror so used produces an enlarged erect image when the face is held between the principal focus and the vertex of the mirThe enlarged image is often a ror. Similarly, a convex distinct advantage. mirror is frequently used as a rear vision mirror instead of a plane mirror. The erect, smaller image produced gives a wider field of view with the same size mirror. Spherical mirrors such as we have been describing have one serious defect, known as spherical aberration. In such mirrors, only those rays parallel to the principal axis and fairly near to it pass through the principal focus on being 193 Chap. 17 LIGHT are axis (Fig. 17:19). reflected from the mirror. Rays farther from the principal reflected through points some distance from the Conversely, if a focus source of light is placed at the principal focus, the reflected rays will not form a parallel beam, but the outer ones will be scattered and hence the light is weakened. Fig. 17:19 Spherical Aberration in Curved Mirrors. To overcome this parabolic rather than spherical mirrors are used defect, 17:20). A parabola is a section (Fig. from a cone obtained by cutting the cone Fig. 17:20 A Parabolic Mirror Used to Prevent Spherical Aberration. by a plane parallel to a line from the apex of the cone to any point on the circumference of the circular base. All rays which emanate from the principal focus of such a mirror, after reflection are parallel to the principal axis, no matter how great the aperture is. As a result parabolic mirrors are used as reflectors for searchlights, car headlights and in the reflecting telescope. IV : 20 QUESTIONS A 1. (a) Define: angle of incidence, angle of reflection. (b) State the two laws of reflection and indicate how you could test them. 5. 2. (a) Distinguish between diffuse and locate the image of an inclined arrow in a plane mirror. (b) Show how an eye sees the image. (a) How many images will be observed in 2 mirrors inclined at 45°? (b) Draw a diagram to locate these images. glare de- 6. (a) Define: principal axis, principal regular reflection. (b) How is undesirable creased? 3. (a) State the rule for determining the position of an image in a plane mirror. (b) Describe the Image. between (c) Distinguish real and virtual images. (a) Show by a diagram how you can 4. 194 focus, focal length. (b) By means of accurate construction show how the of (i) a concave mirror, and (ii) a convex mirror may be located. principal focus 7. (a) Under what conditions does a concave mirror form (i) a real image, (ii) a virtual image? REFLECTION OF LIGHT-MIRRORS Sec. IV: 20 (b) Describe the size and location of all the real images produced by a concave mirror. (c) When a concave mirror is used in shaving, where is the person’s face in relation to the principal focus of the mirror? What kind of image is seen? 8 . An object is located just beyond the centre of curvature of a concave mirror. (a) By means of a diagram locate its image. (b) Show how the eye sees the image. (c) State the characteristics of the image. 9 . (a) State uses of plane, concave and convex mirrors. (b) How does a searchlight produce a narrow beam of light? B 1. A watch seen in a mirror seems to read (a) 2 o’clock, (b) 6.15 o’clock. What does it actually read? Explain. 2 . Support a mirror vertically on a desk and stand a book in front of it to serve as a screen. Draw a triangle on a piece of paper and lay it flat between the book and the mirror. Watching the image in the mirror, but not the triangle itself, try to retrace the triangle. Why do you have difficulty? 3 . What changes would you observe in your image as you walked toward (a) a plane mirror, (b) a concave mirror, (c) a convex mirror? the position grams locate of the image for each position of the object, (b) State the characteristics of each image. accurate 5 . By means of an scale diagram locate the image produced by a convex mirror, whose focal length is 15 in., of an object 6 in. high and 2 ft. distant. State the characteristics of the image. 6 . An object is 4 in. tall and its image is 6 in. tall when the object is placed 2 ft. from the mirror. How far is the image from the mirror? 7 . A 6 in. pencil is ft. in front of a curved mirror. Find the length of the image if it is 8 in. from the mirror. 1 8. How tall is an object if it produces an image 3 in. tall located 7 in. behind a convex mirror when the object is located 1 0 ft. in front of the mirror? 9 . An object 4 in. in front o
f a concave mirror produces an image 1 2 in. behind the mirror. Find the focal length of the mirror. 10. An object is 1 8 cm. in front of a concave mirror which has a focal length of 1 2 cm. How far is the image from the mirror? 11. An object 9 in. from a convex mirror produces an image 3 in. behind the mirror. Find the focal length of the mirror. 12 . An object is 15 cm. from a convex mirror of focal length 20 cm. Calculate the image distance. 4. An object 15 cm. high is located (i) 75 cm. (ii) 60 cm. (iii) 45 cm. (iv) 30 cm. (v) 20 cm. from a concave mirror whose focal length is 30 cm. 13. The focal length of a concave mirror is 12 in. How tall is the image of a 5 in. candle standing 15 in. from the mirror? 14. Repeat question 13 using a convex (a) By means of accurate scale dia- instead of a concave mirror. 195 CHAPTER 18 REFRACTION OF LIGHTLENSES phate (hypo) and a few drops of hydrochloric acid. These materials will render the water slightly turbid and make Incident Mirror Refraction of Light on PassFig. 18:2 ing from Air to Water or from Water to Air. visible a light beam projected through Shine a beam obliquely upon the it. surface of the water (Fig. 18:2). Part of this beam will be reflected at the surface of the water, and part will be If a refracted as it enters the water. normal (perpendicular) is placed at the point of incidence, it will be observed that the light beam is bent toward the normal. Similarly, if we shine the beam of light obliquely up through the water and into the air, it will be observed to bend away from the normal. Both effects can be shown simultaneously by placing a plane mirror on the bottom of the tank Note that no refraction (Fig. occurs if the light enters or leaves the water at right angles to the surface. 18:2). We may also show that a beam of light is refracted on passing obliquely IV; 21 MEANING OF REFRACTION Water frequently appears to be much shallower than it actually is. An oar or stick when only partly submerged appears to be bent upwards at the surface of the water (Fig. 18:1). These phenomena, and many others similar to them, are due to the bending of light rays as they pass obliquely from one medium into another of different optical Fig. 18:1 Partly Immersed Stick Ap- pears Bent. density (Sec. IV: 22). Such bending of the light rays is called refraction. Refraction of light as it passes from air into water, or from water into air, can be illustrated Fill a tank with water containing a small amount of fluorescein, or a little sodium thiosul- easily. 196 REFRACTION OF LIGHT-LENSES Sec. IV: 23 from air into glass, or from glass into Place a semicircular block of glass air. on the optical disc, so that its flat edge is bisected at right angles by the 0°-0° Shine a ray axis of the optical disc. of light so that it strikes the surface at this central point where the axis crosses it. The axis thus is made the normal to the refracting surface. The light will Fig. 18:3 Refraction of Light on Pass- ing from Air to Glass. be refracted as shown in Fig. 18:3. Why is the light not refracted on leaving, or on entering the circular surface of the glass ? From the preceding observations we may summarize the behaviour of light rays in passing from one medium into another of different optical density, as follows: 1. When a ray of light passes obliquely from one medium into another of greater optical density, it is refracted toward the normal. 2. When a ray of light passes obliquely from one medium into another of less optical density, it is refracted away from the normal. It follows from the above that when a ray of light enters a new medium at right angles to the surface, no refraction occurs. IV: 22 EXPLANATION OF REFRACTION We may ask what causes refraction of light. The following illustration should help us to understand it. Suppose the brakes of an automobile are improperly adjusted, those on the right wheels holding better than those on the left. When the brakes are applied, the car will swerve to the right, that is, to the side that is slowed up most. Similarly, the bending of the light beam, or refraction, is caused by the change in the velocity of light as it passes from one medium into another of different optical density. The greater the optical density of a substance the more slowly light will travel through it. According to the wave theory, light travels out from the source as spherical waves. When a wave-front enters an optically denser medium obliquely (Fig. 18:4), that part of it that enters the new medium first will be slowed, while the rest of it continues to advance at the same speed as before. Consequently, the wave-front swerves toward the normal as it enters the denser medium. Conversely, the wave-front would swerve away from the normal as it speeds up on passing into an optically less dense medium. No bending occurs if the light enters the new medium at right angles to the surface, for the entire wave-front would be slowed down or speeded up at the same instant. IV : 23 INDEX OF REFRACTION From Fig. 18:4 it is evident that the light is travelling the distance PR in one medium (air), while it travels the distance OQ in the other medium. Hence these distances must be proportional to 197 Chap. 18 LIGHT the velocities of the light in the two media. Also, it should be evident that the amount of refraction is governed by these In the study of refraction only acute angles are involved. When the angle is contained in a right-angled triangle, its sine is a constant quantity found by dividing the length of the side opposite A the angle by the length of the hypotenuse. For example, in A ABC, Fig. 18:4 Refraction of Wave Front on Entering an Optically Denser Medium. relative distances or these relative velocities, and would be constant for any two given media. This ratio is called the index of refraction, and may be defined as follows: Index of Refraction of a medium (yu,) Velocity of light in air Velocity of light in the medium V (air) V (medium) and AB Sin Z ACB = Sin Z CAB = — AC BC AC To prove that: Index of Refraction (u) = Sin Z i Sin Ir PR OQ PR OQ V (air) ^ Sin Z i V {medium) PR/ OR Sin Ir OQ/OR P ~ Sin Z i TZ Sin Z ^ In experiment 8, chapter 21, by means of a simple geometric construction it shown that Index of Refraction is Sine of angle of incidence Sine of angle of refraction The sine of an angle ( abbreviated sin Z ) is a property of an angle found useful in calculations in mathematics and science. The index of refraction is an important physical property of a transparent medium. An instrument, called a refractometer is used to measure' its value quickly and accurately. This enables scientists to identify substances and to If the index of recheck their purity. fraction of a substance is known the velocity of light may be determined for that substance. A substance with a large index of refraction is said to have a REFRACTION OF LIGHT-LENSES Sec. IV: 25 high optical density, because it permits light to travel through it at a relatively Such substances refract slow velocity. light to a greater extent than do those with a smaller index of refraction. The brilliance of diamonds, and other precious stones is largely due to this fact (Sec. IV: 26). The index of refraction of a substance varies with the colour of the light that passes through it. The rainbow and the beautiful colours obtained when light is refracted through cut glass and precious stones are due to Temperature, too, this has an effect on the index of refraction because it alters the optical density of The waviness observed a when light rising above a hot object is caused by the differing .refractive indices of various layers of hot and cold air. passes through air (Sec. IV: 37). substance. The Index of Refraction of Some Common Substances W ater Crown Glass Flint Glass 1.3 1.5 1.7 Quartz Zircon Diamond 1.5 1.9 2.4 IV : 24 REFRACTION THROUGH A GLASS PLATE When a ray of light passes obliquely through a glass plate with parallel sides (Chap. 21, Exp. 9), it is refracted both on entering and on leaving the glass. On entering, the light is slowed down and Fig. 18:5 Refraction Through a Glass Plate with Parallel Sides. is, therefore, refracted toward the normal. On emerging, speeds up again, and hence is refracted away from the normal. Since this second effect exactly counteracts the first the speeding up exactly compensates for the light the (i.e., Fig. 18:6 Why an Object Appears 'Closer When Viewed Through a Glass Plate. original slowing down), the emergent ray will be parallel to the incident ray but laterally displaced (Fig. 18:5). The amount of this lateral displacement depends on the refractive index of the glass, on the angle of incidence, and on the thickness of the glass. On looking' at an object through such a plate, it will be slightly displaced in position and will appear nearer to the eye than it is in fact (Fig. 18:6). The same thing occurs in water and this accounts for the diffi-culty in locating the exact position of an immersed object (Chap. 21, Exp. 10). IV: 25 REFRACTION THROUGH PRISMS A prism consists of a wedge-shaped portion of a refracting substance, bounded by two plane surfaces inclined at an angle to each other, this angle being called the refracting angle. When a ray of light enters such a prism it is slowed down, and therefore refracted toward the normal. On leaving, it speeds up, and therefore bends away from the normal. This can be shown by mounting a 60° prism on an optical disc and shining a ray of monochromatic light, e.g., red, through it. From Fig. 18:7 it is seen 199 . Chap. 18 LIGHT that both these refractions are toward the thick base of the prism. Hence the light is bent, or deviated, quite consid- A prism when the refracted ray passes through the prism parallel to the base. Finally, the amount of deviation depends upon the colour, or the wavelength, of the light used, which is why monochromatic light was used in these experiments. gives prisms one of their large
areas of usefulness. More discussion on this follows in the chapter on colour, page 213. factor This latter Fig. 18:7 Deviation Through a Glass Prism. erably from its original path. The angle of deviation, D, is obtained by extending the incident and emergent rays to meet (Chap. 21, Exp. 11 ) The amount of deviation produced by a prism depends upon a number of facFirst, the material comprising the tors. prism—^the greater its index of refrac- tion, the greater will be the deviation produced. Secondly, the shape of the prism—^^the greater its refracting angle, the greater will be the angle of deviation. Thirdly, the angle of incidence at which the light meets the prism. This can be shown by rotating the optical disc and observing the amount of deviation for various angles of incidence. Minimum deviation is obtained from an equilateral IV : 26 TOTAL REFLECTION It is more usual to consider light passing from air into an optically denser medium; however, when it goes from the optically denser medium into air, or a less dense medium such as from water or glass into air, a peculiar phenomenon occurs. As the light speeds up on entering the air obliquely it bends away from the normal. As the angle of in the denser medium inincidence creases, we finally come to a position when the refracted ray just grazes the surface, that is, the angle of refraction equals 90°. This angle of incidence is angle. When the called the critical incident angle exceeds the critical angle the light is completely reflected, a phenomenon known as total reflection (Fig. 18:8). Using the semicircular block of glass on the optical disc as in section IV:21, shine a ray of light through it. Rotate the disc to increase the angle of Fig. 18:8 Total Reflection of Light (a) On Passing from Water to Air. (b) On Passing from Glass to Air. 200 REFRACTION OF LIGHT—LENSES Sec. IV: 26 incidence and note the size when total This angle, the reflection first occurs. critical angle for glass, is about 42°. Fig, 18:9 An Illustration of Total Reflection. V An example of total reflection is seen when an empty test-tube is placed in a Canadian Industries Ltd. An Interesting Example of Light Being ''Bent" in a "Perspex" Rod. beaker of water (Fig. 18:9). On looking down into the water the sides of Applications of Total ReFig. 18:10 flection Prisms (a) A Simple Periscope. (b) Field Glasses. 201 Chap. 18 LIGHT, is very Total reflection test-tube appear silvery. This is the due to the light striking the surface of the tube at an angle greater than the This light is critical angle for glass. totally reflected to the eye as shown. Similarly a sooted ball appears silvery on being lowered into a beaker of water. A layer of air entrapped by the coating of soot produces a water-air -boundary at which reflection occurs. a useful phenomenon. Total reflection prisms are usually right-angled prisms with wellpolished faces. Light enters such a prism at an angle of incidence of 45°, which is greater than the critical angle for glass (p. 201), and hence total reflection occurs. Such prisms are used in periscopes (Fig. 18:10a), range finders, field-glasses (Fig, 18:10b), and reflecting telescopes. They are much more efficient reflectors than mirrors, since mirrors reflect only about seventy per cent of the light they receive, whereas prisms reflect a much greater proportion. In addition, prisms are more robust, there is no silvering to tarnish and they give rise to a single well-defined image. The brilliancy of diamonds, brilliants, and cut-glass dishes is due to total reflection. The greater the index of refraction of a substance the smaller is its critical angle. Diamonds have a large index of refraction (Sec. IV: 23) and consequently have a small critical angle. The surfaces of the diamonds meet each other at such angles that much of the light entering a diamond is totally reflected a number of times internally, before eventually being refracted out. This lights up many surfaces, and gives the diamond its sparkle. Brilliants, and cut-glass articles are often made of leaded The addition of lead increases the index of refraction, and consequently the cut faces are able to cause much total reflection. glass. 202 IV : 27 ATMOSPHERIC REFRACTION Light travels faster in a vacuum than in air. Therefore, light reaching us from the sun and stars will slow up and be refracted as it enters the atmosphere of the earth. Since the atmosphere gradually becomes denser as the altitude decreases, the light will be refracted more and more as it passes through successive layers of denser air nearer the surface. Consequently, when light comes to us obliquely from the sun and stars these S' A' Fig. 18:11 Atmospheric Refraction (a) The Sun Low on the Horizon. (b) A Mirage. the distances increased bright objects appear higher than they really are (Fig. 18:11a). This effect is most pronounced at low altitudes because of the larger angles of incidence, and be travelled through the successive layers of air. As a result the sun appears to set several minutes after it has actually passed below the horizon. The enlarged and sometimes elliptical appearance of the sun and moon when near the horizon is due to the fact that rays from the to REFRACTION OF LIGHT—LENSES Sec. IV: 28 lower edge are refracted more than those from the upper edge. A mirage is a well-known optical illusion caused by refraction and sometimes total reflection of light as it passes through layers of atmosphere of varying density (Fig. 18:11b). Objects in the distance may be raised above or depressed below their normal position and may be distorted into irregular fantastic shapes. The most commonly observed mirage is that of an apparent layer of water over a hot level sandy surface, or paved roadway. The mirage in these cases sky, produced by total reflection from layers of air near the ground. really an image of the is IV: 28 LENSES A lens is a piece of transparent refracting medium, usually glass, bounded by two spherical surfaces, or by a plane and a spherical surface. There are two main types: (a) Converging or Convex Lenses are thicker at the centre than at the outer edge. Such lenses always refract rays of light so as to converge them. They therefore collect light. CONVERGING or CONVEX DIVERGING or CONCAVE Fig. 18:12 Kinds of Lenses. (b) Diverging or Concave Lenses are thinner at the centre than at the These lenses cause outer edge. They to diverge. light rays therefore scatter light. Fig. 18:13 Comparison of Lenses and Prisms, (a) Action of a Converging Lens. (b) Action of a Diverging Lens. 203 Chap. 18 LIGHT These two types of lenses may be of varying shapes as shown in Fig. 18:12. Each shape is devised for a specific purpose. The action of a lens is similar to that o-f two prisms base to base (Fig. 18:13). As discussed in section IV: 25, the light is bent toward the base of the prism both on entering and leaving the prism. Similarly, the light is bent toward the thicker part of the lens both on entering and leaving it. These examples explain the converging action of a convex lens, and the diverging action of a concave lens. Fig. 18:14 Terms Pertaining to Lenses. The study of lenses makes use of a new vocabulary. The terms used are explained below (Fig. 18:14) : Centre of Curvature C, In most lenses there are two centres of curvature. They are the centres of the spherical surfaces that bound the lens. Principal Axis of a lens is the line passing through the centres of curvature of the two faces, or, in the case of a lens which has one face plane, it is the line passing through the centre of the curved face and curvature of which is normal to the plane face. Optical Centre O, is the point on the principal axis midway between the two surfaces of the lens. All distances along the principal axis are measured from this point. Principal Focus F, of a convex lens is that point on the principal axis to which a beam of light which is parallel to the principal axis converges 204 after refraction through the lens. (In a concave lens, the principal focus is a virtual point, and is that point on the principal axis from which a beam of light, which is parallel to the prindiverge on cipal being refracted through the lens.) axis, appears to Focal Plane is a surface that passes through the principal focus perpendicular to the principal axis of the lens. Focal Length is the distance of the principal focus from the optical centre of the lens. Note 1: In Fig. 18:15 (see below) a line has been drawn through the optical centre of the lens perpendicular to the principal axis. For simplicity, we may represent the entire refraction of light as occurring at this line. Actually, of course, a light ray will be refracted both Fig. 18:15 Represented and Actual Paths of Light Through a Lens. on entering and on leaving the lens. This actual path of the light through is shown by drawing a line the lens between the point where the incident ray enters the lens, and the point where the refracted ray leaves the lens. Note 2: A ray through the optical centre of a lens may be considered as passing straight through the lens. This lens is through which the light is travelling are almost parallel. Therefore, the light will be refracted on entering and on leaving because surfaces the the of REFRACTION OF LIGHT—LENSES Sec. IV: 29 in such a way that the emergent ray will be parallel to, but laterally displaced from, the incident ray (Sec. IV: 24). We shall consider all our lenses to be very thin, so that the amount of lateral displacement is negligible. Increasing either or both of these increases -.the amount of bending of the light and so shortens the focal length. The powers of lenses used in most optical instruments are usually expressed in terms of their focal lengths. IV: 29 FOCAL LENGTH OF LENSES To determine the focal length of a lens the position of the principal focus In optometry it is more usual to deal with the power of a lens rather than its focal length. The unit of
power is the dioptre, which is the power of a converging lens of focal length one metre (100 centimetres). The shorter the focal the greater the power of the length, lens, and accordingly the power can be the focal related length the by to formula: r, , P (dioptres) X 100 / (cm.) According to our convention of signs (Sec. IV: 18, and IV: 32), a convex lens with a real principal focus has a positive focal length and a 4- power; a concave lens with a virtual principal focus has a negative focal length and a — power. If a lens is used in a different medium, its focal length and therefore its power will change. This can be shown by placing a lens in a tank of turbid water (Fig. 18:17). Shine a parallel beam of Comparison of Focal Lengths Fig. 18:17 of Lens (a) In Air (b) Immersed in Water. 205 Concave Fig. 18:16 Principal Focus of a Convex Lens and a Concave Lens must first be determined and then its distance from the optical centre of the lens measured (Chap. 21, Exp. 12). This may be demonstrated by placing a convex lens on an optical disc. Shine a number of rays, parallel to the principal axis, through the disc and note the point through which they converge. This point is the principal focus (Fig. 18:16). The focal length depends upon the index of refraction of the lens and its thickness. Chap. 18 LIGHT : For example, a glass globe of water in sunlight could focus the sun’s rays. If flammable material should be located at the principal focus of this lens a fire could easily result. Use is made of this very fact in some types of sunlight recorders used by weather bureaus. A sunlight recorder is a device used to determine the number of hours of bright sunshine. In Ontario, the meteorological department uses Campbell-Stokes Sunshine Recorder consists essentially of two parts 18:18). (Fig. the It ( 1 ) A glass sphere which brings the sun’s rays to a focus. (2) An approximately spherical metal light through the water and the lens, and compare the focal length in water with that previously obtained in air. It will be found to be longer. This is to be expected as there is a smaller decrease in velocity of the light going from water to glass than when going from air to glass. Consequently there is less bending of the light, and therefore a longer focal length. A convex air lens in water would function as a diverging lens. ExIt can be constructed by plain why. cementing watch glasses together using a waterproof cement. Care should be observed in the use and location of spherical transparent objects. 206 ; REFRACTION OF LIGHT—LENSES Sec. IV: 31 bowl carries cards which form a belt on which the sun burns a record. lens. This ray may be considered straight through the as lens (Sec. IV; 28). passing recorder. Real care must be observed in setting up There must be no obthe structions that would shield the recorder It must be placed from the sun’s rays. table, and made perfectly on a rigid Instructions are always provided level. for adjusting for latitude, and for adjusting for the time meridian. The glass ball must be kept perfectly clean at all times. IV: 30 IMAGES IN CONVEX LENSES Previously we saw how images were ( Sec. IV formed in concave mirrors 15). Convex lenses form a very similar series of images by refracting the light that passes through them. In experiment 13, chapter 21 the method of studying the characteristics and position of the images formed is given. It will be found that as the object approaches the lens up to the focal plane the image formed on the opposite side of the lens is real, inverted, and gradually becomes larger in size as it moves farther from the lens. When the object is located inside the focal plane, the image becomes virtual, erect, and is located on the same side of the lens as the object. As in mirrors, so in lenses, it is possible to verify these results by means of simple geometric diagrams. To do so it is necessary to draw two rays from any point on object, and determine through what point they are focused by (Fig. 18:19). The two most the lens the suitable rays are; ( 1 ) A ray from the tip of the object parallel to the principal axis. This ray on passing through the lens is refracted principal through the focus. , (2) A ray from the tip of the object through the optical centre of the Fig. 18:19 To Locate a Real Image in a Convex Lens. The point where these two refracted rays cross locates the tip of the image. A similar construction for other points on the object will locate the corresponding points on the image. Fig. 18:20 shows how to locate the virtual image obtained when the object is inside the focal plane of the lens. Fig. 18:20 To Locate a Virtual Image in a Convex Lens. The construction is identical to that just Because the refracted rays described. diverge from each other, it is necessary to produce them back to where they appear to meet. IV; 31 IMAGES IN CONCAVE LENSES By referring to experiment 14 chapter 21 it will be observed that concave lenses can form virtual images only. These are 207 Chap. 18 LIGHT always erect, smaller than the object, and located at less than the focal dis(Fig. 18:21). A tance from the lens close similarity will be noted between the images formed by concave lenses and those formed by convex mirrors (Sec. IV:16). Fig. 18:22 shows how the eye sees the Fig. 18:21 To Locate the Image of image produced by a lens. an Object in a Concave Lens. (b) Virtual Images. IV: 32 THE LENS FORMULAE The formulae we obtained for curved mirrors (Sec. IV: 18) also apply to lenses, that is, (a) Magnification Formula Height of Image Distance of Image FI eight of Object Distance of Object A fio~~ Do (b) Distance Formula Distance of Object Distance of Image Focal Length 1 + 1 = 1 D. D< f The same convention of signs must be followed, namely: real is posi- tive, virtual is negative. 208 REFRACTION OF LIGHT—LENSES Sec. IV:32 Examples 1. An object 5 cm. tall is placed 30 cm. from a convex lens whose focal length is 10 cm. (a) By means of an accurate scale diagram locate the image, and state its characteristics. Fig. 18:23 Scale 5 cm. = ^ in. (b) By using the lens formulae, determine the position of the image, and its size. How could you tell from your answer whether it is real or virtual? Do — 30 cm. Di — ? 10 cm. / Do — + 1 , 30 Di 1 / 1 Di~ 10 1 1 D~ 10 1 30 _ 3 — 1 ~ 30 .'. Di=: 15 Note: Since the image distance i; + 15 cm., therefore the image is real. .*. Imaae distance is 15 cm. Do = 30 cm. Di = 15 cm. Ho = 5 cm. H,= ? . .^_D, ’ Ho~ Do ‘ . Hi _ 15 ~ 30 Hi = 2.5 5 Height of image is 2.5 cm. 209 Chap. 18 LIGHT 2. A concave lens has a focal length of 4 in. An object 1 in. high is 12 in. from the lens. (a) Determine the position and size of the image. (b) Verify your answer by making an accurate scale diagram. / = — 4 in. Do = ( — since / is virtual) 12 in. ..Ill D~ f .•. 1 + '- = -1 12 Di 4 . 1 _ 1 1 _ —3 —1 .\Di = -3 Image distance is 3 in. Note: Since the image distance is negative, therefore, the image is virtual. Hi = ? Ho = I in. Di = 3 in. Do= 12 in. . .Hi _1^ ‘ Ho~ Do ‘ ^- 1. ' 'T~ 12 .\Hi = . 25 Height of image is .25 in. IV: 33 APPLICATIONS OF LENSES part As lenses are an essential of almost every optical instrument, a discussion of their major applications is reserved until chapter 20 where several optical instruments are described. However, to relate specifically to the facts learned in the preceding sections we will mention a few simple uses here. The camera (Sec. IV; 44), (Sec. IV: 45), IV; 48) the telescope and the projection lantern (Sec. IV: 49) all contain a convex lens as an essential part of their construction. All these in- the eye (Sec. struments produce real, inverted images because the object viewed in each case is beyond the principal focus of the lens. (Sec. IV;46) In the magnifying glass and microscope (Sec. IV; 47) an enlarged, erect, virtual image is obtained because the object is inside the principal focus of the convex lens. Concave lenses are used along with convex lenses in many optical instruments to overcome certain defects, e.g., chromatic aberra(Sec. IV; 42), that would be apthe convex lens were used parent if tion alone. 210 REFRACTION OF LIGHT—LENSES Sec. IV: 34 IV : 34 QUESTIONS 1. 2. 3. A (a) Define refraction of light. (b) Describe and explain fraction of light as it passes obliquely from one medium into another of different optical density. the re- (a) Define index of refraction. (b) Draw a diagram showing how you can see a coin lying on the bottom of a dish filled with water, though the coin would be hidden if the dish contained no water. (c) What is the velocity of light in quartz? (See table p. 1 99 for index of refraction of quartz). (d) The velocity of light in a diamond is 75,300 miles per second. What is its index of refraction? (a) How do you account for shimmering effect seen in the above a hot radiator? (b) How does the same principle account the twinkling the the for of air stars? 4. Place a thick glass plate on a line drawn on a piece of paper. View the line obliquely. Describe and explain what is observed. 5. aid of a diagram (a) With the explain why an oar appears bent when partly immersed in water and viewed obliquely. (b) Is the index of refraction greater for glass in which the velocity of light is 1 24,000 miles per second, or for water, in which the velocity is 1 40,000 miles per second? Why? 7. (a) Define critical angle. Illustrate your definition with a labelled diagram. (b) Would the be greater for water (Index of refrac- angle critical 8. 9. 1.33) or for glass (index of tion refraction 1.5)? Why? (a) Draw a diagram to show how a right-angled prism may be used to secure (i) one total internal reflection, (ii) two total internal reflections. (b) Explain why total reflection occurs in these two cases. (c) Why are total-reflection prisms preferable to mirrors in many optical instruments? (a) Compare the action of lenses to that of two prisms. (b) Define principal focus of a lens. (c) How can you determine exper
imentally the focal length of a lens? (d) Calculate the power of a lens whose focal length is 0.15 metres. 10. (a) What is the purpose of a sunlight recorder? (b) What approximate position relative to the glass sphere should the recording-belt occupy In the sunlight recorder? 11. (a) Distinguish between convex and concave lenses. (b) Summarize the types of images possible with both types of lenses and the conditions under which each is obtained. 12. An object is located at a point more than twice the focal length from a convex 6 . (a) Define angle of deviation. lens. (b) State four factors that govern the amount of deviation produced by a prism. (c) Is the index of refraction of glass (a) By means of a diagram locate its image. (b) State the characteristics of the image. constant for all colours of light? 13. (a) What do we mean by the magni- Explain your answer. fication produced by a lens? 211 Chap. 18 LIGHT 9. (b) How does the magnification depend on image distance and object distance? B 1. What is the index of refraction of a liquid in which the speed of light is 1 55,000 miles per second? 2. The index of refraction of diamond is 2.47; that of window glass is 1.51. How much faster does light travel in the glass than in diamond? 3. Light surface strikes the of glass making an angle of incidence of (a) 60°, (b) 45°, (c) 30°. The index of refraction of 1.5. By means of accurate geoglass is metric diagrams draw the refracted ray for each case. Using a protractor, measure the angles of refraction. 4. By means of an accurate construction determine the size of the angle of incidence when the angle of deviation is a minimum in an equilateral crown-glass prism. Measure the angle of deviation. 5 . In which material does light travel faster, one with a critical angle of 25° or one with a critical angle of 30°? Explain, using appropriate diagrams. 6. An object 1 5 cm. high is located (i) 75 cm. (ii) 60 cm. (iii) 45 cm. (iv) 30 cm. (v) 20 cm. from a convex lens whose focal length is 30 cm. (a) By means of accurate scale diagrams locate of the image for each position of the object. (b) State the characteristics of each position the image. 7 . By means of an accurate scale diagram locate the image produced by a concave lens whose focal length Is 1 5 in. of an object 6 in. high and 2 ft. distant. State the characteristics of the image. 8 . A camera forms an image 8 cm. from the lens. If the object is 400 cm. away and 250 cm. tall, what is the height of the image? 212 The image of a tree in a miniature camera is 50 mm. from the lens and 30 mm. high. The tree is 1 5 metres away. How tall is the tree? 10 . The image of an object 3 in. from a lens is formed 20 ft. from the lens. How many times is it magnified? 1 1. The image of an object 24 ft. from a lens is focused clearly on a screen 3 ft. from the lens. What is the focal length of the lens? 12. A convex lens forms a virtual image at a point 1 2 cm. from the lens. The object distance is 8 cm. Find the focal length of the lens. 13. A tree 1 00 ft. from a camera lens has its image very close to the principal focus of the lens. If the tree is 66 ft. tall and the image is 4 in. tall, what is the focal length of the lens? 14. A candle is 1 2 cm. from a convex lens of focal length 8 cm. What is the distance from the lens to the image of the candle? 15. A student uses a convex lens to look at an object held 4 cm. from the lens. If the focal length of the lens is 5 cm., how far is the image from the lens? What kind of image is it? 16 . The image in a camera is 1 0 cm. high and 1 4 cm. from the lens. If the object is 100 cm. tall, what is the focal length of the lens? 17 . A jeweller uses a converging lens of in. to examine a diamond. focal length 1 The virtual image is 10 in. from the lens. Find (a) the object distance, (b) the magnifi- cation. 18 . When photographing a scene at a distance of 6 ft. from the lens, you find that the distance between the lens and the film is 6 in. (a) What is the focal length of the lens? (b) What is the actual size of a portion of a scene which occupies a space 3 in. X 5 in. on the film? CHAPTER 19 COLOUR IV: 35 INTRODUCTION TO COLOUR Imagine how drab and uninteresting the world would be if there were no pleasure and colour. stimulation we receive from the colours of nature—the blue sky, the green grass, Think of the the beautiful flowers, the gorgeous hues of the sunrise and sunset. The use of colour in photography, in movies, and in book illustrations has added tremendously to our enjoyment of these things. How appealing and satisfying are some of the beautifully coloured mastei-pieces of art! Economically too, colour plays a very important role, as shown by the varied colours used in home decorating, clothing, advertising, and the like. For thousands of years men have known that colourless glass of certain shapes, as well as frost, diamonds and other crystals, produce light of many colours when illuminated by white light. Sir Isaac Newton Until the time of everyone supposed that or crystals produced the light by giving something to the light as it was reflected by, or transmitted through, them. It was he who, after thorough scientific investigation arrived at the true explanation of the nature of colour and the character of white light. glass the IV: 36 COMPOSITION OF WHITE LIGHT (a) Dispersion In 1666, Newton permitted a beam of sunlight, passing through a circular hole in a window blind, to fall on a triangular glass prism. He found that the light was refracted or deviated, from its original Instead of obtaining a simple path. image of the hole, he obtained a band of colours which he called a spectrum. Its colours were the same as those found in the rainbow— red, orange, yellow, green, blue and violet, with each colour merging imperceptibly into the ne.xt (Fig. Red •Orange Yellow Green Blue Indigo Violet Fig. 19:1 Dispersion of White Light into its Spectrum. that Newton reasoned white 19:1). light must be composite, that is, made up of a combination of the above colours. The separation of the colours by the prism he called dispersion. A further study of dispersion is made in experiment 15, chapter 21. 213 Chap. 19 LIGHT . ' As shown in Fig. 19:1 this dispersion occurs because the different colours are refracted different amounts by the prism and are deviated different amounts from their original direction. Red is always bent the least from its original direction and violet the most, with the other colours intermediate between these. Red light has the longest waves and violet the shortest, the wave-lengths of the other colours being between these two. the rays encounter less opposition to their passage through the glass prism and, therefore pass through more rapidly than do the violet rays. longer red Evidently, As we learned earlier (Sec. IV; 22), refraction is caused by a change in velocity of light on passing from one medium to another. In free space, or in air, light of all colours travels at the same rate, 186,000 miles per second. On entering an optically denser medium such as glass the light is slowed down, and on leaving this medium it speeds up to regain its original velocity in air. Since the different colours are bent different amounts by the prism, it follows that they must be velocities through the prism. Red light, bent the least, must slow down the least on entering the prism and therefore speeds up the least on leaving. In contrast, violet light must slow down the most on entering, and speed up the most on leaving the prism. travelling different at A question that immediately arises of course is, “Why does the prism have this different effect on the different colours?” The answer relates back to our wave theory of light (Sec. IV: 7). The different colours of light result from different wave-lengths. Table of Wave-Lengths* (1 Angstrom (A) = 10'® cm. Infra-red above 7000 A Red Orange Yellow Green Blue ^ Visible Spectrum 6500A 6000A 5800A 5200A 4700A 4100A ^ Violet Ultra-violet below 4000 A [ * The wave-length shown for each colour is representative only. Each colour consists of wave-lengths that merge into those of the colours adjacent to it. For example, the wave-lengths of red lie between 6470 A and 7000 A. Colour bears the same relation to light that pitch does to sound. The pitch of a sound depends upon the number of vibrations per second that reach the ear Similarly the colour of (Sec. 11:12). light depends upon the number of vibrations per second that reach the eye. In light, however, since the frequency is so great, it is customary to describe the colour in terms of wave-lengths, rather than in terms of vibration frequency. Example Calculate the vibration frequency of red light. Velocity = 3X1 0^® cm. per sec ( Sec. IV : 6 ) Wave-length = 6500 A — 6500 X 10"® cm. Velocity = Frequency X Wave-length (Sec. 11:5) Frequency = Velocity Wave-length _ 3 X 1010 - 6500 X 10-8 = .46 X 1013 The vibration frequency of red light is .46 X lOi^ vibrations per second. 214 COLOUR Sec. IV: 36 Fig. 19:2 Recomposition of the Spectrum into White Light, (a) By Reversed Prisms, (b) By a Converging Lens, (c) By Newton's Disc. (b) Recomposition Newton further supported his theory concerning the composite nature of white light by showing that the colours of the spectrum could be recombined, giving white light (Chap. 21, Exp. 16). He first arranged two prisms with their refracting edges in opposite directions. On passing white light through these reversed prisms, white light was obtained through Fig. 19:2a shows that the first them. prism disperses the colours, while the second prism recombines them by reversing the original refraction and causing the light waves to be superimposed on each other. A similar effect can be secured by using a converging lens to catch the dispersed coloured light from If this light is brought to a a prism. focus on a screen a spot of white light will be obtained (Fig. 19:2b). If the screen is moved beyond the focus, a reth
e original colours will be versal of obtained. Newton also prepared a colour disc on which were coloured sectors whose sizes and colours corresponded fairly closely to the coloured bands obtained in a pure spectrum of white light (Fig. 19:2c). If this disc is strongly illuminated and rapidly rotated it will appear white. This phenomenon is due to what is called ‘the persistence of vision”. Any visual impression on the retina of the eye persists for a short period of time after the It is on this cause has been removed. principle that movies are made to appear continuous. In reality, each picture is thrown on the screen for a fraction of a second, its image persisting in our vision until the next appears. Similarly, if the coloured disc is rotated rapidly enough, the impression produced by one colour persists, while impressions produced by all the other colours are received on the same portion of the retina. Thus all the colours of the spectrum will be superimposed on the retina, and will give the sensation of white light. 215 Chap. 19 LIGHT IV : 37 THE RAINBOW The rainbow is a spectrum of sunlight formed by water droplets. A ray of sunlight entering a drop of water is refracted at A (Fig. 19:3a), the violet rays being refracted more than the red rays. Fig. 19:3 The Rainbow (a) Refraction and Total Reflection in a Raindrop. (b) The Primary Bow. (c) Double Reflection to Produce the Sec- ondary Bow. The refracted light is totally reflected at B and is again refracted at C so that the different colours are dispersed. Each drop of water forms its own little specIn the actual bow which the trum. observer sees, the red rays come at an angle of 42° from drops of water higher 216 (Fig. in the sky, and the violet rays come at an angle of 40° from drops of water lower 19:3b). The other in the sky colours come from drops between these angles. The rainbow has the shape of a bow, since the eye of the observer is at the apex of a cone from which he sees the coloured rays refracted from drops, all of which must subtend approximately the same angle at the eye (between 40° and 42°). Sometimes a larger, but fainter secondary bow is seen above the primary. The colours in it are reversed, the violet being on the outside. The light enters the lower part of the water drops, is refracted, and twice totally reflected before it leaves the drop (Fig. 19:3c). The light is refracted from the drops of water at angles of from 51° to 54°. The double reflection not only reverses the colours, but also absorbs more light thus causing the secondary bow to be fainter than the primary bow. IV : 38 BEYOND THE VISIBLE SPECTRUM So far in our study of the spectrum we have considered only those radiations to which the eye is sensitive. Actually the spectrum of sunlight extends well beyond Sir William Herschel, its visible limits. in 1800, on placing the blackened bulb of a thermometer in the various parts of the spectrum, discovered that the heating effect observed at the red end was continued when the thermometer was placed well beyond the visible limit. He thus indicated the existence of a wide range of invisible radiation beyond the red end of the spectrum (Sec. 111:15). These infra-red radiations, as they are called, convey almost half of the sun’s total outpouring of energy into space. They can penetrate mist, smoke and haze and hence are very suitable for distance photophotography, graphy in the dark, and detection of reconnaissance, ^ <c ^ III 3 \| i 0) -c ^ ^ I <D "D Q- c s D I c c O (D o O) S •4^ Q. O O "D ^ ^-2 < fi-g O O O _C CO — CO u -tu C <13 1 °? CO Q. to LU O Frequency 10 H lo' lO' 10- lo"- 10 lO'-l 10 lo'H - 12 10 10 - 10 10 * — lo' lo' lo' lo' *- 10 lo'- lo'- lo’ 10 10 COLOUR Sec. IV: 38 Electromagnetic Wave Spectrum Generation (Uncertain) Disintegration of atomic nuclei Sadden stopping of fast-moving electrons Radiated by very hot bodies and ionized gases Radiated from hot bodies Heat radiations from hot bodies Spark-gap discharge Oscillating three-element vacuum tube ^ Used in Radio Very long waves Coil rotating in a magnetic field Fig. 19:4 217 Chap. 19 LIGHT (a) American Optical Company (Canada) Ltd. value of a holiday away from the city. For ultra-violet treatment one must be exposed directly to the sun’s rays, or use windows of special “vita glass” which is transparent to some ultra-violet rays. Proper precautions must be taken as overtreatment can produce severe sunburn, as well as damage to the retina of Ultra-violet lamps simply conthe eye. sist of carbon or mercury electrodes between which an electric current produces an arc. Such lamps are usually fitted with quartz lenses which are transparent to the ultra-violet rays. The presence of ultra-violet rays beyond the visible spectrum of white light may be shown by using a quartz prism in place of the glass prism in Fig. 19:1. If a sample of anthracene is placed beyond the violet it will be seen to fluoresce. In addition to the invisible radiations forgeries or substitutions. Under these conditions special photographic film sensitive to infra-red must be used. These rays also have therapeutic value such as the treatment of rheumatism and in other aches and pains. The year following Herschel’s discovery of the infra-red, Ritter discovered, by its action in blackening certain silver salts, a band of invisible radiation beyond the violet, called the ultra-violet. Ultra-violet rays do have special photographic effects, and also make some substances such as vaseline and certain minerals “fluoresce” or give out “cold light”. They have a highly beneficial effect on health, accelerating the manufacture of vitamin D under the skin. These ultra-violet rays are easily absorbed by mist, smoke and ordinary glass. Consequently, one readily the recognizes 218 COLOUR Sec. IV: 39 discussed above in connection with the solar spectrum, there exist other invisible radiations such as radio waves, X-rays, and gamma rays (rays emitted by radioAll these different active substances). radiations are propagated as waves with the same velocity as light, the difference in their properties being due to their different wave-lengths. The relationship between these various electromagnetic is shown waves, diagrammatically in Fig. 19:4. as they are called, IV: 39 SPECTRUM ANALYSIS The spectroscope is an important instrument by means of which the spectra of luminous bodies can be produced and measured. It consists of a tube called the collimator (Fig. 19:5), at one end of which is a convex lens and at the other end an adjustable vertical slit. The length of the tube is equal to the focal length of the collimator lens, so that when the slit is illuminated by light from the source, S, under examination, a parallel beam can be directed onto a prism situated on a small turn-table at the centre of the instrument. The dispersed beams produced by the prism are received by a telescope focused for parlight and fitted with cross-wires. allel This telescope can be moved round a circular scale against which the deviations of the constituent parts of the spectrum formed can be measured. The this way of the light examination in from different luminous bodies reveals that each spectrum is characteristic of the source, and a study of these spectra yields much valuable information to the physicist and astronomer. Accordingly we shall briefly consider here some of the more important spectra and their significance (Fig. 19:6), Continuous Spectra: These consist of a number of coloured bands each shading off gradually into the next. They are produced by incandescent solids, e.g., arc-lamps, white-hot iron, etc. Incandescent gases and Line Spectra: vapours emit light which when analysed produces spectra consisting of a number of well - defined coloured lines or col oured images of the slit. Each element has its own characteristic line spectrum which provides a certain and accurate means of identifying the element (Exp. 17, Chap. 21). The presence of even a minute quantity of an element in a mixture can be detected by spectroscopic analysis, and it is interesting to note that the method has been the means of discovering new elements. If the spectrum of a substance contains lines which do not correspond with those of any known element, the obvious conclusion is that there is present an element as yet not known. In this way the elements caesium and rubidium were discovered by Bunsen. Line spectra can be produced by the Bunsen flame only in the case of subvaporized. stances Other substances must be vaporized, using an electric arc or a spark dis- which easily are Fig. 19:7 An Electrical Discharge Tube. The spectra of incandescent charge. gases are usually obtained by means of special tubes containing a sample of the gas at low pressure through which the discharge from an induction passed (Fig. 19:7). coil is Absorption Spectra: If a sodium flame is set up between the slit of a spectroscope and a high-power electric lamp (Fig. 19:8), a dark line will be observed in the continuous spectrum of the light from the relatively hotter lamp in a position corresponding to that of the 219 Chap. 19 LIGHT yellow line of the incandescent sodium vapour. With other vapours dark lines would be formed in positions corresponding to their line spectra. A spectrum Fig. 19:8 The Production of an Ab- sorption Spectrum. crossed by dark lines in this way is called an absorption spectrum, and it was found that all substances when interposed in the path of light originating from a higher temperature source absorb from it light of the same wave-length that they themselves emit. Most of the The Solar Spectrum is an absorption spectrum containing hundreds of dark lines carefully studied by Fraunhofer in 1814. These lines are due to selective absorption from the radiation emanating from the extremely hot core of the sun by the relatively cooler gases in the sun’s atmosphere. lines have been found to correspond to lines in the spectra of elements present on the earth,
and we have thus good grounds for believing that the chemical components of the sun and the earth are similar. A certain group of lines did not correspond with those of any known element. They were accordingly attributed to an element which was named helium, and which was ultimately discovered on the earth twenty-six years later. The constitution of the stars is determined in a similar way from an examination of their absorption lines. IV : 40 NATURE OF COLOUR (a) Colour We have that colour is a property of light waves (Sec. established already 220 It may be defined generally, IV; 36). as the response of vision to different wave-lengths of light. Colour is the alphabet in our visual language, through which we make interesting our description of things and ideas which otherwise would be dull and prosaic. Our present knowledge and use of colour is based largely on the sciences of psychology, chemistry, and physics. Psychology considers colour through the physical operation of the eye and the mental impression created by colour on the brain. As a result, the actual perception of colour is a highly personal experience which may be influenced by such factors as health, fatigue and response of the eye to colour (colourAccording to psychologists blindness). the average person recognizes four distinct primary colours, red, yellow, blue and green. Colours do have a strong influence on people. Tensions can be heightened or relaxed through a change in colour harmonies; a feeling of warmth or coldness can be obtained by colour schemes used; colour can be stimulating or restful, it can attract or it can repel. Chemistry deals with the production of materials which have the ability to absorb or reflect part of the light which falls upon them. These are chiefly dyes and pigments. The explanation for the colours of these materials is given in part (b) below. Physics deals with colour from the standpoint of differences in wave-lengths of the coloured lights, and the effects of superimposing these on one another. The additive theory of colour to explain (c) these effects is discussed in part below. These three colour theories are the basis for our modern usage of colour. Though separate, they are also interdependent. The range and intensity of the colours we see depend upon the quality and quantity of light, the nature of the . COLOUR Sec. IV: 40 surface visible, and our ability and interpret what meets our eyes. to see (b) Colour of Objects— The Subtractive Theory There are three colours, yellow, red, and blue, which in pigments are the source of all other colours. By mixing these three colours in the right proportion, all other colours may be obtained. No mixture of other colours will produce any of these three colours. For this reason yellow, red, and blue are called primary colours. A colour chart has been prepared (Fig. 19:9) in which these primary colours are placed the same distance apart along the edge of the circle. Mixing any two primary colours produces a third colour called a binary colour which is intermediate between those primaries, e.g., yellow and red produce orange. Various hues can be produced by mixing a binary colour with a primary colour used in making the binary, e.g., yellow and orange produce yellow orange. Certain colours seem to strengthen each other when they are seen together, and hence are said to be complementary to each other. In the chart the colours on the opposite sides of the circle are complementary. White is the total addition of colour, is of then, the result spectrum, and reflects and not, as often believed, the absence of colour. It is produced when a surface reflects all colours equally. Black, on the other hand, is perceived when a surface absorbs all colours and reflects none. Colour, partial absorption, and consequent subtraction, of a band of colour from the spectrum. The surface of an object will appear red when it is of such nature that it absorbs all but the red wave-lengths of red the wave-lengths (Chap. 21, Exp. 18). When we mix pigments, each one subtracts certain colours from white light, and the resulting colour depends upon the light that is not absorbed. For example, if we mix a blue pigment with a yellow one we get a green colour. The blue colours from subtracts or absorbs all white light, except green, blue and violet, while the yellow subtracts all except green, yellow and orange. Green is the only colour not subtracted by either pigment. For this reason the two pigments produce green. these Similarly, transparent objects are able certain colours to absorb or subtract from white light. If white light is passed through a piece of red glass, the glass absorbs all light except red, and a little orange, and we obtain red light (Fig. 19:10a). Objects that are transparent to only one colour are called colour filters. The combination of a yellow and a blue filter placed over a single white light source will result in a green light because all except the green wave-lengths of the spectrum have been absorbed a combination of three filters, each corresponding to a primary colour, will absorb all the colours and allow practically no light to pass (Fig. 19:10c). (Chap. 21, Exp. 18) 19:10b). Similarly (Fig. The light which falls upon an object can also determine its colour. If the light does not contain all the colours of 221 Chap. 19 LIGHT RED YELLOW BLUE (a) (b) Fig. 19:10 Colour Filters (The Subtractive Theory) (a) Transmission of Light by Different Coloured Filters. (b) Combination of Yellow and Blue. (c) Combination of Red, Yellow, and Blue. o the spectrum then the true colour of the object will not be observed. The light given off by a light bulb does not contain all the colours found in sunlight. For this reason an article of clothing may appear quite different in the sunlight than when seen under artificial light. (c) Coloured Lights— The Additive Theory Having produced different coloured lights by the use of filters, two or more of these may be added together to produce still other colours. The three colours, red, green and blue, which combine to produce white are called the three primary colours for lights. None of these primary colours can be produced by adding other colours of lights. However, any other colour can be pro- 222 19:11). duced by a proper combination of them For example, red and (Fig. blue light when combined produce violet; red and green light added together Fig. 19:11 Coloured Lights (The Ad- ditive Theory). COLOUR Sec. IV: 41 will produce yellow, and so on. The colour resulting from the simple combination of each pair of primary colours is complementary to the third primary; for example, violet is complementary to green, and yellow is complementary to If we superimpose these comblue. plementary coloured lights we again get white light. Contrast how colour is obtained in mixing pigments with how it is obtained in mixing coloured lights. In the former the resulting colour is the one not absorbed or subtracted by either pigment in the latter the (the subtractive effect) ; resulting colour is obtained by adding the effect of the individual colours (the additive effect). Both methods are widely used in industry. The mixing of pigments to produce our multi-coloured paints, the many vegetable and synthetic dyes for treating fabrics, etc., and the use of filters in photography, all are applications of the subtractive effect mentioned above. The additive effect is applied in the use of coloured spotlights to illuminate actors on stage, or participants in a carnival or pageant, as well as to produce changing and spectacular effects in the illumination of Niagara Falls and in some of the coloured advertising that is so common nowadays. textiles, Colour is of tremendous importance Its use in advertising has in industry. been mentioned above. Bright contrasting colours of paint on walls, floors, and machinery not only provide better visual conditions but also boost morale, speed up production, and affect safety re- cords. Colour codes are used in busi- ness for invoicing, in electrical indus- tries to denote polarity, voltage, resistance, and the like, and in shipping to encourage due precautions in the han- dling of dangerous articles. All, of course, are familiar with its use in traffic control. (d) Colour Vision are there theory, widely The way in which light is changed into the sensation of sight is not definitely The most known. accepted the Young-Helmholtz theory, of colour three states nerves in the normal eye. One of these sets of nerves responds only to blue light, another set to green light, and the third set to red light. The colour of an object is determined by the relative degree of stimulation of each type of nerve. sets A small proportion of men (about 3/2%) and women (about ^2%), do not have all three of these sets of colour nerves and hence they do not receive the same sensation of colour from an object as a person with all three sets of colour nerves. Such people are said to be colour blind. By this we do not mean that they do not see colour in objects, but that they do not see the same colours as a person with normal eyes. This defect through inherited is usually exists from birth; and there is no known cure for it. it mother; the pattern Because a colour-blind person sees a different colour pattern in a landscape seen by a person from the having normal vision, colour-blind observers are very efficient in detecting camouflage and are so used during wars. Some colour-blind persons are unable to distinguish reds and greens from each other and, consequently, have real difficulty with traffic signals. IV: 41 COLOUR PRINTING Most of the coloured pictures in books and magazines are made by using four separate printing-plates, each plate printing a different coloured ink on white paper—yellow, red, blue and black, in that order. The plates are made from four photographic negatives of the same are made negatives subject. through These fine-meshed coloured
filters. 223 Chap. 19 LIGHT the the creation of each stained with one of three primary colours (the black one is not the actually used in Fig. 19:12). The coloured picture in fine-meshed filter breaks the subject into numerous small dots on the negative. The filter determines what colours can pass through the camera and become recorded on the negative. Thus dots representing only certain colours are recorded on each negative. The first plate reproduces the dots of the first negative in yellow ink on white paper. The second plate reproduces those of the second negative in red ink on or between the yellow dots, etc. The final effect is one of individual and overlapping dots of colour. Where a dot of one colour overlaps a dot of another colour a new colour may be obtained. The eye blends these dots of the four original and the mixed colours to give the total effect. IV: 42 CHROMATIC ABERRATION in contact (Sec. IV: 28), Since the action of a convex lens is similar to that of two prisms with their bases dent rays, in addition to being deviated to form a convergent beam will be dispersed on passing through the lens. The various colours are brought to diffoci on the principal axis, the ferent focus for violet rays being nearer the inci- than red rays that for (Fig. lens 19:13a). In consequence it is impossible to obtain a well-defined image when white light is used, the image received on a screen having a red fringe when placed at A and a blue fringe when placed at B, there being no position be- (a) Cr. FI. V R F fb) Fig. 19:13 Chromatic Aberration. (a) Its Cause. (b) Its Correction—An Achromatic Lens. tween A and B at which the image is This non-focusing of sharply in focus. light of different colours is known as chromatic aberration. By using two reversed prisms, one of crown glass and the other of flint glass which has a higher index of refraction, it is possible to obtain a non-dispersive combination which still has the power of deviating the light (Fig. 19:13b). is possible to combine a convex lens of crown glass with a weaker concave lens of flint glass so that the dispersion produced by the two lenses is equal and opposite. A lens combination of this type is called an achromatic (without colour) lens. These achromatic combinations are still converging and accordingly can be used as objective in field glasses, cameras, and other optical instruments. Similarly it telescopes, lenses 224 Red, Print- green- a Yellow, through 5. made filter; Four-Colour In Note; negative red-yellow print. Plates. Blue and Plate, a Red finished Red 2. through the Yellow, made filter; in the shades to negative and Addition violet-green Plate, a through Blue 4. made yellow; on negative colours in Used various is Plate the Block Notice a ing Red 3. combined. filter; Blue yellow and Plate, Yellow L IV : 43 COLOUR QUESTIONS Sec. IV: 43 1. (a) Define: spectrum, dispersion. (b) Why does a prism deviate violet light more than red light? (c) Calculate the vibration frequency of violet light (See table, p. 214, for wave-length of violet light). 2. Describe and explain three methods for recombining the spectral colours into white light. 3. Explain how refraction produces a solar spectrum (rainbow). 4. Write a note on (a) infra-red rays, (b) ultra-violet rays, indicating, (i) how each differs from visible light. (ii) how each can be detected. (iii) how each is used. (a) What is the purpose of a spectroscope? 5. 8 . (i) a continuous (b) Distinguish a line spectrum from an absorption spectrum. How is each produced? spectrum, (ii) (c) Why does red glass appear red, blue glass appear blue, and ordinary glass appear colourless, white light? State a general rule for the in colour of transparent objects. (d) What would be the colour of the objects in (b) when viewed through (i) a red-glass filter, (ii) a blue-glass filter? (e) Why is it unwise to buy coloured clothing by artificial light? (a) How does the additive theory of subtractive light differ from the theory? (b) In the additive theory what are (i) the primary colours, (ii) complementary colours? (c) If coloured spotlights are shone on a white screen, (i) what colour will be obtained if red and blue lights are used? (ii) what coloured used to produce yellow? lights should be 6 . What is the physicist’s explanation of (a) coloured light, (b) white light? 7. (a) In the subtractive theory of light what are (i) the primary colours, (ii) complementary colours? (b) Why does a rose appear red, grass green and a dandelion yellow when seen in sunlight? State a general rule for the colour of an opaque object. 9. How does the physicist account for the difference between mixing coloured lights and coloured pigments? 10. What is colour-blindness? How does the Helmholtz theory explain it? 11. Describe the process of colour print- ing. 12. (a) What is chromatic aberration? (b) What is an achromatic lens? 225 CHAPTER 20 OPTICAL INSTRUMENTS 1553, Porta, Baptista and camera. Johann Kepler, 1604, later added a lens and then a combination of lenses. From these simple beginnings have come many types of cameras ranging from the relatively inexpensive family-type cameras, through the more expensive press and movie cameras, up into the highly specialized cameras used in television and in aerial photography. (b) Structure All cameras are constructed on the same basic principles as were the first crude, wooden, box-cameras. The essential parts of a camera (Fig. 20:1) are: 1. A light-tight box or bellows. 2. A lens to form the image on the film. 3. A shutter (and diaphragm) to con- IV : 44 THE CAMERA The photographic camera is essentially an enclosed box or bellows chamber with an opening or aperture in the front through which light from an object The image is recorded on a passes. light-sensitive material at the back. (a) History The evolution of the camera is credited to men like Roger Bacon, 1267, and Leonardo da Vinci, 1519, who dedeveloped pin-hole scribed and the 226 OPTICAL INSTRUMENTS Sec. IV; 45 the amount of trol admitted to the film. light that is 4. A negative holder or carrier to hold films in position during exposure. 5. A view-finder or ground glass to determine picture area. By the addition of special accessories and other refinements, the utility and scope of the camera may be extended to meet almost any photographic requirement. (c) Focusing In order to focus objects at different distances from the camera, the lens is often attached to a bellows so that it may be moved nearer to the light-sensitive plate or film, or farther away. In section IV : 30 we learned that as the object is moved nearer to a converging lens, the real inverted image gradually moves farther away from the lens. This should indicate how to focus the camera. For distant objects the lens is moved so that the sensitive plate is located at its principal focus. For close objects, the lens is moved away from the sensitive film, for now the image is focused beyond the principal focus of the lens. Most cameras include a focusing scale on the lens mount or on the bed of the camera and many have built-in range finders to assist in focusing. (d) Light Control As has been indicated previously the film or plate in a camera contains materials which are sensitive to light. The plate is coated with an emulsion consisting of gelatine and a silver salt. Light decomposes this silver salt so that when “developed” and “fixed” a negative is obtained consisting of a dark deposit of silver in places corresponding to the light parts of the original image, and light transparent portions in positions corresponding to the darker parts of the original image. One sees, therefore, that it is necessary to be able to control the amount of light that enters the camera. This is done in two ways : ( 1 ) by controlling the size of the aperture or opening through which the light enters; and (2) by regulating the length of time that the light is permitted to enter. The former is accomplished by using a variable diaphragm behind the lens which enables the aperture to be varied, and the latter by means of a shutter, which is a delicate mechanical device, often consisting of a number of overlapping thin metal blades, held in position by means of springs. On being released or tripped, when taking a picture, the shutter opens and closes for a measured time interval. When using a small aperture a longer exposure is required than with a large It is wise practice to use as small one. an aperture as possible as this permits using the central part of the lens only, and prevents errors due to inaccuracies near the edges of the lens. Also greater depth of focus is obtained, that is, objects relatively smaller and greater disstill be at tances from the camera will sufficiently in focus to look clear in the print. IV: 45 THE HUMAN EYE to be perceived. The eye is the sense organ enabling This, the most light wonderful of all optical instruments, can be compared in many respects to a camera (Fig. 20:2). (a) Structure and Action The eyeball is approximately spherical in shape and slightly less than one inch in diameter (Fig. 20:3). The wall of this sphere is composed of two major layers: ( 1 ) the outer covering, the sclerotic coat, substance forming the white of the eye. The front portion of this sclerotic coat forms a transparent, curved section called the tough opaque white a is 227 Chap. 20 LIGHT Fig. 20:2 Comparison of the Eye and the Camera. cornea that protects the eye and aids in refracting light. (2) The inner layer, the choroid coat, is black to prevent internal reflection and to protect the light-sensitive parts of the eye. is also largely nutritive in function. The aqueous and vitreous humours are jellylike materials filling the spaces within It the eyeball, which help to keep its spherical shape. Behind the cornea is a coloured diaphragm called the iris. This is really an Fig. 20:3 The Structure of the Eye. 228 extension of the choroid coat, and its colour is mainly due to the
variable amounts of pigment in it. In the centre of the iris is a circular aperture, called the pupil, which appears black due to the black interior of the eye. The iris contains muscles, which dilate and contract the pupil adjusting the eye to different amounts of light. The pupil is dilated in dim light, and contracted in bright light. This action occurs involuntarily, i.e., without any control on our part, and hence is called the iris reflex. To note its action, stand in front of a mirror in a dark room for several minutes. Then turn on the light and note the immediate contraction of the pupil. Optometrists use drugs, such as atropine, to temporarily dilate the pupil and permit internal examination of the eye. Behind the pupil and iris is the crystalline lens. This is a transparent structure, made up of numerous concentric layers, increasing in density toward the middle. OPTICAL INSTRUMENTS Sec. IV: 45 it so that tension The curvature of its front surface is considerably less than that of the back Both of these features tend surface. to diminish distortion of the image. The lens is encased in an elastic capsule and is held in place by suspensory ligaments which are attached to the choroid coat. These ligaments hold the lens under a normally certain tends to be flattened and adjusted for distant vision. In the choroid layer near the suspensory ligaments are small ciliary muscles. The contraction of these muscles tension exerted by the suspensory ligaments on the lens which therefore becomes more convex and so adjusted for close vision. The adjustment of the lens to form a sharp image on the retina is called accommodation (Fig. 20:4). The normal eye can accommodate itself to objects ranging from infinity to a point about twenty-five ten inches from the eye. This latter point is known as the nearest point of distinct Objects closer to the eye than centimeters or relieve vision. tends the to • Distant Vision (b) Close Vision. this cannot be clearly focused by the unaided eye. their characteristic The retina is the innermost layer of the eye and is sensitive to light. The retina is composed of visual cells, nerve cells, and the fibres which connect these and conduct the nerve impulses to the brain. The visual cells are of two types, called rods and cones, so named because rod and cone of shapes. The rods are concerned with colourless and dim-light vision, while the cones handle colour and bright-light vision. The nerve cells integrate and relay the impulses which they receive, to the optic nerve, and through it to That part of the retina where the optic nerve enters the eye, is insensitive to light and is known as the Its existence can be readily blind spot. demonstrated as shown in Fig. 20:5. The most sensitive part of the retina is a small yellow spot located on the principal axis of the lens. An object being viewed is automatically focused on this brain. the area. X Fig. 20:5 The Blind Spot. Close one eye. Hold the book at arm's length. Stare at the X. Move the book toward you. At a certain distance the circle will disappear, for its image is formed on the blind spot. 229 . Chap. 20 LIGHT (b) Defects of Vision and Their Correction In view of the delicate adjustments made by our eyes and the way we abuse them these days with excessive reading, watching movies, television, only natural that many abnormalities or defects develop. Here, however, we are only concerned with defects or errors caused by refraction of light. etc., it is is rest, parallel When an eye with no refractive error rays of light are at focused exactly on the retina. Many eyes, owing to abnormalities of the refracting mechanism, do not focus the light exactly on the retina. Such defects of the eye can in large measure be overcome with the (spectacles) of suitable lenses aid /. Near-sightedness to close Persons suffering from this defect are unable to focus on distant objects distinctly, although objects the eyes are clearly seen. This is due to the axis of the eyeball being too long or the crystalline lens being too powerful, so that rays from a distant object tend to be focused in front of the retina (Fig. This defect is largely heredi20:6a). also aggravated by too strenuous use of the eyes during the early years. Concave lenses placed in front of the eye will make the light more divergent and bring the image to a focus on the retina. but is tary, 2. Far-sightedness A far-sighted person cannot see clearly objects close to the eye although distant objects may be distinctly The defect is due to the axis of the eyeball being too short or the crystalline lens not sufficiently converging so that rays from a nearby object tend to converge seen. to a point behind the (Fig. It is the most common of re20:6b). fractive errors of the eyes and is present retina 230 to some degree in about two-thirds of all adults. Convex lenses placed in front of the eye will reduce the divergence of the rays which can then be focused on the retina. (c) Fig. 20:6 Defects of Vision. (a) Near-sightedness, Corrected by Con- cave Lens. (b) Far-sightedness, Corrected by Con- vex Lens. (c) Test for Astigmatism. the lines appear equally distinct to you? Do all 3. Astigmatism As a result in different planes are brought to different foci and in consequence one set of lines of this defect lines OPTICAL INSTRUMENTS Sec. IV: 46 (Fig. 20:6c). will be in sharp focus, while others inclined to them are blurred and indisFor example, an tinct astigmatic eye may clearly discern all the vertical lines of a building, while those in a horizontal plane are scarcely perceived. The main cause of this defect is the lack of sphericity of the cornea, in which the vertical section is frequently more curved than the horizontal. To overcome the defect cylindrical lenses are used to assist refraction in the plane of least curvature of the cornea. Astigmatism is present to varying degrees in most eyes. ing rays from the object without undue With the use of a short-focus strain. convex lens, however, it is possible to bring the object even closer, thereby enabling it to be seen still more distinctly without straining the eye. The object is placed inside the focus of the lens at such a position that the eye focuses the magnified virtual image formed at the nearest point of distinct vision. The path of the rays by which the eye sees the image is shown in Fig. 20:7. The magnifying power of such a lens the size of image on retina when is object is viewed through the lens divided 4. Presbyopia It loss This is a defect of age resulting from of accommodation brought the about by hardening of the crystalline lens, or of the ciliary muscles. is evidenced by both near and far objects being indistinct. Clear vision by the unaided eye is possible over only a very restricted range. Two pairs of spectacles are required by such people, one pair (convex) for reading, and another pair (concave) for distant vision. Both purposes may be served by a pair of bifocal lenses, the top part being used for distance, and the lower part for reading. IV : 46 THE MAGNIFYING GLASS The apparent size of an object depends on the angle it subtends at the eye. This angle is known as the visual angle of the object, and in order to see a small object clearly one instinctively brings it close to the eye, with consequent increase in its visual angle. With the unaided eye there is, however, a limit to which this process can be carried. This is reached when the object is at the nearest point of distinct vision (normally 25 cm. or 10 in.) at which position the object is seen most clearly. If brought within this natural limit the eye cannot accommodate itself to the widely diverg- The Action of a MagnifyFig. 20:7 ing Glass. The image is focused at the least distance of distinct vision. by size of image on retina when object is viewed directly at the least distance of distinct vision least distance of distinct vision focal length of the lens 10 in. 25 cm. In using a magnifying glass we focus the image at the least distance of distinct vision by moving the lens back and forth from the object. A near-sighted person might focus on an image 7 inches from the eye, for example, whereas a far-sighted person would change the lensobject distance so that the image might be 13 inches from the eye. If / 1 inch., the former person would only get a magnification of 7 X, whereas the farsighted person would get a magnification of 13 X. To save argument the average value for least distance of distinct vision has been set at 10 inches. 231 a as simple magnifying principal focus, F' , of the eyepiece which acts glass. Through this lens the eye sees a large virtual image, pq, when the length of the tube is properly adjusted so that the image is sharply in focus at the nearest As with the point of magnifying glass, position of the the virtual image depends upon the accommodation of the observer’s eye. distinct vision. The magnification produced by the compound microscope is obtained in two stages. Both lenses individually produce a magnified image, so that the total magnification is the product of the two. Microscopes for visual use can be constructed to give magnifications of 2000 and above, the higher magnifications being obtained by using objective lenses of very short focal length. Often such lenses are designed to employ a medium of cedar oil between the specimen and Microthe objective (Oil-Immersion This medium increases the scopes). light-gathering power and the resolving power of the objective by reducing reflection and scattering of light at the surface Single lenses of very short of the lens. images which are length give focal LIGHT Chap. 20 Note: Magnification = — := — (Sec. IV: 32) Ho and Di — least distance of distinct vision Do Hi Di r= 10 in. or 25 cm. and Do = focal length (/) approximately Magnification 10 in. ~T~ or 25 cm. / of short focal length, IV : 47 THE COMPOUND MICROSCOPE The compound microscope is very widely used to produce great magnificat
ions of small objects. The instrument (Fig. 20:8a) consists mainly of a convex called the lens objective, O, and a second convex lens, called the eyepiece, E. These lenses are mounted in a tube of adjustable length. The object, PQ, to be viewed is placed just outside the principal focus, F, of the objective (Fig. 20:8b). This produces a real magnified image of the object at P'Q,', and this image falls inside the American Optical Company (Canada) Ltd. (a) Fig. 20:8 The Compound Microscope. 232 OPTICAL INSTRUMENTS Sec. IV:48 Virtual Image Fig. 20:9 The Refracting Astronomical Telescope. greatly distorted, and which have coloured edges. To avoid this the objectives of actual microscopes are usually compound IV:42). achromatic lenses (Sec. of Microscopes generally have a concave mirror, and frequently a condensing lens above it, to focus sufficient light on the object being examined. The inventhe microscope in 1590 by tion Zacharias Janssen, with the many improvements added since then, has contributed advances made in medicine and public health, as well as revealing many marvels of plant and animal life, and hidden secrets of the mineral kingdom. immensely the to IV: 48 THE TELESCOPE (a) Refracting Telescope The method by which the telescope works is similar to that of the microscope. The refracting telescope consists be the focal larger length of an arrangement of two convex lenses for viewing very distant objects (Chap. 21, Exp. 19), The objective should have a very great focal length, for the greater the will the image of an object at infinity. It should also be of large diameter to enable a maximum amount of light to be collected, thereby producing a brighter image. This lens will produce a real inverted image at its principal focus. This should be located just within the principal focus of the eyepiece through which the eye sees a magnified virtual image (Fig. 20:9). The magnifying power of the telescope is approximately equal to the focal length, F, of the objective divided by the focal length, /, of the eyepiece. Magnifying power = — Fig. 20:10 Two Forms of the Reflecting Telescope. 233 Chap. 20 LIGHT Palomar Mountain Observatory, California. Star Newspaper Service (h) Reflecting Telescope Reflecting telescopes are used for astronomical purposes. The objective is a parabolic mirror (Sec. IV; 19) which light from the distant object. collects Since the light is reflected back toward the object, a means of viewing the image with an eyepiece necessitates the use of an additional mirror or prism 20 : 10 ). (Fig. Large reflecting telescopes include the 74 inch diameter telescope at the David Dunlap Observatory at Richmond Hill, Ontario, and the 72 inch telescope at the Dominion Observatory near Victoria, British Columbia. Larger Astro-Physical telescopes of this type are located at Mount Wilson, California (100 in.) and at Palomar Mountain, California (200 in. ) . This latter is the largest telescope in the world, having one million times the light-gathering power of the human eye. (c) Terrestrial Telescope The drawback to the astronomical telescope is that the image is inverted. For observing familiar terrestrial objects a telescope requires an image-erecting system. The simplest means of obtaining this is to insert a third convex lens to re-invert the image (Fig. 20:11). Such terrestrial telescopes are used by surveyors, by mariners and by military men. Final Image First Image 234 OPTICAL INSTRUMENTS Sec. IV; 50 Fig. 20:12 The Slide Projector. IV: 49 THE SLIDE PROJECTOR The purpose of the slide projector is to cast a magnified image of a slide or transparency upon a screen coloured some distance away. Fig. 20:12 shows the essential structure of such a projector. A study of the paths of the light rays shown in the diagram will serve as an excellent review of some of the work covered earlier in concave mirrors and converging lenses. For this instrument a very powerful light source is needed. This is placed at the centre of curvature of a concave reflector (mirror) M, so that many rays that tend to be lost, on hitting this mirror are reflected back on themselves thereby intensifying the light source. It is also placed at the principal focus of is this condenser a condensing lens system, C. As a result the diverging cone of rays from the light source on striking concentrated into a parallel beam of light and sent through the slide, O. The slide must be located outside the principal focus of the objective, L. This lens as a result produces a real, enlarged, and inverted image on the screen. The slide is always inserted upside down in the carrier, so that its image will be erect on the screen, /. The size of the image may be increased by increasing the distance between the projector and the screen. The objective is contained in a moveable mount, to permit focusing the image for different distances of projecOptically the motion-picture pro- tion. jector and slide projector are similar. 3. IV: 50 QUESTIONS 1. Compare lens camera and the human eye under the following headings: contrast the and (a) structure. (b) focusing. (c) light control. (a) an object close to the camera. (b) an object distant from the camera. (c) an object on a cloudy day. (d) an object on a sunny day. (e) a moving object. The focal length of a camera lens is 2. What adjustments would you make in your camera for taking pictures under the following conditions: 3.0 in. (a) How far from the film is the lens when a man standing 1 2 ft. from the 235 (a) Make a sketch of a compound microscope locating the two images, (b) If the first (objective) lens magnifies the object 8 times and the eyepiece magnifies the real image 20 times, how many times is the final image larger than the object? 8 . Discuss (a) the refracting telescope, (b) the reflecting telescope and (c) the terrestrial telescope, as to: 9 . (i) essential parts. (ii) how each works. (iii) types of images obtained. In a slide projector: (a) Where is the light source relative to (i) the concave reflector, (ii) the condensing lens? Why is it placed in this position? (b) Where is the slide relative to the principal focus of the objective lens? In what relative position is the slide inserted in the carrier? Why? (c) If the image formed by the projector is too large for the screen should you move the projector closer to the screen or farther away from it? Why? What further adjustment would be necessary? (d) The object is 16 mm. high and 50 mm. from the lens. If the screen is 6 metres away, what b the height of the image? Chap. 20 LIGHT camera is brought into focus upon the 7. film? (b) How tall will be the man’s image on the unenlarged picture his if 4 . iris reflex, height is 6 ft.? (a) What is meant by (i) (ii) lens accommodation? (b) Why do your eyes become more sensitive when you go into a darkened room? (c) By means of a diagram show how the lens of the eye changes when the object viewed is moved closer to the eye. 5 . Describe, with the aid of diagrams, (a) near-sightedness, (b) far-sightedness, (c) astigmatism, under the following head- ings: (i) cause. (ii) effect. (iii) correction. 6 . (a) What is the purpose of the magnifying glass? (b) Describe the image you see through a magnifying glass. (c) Calculate the magnifying power of such a glass having a focal length 1 ,5 in. of (i) 3.0 cm, (ii) (d) What would be the size of the image of a crystal 3 mm. long when viewed through each of the magnify- ing glasses in 6 (c)? 236 CHAPTER 21 EXPERIMENTS ON LIGHT EXPERIMENT 1 To study the characteristics of an image produced by a pin-hole camera. (Ref. Sec. IV: 4) Apparatus Pin-hole camera (Fig. 16:3), one large candle, one small candle. Method 1. With the adjustable section of the camera held stationary so that the ground-glass screen is a fixed distance from the pin-hole, observe the size and other characteristics of the image of the candle placed at some distance from the camera. Move the camera closer to the candle and note the size of the image. 2. Holding the camera a fixed distance from the candle, note the size of the image. Move the adjustable section thereby decreasing the distance of the screen, and hence of the image, from the pin-hole, and again note the size of the image. 3. With the adjustable section held stationary as in step 1, observe the size of the image produced by first the large, and then the small candle at a fixed distance from the camera. 4. If possible, enlarge the pin-hole and note the effect on the brightness and sharpness of the image. Observations Record carefully the observations for each of the previous steps. Conclusions 1. State the characteristics of the image. 2. What effect did each of the following have on the size of the image: (a) decreasing object distance? (b) decreasing image distance? (c) decreasing size of object? 3. Establish the formula: Hi Di Ho~~Do 4. On what does the brightness of the image and its sharpness of defini- tion depend? Explain fully. 237 Chap. 21 LIGHT Questions 1. Why do we get an image of the candle and not an image of the pin-hole ? 2. What image would be obtained if the hole were increased in size until it has a diameter of 1 in. or more? Why? EXPERIMENT 2 To establish the laws of reflection. (Ref. Sec. IV: 9) Apparatus Plane mirror, pins, sheet of paper, tack board, ruler, protractor. Method A. Demonstration Experiment—Optical Disc Method (Sec. IV: 9). B. Student Experiment^—-Pin Method. 1. Draw a straight line AB (Fig. 21:1). Place the mirror perpen- dicular to the paper so that its back edge is along this line. 2. Place pins at P and Q about 2 in. apart so that the line PQ strikes the mirror obliquely. 3. Look into the mirror with one eye at the level of the paper and line up the images of P and Q. Insert pins R and S in line with these images. 4. Mark the position of all the pins. Remove the mirror and the pins. Draw the lines PQ and RS. Produce these lines until they meet at T. Through T draw TN perpendicular to AB. With a protrac
tor measure angles PTN and STN. 5. Repeat, with other positions of the pins. Observations 1 . Observation No. Angle of Incidence Z PTN Angle of Reflection Z STN 1. 2. 3. 238 EXPERIMENTS ON LIGHT 2. When the incident ray is on the plane of the paper where are the reflected ray and the normal found? Conclusions 1. State the two Laws of Reflection. 2. Define, and label on your diagram: incident ray, reflected ray, point of incidence, normal, angle of incidence, angle of reflection. Questions 1. If a polished piece of metal were used as the mirror in the above experiment, where would T be found? Why? 2. Add an observer’s eye to the diagram and trace the path of the rays to the eye. EXPERIMENT 3 To study the position and characteristics of the image in a plane mirror, (Ref. Sec. IV: 11) Apparatus Plane mirror, two wooden pegs, paper, ruler, protractor. A Mb — izi-:. I o Fig. 21:2 Method 1. Draw a straight line AB in the middle of the sheet of paper. Place a plane mirror perpendicular to the paper, so that its reflecting surface is on AB. 2. Place the object (wooden peg) about 3 in. in front of the mirror. Mark its position O. 3. Look into the mirror to see the image of this object, and using the second wooden peg as a finder place it exactly where the image This can be done quite accurately by using the appears to be. method of parallax. That is, look into the mirror at the image, and over the top of the mirror at the finder. Adjust the position of the latter so that when you move your eye from side to side there is no relative movement between the finder and the image. Mark the position of the image I. Compare the size of the image to the size of the object. 239 Chap. 21 LIGHT 4. Remove the mirror. Join OI and let the line cut AB at M. Measure the lengths of OM and IM, and measure the sizes of the angles at M. 5. Repeat for other positions of the object. 6. Look at yourself in a mirror and raise your left hand. Note what you see. Observations 1 . Observation Number Distance of Object OM Distance of Image IM Size of Angles AT M 1. 2. 3. 2. How does the size of an object and its image compare? 3. Which hand appeared to be raised in the mirror? What term applied to the image describes this phenomenon? 4. What kind of image is it? Conclusions 1. State your conclusion relating to the position of an image in a plane mirror. 2. Summarize the characteristics of such an image. 3. Define or explain; parallax, lateral inversion, virtual image. Questions 1. How could you locate geometrically the position of the image of an object in a plane mirror? Show this by a diagram. 2. Include in the diagram an eye, and show the path of the light rays whereby the eye sees the image. EXPERIMENT 4 To show the location of images produced by two plane mirrors at right angles. (Ref. Sec. IV: 13) Apparatus Two plane mirrors, four wooden pegs, paper, ruler, protractor. Method Repeat parts 1, 2 and 3 of experiment 3 using two plane mirrors placed at right angles to each other. Observations Describe the number and location of the images. Conclusion Construct a diagram to show how the images were obtained. 240 EXPERIMENTS ON LIGHT Questions 1. Add an observer’s eye to the above diagram and trace the path of the rays to the eye. 2. Determine experimentally how many images are obtained when the mirrors are inclined at 60°, at 45°, etc. Use your results to verify the formula; Number of images = 360° —— : ; angle ol inclination 1 EXPERIMENT 5 To study the action of a concave mirror, and to determine its focal length (f), (Ref. Sec. IV: 14) Apparatus Optical bench, concave mirror, candle, screen. V D 1 ' U . Fig. 21:3 J U Method A. Demonstration Experiment—Optical-Disc Method (Sec. IV: 14). B. Student Experiment—Optical-Bench Method. 1. Mount the concave mirror and screen on the optical bench. Hold the mirror so that rays from the sun fall directly upon it. Failing that, hold it so that rays from a candle held at the opposite end of the room fall upon it. Focus the image on the screen. To do this move the screen until the image is clear and sharply defined. 2. Measure the distance of the image from the vertex of the mirror. Observations 1. What is observed when parallel incident rays are reflected from the mirror ? 2. Describe the image obtained when the object is at infinity. 3. What is the distance of the image from the vertex of the mirror? Conclusions 1. What effect does a concave mirror have on light rays incident upon it? 2. Define; principal focus, focal length. 241 Chap. 21 LIGHT Questions 1. Why may a concave mirror be called a converging mirror? 2. Why did we use an infinitely distant object in the above experiment? EXPERIMENT 6 To study the images produced by a concave mirror, (Ref. Sec. IV: 15) Apparatus Optical bench, concave mirror of known focal length, candle, screen, finder. Method 1. Place the candle at more than twice the focal length (/) from the mirror. Locate the image on the screen and state its characteristics. 2. Repeat the above for two or three other positions of the candle as it is gradually moved nearer to the principal focus of the mirror. 3. Place the candle between the principal focus and the mirror. Look into the mirror to see the image. Note its characteristics and locate its position by the method of parallax using the finder. Observations Observation Number Position of Object Position of Image Characteristics of Image Kind Attitude Size 1. 2. 3. Conclusions 1. Where may an object be placed so that a real image of it is produced by a concave mirror? 2. Describe the changes in position and characteristics of the image as the object is gradually moved from infinity to the principal focus of a concave mirror. 3. Describe the position and characteristics of the image when the object is between the principal focus and the mirror. Question Make accurate scale diagrams to locate the images for the above positions of the object. EXPERIMENT 7 To study the images produced by a convex mirror, (Ref. Sec. IV: 16) 242 EXPERIMENTS ON LIGHT Apparatus Optical bench, convex mirror, candle, finder. Method Using the candle at three different positions from the mirror, locate the image each time by the method of parallax, and state its characteristics. Observations Observation Number Position of Object Position of Image Characteristics of Image Kind Attitude Size 1. 2. 3. Conclusions 1. What kind of image is produced by a convex mirror? 2. Describe the position and characteristics of the images produced by a convex mirror for all positions of the object. Questions 1. How would you find the focal length of a convex mirror? 2. Why may such a mirror be called a diverging mirror? 3. Construct a geometric diagram to locate the image of an object produced by a convex mirror. EXPERIMENT 8 To measure the refractive index of glass, (Ref. Sec. IV: 21, 23) Apparatus Rectangular block of plate glass with one pair of opposite edges polished, sheet of paper, tack board, pins, geometrical instruments. M 243 Chap. 21 LIGHT Method 1. Lay the sheet of paper on the tack board. Place the glass block on the sheet of paper. Outline its position carefully. 2. Stick a pin A upright in the paper, near the centre of one edge of the glass and touching it. 3. Stick a second pin B upright in the paper about 2 in. from A so that a line joining the two meets the glass obliquely. 4. Look through the block, line up the images of A and B and insert pin C against the opposite edge of the glass block so that it and the images of A and B appear in a straight line. Mark the position of each pin. 5. Remove the block and pins and draw BA, and AC. Draw the normal MN at A. With centre A describe a circle to cut AB at D and AC at E. From D and E draw perpendiculars DF and EG to the normal MN. 6. Measure the distances DF and EG and calculate the value of DF With a protractor measure the angles of incidence ( Z BAF) and of refraction ( Z CAG). Using a table of sines calculate sin Z i -r- sin Z r. EG. 7. Repeat for several other positions of the pins and arrange your results in the following observation table. Observations Obs. No. DF EG DF/EG Z i Z r sin Z i/sin Z r 1. 2. 3. 4. Average Average Conclusion Define: Index of Refraction. Questions 1. Compare your experimental value with the theoretical value (table p. 199) . Calculate your percentage error. 2. Which way does light bend on entering obliquely a different medium of greater optical density? Why? EXPERIMENT 9 To trace the path of light through a glass plate with parallel sides. (Ref. Sec. IV: 24) 244 EXPERIMENTS ON LIGHT Apparatus Same apparatus as in experiment 8. Method 1. Lay the sheet of paper on the tack board. Place the glass block on the sheet of paper. Outline its position carefully. 2. Stick two pins A and B upright in the paper, about 2 in. apart, so that a line joining them (the incident ray) strikes the glass obliquely. 3. Look through the block, line up the images of A and B and insert pins C and D about 2 in. apart so that they and the images of A and B appear in a straight line. Mark the position of each pin. 4. Remove the block and pins. Join AB and produce it to meet the block at E. Join CD and produce it to meet the block at F. Join EF. On your diagram label the incident, refracted, and emergent rays. 5. Draw the normals at E and F. Measure the angle of incidence at E, and the angle of emergence at F. 6. Repeat the experiment for angles of incidence of different sizes. Observations 1. In what direction relative to the normal does the light ray bend on entering the glass at E and on leaving it at F? 2 . Angle of incidence at F = Angle of emergence at F = 1. 2. 3. Conclusions 1. How does the angle of incidence compare with the angle of emer- gence? 245 Chap. 21 LIGHT 2. What relationship exists between the directions of the incident and emergent rays? 3. Describe the path of a light ray through a glass plate with parallel sides. Explanation Explain the observed bending of the light ray on entering and on lea
ving the glass. Question What factors affect the amount of lateral displacement of the emergent ray? EXPERIMENT 10 To illustrate the apparent change in depth due to refraction of light at a plane surface. (Ref. Sec. IV: 24) Apparatus Tall beaker, two pins, water. r_- Water Real Depth Apparent Depth ^Search Pin S Object Pin O Fig. 21:6 Method 1. Place a pin at the bottom of the beaker and pour in water to a depth of about 15 cm. This pin is the object, O. 2. Hold the search pin, S, horizontally outside the beaker. View from directly above and adjust the position of the search pin until there is no parallax between it and the image of the object pin. 3. Determine the distance of the object pin, and of the search pin from the surface of the water, thus finding the real and apparent depths. 4. Repeat the experiment using different depths of water. 5. Tabulate your observations and determine the average value for real depth apparent depth. 246 EXPERIMENTS ON LIGHT Real Depth Apparent Depth Real Depth Apparent Depth Observations Observation Number 1. 2. 3. 4. Conclusions 1. Construct a diagram to show the path of the light rays from the immersed object to an observer’s eye. 2. Why is the apparent depth always less than the real depth? 3. Compare the value obtained by dividing the real depth by the appar- ent depth to the index of refraction of water (table, page 199). Questions 1. In step 3 of the method, why must the measurements be made on the outside of the beaker? 2. If the water in the above experiment were replaced with an optically denser liquid, e.g., carbon tetrachloride, what effect would this have on the apparent position of the object? Why? Try the experiment. EXPERIMENT 11 To show deviation produced by refraction through a prism. (Ref. Sec. IV: 25) Apparatus A 60° prism, sheet of paper, tack board, four pins, geometrical instruments. A Fig. 21:7 Method A. Demonstration Experiment—Optical-Disc Method (Sec. IV: 25). B. Student Experiment—Pin Method. 247 Chap. 21 LIGHT 1. Lay the sheet of paper on the tack board. Place the prism on the paper. Outline its position carefully. Label its vertices A, B and C. 2. Stick two pins P and Q upright in the paper about 2 in. apart so that Q is touching the prism, and a line joining them (the incident ray) strikes the surface of the prism, AB, obliquely. 3. Look through the prism on the side AC. Line up the images of P and Q and insert pins R and S about 2 in. apart with R touching the prism, so that they and the images of P and Q appear in a straight line. Mark the position of each pin. 4. Remove the prism and the pins. Join PQ, QR and RS. Label the incident, refracted and emergent rays. 5. Extend PQ and SR to meet at D. Measure the size of Z D (angle of deviation). 6. Construct the normals at Q and R. Measure the angle of incidence and the angle of emergence. 7. Repeat on separate diagrams for three other angles of incidence, including the special case where the angle of incidence is equal to the angle of emergence. Observations 1 . Which way does the light ray bend (a) on entering the prism? (b) on leaving the prism? Angle of Incidence Angle of Deviation Angle of Emergence 2 . Obs. No. 1. 2. 3. 4. Conclusions 1 . Describe the path of a light ray through a prism. 2. Describe how the angle of deviation varies as the angle of incidence is changed. 3. What is the relationship of the angle of incidence to the angle of emergence when the deviation is a minimum? Questions 1. Draw a diagram showing a ray of light passing through the prism at the position of minimum deviation. 2. What factors affect the amount of deviation produced by a prism? EXPERIMENT 12 To study the action of a convex and a concave lens, and to determine their focal lengths, (Ref. Sec. IV: 28, 29) 248 EXPERIMENTS ON LIGHT Apparatus Optical bench, convex lens, concave lens, candle, screen, finder. [ 'r'^ I I ' ' Fig. 21:8 Method A. Demonstration Experiment—Optical-Disc Method (Sec. IV: 29). B. Student Experiment—Optical-Bench Method. 1. Mount the convex lens and screen on the optical bench. Hold the lens so that rays of light from the sun, or from a distant candle, fall directly upon it. Focus the image on the screen. 2. Measure the distance of the image from the optical centre of the lens. 3. Repeat using the concave lens. In this case, in order to see the image, it is necessary to look into the lens toward the object. The position of the image can be determined by the method of parallax with the aid of the finder. Observations 1. What is observed when parallel incident rays are refracted through (a) the convex lens, (b) the concave lens? 2. Describe the image obtained when the object is at infinity using (a) the convex lens (b) the concave lens. 3. What is the distance of the image from the optical centre of the lens? Conclusions 1. What effect do (a) a convex lens, (b) a concave lens, have on light rays incident upon them? 2. Define: principal focus, focal length. Questions 1. What other name may be given to (a) a convex lens, (b) a concave lens? Why? 2. Why did we use an infinitely distant object in the above experiment? EXPERIMENT 13 To study the images produced by a convex lens, (Ref. Sec. IV: 30) 249 Chap. 21 LIGHT Apparatus Optical bench, convex lens of known focal length, candle, screen, finder. Method 1. Place the candle at more than twice the focal length (/) from the lens. Locate the image on the screen and state its characteristics. 2. Repeat the above for two or three other positions of the candle as it is gradually moved nearer to the principal focus of the lens. 3. Place the candle between the principal focus and the lens. Look into the lens to see the image. Note its characteristics and locate its position by the method of parallax using the finder. Observations Observation Number Position of Object Position of Image Characteristics of Image Kind Attitude Size 1. 2. 3. Conclusions 1. Where may an object be placed so that a real image of it is produced by a convex lens? 2. Describe the changes in position and characteristics of the image as the object is gradually moved from infinity to the principal focus of a convex lens. 3. Describe the position and characteristics of the image when the object is between the principal focus and the lens. Question Make accurate scale diagrams to locate the images for the above positions of the object. EXPERIMENT 14 To study the images produced by a concave lens. (Ref. Sec. IV: 31) Apparatus Optical bench, concave lens, candle, finder. Method Using the candle at three different positions from the lens, locate the image each time by the method of parallax, and state its characteristics. Observations Observation Number Position of Object Position of Image Characteristics of Image Kind Attitude Size 1. 2. 2 250 EXPERIMENTS ON LIGHT Conclusions 1. What kind of image is produced by a concave lens? 2. Describe the position and characteristics of the images produced by a concave lens for all positions of the object. Questions 1. How would you find the focal length of a concave lens? 2. Why may such a lens be called a diverging lens? 3. Construct a geometric diagram to locate the image of an object produced by a concave lens, EXPERIMENT 15 To study the dispersion of white light into the spectrum. (Ref. Sec. IV:36(a) Apparatus Projection lantern, single slit, converging-lens, prism, white cardboard screen. Method Direct a beam of light from the projection lantern on to the slit. Place the converging-lens in the light that diverges from the slit, and focus these rays on the screen at A. Place the prism in the path of the beam and note what happens to the light. Rotate the prism so that the emergent beam sweeps to and fro. Adjust the position of the prism until the emergent beam is as near as possible to A. The prism is now in the position of minimum deviation. Move the screen into the path of the beam and adjust its position until the light which falls on it is in clear focus. Observations 1. What happened to the beam of light on passing through the prism? 2. Name the colours of the spectrum that you see. 3. Which coloured rays are deviated least, and which most, by the prism ? 251 Chap. 21 LIGHT Explanation 1. Why is light deviated on passing through the prism? 2. Why are different colours deviated different amounts? 3. Which is deviated the least? Why? Which the most? Why? Conclusions 1. What must white light consist of? 2. What is meant by dispersion of white light? Questions 1. What would be observed if a sensitive heat detector were placed just beyond the red end of the spectrum? Why? 2. What would be observed if a fluorescent material (e.g., anthracene), were placed just beyond the violet end of the spectrum? Why? EXPERIMENT 16 To recombine the colours of the spectrum into white light. (Ref. Sec. IV:36(b) Apparatus As in experiment 15, second prism identical to one used there, second converging lens, Newton’s colour disc, rotating-machine. Method 1. With apparatus arranged as in experiment 15, place the second prism with its refracting edge in the opposite position to that of Note what is first. Move the screen to catch the image. the obtained. 2. With apparatus arranged as in experiment 15, place the second converging- lens in the spectrum. Move the lens until the image is brought to a focus on the screen. Again note what is observed. 3. Examine Newton’s colour disc. Strongly illuminate it with white light from the projection lantern. Rapidly rotate it. Note the effect obtained. Place on the rotator. Observations Describe carefully what is observed in each part above. Include diagrams. Conclusion How may the colours of the spectrum be combined to form white light? Explanation 1. Why did the reversed prisms and the converging- lens re-form white light? 2. Why did Newton’s disc appear white when rapidly rotated? 252 EXPERIMENTS ON LIGHT Question How could a concave mirror be used to recombine the colour of the spectrum? EXPERIMENT 17 To study the spectra of a
few common elements, (Ref. Sec. IV: 39) Apparatus Direct-vision spectroscope, Bunsen burner with monochromatic flame attachment, sodium, calcium and lithium salts; electrical discharge tubes containing neon, nitrogen and hydrogen, induction coil, source of direct current. Method 1. Place a sample of the sodium salt in the attachment on the burner. Light the burner and heat the salt. Observe the flame through the spectroscope. Repeat for the other salts. 2. By means of the induction coil send a high-voltage discharge of electricity through the tube containing neon. Observe the discharge Repeat using the tubes containing the through the spectroscope. other gases. Observations Describe the spectra obtained and compare with those given in Fig. 19:6. Conclusion How is the spectrum of an element obtained? Questions 1. What is meant by spectrum analysis? 2. What are some of its advantages? EXPERIMENT 18 To study colour in natural objects (a) opaque (b) transparent, (Ref. Sec. IV: 40) Apparatus As in experiment 15, red, yellow and blue cardboard screens; red, yellow and blue glass filters. Method Produce a pure spectrum as outlined in experiment 15. 1. Replace the white screen by the coloured screens in turn. Note any change in the appearance of the spectrum each time. 253 Chap. 21 LIGHT 2 . With the white screen in place, insert each of the coloured filters in the path of the beam. Note any change in the appearance of the spectrum each time. 3. Repeat 2 using both the yellow and blue filters in the path of the light beam. Note the colour of the light obtained. 4. Repeat 2 using all three filters (red, yellow and blue) in the path of the beam. Note what is obtained. Observations Describe carefully all that is observed. Represent your observations by means of simple diagrams. Explanation Account for each of the observations. Conclusion What is the cause of colour in (a) opaque (b) transparent objects? Questions 1. An object is viewed in white light and appears red. Why? 2. A blue filter is placed between the source of white light and the same object. What will be the appearance of the object? Why? EXPERIMENT 19 To study the action of the astronomical refracting telescope, (Ref. Sec. IV: 48) Apparatus Optical bench, convex lenses of 20 cm. and 5 cm. focal lengths. Fig. 21:10 Method 1. Place the two convex lenses on the optical bench so that the distance between them is equal to the sum of their focal lengths (25 cm.). Using the convex lens of shorter focal length as the eyepiece and the other as the objective, look through the lenses at distant objects. 254 EXPERIMENTS ON LIGHT 2. Construct two similar scales, P and Q, with divisions about 1 in. apart and place them near each other at the far end of the room. (Similar scales may be drawn on the blackboard if this is preferred.) Look through the telescope at the scale Q with one eye, and at the same time look at scale P with your other eye. Have your partner mark the point on scale P which is level with the uppermost division of Q observed through the telescope. Repeat for the lowest division observed. Compare the number of complete divisions, AB, observed through the telescope with the number of Determine corresponding scale divisions, A^B^, observed directly. the magnifying power of the telescope. Observations 1 . What is observed when distant objects are viewed through the telescope? 2. How many complete scale divisions were observed (a) through the telescope? (b) directly? Calculations AB = A^B^ = 1. Determine the magnifying power of the telescope from the above observations. 2. Compare this value to that obtained by dividing the focal length of the objective by the focal length of the eyepiece. Conclusions 1. What is the action of an astronomical telescope? 2. What is the magnifying power of this telescope? Questions 1. By means of a diagram trace the paths of the light rays by which the observer sees the object through this telescope. 2. What disadvantage has this telescope for viewing terrestrial objects? How could this disadvantage be overcome? 255 « ' . ' . H - i • .-'- . V. ' '.. ' i". i' . : hr‘ ' - ' ' ' .V i I };.•• i^. ' 'riV 'la. • e- ': i.-. -'a>^ '-'' . T'/h . r ; '^r •;; . • 1. 1 .'1 '''V'-r,";';' . '.' ' ' ‘ : . ..‘’..a,-.,, -u'-m!' a-.ltl'" •4 . ' '/• ? . .ovv;i j- t, ba:--';-'- '' ' • '"b '.'v ; Jii j'r: , -> ''‘i ' -< >'il'/-’ ' '- ' ‘.‘H : ' ''I?!'' tC'i "' K ' i: : '' j 'I-! -• i'i,- • '- ! . •«"•'( < :ir't ' ' •’'’"''a''' MrbiVi ab lb ^'.v. .••.- 5 .'VV-. / l ^ -ifll ' ^ - f. i ; 't a«ij >" r. . '< ^^^^,,, \|,i' ’ ibwjv I' ' # ' '’ fiij, ."m -fl <:.••./ v ': . :3V.'' •'•<1 mml ’ :4^ . " ' a £ a'" ’ a ; ' ,:./ .; • ''[ivl: .l^itefe '''iifcib"' ,iv a j -(P* • ' '' A' ' ’ 'j,' .-K,:'. f ,' . " W -^»h' a '; ' ''’b^' ''av.-'.iaa'’ .a 5' ' ' '•’ . ti •, "' 'r.'i , "v v'v.ay. ^^ v' • -.::' 'T^::;:]m " .../; I«ic -aba ^ ' .< ';^ n^lJ'^Ts JV ' vi>, 'fl‘'J V? //,;i . . ..... ->': 'Oi!- .ib;i '.'i.yv b"* \-‘v<:.ir.'^i .'I-,. -•;-'/ /'b’v^.a a . ,r,.a'>aa' . •' o , -..^.•jv :i-,.,_ 'W, .yji Si. UNIT V MAGNETISM and ELECTRICITY Happens What Exactly Explain Fire. this CHAPTER 22 MAGNETISM V ; 1 INTRODUCTION TO MAGNETISM The Study of magnetism began with the observation of the power of attracLodestone tion of lodestone for iron. is the mineral magnetite, an oxide of iron, which was first discovered in the province of Magnesia (from which the name magnet was derived) Asia this mineral are Minor. Deposits of found in many other parts of the world including large areas of Canada and the United States. in The original discovery of magnetite is recorded in an ancient legend which tells of a Cretan shepherd whose irontipped crook was so strongly drawn to the earth that he dug to find the cause. Fables tell of magnetic rocks covered on their seaward sides with nails drawn from passing ships, and of iron statues supported in mid air. For centuries men have been intrigued by the phenomena of magnetic propIn the writings of Lucretius, a erties. Roman who died in 52 b.c., the following reference is found: “And iron filings in the brazen bowls furiously; when underneath seethe was set the magnet stone.” It was in the period of the Renaissance physician emphasized when William Gilbert, to Queen Elizabeth I of England, published his famous and hitherto unequalled book entitled De Magnete. In his book Gilbert of scientific method, and of arriving at conclusions only after careful experiment. He dedicated his work “to you alone, true philosophers, ingenious minds, who not only in books but in things themselves look for knowledge.” importance the De Magnete laid the foundation of magnetic science. It described in detail a wide variety of experiments in magnetism. The book proved, however, to be too much in advance of its time and was soon forgotten. For the next hundred and fifty years there was little progress, but in the latter part of the eighteenth century magnetism began to be recognized as the exact science that it has become today. 22:1). V : 2 SOME PROPERTIES OF MAGNETS When an iron bar is stroked with a natural magnet, the iron itself becomes capable of attracting other pieces of iron It has become an artificial or steel. magnet (Fig. this magnet If is brought near to a number of different materials in succession, it is found that only a few of these, called magnetic substances, are attracted. Among these are iron, nickel, cobalt and a few special alloys. The materials which are not attracted are said to be non-magnetic (Chap. 31, Exp. 1). 259 Chap. 22 MAGNETISM AND ELECTRICITY If a bar magnet is rolled in iron filings, the filings are seen to be attracted in large clumps around the ends of the magnet (Fig. These points at which the magnetic force is concentrated are called the poles of the magnet (Chap. 31, Exp. 2). 22:2). Fig. 22:1 Artificial Magnets (a) A Horseshoe Magnet. (b) A Bar Magnet. A magnet which is suspended freely from a non-magnetic stand will always come to rest with its axis along a line running approximately north-south. The same end of the magnet will always point towards the north. This is designated the north-seeking (N-pole) of the Magnetic Force Concentrated ^ near Ends ' Clustered Fig. 22:2 Around the Poles of a Bar Magnet. Filings Iron magnet and the other end is the southseeking pole (S-pole) (Chap. 31, Exp. 3). If the N-pole of the suspended magnet is now approached by the N-pole of 260 another magnet, the poles will be observed to move farther apart (Fig. Fig. 22:3 Demonstrating the Law of Magnetism. of the This motion is caused by a 22:3). force, here one of repulsion. When the suspended magnet is S-pole approached by an N-pole of another magnet they will be attracted to each other. From such observations we may conclude that: like magnetic poles repel; unlike magnetic poles attract. This is called the Law of Magnetism (Chap. 31, Exp. 4). V:3 THE EARTH AS A MAGNET In the previous section it was shown that a suspended magnet always comes to rest along a north-south line. The magnet behaves in this way because the earth itself possesses a magnetic field as if it contained a huge bar magnet lying (Fig. 22:4). nearly parallel to its axis Why is the end of the magnet that points north labelled 6'? The magnetic north pole is located about 1400 miles south of the geographic north pole in the extreme north-central part of Canada. MAGNETISM Sec. V:3 Fig. 22:5 Demonstration Magnetic Compass. consists The modern magnetic compass (Fig. of a magnetized steel 22:5) needle suspended on a point so that it is free to move in a horizontal plane. The N-pole of the compass will point Fig. 22:6 Map of Canada Showing Points of Equal Declination, 1955. 261 Chap. 22 MAGNETISM AND ELECTRICITY always to the earth’s magnetic north pole. The angle between the true north and the direction in which the compass points is called the angle of declination (Fig. 22:6). It is evident that the angle of declination will vary with position on the earth’s surface. A navigat
or must determine its value from charts in order find true north from the compass to bearing at his position. The declination at Toronto, Ontario, was approximately 7° west in 1955. This indicates that the angle between the lines drawn from Toronto to the geographic north pole and to the magnetic north pole is 7°, and the magnetic line is west of the line to the geographic pole. The magnetic north pole shifts its position slowly westward, and its exact location must be resurveyed frequently in order to correct charts for navigation purposes. V : 4 THE MAGNETIC FIELD compass-needle at a number of positions at varying distances from the magnet and observing the direction in which the needle points. The space surrounding a magnet in which its magnetic influence can be detected is known as the magnetic field of the magnet. in as the century, Magnetic fields were studied by Gilbut bert sixteenth Michael Faraday (1791-1867) showed that they could be best described and interpreted by means of magnetic maps made up of a series of lines known as lines of force. A line of force may be the path that would be defined travelled by a free N-pole if at liberty to move in a magnetic field. By specifying the movement of an N-pole in our definition we obtain the conventional direction for the lines of force, i.e., from the N-pole of the magnet, which would repel the free N-pole, to the S-pole of the magnet which would attract it. The effect of a magnet is noticeable over a considerable space surrounding it. This can be demonstrated by placing a An effective method of demonstrating the disposition of the lines of force about a magnet is described in chapter 31, Fig. 22:7 The Magnetic Field about (a) A Bar Magnet (b) Unlike Poles (c) Like Poles. 262 MAGNETISM Sec. V:6 experiment 5. Some examples of magnetic maps obtained in this way are shown in Fig. 22 : 7. The properties of lines of force as shown on these maps may be summarized as follows: the 1. Lines of force form closed curves, the magnet, N-pole of leaving proceeding through space to the S-pole, and completing their path through the The apparently loose ends magnet. shown in the diagrams are actually parts of complete lines linking the poles of the magnets. of a magnetized needle that is free to turn in a vertical plane, and a protractor to measure its inclination from the horizontal (Fig. 22:8). The plane of this needle must be adjusted so that it is parallel to the direction indicated by a compass. The instrument will then indicate the direction of the lines of force. At the magnetic poles the needle will repel each other, 2. Lines of force never cross but tend thus spreading to farther and farther apart as they leave the N-pole of the magnet. 3. Lines of force behave like stretched elastic bands, tending to contract and thus to shorten their paths. 4. The concentration of the lines of force determines the strength of the magnetic field at any point. Thus the field is strongest close to the poles of the magnet. The property of that allows magnetic lines of force to pass through it is called permeability. Substances such as soft iron and permalloy, an alloy of nickel and iron, which allow pass through them lines of readily are said to have a high permeability. For this reason these materials are used where a strong magnetic field is desirable as in the electromagnet (Sec. V:50) V:53). galvanometer substance force and (Sec. the to a V ; 5 THE ANGLE OF INCLINATION In section V : 3 the earth was described as possessing a magnetic field of its own. It is the horizontal component of this field that activates the magnetic compass. The exact direction of the magnetic lines of force may be determined by the use of a dipping needle. This consists Fig. 22:8 Dipping Needle. thus be 90°. stand vertically and the reading on its At the protractor will magnetic equator the needle will be horizontal and the angle of inclination or dip will be 0°. Other locations on the earth’s surface will give readings between these extremes. At Toronto the angle of inclination is about 74°. V:6 INDUCED MAGNETISM An unmagnetized piece of soft iron If, however, will not attract iron filings. we bring it near the pole of a bar magnet, the iron will become a magnet and exhibit powers of attraction. When the bar magnet is removed the soft iron will lose its magnetic properties very rapidly. The iron is said to have been magnetized by induction. If the S-pole of the bar magnet approaches the soft iron it can be shown that the end of the iron 263 Chap. 22 MAGNETISM AND ELECTRICITY A Permanent MagUsed to net Being Separate Metal Sheets. Canadian General Electric nearest the magnet becomes an N-pole and the otlier end becomes an S-pole (Chap. 31, Exp. 6). If a bar of soft iron is held at the angle to the earth’s surface that is indicated by a dipping needle and one end is tapped sharply with a hammer, it can be demonstrated that the iron has become a weak magnet (Chap. 31, Exp. 7 ) . Thus we see that even the weak ships, hulls of etc. may of buildings, become magnetized by induction. The magnetic fields from these structures may seriously affect delicate measuring instruments such as watches and com- passes. In order to protect such instruments they are frequently enclosed in a magnetic shield of some highly permeable metal. The lines of force pass through the shielding material more readily than through the space enclosed by it (Chap. 31, Exp. 8), and so the instruments operate free from magnetic disturbance. The effect illustrated in Fig. 22:9. of magnetic shielding is Induced magnetism plays an important part in many of the devices that will be studied in later chapters of this unit. V:7 A THEORY OF MAGNETISM If a piece of iron wire magnetized as in experiment 9, chapter 31, is subjected to continued division and sub-division, each piece will be found to be a small magnet possessing an N-pole and an 10). When S-pole these pieces finally become too small for further division, we may imagine the process to be continued until finally the (Chap. 31, Exp. Fig. 22:9 Magnetic Shielding. Lines of Force Follow the Soft Iron Ring. Instrument I, in centre, is Unaffected by Magnetic Field. magnetic field of the earth is capable of causing magnetism by induction. It may be readily understood then, that magnetic substances such as the iron parts of machines, steel frameworks 264 MAGNETISM Sec. V:7 molecule of the iron is reached. Thus each molecule would possess its own Nand S-pole of equal strength. We may then reason that all magnetic substances are composed of molecules that are tiny magnets in themselves (Chap. 31, E.xp. 11 )-, Using this theory the unmagnetized state is accounted for by the random EJ m LEi] LXJ 0:1 LX] aa 03 m 03 aJLJJ rm ni] na ra DU [. i£i LL_j rm m aji Id (b) The Theory of Magnetism Fig. 22:10 (a) Unmagnetized State (b) Magnetized State. we can now account for the concentration of magnetic properties at the poles of the magnet. taken across The fact This molecular theory of magnetism provides a simple explanation of many magnetic phenomena. that there are the same numbers of unbalanced N- and S-poles at either end of a magnetized iron bar means that the poles of a bar magnet will be equally strong. A section the magnetized bar will expose an equal number of N- and S-poles showing that equal and opposite poles will be produced on breaking the magnet. Saturai.e., a magnetic substance tion effect, can be magnetized only up to a certain strength is clearly due to the fact that when all molecular magnets have been drawn into line there can be no further increase in magnetic strength. The loss of magnetism due to rough handling or heat is due to the disturbance of the molecules and a reversion to the random arrangement. of of N- arrangement distribution of molecular millions magnets in the bar (Fig. 22:10a). The haphazard and S-poles neutralizes their effect and no external magnetism is evident. When the bar is stroked with the N-pole of a magnet the S-poles of the molecular magnets are attracted in the direction of the stroking magnet and the molecules become aligned as shown in Fig. 22: 10b. At the end of the bar last touched by the N-pole of the magnet there will be many S-poles of the molecular magnets and thus it will act as the S-pole of our new magnet. The opposite end, having many molecular N-poles will be the new N-pole. In the central region of the bar the N- and S-poles of the molecules will again neutralize each other so that Fig. 22:11 Keepers (a) on Bar Magnets, (b) on a Horseshoe Magnet. The mutual repulsion of the similar molecular poles at the ends of a magnet tends to push the molecular magnets out 265 Chap. 22 MAGNETISM AND ELECTRICITY of alignment and so slowly to destroy its magnetism. To prevent this, bar magnets are usually stored in pairs with opposite poles adjacent and a piece of soft iron, called a keeper, placed across each end as shown in Fig. 22:11. The keepers become magnetized by induction and their molecules tend to hold those of the magnet in alignment. A horseshoe magnet requires one keeper as indicated. only A material which is readily magnetized but loses its magnetism rapidly, such as soft iron, is used to form temporary magnets used in electromagnets (Sec. V:50). The molecules of such substances are readily rearranged. Materials which have a more stable arrangement of molecules such as steel of aluminum, and alnico, an alloy nickel, and cobalt, are more difficult to magnetize, but tend to retain their magnetic properties much longer. Such substances are used to make the permanent magnets used in telephone receivers, galvanometers and other instruments. V:8 QUESTIONS A 1. (a) What is a natural magnet? (b) What is the common name of a natural magnet? 2. From the following list select those materials which are magnetic and those which nickel, aluminum, wood, rubber, iron, copper, lead, non-magnetic: glass, are cobalt, steel, zinc, tin. 3. 4. 5. 6 . (a) State the Law
of Magnetism. (b) Explain how, with the aid of a compass-needle, you would determine which end of a magnet is the N-pole. simplest form of (a) Describe the magnetic compass. (b) Explain why the N-pole of the compass points to the magnetic north pole of the earth. (c) Why doesn’t a compass-needle point to the true north pole? Define angle of declination. (a) Define line of magnetic force. (b) Draw the magnetic field about a bar magnet. (c) State three fundamental characteristics of these lines of force. (a) Define angle of inclination or dip. (b) What would be the angle of inclination at (i) the magnetic north 266 pole, (ii) the magnetic south pole and (iii) the magnetic equator? 7. (a) What is induction? meant by magnetic (b) By means of a labelled diagram show how you could induce magnetism into a nail so that the head is an N-pole. 8 . 10. 9. (a) Define magnetic permeability. (b) What is "magnetic shielding”? (c) Where and why would shielding be necessary? such (a) Describe how you could magnetize a steel needle. (b) How could a piece of steel wire be magnetized by a bar magnet without returning the magnet in a wide arc? (a) Outline the theory of magnetism, (b) Use this theory to explain: (i) that a magnet when repeatedly cut gives sections that are magnets also. (ii) polarity of a magnet. (iii) that there is no magnetic power in the centre of a magnet. (iv) that magnetism is lost on heating or jarring a magnet. (v) magnetic induction. (vi) magnetic saturation. MAGNETISM Sec. V:8 (vii) the difference between temporary, and permanent magnets. B 1. A rod of soft iron is held vertically and struck with a hammer. test (a) Describe how you would whether the rod is magnetized. (b) Which end should become an S-pole? (c) Why strike it with a hammer? 2. How could you use a compass-needle to distinguish between a magnet and a magnetic substance? 3. (a) How could the lines of magnetic force of a horseshoe magnet be traced? (b) Show by means of a diagram the pattern that you would expect to see. 4. A steel bar is repeatedly stroked from A to B with the N-pole of a bar magnet. A compass-needle is then placed near the end B of the steel bar. Make a diagram of the steel bar and compass-needle and label all poles. 5. How would you use a compassneedle to distinguish between three bars similar in shape and appearance if one is brass, the second is unmagnetized iron and the third is a magnet? 6. You are given two iron bars — one magnetized, the other not — and nothing else. Describe carefully how you could determine which is the magnet. 267 CHAPTER 23 ELECTROSTATICS 2500 years ago. However, similar discoveries have been made by everyone early in life. What child has not been fascinated to discover that on a cold dry day the comb which has just straightened his hair will now pick up bits of paper? Who has not been startled by a shock after scuffing across a carpet to touch another person or a metallic object? Materials which have gained the property of attracting small bits of paper and other light objects are said to have an electric charge (Chap. 31, Exp. 12). Substances that do not have this charge are said to Since many non-magnetic be neutral. materials are attracted, it is evident that electric charges differ from magnetism. It has been found by many experiments that whenever two different substances are rubbed together each will become this charge electrically charged. that builds up in gasoline tank trucks as they move along the highways. To prevent sparking, which could cause an explosion, a chain drags behind the truck to allow the charge to leak off gradually. Responsibility for many fires has been traced to charges of static electricity and precautions are necessary in many industrial plants to ensure that sparking does It is not occur. V : 1 1 KINDS OF ELECTRIC CHARGES It is a simple matter to prove that a hard rubber or ebonite rod rubbed with cat’s fur or flannel, or a glass rod rubbed with silk, will become electrically V : 9 INTRODUCTION TO ELECTROSTATICS is attributed Greece, about 600 b.c. Every high-school student is familiar with many devices operated by electricity. However, this is electricity in motion, a comparatively recent scientific development. Early experiments in this field were confined to electricity at rest, or static electricity. The discovery of generally electricity to Thales of Miletus, one of the seven wise men of He learned that amber, a yellowish fossilized resin, when rubbed with a fabric would attract light objects. This knowledge an of for over two thousand years. science From the beginning of the seventeenth century electrical science began to grow, and it is from the early experiments of such men as Gilbert and Gray in England, du Fay in France, and Franklin in America that many of our fundamental concepts of electricity have arisen. The study of electrostatics has paved the way for advancement in current electricity. remained isolated fact V: 10 ELECTRIC CHARGES As mentioned above, the phenomenon of electric charges was first noted over 268 ELECTROSTATICS Sec. V:12 charged (Sec. V:10). If we charge an ebonite rod in tliis way, suspend it by a stirrup from an insulated stand, and approach it with another charged ebon- Xt-N. Fig. 23:1 Like Charges Repel. ite rod (Fig. 23:1), the suspended rod is repelled. If the suspended ebonite rod is approached by a charged glass rod, attraction will be observed (Chap. 31, Exp. 13). Thus we can see that these introduced the term negative to describe the charge produced on an ebonite rod rubbed with fur or wool, and positive for the charge on a glass rod rubbed with silk. A simple law of electrical charges may now be stated: like charges repel each other; unlike charges attract each other. V: 12 THE ELECTRON THEORY Our knowledge of the structure of matter has made great progress since Experimental the turn of the centuiy. work beyond the scope of this text has led to a concept of the atom far different from that held until 1897. Until that time the atom was thought to be the Since then smallest particle of matter. our concept of the structure of the atom has been constantly changing. For our is convenient to think of the purposes it atom as follows: 1. It may be pictured as a miniature solar system with a central relatively large mass, called the nucleus corresponding to our sun, and a number of smaller objects, called electrons travelling in their orbits about the nucleus much as the planets move around the sun. 2. The nucleus is composed of two kinds of particles: (a) Protons—which are positively charged and are considered to have unit mass. (b) Neutrons—which have no electric charge and have the same mass as the proton. 3. The electrons are negatively charged and are approximately as heavy as a proton. University of Toronto Benjamin Franklin are two charges. In different kinds electric 1750 Benjamin Franklin of 4. Atoms are neutral normally in charge. Therefore we may conclude that in each atom there must be as many electrons moving around the nucleus as there are protons within the nucleus. 269 Chap. 23 MAGNETISM AND ELECTRICITY Hydrogen Oxygen Neon Magnesium Chlorine Fig. 23:2 Diagrams of Several Atoms— Electrons (•), Protons ( + ), Neutrons (n). 5. There are over one hundred different kinds of atoms known at the Scientists believe that present time. even more will be identified in the near future. The atoms of different elements differ from each other in the numbers of protons and neutrons contained in the nucleus and in the number and arrangements of electrons in their orbits. Some typical diagrams of atoms are shown in Fig. 23:2. An understanding of this theory of atomic structure enables us to explain why an object becomes charged by friction. Tremendous forces are required to separate a proton from the nucleus of an atom. However, it is relatively easy to cause an electron to leave an atom or to move to another atom. When an ebonite rod is rubbed with cat’s fur, electrons are transferred from the fur to the ebonite and the rod becomes negaThe fur, having lost tively charged. electrons, becomes positively charged. Similarly, when a glass rod is rubbed with silk, electrons are transferred from the glass to the silk so that the glass becomes positive and the silk negative. We may now define a positive charge as a deficiency of electrons. A negative charge is a surplus of electrons. A neutral body has equal numbers of protons and electrons. 270 V:13 ELECTROSCOPES Before proceeding further in our discussion of electrostatics it is convenient to study the means by which electric charges may be detected and identified. Instruments used for this purpose are called electroscopes. (a) The Pith Ball Electroscope The simplest form of electroscope consists of two balls of dried elder pith, each about the size of a pea, suspended side by side from an insulated stand by silk threads (Fig. 23:3). When the pith balls are approached by an uncharged object no attraction or repulsion will be noted. When a charged ebonite rod is touched to the pith balls, they will be ELECTROSTATICS Sec. V:13 seen to be attracted, cling to the rod for a moment, and then be violently repelled from it. The pith balls have become negatively charged and they will spring apart due to the repulsion of like charges. The stronger the charge, the greater is the repulsion. When the charged pith balls are now approached by an object bearing the same kind of charge, they will be repelled from it. If the charge on the object is opposite to that placed on the pith balls, they will be attracted to it (Chap. 31, Exp. 14). ball to the rod which, being positively charged, has a deficiency of electrons. Thus, the pith ball, when charged by contact, receives the same charge as the charging object. (b) The Gold-Leaf Electroscope This instrument, shown in Fig. 23:5 is much more sensitive than that described above. It consists of a piece of These observations may be explained electron the readily
by reference The to being rod, negatively theory. charged, has an excess of electrons. Because the rod is charged and the pith balls are not, they are attracted to it. As the pith balls touch the rod some electrons move from the rod to the electroscope. Thus the pith balls become negatively charged and are repelled. This is shown diagrammatically in Fig. 23:4. It is frequently convenient to use an electroscope with only one pith ball as shown for electrostatic experiments. events series of Similarly, the pith ball may be given a positive charge by touching it with a charged glass rod. In this case the electrons will be transferred from the pith Fig. 23:5 Gold-Leaf Electroscope. gold-leaf or thin aluminum-foil attached at its upper edge to a brass plate at the end of a brass rod. This assembly is enclosed in a metal case with glass windows for observation. The rod is in- 271 Chap. 23 MAGNETISM AND ELECTRICITY manner (Sec. V:40, V:77). Substances in which the electrons can not move insulators. A few readily common conductors and insulators are tabulated below. called are V : 15 INDUCED CHARGES Two metal spheres, A and B (Fig. 23:6), supported on insulated stands are placed in contact so that they form a ring causes sulated from the case by a of sulphur, wax, or rubber. The rod generally terminates at the upper end in a knob or flat disc to receive the charge. A negative charge on the electroscope may be obtained by touching the knob with a negatively charged rod (Chap. Electrons will be trans31, Exp. 15). ferred from the charged rod. to the knob, rod, plate and leaf of the electroscope. The similar charge on the plate and leaf leaf diverges. The extent of this divergence is an indication of the strength of the charge. If the knob is now approached by an object bearing the same kind of charge as that on the electroscope, there will be further divergence of the goldIf, however, the object bears the leaf. opposite charge, the leaf will be seen to fall. When touched by a positively charged object, electrons will leave the electroscope and it will be left with a positive charge. repulsion and the V : 14 CONDUCTORS AND INSULATORS In some materials, mostly metals, the loosely bound orbital electrons are so that they will leave the atom readily. In some other substances the electrons are so tightly bound to their nuclei that their movement is slight. Materials in which electrons can move readily are called conductors (Chap. 31, Exp. 16). Conductors may be solids, liquids, or gases, although liquids and gases conduct electricity in a slightly different single conductor. If a negatively charged ebonite rod is now brought near to A as described in chapter 31, experiment 17, and the surfaces of the spheres are examined for charge by means of a charged electroscope and a proof plane (a small metal disc at the end of an insulating-handle used to transfer charges to an electroscope), it will be found that the surface of A near the ebonite rod has acquired a positive charge. A negative charge will be found on the surface of B remote from the rod. If the ebonite rod is removed and the spheres Good Conductors metals graphite water solutions of acids, bases, and salts Poor Conductors dry wood paper alcohol kerosene pure water Good Insulators hard rubber paraffin wax sulphur mica porcelain dry air 272 . ELECTROSTATICS Sec. V:17 will again examined, no charges be This is an example of electrofound. static induction. The charges discovered on the spheres while the charged rod was near are known as induced charges. They are due to the repulsion of electrons from sphere A, which thus becomes positively charged, to sphere B, which thus acquired a negative charge. Induced charges are possible only with conductors since there are no free electrons in insulators to allow the moveIf A and B are separment necessary. ated while the ebonite rod is in position, peiTnanent charges will remain on the spheres (positive on A, negative If the rod is now removed and on B) the spheres allowed to touch, it will be found that no charge remains on either sphere. We may conclude then, that induced charges are equal. still Exp. 18). The leaf diverges as shown in Fig. 23:7 because electrons are repelled to the plate and leaf. Keeping the charged rod near, touch the knob with a finger, that is, “ground” it. The leaf will fall as electrons are repelled from it to “ground”. Next, remove your finger, and then the charged rod. The leaf will diverge again. The remaining electrons are distributed over the electroscope which is left with a deficiency of electrons and thus a positive charge. In order to give the electroscope a negative charge the above procedure must be followed positively charged rod. This method is illustrated Students are advised to in Fig. 23:8. draw the diagrams and make the necessary explanations bearing in mind that “ground” acts a extra source using of as a electrons. V:16 CHARGING BY INDUCTION V: 17 THE ELECTROPHORUS Approach, but do not touch, the knob of a gold-leaf electroscope with a negatively charged ebonite rod (Chap. 31, About 1775 Volta invented the simplest form of machine to produce electric charges by induction, called the electro3) Ground ( 4 ) (5) Fig. 23:7 Charging Gold-Leaf Electroscope Positively by Induction. (1) Uncharged electroscope. (2) Approach rod (— ). with charged ebonite (3) Touch with finger (ground) (4) Remove finger (5) Remove charged rod. 273 Chap. 23 MAGNETISM AND ELECTRICITY Fig. 23:8 Charging Gold-Leaf Electroscope Negatively by Induction. (1) Uncharged electroscope (2) Approach with charged glass rod ( + ) (3) Touch with finger (ground) (4) Remove finger (5) Remove charged rod It consists of a hard rubber or phorus. ebonite cake and a separate metal disc provided with an insulating handle. The ebonite cake is charged negatively by rubbing with fur. The metal disc is then placed on it and touched with a finger. If the disc is then removed by means of the insulating handle it will be found to possess a relatively strong positive charge. The disc has been charged by induction. In order to explain this we must that neither the disc nor the realize cake are perfectly smooth. Thus actual contact between them is made at very few points and we may assume that the two are separated by a thin air space over most of their areas. By induction a positive charge will appear on the under surface of the disc and a negative charge on its upper surface. When the Fig. 23:9 Charging an Electrophorus by Induction. (1) Ebonite rubbed and given neg- (3) Electrons repelled through finger to ative charge. ground. (2) Electrons repelled to top surface of metal disc. 274 (4) Metal disc positively charged. Ebonite cake still negatively charged. ELECTROSTATICS Sec. V:18 finger touches the upper surface, elec- are disc positive repelled from the trons to ‘Aground” and on removal of the finger the disc is left with a deficiency of electrons and a These changes are illustrated in Fig. 23:9. will be noted that very little of the charge is removed from the ebonite cake. The disc, therefore, can be charged repeatedly in the manner described above without recharging the ebonite. charge. It V:18 THE WIMSHURST MACHINE This is a well-known laboratory ma- metal sectors which serve both as inductors and carriers. Wire brushes, supported at the ends of two metal rods placed at right angles to one another across each plate, rub against the sectors as they pass towards collecting combs consisting of a number of sharp metal teeth directed towards the sectors on both front and back plates. These combs are placed at either end of the horizontal diameter of the plates and each is connected to a discharging knob supported above the machine. chine for the continuous production of The action of this machine is beyond electric charge by inductive separation and was first made by James WimsIt consists of two hurst about 1878. circular ebonite or shellacked glass plates mounted together geared close and as to rotate in opposite directions so (Fig. 23:10). On the outer surface of each plate near the rim are fixed an even number of equally spaced thin Fig. 23:10 Wimshurst Machine. the scope of this Interested students are advised to refer to a good text. encyclopedia. is interesting to note that with a 1 cm. air gap, 30,000 volts It are required to cause a spark discharge between the terminals. In spite of the high voltage there is little danger from the discharge because of the small quan- tity of electricity involved. Canadian Laboratory Supplies Ltd. 275 Chap. 23 MAGNETISM AND ELECTRICITY 1. In conductor (a) the leaves of the electroscope show the same divergence matter what part no is touched by the proof plane. 2. In conductor (b) the divergence proves to be a little greater from tests at the ends than at the centre. 3. In conductor (c) the divergence is much greater from tests at the more pointed parts of the conductor than elsewhere. From these tests we may conclude that the charge on a conductor tends to be concentrated at the more pointed parts of the surface. V : 20 LIGHTNING-RODS About the middle of the eighteenth century Benjamin Franklin demonstrated in his classic kite experiment that thunder-clouds carry electrical charges and that lightning is an electrical discharge clouds and objects on the earth. spark between clouds, or or If a metallic point is connected to one of the discharge knobs of a Wims- V: 19 DISTRIBUTION OF CHARGES The location of charges on a charged body may be demonstrated as in chap31, experiment 19A, using Biot's ter Spheres shown in Fig. 23:11. The apparatus consists of an insulated metal over which two tightly-fitting sphere can metal hemispheres be insulated placed to cover it completely. A charge is given to the sphere and then the tightly over it. hemispheres are fitted When the hemispheres are removed and tested using a proof plane and electroscope, each will be found to possess a charge, while the sphere is found to ha
ve lost its charge. The charge, then, must have passed to the hemispheres as they formed the outer surface of the sphere. Charges reside only on the outer surface of a charged conductor. To determine how the charge is distributed over the surface of a charged conductor we can charge several insulated conductors of different shapes as described in chapter 31, experiment 19B. On investigation with a proof plane and electroscope the following observations are made: 276 Fig. 23:12 Action of Points hurst machine, as shown in Fig. 23:12, the streaming of charge away from the point is sufficient to blow aside the ELECTROSTATICS Sec. V:21 flame of a small candle (Chap. 31, Exp. 20). The action of points is as if the point produces a self-discharging action, the charge on the conductor streaming away from it as an “electric wind”. Franklin made use of this action of points as he designed the lightning-rod to protect buildings against the dangers of lightning. As a charged cloud passes its ning with occurs over the earth it induces the opposite charge on objects below. If the charge becomes great enough a flash of lightaccompanying thunder, a shock-wave, caused by the sudden and violent expansion of the air heated by the discharge. The rods are the induced charge so designed escapes from the sharp points and so an accumulation of charge is prevented. If the lightning does strike the rods, a good that conductor leads the electricity safely to ground through a metal rod deeply buried in the damp ground near the building (Fig. 23:13). V : 21 QUESTIONS 1. 2. (a) What is the meaning of static electricity? (b) How is it produced? (c) How is it detected? (a) Name two kinds of static electricity. How is each produced? (b) How would you prove that there are two kinds? (c) State the law for electrical charges. 3. (a) Describe in point form our modern concept of the structure of an atom. (b) In terms of the electron theory, what is (i) a positively charged, (ii) a negatively charged, (iii) a neutral body? (c) Why does (i) ebonite become 5. (ii) glass become negative when rubbed with wool, when rubbed with silk? What is the electrical condition of the wool and silk afterward? positive 4. (a) What are the purposes of electroscopes? (b) Describe and explain what occurs when (i) a positively charged, (ii) a negatively charged object Is brought up to an uncharged pith ball electro- scope. (c) How would you use it to identify an unknown charge? the (a) Describe the gold-leaf electroscope and state the purpose of each part. structure of 277 Chap. 23 MAGNETISM AND ELECTRICITY 8 . Describe, using a series of diagrams to illustrate your answer, how to charge a gold-leaf electroscope (a) negatively by induction. positively, (b) 9. Explain the following: (a) the metal disc of an electrophorus can be repeatedly charged without recharging the ebonite cake. (b) the purpose of the Wimshurst machine. (c) electrical charges reside on the 10. outside of a conductor. (d) lightning-rods are sharply pointed. (e) lightning-rods are connected by a good conductor to ground. A positively charged cloud hovers over a tall church spire. What will be the resulting electrical condition of the spire? What will happen if a large enough charge accumulates on the cloud? 6 . 7. (b) Explain how to charge it positively, (ii) negatively by contact. (c) How would you use it to identify an unknown charge? (i) (a) What is a good conductor of electrical charges? Give examples. (b) What is an insulator of electricity? Give examples. (c) How could you use a gold-leaf electroscope to show that iron is a better conductor than wax? (a) What is meant by electrostatic induction? (b) You have a charged glass rod and two metal spheres mounted on insulated stands. Describe how you would charge one sphere negatively, and the other positively, by electro- static induction. (c) How would you verify that the spheres were oppositely charged? 278 1 ^ . CHAPTER 24 ELECTRIC CURRENT V:22 INTRODUCTION In chapter 23 it was shown that electrons moved between the terminals of the Wimshurst machine and that lightning is actually a movement of electrons between earth and charged clouds. Such movement of electrons is called an electric current. In the eighteenth century Luigi Galvani, an Italian physiologist, discovered that a freshly dissected frog’s leg showed violent muscular contractions when touched by two dissimilar metals. He erroneously attributed this property to the frog, and it remained for Alessandro Volta, an Italian professor of physics, to demonstrate a short time later that the frog’s leg had acted only as a sensitive detector of electricity which had been produced by contact with the dissimilar metals. Volta showed that this electricity was identical with that produced by frictional means (Sec. V:10). From this beginning he constructed the first source of continuous electric current, the voltaic cell (Sec. V:23). From the time of of Volta’s works in 1800, there has been continuous progress in the field of electricity. The sections that follow will publication help to give you an understanding of the fundamentals of electricity that mean so much in our modern way of life. V:23 THE VOLTAIC CELL Every high-school student is familiar with the tiny dry cell that is used to operate a flashlight and the larger storage battery used to start a car; these are just two modern forms of the voltaic cell. A voltaic cell consists of two dissimilar materials immersed in a solution of an acid, called the base, or salt, electrolyte. voltaic The simplest form of cell consists of a copper plate and a zinc plate placed in a vessel containing dilute (Chap. 31, Exp. 21). sulphuric acid As a result of the chemical action in the cell the copper plate becomes posiand plate negatively charged. If the plates are charged tively zinc the i—O— 1 Dilute Sulphuric Acid Flashlight Bulb 1 ^ . 11111 1 1 11 1111111 'HWT ' Copper N. 1 111 II Zinc - Fig. 24:1 Diagram of Simple Voltaic Cell. 279 Chap. 24 MAGNETISM AND ELECTRICITY then connected by a conductor (Fig. 24:1) the potential difference between them will cause electrons to flow from the zinc to the copper. The circuit is As completed through the electrolyte. the current flows, hydrogen is deposited on the copper plate and zinc sulphate dissolves in the electrolyte as the zinc is gradually used up. The potential difference between the plates is approximately 1.1 volts. This P.D. is independent of the size of the plates used, but In all varies with the materials used. such cells chemical energy is transformed into electrical energy. V : 24 DEFECTS OF THE VOLTAIC CELL 1. Polarization If a small flashlight bulb is connected in the circuit of a voltaic cell it will glow brightly for a short time and then fade gradually until no light is seen (Chap 31, Exp. 21). It is apparent that the current delivered by the cell has decreased. This effect is caused by po- METAL CAP, with washer to insulate the cap from the metal cover. EXPANSION SPACE, for expansion of cell contents during use. as the ZINC CAN serves as the negative plate and at the same time container for the cell. ELECTROLYTIC PASTE sal-ammoniac containing and zinc This reacts with the zinc making it electro-negative and producing hydrogen which moves toward the carbon rod. chloride. larization, which is the accumulation of bubbles of hydrogen on the positive plate. The hydrogen deters the acid from reaching the plate and serves as an insulator preventing further transfer of electrons. To restore a polarized cell the bubbles may be removed mechanically by wiping the positive plate, shaking it, or, chemically, by adding an oxidizing agent such as potassium dichromate to the solution. Such a chemical releases oxygen which unites with the hydrogen to form water. Substances added for depolarizing purpose called are this agents. 2. Local Action cell, the If impure or commercial zinc is used the plates are found to in deteriorate rapidly even when the cell is not in use. This is caused by small particles of such impurities as lead, iron, carbon, etc., occurring in the plates used. Many small voltaic cells are set up between these impurities and the zinc, and local action continues until the plate METAL COVER, which closes the cell tightly at the top, making it safe against bulging and breakage. ASPHALT INNER SEAL. CENTERING WASHER. DEPOLARIZING MIX. This contains carbon to provide conductivity, and manganese dioxide which reacts with the hydrogen forming water which keeps the contents moist. CARBON ROD, serves as the positive plate. It is powdered composed of bonded carbon particles together and baked at a very high temperature. COMPLETE CELI.. BOTTOM WASHER. 280 Fig. 24:2 Structure of a Dry Cell. ELECTRIC CURRENT Sec. V:26 is consumed. Thus zinc is used up while delivering no useful current outside the thinner and may eventually break open allowing the paste to dry out. Tliis in- cell. The use of pure zinc, of course, would be a simple remedy, but pure zinc is expensive. A more practical method of eliminating local action is to amalgamate the surface of the zinc plate with mercury. The zinc dissolves in tlie mercury and gradually comes to the surface where it reacts with the acid. The impurities are covered by the mercury and their contact with the electrolyte is prevented. V:25 THE DRY CELL Obviously, the voltaic cell described in section V : 23 has several disadvanThe dry cell tages eliminates many of these undesirable Its structure is shown in Fig. features. practical use. in 24:2. creases the internal resistance of the cell to such an extent that current will flow in the circuit. if any, little, The potential difTerence of a fresh dry cell is approximately 1.5 volts regardless of size. Large dry cells are able to produce current over longer periods than smaller cells. Dry cells of various sizes and shapes have been made to fill many needs. They are widely used for flashlights, portable radios, telephones, hearand
many other ing devices used by the armed forces and industry. lanterns, aids, V:26 THE ELECTRIC CIRCUIT The flow of electrons in a wire is so similar to the flow of water in a pipe a comparison may be made as that As the cell is used the zinc becomes follows: The Water System The Electric Circuit 1. Water current is the flow of water through the system. 1. Electric current is the flow of electrons through the wires. 2. Water flows through a closed system of pipes between source and user. 3. The quantity of water in a system is measured in gallons or similar units. 4. The current of water is measured in gallons moving through the pipe in a unit of time. 5. Water current is reduced by the friction between the water and the pipes. 2. Electrons flow through a closed path of wires called a circuit. 3. The quantity of electricity is measured in coulombs. A coulomb is equal to approximately 6.28 billion billion electrons. 4. The electric current is measured in amperes. An ampere is the current when one coulomb passes a point in the circuit in 1 second. 5. Electric current is decreased by the resistance of the conductor. The unit of resistance is the ohm. 281 : Chap. 24 MAGNETISM AND ELECTRICITY The Water System 6. Water is driven through the pipes by pressure which is measured in pounds per square inch or related units. Pressure is provided by increasing the potential energy of the water by high tanks. The force so created pushes the water to its destination although the water pressure decreases as the length of pumping into it pipe increases. The Electric Circuit 6. Electrons move through a circuit because of a difference in electrical pressures or potential difference (P.D.) between points in the wire. This potential difference is measured in volts. The pressure difference is maintained by the use of batteries or cells which provide a source of electrons by means of the transformation of chemical energy. This potential difference is the electromotive force necessary to move the electrons through the wire. Resistance in the circuit causes a lowering of potential ence between points the conductor. This is referred to as a voltage drop. differ- in The units of electricity in common use and their symbols are sum- marized below for convenient reference (b) Current of electricity (a) Quantity of electricity (Q) —-unit is the coulomb (/) — unit is the ampere {V ) — unit is the volt (R) — unit is the ohm Potential difference (d) Resistance (c) Smaller or larger units are obtained by using submultiples or multiples of these as: 1 milliampere = 1/1000 ampere 1 microampere = 1/1, 000, 000 ampere 1 millivolt = 1/1000 volt 1 megohm V : 27 ELECTRIC CURRENT AND ELECTRON FLOW Electric current has been defined as the flow of electrons through the wires of a circuit (Sec. V:26). From the beginning of electrical theory, the current has been considered to flow, as is the case with water, from the place of high potential to the place of lower potential. Early scientists considered the positive side of a source of electricity to be at 282 1 microvolt 1 microhm 1 kilovolt = 1,000,000 ohms etc. = 1/1,000,000 volt = 1/1,000,000 ohm = 1,000 volts a higher potential than the negative side. Thus, the convention that electric current flows from positive to negative came into common use. We now know that the negative side of a source has a surplus of electrons and the positive side a deficiency of a electrons fundamental law of nature, electrons move from where there are more to where there are fewer. Electrons, there- (Sec. V:12). Following fore, actually move from negative to (b) Parallel Circuit ELECTRIC CURRENT Sec. V:29 positive. In all work on electronics, including radio, it is necessary to refer to electron movement in order to understand the operation of modern devices. However, in most other electrical work, unless definitely specified, references have been confined in the past to the conventional current flow, i.e., positive to negative. In most modern writings, including this text, all references are to the electron flow. No difficulty need arise if the difference between current flow and electron flow is recognized. V : 28 TYPES OF CIRCUITS Two main types of connection are in common use in electric circuits and all others are merely combinations of these. (a) Series Circuit In this arrangement the current goes from one terminal of the source to one terminal of the first appliance. On passing through this, it flows to the next and the next until it reaches the other terminal of the source (Fig. 24:3). All In the parallel or shunt type of connection the conductor from the positive terminal of the source is connected to one terminal of each appliance and the negative terminal of the source is connected to the other terminal of each Electron Flow 4- Battery Appliance = © (I) (£) (p Fig. 24:4 Parallel Circuit. appliance (Fig. 24:4). This is typical of wiring circuits used in your home. Each appliance receives current independently from the others and thus a break in one does not interfere with the operation of the rest. V:29 ARRANGEMENT OF CELLS (a) Series If the resistance in a circuit is so great that the P.D. of one cell is not sufficient to provide the required current, a number of cells may be connected in series. The positive terminal of the is connected to the negative of the second, the positive of the second cell first the current must pass through each appliance, causing a voltage drop (Sec. V:26) in the circuit. A break in any part of the circuit will stop all flow of current. Series circuits are frequently found in Christmas-tree lights, but are seldom used for other purposes. Fig. 24:5 Battery of Cells Connected in Series. 283 Chap. 24 MAGNETISM AND ELECTRICITY to the negative of the third, etc. Such an arrangrnent of two or more dry cells is called a battery. When cells are connected in series the total P.D. is equal to the sum of the voltages of the indiradio vidual B-battery, for example, is composed of thirty dry cells connected in series (Fig. cells. A forty-five volt 24:5). (b) Parallel If the resistance in a circuit is low and a large current is to be used, dry cells are connected in parallel. All the positive and negative terminals are connected as shown in Fig. 24:6. The P.D. of such a battery remains the same as that of a single cell. Since the current is provided equally by all cells the parallel arrangement gives a large current for a relatively long period of time. V : 30 SWITCHES Just as taps are inserted in a pipe line to stop the flow of water, so switches are used to interrupt the flow of electric current in a circuit. These are devices which break or complete the conducting path. Many types of switches are in common use, but only two will be described here. erally of the type shown in Fig. 24:7a. Wall switches are smaller variations of the same type. (b) Push Button Switch This familiar type is shown in Fig. 24:7b. It is used to control low-voltage Fig. 24:7 Switches (a) Main House Switch (b) Push Button Switch circuits in which the current is required for short intervals only. Electric doorbells, signal buzzers, etc. make use of this type of switch. at this their description Many newer types of switches have been devised for specific purposes, but the scope of this text does not allow The interested student will find information on mercury switches, automatic circuit breakers, and special switching arrangements such as those used to control a stairway light from either upstairs or downstairs in any good wiring handbook. time. V:31 USE OF SYMBOLS (a) Knife Switch Modifications of the common knife switch are used to control many household circuits. The main switch is gen- In order to represent the many pieces of equipment encountered in electrical drawings and circuit diagrams, a number of conventional symbols have been de- 284 ELECTRIC CURRENT Sec. V:31 vised. A few of the most common symbols are introduced at this point so that the student may use them in the work which follows. SOURCES OF •potential DIFFERENCES + Cell —wwvw Fixed Resistors WIRING I Battery (Cells in Series) RESISTORS AWWW- L T T T Battery (Cells in Parallel) METERS Wires Joined Wires Crossing Not Joined RADIO Voltmeter Ammeter Fig. 24:8 Common Electrical Symbols 285 Chap. 24 V : 32 MAGNETISM AND ELECTRICITY QUESTIONS 5. Define the following: electric circuit, ampere, coulomb, electrical resistance, potential difference. 6 . (a) Name the two principal types of circuits and define each. electrical (b) State the advantages and disadvantages of lights each way. (c) Make a diagram showing three lights connected in each of these ways. connecting electric 7. (a) Distinguish between the arrangement of cells in series and in parallel. (b) What is the purpose of each of these arrangements? (c) What would be the difference of three dry cells connected potential (I) in series (ii) in parallel? 8. Make a diagram of a circuit containing a cell, a fixed resistor, two wires joined, two wires crossing but not joined, a fuse, a lamp, a switch. 1. 2. 3. (a) What is an electric current? (b) How is an electric current produced? (a) Make a diagram of a simple voltaic cell. Mark the direction of the electric current (electron flow). (b) What determines the voltage of such a cell? (c) What transformation of energy takes place in a voltaic cell? (a) What are defects in a voltaic cell? (b) How are the defects overcome? two the principal 4. (a) Make a labelled diagram of a dry cell. (b) State the purpose of each struc- ture or material labelled in part (a). (c) What is the voltage of a dry cell? (d) Why are different sizes? dry cells made in 286 ; CHAPTER 25 OHM’S LAW AND RESISTANCE V:33 OHM'S lAW relationship The between current strength, potential difference, and resistance of the circuit is of the utmost importance in practical electricity (Chap. In such an experiment 31, Exp. 22). the potential difference across the revaried by connecting sistance different
numbers of dry cells coil is Dry Cells ment is shown in Fig. 25:1. Let us consider some sample results of such an experiment No. of Cells Galvanometer deflection No. of Cells (divisions) Deflection 1 2 3 4 5 4.9 9.8 14.8 19.7 24.5 0.204 0.204 0.203 0.203 0.205 Within the limits of experimental error. Number of cells in the Galvanometer deflection And, since the P.D. across the resistance coil varies directly as the number of cells used, and the galvanometer deflection is proportional to the current through the coil, we have Potential Difference ~ a constant Current Strength or Current Strength is proportional to Potential Difference The strength current circuit. flowing through the coil is gauged by the deflection of a galvanometer needle ( Sec. V : 53 ) . A circuit for this arrange- the of in the This relationship was first investigated by nineteenth early George Simon Ohm. The results of his work, published in 1826, are embodied in the law bearing his name. Ohm’s Law may be stated as follows: century The current through a given conductor is proportional to the potential difference between its ends. It must, however, be emphasized that 287 . : Chap. 25 MAGNETISM AND ELECTRICITY this law is true only if the temperature remains constant (Sec. V:34). The value of the constant derived above is known as its resistance {R) The unit of resistance is the ohm which may be defined on the basis of Ohm’s Law as being the resistance of a conductor which permits a current of one ampere to flow when a potential difference of one volt is applied across it. Ohm’s Law may be written, then, as Potential Difference „ = Resistance . or = Ohms Current Volts Amperes V I This formula may be rearranged to read V = IR or I = V/R All of these forms will be found useful in numerical work. Problem examples are given in sec- tion V:36. FACTORS AFFECTING >f:34 RESISTANCE The resistance of a conductor depends on four main factors; (a) Length—The resistance of a con- ductor varies directly as its length i.e., twice the length gives twice the resistance; half the length gives half the resistance; etc. (b) Cross-section—The resistance of a conductor varies inversely as the area of twice the cross-sectional area gives half the cross-section i.e., its resistance; half the cross-sectional area gives twice the resistance; etc. (c) Temperature—In most metals an increase in temperature causes an inin some crease resistance, but in 288 substances such as glass, carbon and electrolytes an increase in temperature causes a decrease in resistance. (d) Material—Materials differ widely in their resistances. Good conductors such as copper and aluminum have very low resistances. Poor conductors such as nichrome and manganin have much higher resistances. Good insulators, such as glass, mica, etc., have very high resistances. Relative Resistances of Some Common Materials (in ohm-cms at 20°C.) Aluminum Copper Iron Mercury Nichrome Platinum Silver Tungsten X 10-6 2.83 1.72 10.0 95.8 100 10.0 1.63 5.51 V : 35 RESISTORS IN SERIES AND IN PARALLEL The potential difference which causes the current to flow between the ends of a resistor is frequently called the voltage drop across the resistor. The value of a voltage drop (F) may be readily calculated by using Ohm’s Law. (a) Resistors in Series If resistances of r^, and ohms are AVWV-t-AAAA/V Fig. 25:2 Resistors in Series. joined end to end as shown in Fig. 25:2, and a current of 1 amperes is passed OHM’S LAW AND RESISTANCE Sec. V:36 through Ts- The potential difference beV volts and the total tween A and B effective resistance in the circuit is R ohms. through them, the voltage drop across V — IR) will be respectively: each Vi — hi, Vo z=z lr2. Vs = hs will be The total voltage drop Vi h Vo + Vs or I {ri + rg + rs). But V = IR, where R is the total re- sistance in the circuit. IR — I [ti + rg + Ts) and R — Ti r 2 h Ts (b) Resistors in Parallel By connecting resistors in parallel, • alternative paths are provided for the --I . Ij — — , 12 — — , 13 — V . V . . ri rg But I = ii is + is V rs + L + ri rg R And — = — + — + — Ts Ti rs rs R reciprocal of the total Thus we see that when resistors are in series, the total resistance of the circuit equals the sum of the individual resistances. When resistors are in parallel, resistance the equals the sum of the reciprocals of the individual resistances. Note The reciprocal of the resistance is known as the conductance of a conductor. The unit of conductance is the mho. Thus a wire with a resistance of 2 ohms will have a conductance of half a mho. current through the circuit (Fig. 25:3). The current, /, divides at A so that ii passes through r^, fg through rg, and is V : 36 PROBLEM EXAMPLES INVOLVING OHM'S LAW Example 1 Calculate the resistance of a light bulb which carries a current of 2 amperes when connected to the 110 volt circuit. T = 1 10 volts 1 = 2 amp. R = ? / no . „ V ^ ' R = 55 Resistance of the bulb 55 ohms. Example 2 If two lamps Li, Lg, of resistance 30 ohms and 20 ohms are connected in the total resistance in the series in a 110 volt circuit, determine (a) 289 Chap. 25 MAGNETISM AND ELECTRICITY circuit (b) the current in the circuit (c) the voltage drop across each lamp. fi = 30 ohms t2 = 20 ohms (a) (b) ri = 2)0 ohms 72 = 20 ohms R = ? v = no volts /? nr 50 ohms 1 = ? / = 2.2 amp. 7i = 30 ohms rg rr 20 ohms 1^ V =110 volts H ' R = Ti + 72 (series connection) i? = 30 + 20 = 50 total resistance in circuit = 50 ohms I = — (Ohm’s Law) R 7 = 112 = 22 50 current in the circuit = 2.2 amp. Vi — hi (Ohm’s Law) vi = 2.2 X 30 = 66 voltage drop across Li = 66 volts V2 = Ir2 .\v2 = 2.2 X 20 = 44 voltage drop across = 44 volts. Example 3 If two lamps Li, Lg, of resistance 30 ohms and 20 ohms are connected in parallel in a 110 volt circuit, determine (a) the effective resistance of the circuit (b) the current through each lamp. 290 OHM’S LAW AND RESISTANCE Sec. V:37 :a) Ti — 30 ohms r =20 ohms R = ? (b) = 110 volts = 30 ohms = 20 ohms _ p — ? OF RESISTORS Ti 72 — = — + — (parallel connection) R 1 _ 1 “so R=\2 effective resistance of the circuit =12 ohms. 1 _ 5 _ 1 20 ~~60 ~ 12 V I j =z — (Ohm’s Law) ri 110 II 11 CO 30 current through Li — 3.7 amp. rz .•./5 = — = 5.5 current through Lg = 5.5 amp. 20 be made from special constantan or manganin (alloys of copper and nickel), having an accuracy within 1 % . These are in general use laboratory work and measuring- for instruments. A number of coils of fixed resistance are frequently arranged together in a Resistors are used primarily to control current and potential difference in eleccircuits. They are constructed in tric many forms and sizes varying from those which have resistances of only a fraction of an ohm, as used in some measuring instruments, to those having resistances of many megohms as found in radio receivers. In general, however, resistors may be classified or variable. either fixed as (a) Fixed Resistors These may be constructed of short lengths of metal strip for very low resistances. Carbon or wire coils will provide higher resistances. Carbon resistors are frequently used in radios and other equipment in which some variation in resistance is permissible since such resistors may have an error tolerance of up to 20%. Wire-wound resistors may Fig. 25:4 Section of Resistance Box. Details of construction resistance box. are shown in Fig. 25:4. The ends of the 291 Chap. 25 MAGNETISM AND ELECTRICITY the two (Chap. 31, Exp. 23 and 24) which follow: (a) Voltmeter—Ammeter Method The unknown resistance, R, is connected in the circuit shown in Fig. 25:6. The rheostat is adjusted until the amme(Sec. V:54) records any suitable ter coils are attached to brass blocks on the top of the box and the current passes through the blocks when the shortingplugs are placed in position. Pulling out the plug puts the resistor below it into the circuit. Such a box provides a large range of standardized resistors for laboratory use. To avoid errors in using the plugs must be a resistance box, Plugs and sockets must inserted firmly. be kept clean at all times. (h) Variable Resistors (Rheostats) Such resistors are usually continuously variable between certain limits. A common form of rheostat is shown in Fig. It consists of a number of turns 25:5. Fig. 25:6 for Determining Voltmeter-Ammeter Method the Value of an Unknown Resistance. value for the current passing through the resistor. The voltmeter (Sec. V:54) across the resistor indicates the potential difference between the ends of the resistor. From this information the resistance may be calculated by using Ohm’s Law. V = LToUs R= ? Example = J .•./e = l^ = 4 3 .'. The resistance is 4 ohms. (b) Substitution Method The unknown resistance may be connected in a circuit as shown in Fig. 25:7. The rheostat is then adjusted until a large deflection is shown on the ammeter, or galvanometer (Sec. V:53). The meter reading is noted and the un- Fig. 25:5 Rheostat. of wire wound on a porcelain tube. Above the coil is supported a metal bar along which moves a metal spring which can make contact at any turn of the coil. Connections are made to the end of the bar and to the end of the coil to enable the resistance in the circuit to be readily adjustable by moving the spring. Variable resistors such as this are of various sizes and shapes and are used in radio volume controls, for dimmer lights in theatres, starter boxes for heavy electric motors, etc. V:38 METHODS OF MEASURING RESISTANCE Although there are several methods the scope of of measuring resistance, this text permits the discussion of only 292 OHM’S LAW AND RESISTANCE Sec. V:39 known resistance is replaced by a rePlugs are removed until sistance box. the meter reading is the same as before. The resistance of the box is now equal to the unknown resistance. WAW\/\ Rheostat Substitution Fig. 25:7 for Determining the Value of an Unknown Resistance. Method Battery Galvanometer O Ammeter or —WWW Unknown Resistance or Resistance Box V : 39 QUESTIONS 1. 2
. (a) State Ohm’s Law and explain how it may be determined. (b) Calculate resistance the of a light bulb which carries a current of 5.0 amperes when connected in a 110 volt circuit. (a) In terms of Ohm's Law define: ohm, volt, ampere. (b) What weakness in these definitions is apparent? 3. (a) Explain electrical resistance. (b) List four factors affecting re- sistance. State the effect of each. 4. (a) What is meant by voltage drop across a resistor? (b) What is the effect on the total resistance of connecting a number of resistors (i) in series, (ii) in paral- lel? If two lamps of resistance 50 (c) ohms and 40 ohms are connected in series in a 110 volt circuit calculate (i) the total resistance of the lamps (ii) the current in the circuit (iii) the voltage drop across each lamp. (d) If two lamps of resistance 20 ohms and 40 ohms are connected in parallel in a 110 volt circuit, determine (i) the effective resistance of the circuit. (ii) the current through each lamp. 5. 6 . (a) Describe a resistance box and a rheostat. State the purposes of each, (b) What is the effect of removing a plug from a resistance box? (a) Describe two methods for finding the resistance of a conductor. (b) In an electrical circuit, the ammeter reading is 5.5 amperes and the is 77 volts. potential What is the resistance? difference B 1. What is the resistance of a flashlight bulb which carries a current of 0.50 ampere when connected in series with a 6.0 volt battery? 2. The resistance of a conductor is 25 ohms and it carries a current of 8.5 amperes. What is the potential difference? 3. How much current does a 36 ohm resistance draw when operated on a 120 volt line? 4. (a) What current flows through a 99.5 ohm resistance connected in series with a 1 2 volt battery having an internal resistance of 0.50 ohm? (b) What is the voltage drop across the resistance and across the battery? (a) What is the electric toaster which passes a current of 5.0 amperes when connected in alio volt circuit? (b) What resistance must be placed resistance of an 5. 293 Chap. 25 MAGNETISM AND ELECTRICITY in series with the toaster to lower the current to 4.0 amperes? ference of 20 volts is applied between points A and B. 6 . Eight lamps, each having a resistance of 330 ohms, are connected in parallel in the 110 volt circuit. Calculate (a) the effective resistance of the circuit. (b) the current in each lamp in the circuit. (c) the total current in the circuit. 7 . A battery of resistance 0.10 ohm is connected in series with an ammeter of resistance 0.5 ohm. The ammeter registers 1 0 amperes. When a resistance is placed in series with the ammeter the current drops to 4 amperes. (a) Draw a diagram of the circuit described. (b) Find the voltage of the battery. (c) Find the value of the added resistance. 8 . The difference of potential between the ends of a certain resistance coil when a current of 0.36 ampere is passed through is 1.2 volts. What resistance must be connected in parallel with the coil so that, with the same total current, the difference of potential will be only 1.0 volt? 6 ohm the total (a) Calculate resistance between A and B when the switch is closed. (b) Find the current in the 6 ohm resistance with switch open. (c) Find the current in the 1 2 ohm resistance with switch closed. 10 . A cell with a potential difference of 2 volts and negligible resistance sends a current through two resistances of 6 ohms and 9 ohms connected in parallel. In series with the cell is a third resistance of 2 ohms. Calculate the current in the 6 ohm resis- tance. 11. A current of 4.0 amperes is passed through a resistance of 2.0 ohms in series with a parallel combination of 4.0 ohms and 6.0 ohms. Calculate (a) the current in each of the parallel resistances. 9 . Three resistances are connected as shown in the diagram and a potential dif- (b) the potential difference across the whole circuit. 294 CHAPTER 26 CHEMICAL EFFECTS OF ELECTRIC CURRENT reason that one can receive a severe electric shock while standing in water or on a damp floor. V : 40 ELECTROLYSIS others All chemical compounds are composed of two or more elements. The elements are composed of atoms as described in Section V:12. Frequently when atoms unite, electrons are transferred from one kind of atom to the other. The atoms electrons become charged which lose positively and the negatively. These charged particles are called ions. The oppositely charged ions attract each other to form ion-pairs. However, the total number of positive and negative charges has not been altered, so that the ion-pairs formed are neutral. When substances formed in this way are dissolved in water, some of these particles and separate dissociate negative ions. Compounds which form ions in solution are capable of carrying electric current and are called electrolytes. Examples of good electrolytes are solutions of acids bases and salts. Many other substances do not ionize when in solution and so do not conduct electricity. These are classed as non-electrolytes. Sugar, alcohol and distilled water are examples of these. Ordinary water salts which usually contains dissolved make it a weak electrolyte. It is for this positive into the water, decomposition of Shortly after the discovery of current electricity by Galvani and the construction of the first voltaic cells, experimenters like Sir Humphrey Davy of England and Svante Arrhenius of Sweden investigated the decomposition of water by electricity, or as we speak of it now, the electrolysis of water. Davy showed that the in volume of hydrogen produced is double His most striking disthat of oxygen. coveries were the breaking apart by electricity of the alkalis, caustic soda and caustic potash. Arrhenius explained these phenomena by his theory of ionization which assumes that in solutions of electrolytes there is at least a partial dissociation of the dissolved substances into It was not until 1834 that separate ions. Faraday introduced the term anode for the positive plate and cathode for the negative plate cell (Fig. 26:1). The negatively charged ions which are attracted to the anode became known as anions, and the positively charged ions which are attracted to the cathode as cations. When compounds dissociate in solution, metals and hydrogen form cations while non-metals radicals become and in an electrolytic chemical most anions. A typical electrolytic cell consists of two conducting-plates or electrodes immersed in an electrolyte. The anode is 295 Chap. 26 MAGNETISM AND ELECTRICITY the negative connected to the positive terminal of a battery or generator and becomes positively charged while the cathode is conterminal and nected to becomes negatively charged. Anions will move to the anode where they give up their surplus electrons, are neutralized, and released as neutral atoms. These electrons flow from the anode, through the connecting wire to the battery terminal. At the same time cations move to the cathode where they receive electrons and became neutral atoms. Electrons move from the negative battery terminal to the cathode to replenish the to the positive terminal of the battery. The electrolyte is decomposed by the electric current. This process is known as electrolysis. V : 41 ELECTROLYSIS OF WATER Water may be decomposed readily in a simple electrolytic cell (Fig. 26:1), or in the HoflFman water voltameter (Fig. 26:2), by the passage of a direct current through it (Chap. 31, Exp. 25). Fig. 26:2 Hoffman Water Voltameter. A small amount of sulphuric acid (about 10% of the total volume) is added to the water to make it a better electrolyte. The anode and cathode are made of platinum which is unaffected by the electrolyte and is a good conductor of electricity. In solution the acid is highly dissociated into positive hydrogen ions (H"^) and negative sulphate ions (S04““). H2S04^2H^ + SO4-- The water is slightly dissociated into positive hydrogen ions and negative hydroxyl ions (OH“) : H 2O + OH-. Fig. 26:1 Electrolytic Cell as Used in the Electrolysis of Water. supply. Thus we have electrons moving from the negative battery terminal to the cathode, through the electrolyte by way of the ions to the anode, and back 296 . CHEMICAL EFFECTS OF ELECTRIC CURRENT Sec. V:42 When the voltage is applied to the electrodes the following reactions occur: within the electrolyte completes the circuit (Sec. V:40) At the Cathode 1. The positive hydrogen ions are attracted to the negative cathode. 2. Each hydrogen ion is neutralized by gaining an electron from the cathode and becomes a hydrogen atom: + le-»H° 3. The hydrogen atoms immediately combine in pairs to form molecules of hydrogen gas which escape as bubbles: H° + H° ^ H 2 . At the Anode 1. The negative sulphate and hydroxyl ions are attracted to the positive anode. 2. The hydroxyl ions are discharged in preference to the sulphate ions. Each hydroxyl ion gives up an electron to the anode to become a neutral hydroxyl group. OH--le-^OH° 3. Pairs of hydroxyl groups then combine to form water and an atom of oxygen. pairs OH° + OH° ^ H 2O + 0 ° 4. The oxygen atoms immediately comto form molecules of bine in oxygen ’gas which escape as bubbles. 0° + 0° ^ O2 As the ions are discharged more water dissociates into ions. The sulphuric acid remains in the electrolyte throughout the process and only the water is decomposed, producing two volumes of hydrogen for every one volume of oxygen. The electrons given up by the hydroxyl ions at the anode are propelled through the external circuit by the battery to the cathode where an equivalent number of electrons is taken up by the hydrogen ions. This constitutes the current in the external circuit. The movement of the ions to the oppositely charged electrodes V:42 ELECTROLYSIS OF COPPER SULPHATE SOLUTION If copper sulphate solution is used as the electrolyte and carbon rods as the anode and cathode, copper will be deposited at the cathode soon after the voltage is applied (Chap. 31, Exp. 26). the copper sulphate
is highly dissociated into positive copper ions (Cu^^) and negative sulphate ions (SO4--). In solution Cu SO 4 Cu"" + SO 4 - - The water is slightly dissociated into positive hydrogen ions (H"^) and negative hydroxyl ions (OH“). H 2O + OHis applied to When the voltage electrodes the following reactions occur: the At the Cathode 1. The copper ions and the hydrogen ions are attracted. 2. The copper ions are discharged in preference to the hydrogen ions. Each copper ion takes two electrons from the cathode and becomes a neutral copper atom. Cu"" + 2e ^ Cu° 3. The copper atoms are deposited on the carbon rod. At the Anode 1. The sulphate ions and the hydroxyl ions are attracted. 2. The reactions are exactly the same as in the electrolysis of water (Sec. V:42), in which oxygen bubbles were formed. Commercial use has been made of such a process in the electroplating industry (Sec. V:45), in which copper is used as the anode and the object to be plated as the cathode. Also, in the purification of various metals, a bar of impure metal 297 Chap. 26 MAGNETISM AND ELECTRICITY is used as the anode, pure metal as the cathode and a solution of a salt of the same metal as the electrolyte. Only the pure metal is transferred from the anode to the cathode where bars of pure metal are obtained. The impurities are left in the electrolytic cell. V:43 LAWS OF ELECTROLYSIS From your experimental work on electrolysis it will be apparent that there is a close connection between the current strength used and the amount of material that is decomposed (Chap. 31, Exp. It will also be readily noted in 27). the previous examples that the greater the time that the current flows, the greater the amounts of water decomposed or copper deposited. Further, it is not difficult to show that a greater mass of silver than of copper will be deposited using the same current for the same length of time. These observations were made as long ago as 1834 by Faraday who formulated Faraday’s Laws of Electrolysis as follows: 1 . The amount of chemical change produced (i.e., the amount of any sub- 298 stance deposited) by an electric current is proportional to the quantity of electricity passed (Quantity = Current X Time.) 2. The amounts of different substances deposited by the same quantity of electricity are proportional to their equiva(The equivalent weight lent weights. of a substance is the weight of the substance deposited by 96,400 coulombs.) The number of grams of various elements liberated by one ampere in one electrochemical second equivalent of the element. called the is Electrochemical Equivalents Aluminum Chlorine Copper Hydrogen Magnesium Oxygen Potassium Silver Sodium Zinc 0.000093 0.000368 0.000329 0.0000105 0.000126 0.0000829 0.000405 0.001118 0.000238 0.000339 V ;44 THE COPPER VOLTAMETER The internationally accepted definition of the ampere is based on the deposition of metal at the cathode of an electrolytic cell or voltameter. Silver was selected as the standard, since the relatively large amount of silver deposited in a short time enables a greater degree of accuracy in weighing, and because silver is very resistant to oxidation. The international ampere is defined as that current which, when passed through a solution of silver nitrate in accordance with given specifiat the rate of cations, 0.001118 gm. per second. deposits silver While silver is the best substance for accurate results, a copper voltameter is current satisfactory strength in elementary work (Chap. 31, Exp. 28) . In this experiment both anode determining for CHEMICAL EFFECTS OF ELECTRIC CURRENT Sec. V:45 and cathode should be copper and the electrolyte copper sulphate solution. The circuit is adjusted so that a small current flows. The cathode should be thoroughly cleaned with fine emery paper, washed and dried. It is then carefully weighed. Next the is connected, as circuit in Fig. 26:3, and the current is allowed to flow for exactly twenty minutes. The cathode is removed, dipped in alcohol or ether and allowed to dry by evaporaIt is then carefully reweighed. The this tion. current may be calculated as example. in Example Initial weisfht of cathode = 5.52 gm. = 5.71 gm. =0.19 gm. = 20 min. = 1200 sec. .‘. Weight of copper deposited Time of current flow Electrochemical equivalent of copper = 0.000329 In 1200 sec. weight of copper deposited =0.19 gm. .'. In 1 sec. weight of copper deposited = 0.19 = 0.000158 srm. Current flowing when 0.000329 gm. copper is deposited in 1 sec. = 1 amp. Current flowing when 1 gm. copper is deposited in 1 sec. = amp. 000329 1200 Current flowing when .000158 gm. copper is deposited in 1 sec. = 0.000158 X = 0.48 amp. 0.000329 .'. The current strength is 0.48 amp. V:45 ELECTROPLATING Electroplating is an important industrial process designed to improve the appearance, or resistance increase to corrosion, of various metals. It consists of covering the metal with a thin layer silver, chromium, copper, etc., by of electrolysis. The article to be plated serves as the cathode and a solution of a compound of the metal to be deposited makes up the electrolyte (Chap. 31, Exp. 29). Before plating, the object must be thoroughly cleaned of rust, grease, etc. This is usually accomplished by treatment with abrasives followed by rinsing in strong acid, then strong alkali, then weak acid again, and finally distilled In order to obtain a fine longwater. lasting deposit a small current over a long time interval must be used. Special platings require modifications in the process. For example, silver plating is usually preceded by copper plating to give a suitable base. In copper plating better results are obtained if a little sulphuric acid is added to the electrolyte. Chromium plating directly on another metal leaves small openings like pin-holes in the surface so that corrosion may begin below the chromium layer. To avoid this the base metal is often lightly coated with nickel before the chromium is applied. Electroplating has also been applied to the printing industry in the reproduction of pages of type and illustrations, a process called electrotyping. An impression of the original is made in wax or plastic. This mould is covered with graphite or metallic powder to serve as a conductor 299 Chap. 26 MAGNETISM AND ELECTRICITY Industrial Electroplating. By Electrolysis These Metal Plates are Being Coated with a Thin Layer of Copper. International Nickel Co. of Canada and is then used as the cathode in a copper or nickel electrolytic cell. The metal deposited on it is stripped off, backed with a metal of low melting-point to strengthen it and is then ready for use. This reproduction can be used over and over again and is readily stored for later editions. V:46 THE LEAD-ACID STORAGE BATTERY When the simple voltaic cell (Sec. V;23), and the dry cell (Sec. V:25), become discharged they must be discarded and replaced by new cells. Cells of this type are classified as primary cells. The lead storage cell differs from these in that once its chemical energy 300 has been exhausted, the cell may be restored to its original condition by electrical means. called secondary cells. The familiar lead-acid storage battery used in the electrical system of many automobiles consists of six secondary cells connected in series Such cells are 12 volts. provide approximately to Chemical energy is stored in the active materials of the plates and in the electrolyte when current is passed through during charging. This chemical energy is transformed into electrical energy as the cells are discharged in providing current in a circuit (Chap. 31, Exp. 30). (a) Structure of a Cell Several positive plates, consisting of CHEMICAL EFFECTS OF ELECTRIC CURRENT Sec. V : 46 VENT PLUGS SEALED COVER ELEMENT PROTECTOR SEPARATOR NEGATIVE PLATE PLATE HARD RUBBER CONTAINER SEDIMENT SPACE Exide Automotive Division Fig. 26:4 Structure of a Lead-Acid Storage Battery. lead peroxide forced into a strong grid of lead alloy, are joined to each other by metal strips (Fig. 26:4). Between each of these, and at the ends, are negative electrodes of spongy lead. These are connected by metal strips and kept plates by from touching the positive wood separators. The plates are contained in a moulded, hard rubber case holding the dilute sulphuric acid which serves as the electrolyte. (b) Discharging the Cell As the cell is used to provide the E.M.F. which causes current to flow in a circuit, it becomes discharged. The sulphuric acid combines with the porous materials of the plates and the following chemical reaction occurs: negative positive diluted plates electrolyte lead + lead peroxide + sulphuric acid -» lead sulphate + water both plates plates electrolyte As the two plates become coated with the same material the potential difference decreases. Simultaneously, the removal of sulphuric acid from the electrolyte and the formation of water causes dilution of the acid in the cell. Thus, discharged cells are indicated by a lowering of the P.D. of the cell and a decrease in the specific gravity of the electrolyte. A fully charged cell should have a P.D. of approximately 2.2 volts, and 301 Chap. 26 MAGNETISM AND ELECTRICITY a specific gravity of 1.275 to 1.300. specific gravity falls to 1.185. (c) Recharging the Cell It is ready for recharging when the In order to recharge the cell direct current from a rectifier or some other source must be passed through the cell in the direction opposite to that of discharge. The chemical reactions in the cells are reversed: diluted negative both plates lead sulphate + water electrolyte positive plates plates —» lead peroxide + lead + sulphuric acid electrolyte Thus, the plates are again made dissimilar and the P.D. increases to 2.2 volts. At the same time water is used up and sulphuric acid is produced by the reaction, so the specific gravity of the electrolyte increases. The charging and discharging reactions may be summarized as: negative positive electrolyte both plates diluted plates
plates lead + lead peroxide + s (d) Care of Storage Batteries 1. Keep the battery clean and dry. 2. Do not allow the battery to dis- 3. Keep the charge below specific gravity 1.185. above the level of the wood separators by addition of distilled water. electrolyte 4. Do not allow a fully discharged cell to stand in this condition for more than a few days before charging. (e) Battery Rating The life of a battery is rated in ampere-hours, i.e., amperes X hours. For example, a battery rated at 100 ampere-hours would be capable of de- discharge iric acid ^ lead sulphate + water electrolyte charge livering 5 amperes for 20 hours, or 12.5 amperes for 8 hours, etc. In general, the greater the number of plates per cell, the greater the ampere-hour capacity of the 3. battery. (f) Uses of Storage Batteries Storage cells are frequently used to provide a source of E.M.F. for extended periods of time, as in some radio transmitters and receivers. They are used in the starting circuits of automobiles and aircraft, for auxiliary power supplies for telephones, trains, submarines and ships and for emergency lighting-systems. V : 47 Q U E S noNS A 1. (a) Define: ion, electrolyte, non-elec- trolyte, electrolytic cell, anode, cath- ode, anion, cation, electrolysis. (b) Outline Arrhenius’ theory of ion- ization. 2. Explain how the electrolyte completes In the electrolysis of water describe: (a) the condition of the water and the sulphuric acid before the voltage is applied. (b) what happens in the electrolyte at the moment when the voltage is the circuit in electrolysis. applied. 302 CHEMICAL EFFECTS OF ELECTRIC CURRENT Sec. V:47 (c) the reactions that occur at (i) the cathode (ii) the anode. lysis of water by a current of 2.0 amperes flowing for 4 hours? 4 . Explain the electrolysis of copper 4 . What current would deposit 0.1 sulphate solution. gm. of silver in 8 hours? 5. Describe how electrolysis may be used in (a) silver plating (b) refining of copper 5. A constant current is passed through 20 copper voltameter a minutes and it is found that 0.500 gm. of copper is deposited on the cathode. exactly for 6 . 7 . (c) electrotyping. (a) State Faraday’s Laws of Electro- lysis. (b) Define electrochemical equiva- lent. (a) Define an ampere in terms of the deposition of metal at the cathode of an electrolytic cell or voltameter. (b) Why Is acceptable Ohm’s Law? (c) Why was silver selected as the international standard rather than more on definition based than that this copper? (a) Describe the construction of a 8 . lead-acid storage cell. (b) What transformations of energy (i) charging, (il) distake place in charging a lead-acid cell? (c) Give word equations describing the reactions as a lead-acid cell is 9 . (i) charged, (Ii) discharged. (d) How may the condition of charge of such a storage battery be determined? (a) Define ampere-hour. (b) What determines the amperehour capacity of a battery? (c) What precautions should be obstorage served caring for a in battery? B Calculate the strength of the current. 6 . An ammeter connected in series with a silver voltameter reads 1.50 amperes. In 50 minutes the increase in weight of the cathode is 4.947 gm. What Is the error in the ammeter reading? 7. A copper voltameter and a water voltameter are connected in series with a direct current supply. In 25 minutes 0.09 gm. of copper is deposited on the cathode of the copper voltameter. . (a) What current is flowing in the circuit? (b) What mass of hydrogen is liberated in the water voltameter? 8 . A current of 2 amperes is passed through a copper voltameter. Copper is deposited evenly on the cathode which has an area of 66 sq. cm. Find the thickness of the layer when the current has been flowing for 30 minutes. (Density of copper = 9.0 gm. per c.c.) 9. The anode of a copper voltameter is made of impure copper, and the particles of impurity detach themselves and fall to as cell bottom of the the electrolysis continues. A current of 2.0 amperes flows for 1 hour and 40 minutes. At the start the anode weighs 85.69 gm. and at the end 81.37 gm. Find the mass of copper dissolved from the anode and the mass of impurity released. 10 1. What weight of (a) copper (b) silver will be deposited in 3 hours by a current of 1 ampere? 2 . What weight of (a) copper (b) silver will be deposited in 4 hours by a current of 2.0 amperes? 3. What weight of (a) oxygen (b) hydrogen would be liberated in the electro- series connected (a) If four 2 volt lead-acid cells, each of 0.1 ohm internal resistance, a are in conductor of 3.6 ohms resistance, what current would flow? (b) How long continue to flow if the battery has a capacity of 30 ampere-hours? would current with this 303 CHAPTER 27 MAGNETIC EFFECTS OF ELECTRIC CURRENT through a piece of cardboard, as shown in Fig. 27:1. When the switch is closed and iron filings are sprinkled lightly over the cardboard, gentle tapping will cause the filings to arrange themselves in concentric circles with the wire as centre. Thus the pattern of the lines of magnetic force becomes apparent. The direction of these lines of force may be determined Left Hand Rule to DeterFig. 27:2 mine the Direction of the Magnetic Field about a Conductor. by placing a number of small magnetic compasses on the cardboard. The N-poles of the compasses indicate that the lines go in a counter-clockwise direction about the wire (Chap. 31, Exp. 31). It is apparent, then, that if the direction of current is known, the direction of the lines of force about a conductor may be determined as follows: Grasp the conductor with the left hand with the thumb extended in the direction of the electron flow ( — to + ) . The fingers V:48 ELECTROMAGNETIC EFFECT In 1820, Hans Christian Oersted, professor of Physics at the University of Copenhagen, made a discovery that has made possible most of the modern advancements in electrical knowledge. He observed that a wire carrying an electric current deflected a compass-needle. Further that whenever a current was passing through a conductor a magnetic field was set up around it. experiments established It is possible to investigate this magconductor passing field by a netic Fig. 27:1 Magnetic Field about a Conductor. 304 MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V:49 When considering the magnetic field about a helix, it is convenient to represent it as in Fig. 27:5. The helix is shown in cross-section with the front half removed. The electron flow is up the back section of the turns and down the front. In the diagram, electron flow is indicated by a dot (.) representing the head of an arrow and a cross ( + ) representing the Fig. 27:6a shows the lines of magtail. netic force around a few turns of wire. The Left-Hand Rule will confirm the will now point in the direction in which the lines of magnetic force encircle the is This called statement the wire. Left-Hand Rule (Fig. 27:2). Conversely, this rule may be used to determine the direction of the electron flow if the direction of the lines of force is known. V : 49 MAGNETIC FIELD ABOUT A HELIX A single turn of wire (Fig. 27:3) is called a loop or coil. A series of such Fig. 27:3 Magnetic Field about a Loop Carrying Electric Current. loops in a wire (Fig. 27:4) is referred to as a helix or solenoid. If a conductor is formed into a single loop, the lines of Fig. 27:4 A Helix or Solenoid. magnetic force which surround the wire will pass through the centre of the loop as shown. Note that the magnetic field inside the loop will be more dense than outside since the lines of force are crowded into a smaller area. The Lines of Magnetic Force Fig. 27:6 Around the Individual Turns of Wire in a Helix as Shown at (a). Combine to Form the Resultant Field Shown at (b). Fig. 27:5 Cross-sectional View of a Helix with the Front Half Removed. The Direction of Electron Flow is Indicated. direction of these lines. Between the turns the forces are in opposite directions and so tend to cancel each other out. In the centre of the coils the forces act in the same direction and are very concentrated. On the outside of the helix the forces also reinforce each other to give us Fig. 27:6b. Note the similarity between this field represented in the 305 Chap. 27 MAGNETISM AND ELECTRICITY pattern and that formed by a bar magnet (Sec. V:4). The existence of field may be readily proved by inserting a helix, carrying current, in a slit in a piece of cardboard Iron filings 27:7). (Fig. will this Apparatus Used to DemonFig. 27:7 strate the Field of Force about a Helix Carrying Electric Current. the pattern of the magnetic indicate field (Chap. 31, Exp. 32). The polarity of the field may be determined by testing with a compass-needle. From this information a simple rule may be confirmed: Grasp the helix in the left hand so that the fingers extend in the direction the electrons are flowing around the turns. The extended thumb will then point towards the N-pole of the helix This may be called the (Fig. 27:8). Left-Hand Rule for the Helix or, to avoid confusion with other rules, simply the Helix Rule. V:50 ELECTROMAGNETS If a bar of iron is placed in the centre of a coil of wire through which a cur- 306 rent is flowing, the iron will become magnetized. The elementary magnets of the iron become aligned in such a direction as to reinforce the magnetic field of the coil. Thus the iron core greatly strengthens the magnetic field. Such an arrangement is known as an electromagnet. Soft iron makes the most satisfactory core, since a temporary magnet is produced. When the current in the is stopped the magnetism of the coil core is lost in a short time. An increase in the number of turns in the coil or an increase in the strength of the electric current flowing through it will result in a more powerful electromagnet (Chap. 31, Exp. 33). Electromagnets have been designed in a variety of shapes to meet many needs. The iron clad type is widely used where strong magnetic fields are desirable as Iron Core * ^ 1 \ Electron Flow 1 ) ) )
) ) ) ..--1 1 I . ^ ^ 2 r 1 Fig. 27:9 Iron-Clad Electromagnet. for a lifting magnet for scrap iron or in radio loudspeakers. The iron core not only passes through the centre, but almost completely surrounds it as shown in Fig. 27:9. The horseshoe electromagnet (Fig. 27:10) is used in such devices as earphones, telephone receivers, electric bells and electric buzzers. The electric bell (Fig. 27:11) consists of a gong, a horseshoe electromagnet, soft iron armature and contact screw arranged in a circuit as shown. When the switch is closed MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V:50 loses The electromagnet broken. its magnetism and the armature is pulled back to the contact point by the spring. This completes the circuit and the action Thus, the hammer will is vibrate against the gong as long as the switch is depressed and a continuous ringing is produced. repeated. Electromagnets are being used more and more to protect circuits carrying large currents. Magnetic circuit breakers employ a switch which is opened by the magnet if current becomes too strong. A simple circuit is shown in Fig. 27:12. The student should trace the the Fig. 27:10 The Horseshoe Electromagnet. current flows in the winding ot the electromagnet, causing the armature to be attracted. As the armature moves away from the contact screw the circuit is Magnetic Circuit Breaker. Fig. 27:12 If current becomes too great, core is pulled upwards by magnetism of the coil, tripping the catch held by Si. pulls up the knife switch Spring S 2 breaking the circuit. path of the electrons and explain the action of the circuit breaker. The automobile generator cut-out relay is a magnetic switch which opens and closes the circuit between the genbattery erator (Sec. V:63) and the 307 Chap. 27 MAGNETISM AND ELECTRICITY An Electromagnet Used for Lifting Scrap Iron. steel Co. of Canada Ltd. (Sec. V:46). It serves to connect the generator to the battery when the generator is operating at charging speeds and to open the circuit when the generator stops or slows down to prevent back the through the generator. A simple circuit is shown in Fig. 27 : 13. discharging battery from The cut-out relay has two windings assembled on the same soft iron core. The shunt winding (dotted) is connected in parallel with the generator so that, when the generator starts to operate, the potential difference created causes a current to flow through the winding. This 308 produces a magnetic field strong enough to pull the armature toward the core, and the circuit is closed as the contact points meet. Current then flows from the and back generator through “ground”, passing through the series winding in such a direction as to add armature down. the magnetism holding battery the the to to When the generator slows down, the voltage produced by the generator becomes less than the voltage. Thus, a current begins to flow in the reverse direction through the series winding but continues to flow in the same battery MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V:52 the shunt winding. The direction in magnetic fields of the two windings are now in opposite directions and so tend to cancel each other. The armature is pulled upward by spring tension to open Fig. 27:14 A Simple Galvanoscope. parison of current strengths may be made. For weak currents the coils with many turns will be used and for strong currents the coils with few turns are satisfactory (Sec. V:50). This instrument has been largely replaced by more sensitive and more accurate current-measuring devices. Fig. 27:13 Automobile Generator Cut-Out Relay. V:52 THE MOTOR PRINCIPLE the contact points and the broken between the is battery and the circuit generator. V;51 THE GALVANOSCOPE The galvanoscope is a simple device used to determine the directions and comparative strengths of electric curIt consists of several coils having rents. varying numbers of turns of wire (Fig. 27:14). The current is made to pass over a compass-needle in one direction and under the needle in the opposite In this way the magnetic direction. effect of the current is magnified and the compass-needle will be deflected. By Left-Hand applying (Sec. V:48), the direction of current may be determined. By noting the amount of deflection of the needle a rough com- Rule the In experiment 34, chapter 31, a conductor AB was suspended between the poles of a strong permanent magnet as shown in Fig. 27:15. When a heavy current was sent through the conductor from a storage battery or similar source, the conductor was pushed aside. When the current was reversed, the conductor was thrust in the opposite direction. Thus we see that a conductor carrying current in a magnetic field is acted upon by forces which cause the conductor to move. The magnetic field between the poles of the permanent magnet may be represented as shown in Fig. 27: 16(a). The current in the conductor sets up a magnetic field as in Fig. 27: 16(b). A crosssection of the conductor is shown in Fig. 27: 16(c) . When the conductor is in the permanent magnetic field, Fig. as in 309 Chap. 27 MAGNETISM AND ELECTRICITY 27: 16(d), the lines of force from both the conductor and the permanent magnet are in the same direction on the force are like stretched elastic bands (Sec. V:4), they try to straighten themselves and consequently, push the conductor to the left. Similarly, if the electron flow in the conductor is reversed, the motion will be in the opposite direction. If now conductor AB (Fig. 27:15), is replaced by a single loop ABCD (Fig. 27:17), section AB will be moved toward the left by the combined forces of the permanent magnetic field and the electromagnetic field, while section CD will be thrust toward the right. Thus, the loop will set itself so that the plane of the coil will become perpendicular to the direction of the lines of force of the permanent magnet. This behaviour may be more simply explained if we consider of a Conductor Fig. 27:15 Carrying Electric Current in a Mag- Action netic Field. right and in opposite directions on the left. Thus the lines tend to be crowded on the right side, while on the left they cancel each other. Since lines of magnetic Fig. 27:17 Action of a Loop CarryElectric Current in a Magnetic ing Field. the loop to be a form of helix. The Helix Rule (Sec. V:49), indicates that the loop, when carrying current, will have an N-pole and an S-pole. Following the Law of Magnetism (Sec. V:2), the N-pole of the loop will be attracted to the S-pole of the permanent magnet while the S-pole of the loop will face the N-pole of the magnet. V:53 THE D'ARSONVAL GALVANOMETER Fig. 27:16 The Motor Principle. The D’Arsonval Galvanometer is a 310 MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V;54 sensitive instrument designed to detect the presence of small direct currents of It also serves to determine electricity. Fig. 27:18 A Simple Galvanometer. the direction and compare the strengths of such currents. The simplest form of this type of gal- vanometer is shown in Fig. 27:18. It consists of a permanent horseshoe magnet and a coil of wire which is free to swing between the poles of the magnet. A stationary soft iron core is mounted inside the coil to concentrate the lines of force. Springs attached to the coil serve to conduct the electrons into and out of the coil and also to bring the coil to rest in a position where the plane of the coil faces away from the poles of the magnet, as shown in the diagram. As current passes through the coil it will turn, so that the N-pole of the coil is toward the S-pole of the magnet, and is toward the the S-pole of the N-pole of the magnet (Sec. V:52). This movement is restricted by the springs. The amount of turning is indicated by the attached pointer moving across a calibrated scale. The stronger the electric current through the coil, the greater the magnetic field produced, and so the coil higher the reading on the scale. This moving-coil type of meter is the basis of most standard electrical measuring instruments in common use. V : 54 AMMETERS AND VOLTMETERS Ammeters and voltmeters are instruments designed to measure electric current and potential difference respectively. They are essentially galvanometers of the moving-coil type (Sec. V:53), modified by the addition of suitable resisTheir scales are calibrated to tances. read directly in amperes or volts. (a) Ammeters The wire winding on the moving coil of an ammeter must be very fine and light to allow the necessary freedom of movement. Such a conductor cannot carry a large current without undue (Sec. V:71), and consequent heating melting of the wire. The movable coil of such an instrument is rarely allowed to carry more than 0.05 amperes. If the ammeter is to be used to mea- Fig. 27:19 Ammeter Connected in Series. 311 ) Chap. 27 MAGNETISM AND ELECTRICITY sure higher currents than this the current must be divided so that no more than 0.05 amperes will flow through the moving coil, and the rest of the current will be carried around the coil through a shunt connected in parallel with the instrument (Fig. 27:19). Such a shunt must have a very low resistance and be accurate over a wide range of temperatures. To measure the current in a circuit the ammeter must be connected in series. As a result, all the current, or a known fraction of the total current, will pass through the instrument coil. In order to avoid affecting the total current in the circuit, the resistance of the ammeter must be very low. If the ammeter were accidentally connected across the circuit, a high current would flow through it and the instrument would be “burned out”. To avoid this a fuse (Sec. V:72) should be connected in series with the ammeter. Example A galvanometer has a resistance of 5 ohms and gives a full-scale deflection with a current of 0.05 amperes. What value of shunt resistance must be used to convert it to an ammeter reading up to 5 amperes? Current to be carried by ammeter = 5 amp. Current to be carried by galvanometer = 0.05 amp. Cu
rrent to be carried by shunt = 5 — 0.05 — 4.95 amp. Resistance of galvanometer = 5 ohms. Potential difference across galvanometer — 0.05 X 5 = 0.25 volts {V = IR) Potential difference across shunt = 0.25 volts (parallel connection) Resistance of shunt to be used = ^ 4.95 0.05 ohms {R — — I Fig. 27:20 Voltmeter Connected in Parallel. 312 MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V:55 (b) Voltmeters potential Since a voltmeter is used to measure difference between two the points in a circuit, it must be connected circuit between the in these points. Therefore, a voltmeter must have a very high resistance to avoid parallel with drawing a large current. As a result the total current in the circuit will not be affected significantly. A galvanometer may be converted to a voltmeter by the addition of a high resistance in series with the moving coil (Fig. 27:20). Example A galvanometer has a resistance of 5 ohms and gives a full-scale deflection with a current of 0.05 amperes. What value of series resistance must be used to convert it to a voltmeter reading up to 5 volts? Potential difference across voltmeter = 5 volts Current through voltmeter = 0.05 amp Resistance of voltmeter = 5 =100 ohms (R 0.05 Resistance of galvanometer = 5 ohms Series resistance to be added = 100— 5 = 95 ohms. V:55 THE ELECTRIC MOTOR The electric motor is studied in experiment 35, chapter 31, using a St. Louis motor (Fig. 27:21). In order to understand its consider an electromagnet mounted on an axis so that it is free to rotate between the poles of a magnet (Fig. 27:22). If a current is passed through the electromagnet so as to produce an S-pole at the top, as operation shown, the S-pole will be attracted by the N-pole of the stationary field. Similarly the N-pole of the electromagnet will be attracted by the S-pole of the These forces cause stationary magnet. the electromagnet, which serves as the motor armature, to rotate. The momentum of this armature causes it to rotate so that the S-pole of the armature passes slightly beyond the N-pole of the magnet. Commutator Brushes Commutator Plates Armature — Field Magnet Fig. 27:21 St. Louis Motor. Central Scientific Co. of Canada Ltd. 313 Chap. 27 MAGNETISM AND ELECTRICITY ments of the commutator also rotate. Connection to the commutator segments is made by a pair of brushes, one leading the electrons in ( — ) and the other out ( + ) . At exactly the correct instant, segment R makes contact with brush Bi causing the electrons to flow through the armature. One-half turn later, segment R makes contact with brush B 2 as segment S contacts brush Bi, etc. Thus the commutator serves to reverse the direction of electron flow with every 180° rotation of the armature. The magnetic field in which the armature turns is usually provided by an Fig. 27:24 Electric Motor. The Commutator Reverses the Direction of Electron Flow Every 180°. electromagnet, since such a magnet may be made more powerful than a permanIts windings are called the ent type. field coils to difTerentiate them from the armature coil. The field coils may be in series or in parallel with the armature coil, depending on the motor characteristics desired. A more advanced text should be consulted if greater detail is required. The motor described here operates on direct current only, and thus is seldom used. However, it serves to illustrate the principle electric operating on of operation of including motors, those all alternating current. S-poIe Electric Motor, The ArFig. 27:22 mature is Free to Rotate Between the Poles of the Magnet. If the current in the armature windings could be reversed at this instant, the polarity of the armature would change and repulsion and attraction would carry the armature through another 180° (Fig. Successive reversals of current 27:23). Electric Motor. The Poles Fig. 27:23 Reverse in this Position so Armature Continues to Rotate. and consequent changes of armature polarity cause the rotation to continue. Such a reversal of current can be caused of a commutator (Fig. by the 27:24). The armature wires are attached to two segments of good conducting material which are insulated from each other. As the armature rotates the seg- use 314 MAGNETIC EFFECTS OF ELECTRIC CURRENT Sec. V:56 V : 56 QUESTIONS A (b) State the Left-Hand Rule. 1. 2. (a) Draw diagrams showing the lines of magnetic force for (i) a straight wire (ii) a single turn of wire when carrying an electric current. Indicate clearly on each diagram the direction flow and of the electron of the magnetic field. (a) On a diagram of a solenoid indicate electron direction flow and the magnetic field rounding it. the sur- of the the (c) Indicate current in wires (1), (2) and (3) in which the direction of the lines of direction of force is shown. (d) Indicate the direction of the lines (5) and of force around wires (4), (6) carrying a current. (b) State the Helix Rule. (c) On the following diagrams mark the direction of the current and the polarity of the solenoids. 315 Chap. 27 MAGNETISM AND ELECTRICITY 3. (a) Why is soft iron, rather than steel, used as the core in most electromagnets? (b) How may the strength electromagnet be increased? of an 8. (d) How may a galvanometer be (i) an modified to convert it into ammeter (ii) a voltmeter? Make a labelled diagram of a simple D.C. motor and describe its opera- 4. (a) Make a fully labelled diagram of an electric bell connected in a tion. B circuit. (b) With the aid of this diagram describe the operation of an elec- tric bell. 5. 6 . (a) What is the purpose of an automobile generator cut-out relay? (b) Describe its operation. (a) What is the purpose of a galvanoscope? (b) Describe and explain the operation of a galvanoscope. 7. (a) State the motor principle. (b) Make a labelled diagram of a simple moving-coil galvanometer and explain how its operation employs the motor principle. (c) What precaution should be obobserved when a galvanometer is connected in a circuit? 1. A galvanometer has a resistance of 1 0 ohms and gives a full-scale deflection with a current of 0.04 ampere. What value of shunt resistance must be used to convert it to an ammeter reading up to (a) 5 amperes (b) 50 amperes (c) 500 amperes. 2. A galvanometer has a resistance of 2 ohms. It gives a full-scale deflection with a current of 0.01 ampere. What value of shunt resistance must be used to convert it to an ammeter reading to 10 amperes? 3. What value of series resistance must be added to convert the galvanometer of question 1 to a voltmeter reading up to (c) 500 volts? (a) 5 volts 4. What value of resistance must be connected in series with the galvanometer (b) 50 volts of question 2 to convert it to a voltmeter reading up to 100 volts? 316 CHAPTER 28 ELECTROMAGNETIC INDUCTION V:57 THE STORY OF FARADAY Michael Faraday was one the greatest experimental scientists the world has ever known. He was born in England in 1791 and, after a very ordinary education, became an apprentice to a book- of pressed the great chemist. Sir Humphrey Davy, so much that at the age of twentyone he became his assistant at the Royal Institution. Within twelve years he was made a director of the Institution. From the time that Oersted demonstrated the magnetic effect of an electric current Faraday dreamed of the converse effect, the production of electricity from magnetism. After many unsuccessful experiments he finally produced a momentary current in a coil of wire wound on an iron ring by starting and stopping a current in another coil wound on the same ring (Fig. 28:1). With this small beginning Faraday began a series of experiments on electro- University of Toronto Michael Faraday binder. He read every scientific book which passed through his hands and quickly became well educated. He im- Fig. 28:1 Electricity from Magnetism. to magnetism that eventually led his invention of the dynamo. The services of Faraday were in great demand by firms but he refused great industrial wealth in order to devote his talents to scientific discovery. Honours of all kinds were showered upon him but he remained the humble scientist and experimenter to the end, devoting his time and 317 Chap. 28 MAGNETISM AND ELECTRICITY energy to discovery. When all else is forgotten, he will be remembered as the father of the electric current that serves our homes, offices, factories and communities. Some of his other discoveries, such as the methods of liquefying gases, making optical glass, of benzine, and the like, are not remembered so well even though of great importance. the isolation Although he was never one to seek publicity his name has become immortal because of his contributions to humanity through physics and chemistry. V:58 CAUSE OF AN INDUCED CURRENT It is a simple matter to demonstrate, as Faraday did, that, just as electrons in motion can set up a magnetic field, a magnetic field in motion can cause a Three such experiflow of electrons. ments are described in chapter 31, experiment 36. is i.e., no coil, (Fig. current If a galvanometer is connected to the terminals of a solenoid and a bar magnet or an electromagnet is held above the solenoid 28:2), obtained. When the magnet is thrust into the centre of the the strength of the magnetic field is increased, the galvanometer needle is momentarily deflected from its centre position. The galvanometer needle returns to the zero position when the magnet remains stationary in the solenoid, i.e., while the strength of the magnetic field remains constant. When the magnet is withdrawn quickly from the solenoid, the strength of the magnetic field decreases, and the galvanometer needle is again deflected, but this time in the opposite direction. It is apparent, then, that a flow of electrons originated each time the magnet was moved. This is called induced current, and the effect caused by the changing magnetic field around a conductor is called electromagnetic induction. 318 For an induced current to flow in the circuit there must, of course, be an electro
motive in (Sec. V:26). This induced E.M.F. is caused by the changing magnetic field about the conductor. circuit force the Similarly, in the circuit used by Faraday (Fig. 28:1), when the switch in circuit containing the battery and the primary coil (C) was opened, the electron flow was stopped. Thus, the magnetic field about the primary coil deThis changing creased rapidly to zero. magnetic field the the secondary coil in the circuit and an E.M.F. was induced. When the switch was closed, current flowed in the primary circuit and as the magnetic field increased from zero to a maximum, an E.M.F. was induced again in the secondary circuit, causing a current to flow in the turns cut (5') of opposite direction. From these experiments we may conclude that an induced E.M.F. is caused by the changing magnetic field about the conductor. V:59 MAGNITUDE OF INDUCED E.M.F. In experiment 37, chapter 31, it will be observed that, if the magnet is slowly plunged into and withdrawn from the solenoid, there will be slight deflection of the galvanometer needle. If the action is repeated more and more rapidly, it will be seen that the amount of deflection is directly proportional to the speed of movement of the magnet. Thus, we may conclude that the strength of the induced current is proportional to the speed at which the magnet is moved or the E.M.F. induced in a circuit is proportional to the rate of change of the magnetic field cutting the conductor. If a magnet is thrust first into a coil of few turns and then, at the same speed, into a coil of many turns, and the deflection of a galvanometer in the ELECTROMAGNETIC INDUCTION Sec. V: 60 circuit is noted, it will be found that a stronger induced current flows in the coil having the larger number of turns. Therefore, it is evident that the E.M.F. induced in a circuit is proportional to the number of turns of the conductor cut by the varying magnetic field. N-pole at the end of the solenoid where the magnet enters. This N-pole repels the N-pole of the magnet and opposes the motion of the magnet. As the N-pole of When a powerful electromagnet with an iron core is substituted for the bar magnet in the experiment described above it will be observed that, with the rate of movement and the number of turns in the coil remaining constant, the deflection of the galvanometer needle is greatly increased. It is apparent, then, that the E.M.F. induced in a circuit is the proportional the changing magnetic field. strength of to V:60 DIRECTION OF AN INDUCED CURRENT The direction of an induced current is studied in experiment 38, chapter 31. Here, the direction of the current is noted when (a) the N-pole of a magnet is thrust into the coil; (b) the N-pole is Induced Current as the Bar Fig. 28:3 Magnet is Withdrawn from the Solenoid. the magnet is withdrawn from the coil (Fig. 28:3) the induced current is in such a direction as to produce an S-pole at the same end of the solenoid. The attraction of this S-pole of the solenoid for the N-pole of the magnet tends to oppose the withdrawal of the magnet. Similar results may be found when the electromagnet is used. These eflfects are summarized in Lenz’s Law which states: The direction of an induced current is such that the magnetic field which it produces opposes the motion or change that induces the current. Induced Current as the Bar Fig. 28:2 Magnet is Plunged into the Solenoid. withdrawn; (c) the S-pole is inserted; (d) the S-pole is withdrawn. As the Npole enters the solenoid (Fig. 28:2) a current is induced which produces an Because of their importance, primary and secondary currents require special It was noted (Sec. V:58), treatment. that secondary currents are in opposite directions when the primary is opened compared to when it is and demonstrable secondary currents are in the same direction when the primary circuit is opened (at the “break”) and in the opposite direction when it the “make”). The idea of opposing mag- primary is closed. closed that the (at It is 319 Chap. 28 MAGNETISM AND ELECTRICITY As it nears the position at right angles to the first, much cutting of the lines of force netic fields is carried out here in that the induced current, while flowing in the same direction as the primary, attempts to maintain the magnetic field that is Similarly, at the completion collapsing. of the primary circuit, the induced current is attempting to oppose a magnetic field which is being built up. It is further to be noted that the induced E.M.F. is greater at the “break” because the induced current tends to strengthen the field of force of the primary. V:61 THE EARTH INDUCTOR The earth inductor is a coil with a diameter of about 18 in., made of several hundred turns of fine insulated copper wire. When the two end leads of the coil are connected to a galvanometer and the rotated through 360°, the galvanometer needle will be seen to swing first to one side and then to the other (Chap. 31, Exp. 39A). This action continues as long as the inductor is rotated. inductor rapidly is In Fig. 28:4, as the coil rotates clockwise, A downward and B upward, little cutting of the lines of force occurs and a small E.M.F. will be developed at first. occurs, and the E.M.F. produced will be larger. Throughout this 90° of rotation the induced E.M.F. has been built up from zero. It will be obvious that the E.M.F. will gradually decrease to zero in the next 90°. Throughout this rotation of 180°, the E.M.F. has gone from zero to a maximum and back to zero again (Fig. 28:5). The direction of the current has been constant. All this time DIRECTION OF CURRENT 320 ELECTROMAGNETIC INDUCTION Sec. V: 62 Fig. 28:6 Simple Alternating Current (A.C.) Generator. the magnetic field of the induced current has been opposing the motion causing it (Lenz’s Law). Rotation through the next 180° in which A will move up and B down will result in the same cycle of events as before except that the direction of the current will be reversed. The energy required to overcome the opposition of the fields of force is transformed into electrical energy. While the earth inductor itself is of no practical value it does permit a preliminary study of the generation of electricity by induction. This simple knowledge has made possible the development of all the many forms of electric generators which supply power for modern living. V:62 THE ALTERNATING-CURRENT GENERATOR As indicated in section V:61, mechanical energy may be used to produce electrical energy. A device used for this purpose is called a generator or dynamo. A simple generator (Fig. 28:6) consists of a coil of wire called an armature which is made to rotate between the poles of a magnet (Chap. 31, Exp. 39B). To facilitate connections to the rotating armature, slip rings {Ai, A 2 ) are connected to each end of the coil and rotate with it. Contact is made by metal or carbon brushes {Bi, B2 ) which connect the coil with the external circuit containing the lamp (L). As the coil rotates in the direction shown in the diagram, it cuts the magnetic field between the pole-pieces {S, N) of the magnet and a current is induced In position (1) the in the conductor. sides ab and cd of the conductor are momentarily moving parallel to the lines of magnetic force so there is no cutting and no induced E.M.F. Therefore no electrons will flow. In position (2) the coil is cutting an increasing number of lines of force so the induced E.M.F. is Electrons flow in the direcincreasing. tion abed so as to produce an N-pole and an S-pole in the coil which will oppose its motion (Lenz’s Law). Electrons are now leaving the slip ring B, passing through the lamp, and returning to A. In position (3) the conductor is cutting the maximum number of lines of force and the induced E.M.F. and induced electron flow are at a maximum. As the 321 Chap. 28 MAGNETISM AND ELECTRICITY (4) coil reaches position the electron flow is decreasing and, in position (5) as the motion is parallel to the lines of force once more, it ceases entirely. In the electron flow builds up again (6) but this time in the direction dcba so that electrons leave slip ring A, pass through the lamp, and return through B. The current again reaches a maximum (8), and falls to in zero in position (9). Thus, during one complete rotation of the coil, the induced current starts at zero, increases to a maximum, falls to zero, increases to a maximum in the opposite direction, and again falls These (7), decreases in zero. to changes are summarized in Fig. 28:7. A current with such characteristics is said to be an alternating current (A.C.) and one cycle has been described above. Alternating current is provided by all hydro-electric installations. Those who have lived in, or visited, 25-cycle power areas will have noticed the rapid flickering of the lights. This is caused by the reversal of current flow occurring 25 times per second. When the number of reversals is increased to 60 times per second as in 60-cycle power, this flickering is not noticeable. Very few areas 25-cycle of power. the world use this still 322 ELECTROMAGNETIC INDUCTION Sec. V: 64 V:63 THE DIRECT-CURRENT GENERATOR When electrons flow continuously in one direction it is said to be a direct current. Such current is required for special installations such as electroplatIn order to obtain direct current ing. from an A.C. generator, a commutator must be used in place of the slip rings (Chap. 31, Exp. 39B). For a single-coil generator as described in section V:62, the commutator consists of a collectingring made of two segments or com- Brush trons are about to reverse direction. Thus the commutator bars change brushes just as the electron flow reverses so that the current flowing through the external circuit is always in the same direction (Fig. 28:9). The current from a single turn of wire is clearly a pulsating current. To eliminate this the armature of a commercial generator consists of many such coils each with its commutator arranged around the shaft
(Fig. 28:10). Such an arrangement will give a more continuous flow of current, since some of the coils will always be cutting the magnetic lines of force. Such generators are used in autoremote mobiles, places, direct-current generating systems in schools, hospitals, etc., and for electroplating and electro-purification of metals. electrical systems in Segments Fig. 28:10 Armature of a Commer- cial D.C. Generator mutator bars insulated from each other. Each bar is connected to a terminal of the coil (Fig. 28:8). The brushes are so placed that they rest on the insulating material between the bars at the instant when the elec- V : 64 TRANSFORMERS Transformers are used to change the voltage in an A.C. circuit as required. A transformer consists of two separate coils of insulated wire, the primary and secondary coils, wound on the same soft iron core (Fig. 28:11). Alternating currents from a generator flow through the windings of the input or primary coil. As the current changes in intensity and direction, the magnetic field surrounding the coil changes as well. Thus lines 323 — Chap. 28 MAGNETISM AND ELECTRICITY of magnetic force of changing strength and direction are made to cut the turns of the output or secondary coil. An alternating E.M.F. is induced in the secondary coil. Since the magnitude of Fig. 28:11 Step-up Transformer. the induced E.M.F. is proportional to the number of turns of the conductor cut by the changing magnetic field, the strength of the induced voltage may be varied by using a different number of the primary and secondary turns windings. For any transformer in Output Voltage _ No. of turns on secondary coil Input Voltage No. of turns on primary coil A step-up transformer has more turns on the secondary coil than on the primary coil and so serves to increase or “step up” the voltage. A step-down transformer has more turns on the primary coil than on the secondary and so serves to decrease or “step down” the voltage. In section V : 75, it will be established that the power of an electric current = the number of volts X the number of Obviously the power of the amperes. secondary current must equal the power of the primary current (except for a very small decrease owing to conversion to heat). Thus, if a step-up transformer causes a voltage rise, there must be a corresponding fall in the current. Students will be familiar with the small step-down transformers used to operate electric trains, electric bells and electric janitors around the home. Some other uses of the transformer will be described in the sections that follow. Example A transformer is to be used to provide 6 volts to operate an electric If the primary coil has 2000 turns, how door-bell on a 120 volt circuit. many turns should be on the secondary coil? Output voltage ~ 6 volts Input voltage — 120 volts Number of turns on primary = 2000 Number of turns on secondary = x Output voltage Number of turns on secondary coil Input voltage Number of turns on primary coil 6 X 120 2000 6 .•.x=r—X 2000 =100 .’. there should be 100 turns on the secondary coil. 324 ELECTROMAGNETIC INDUCTION Sec. V:65 V:65 THE TELEPHONE The first telephone was invented by Alexander Graham Bell, a Scottish emigrant to the United States, 1875, and first used between Brantford and Bell’s main interest was Paris, Ontario. he had deaf-mutes, but in many other interests and the telephone was one of the many products of his fertile mind. From his simple instru- teaching in Fig. 28:12 Telephone Transmitter. vibrate. When the vibrations are transmitted to the carbon granules, they are compressed to a vaiying degree. This, in turn, causes their electrical resistance to vary and fluctuations in the current strength are produced at the same freThis fluctuatquencies as the sounds. ing current is transmitted to the re- ceiver. into sound again. As we know so well, the purpose of the receiver is to convert these fluctuatThe ing currents receiver (Fig. 28:13) consists of a permanent magnet (M) having an electromagnet at the poles. As the fluctuating current caused by the transmitter passes through the coils {CC) of the electro- Diaphragm (D) Electromagnet (C) Star Newspaper Service Alexander Graham Bell. ment has been developed the worldthat we wide communication system know to-day. The essential parts are the transmitter and the receiver. The transmitter (Fig. 28:12) consists of a capsule containing carbon granules (C) connected the to diaphragm (D) on one side and to a carbon block (B) on the other. A current from a battery passes through the granules. metal called sheet thin a It responds to sound waves as follows: the condensations and rarefactions falling on the diaphragm cause it to Fig. 28:13 Telephone Receiver. 325 Chap. 28 MAGNETISM AND ELECTRICITY Transmitter Ground Fig. 28:14 Simple Telephone Circuit. magnet, the magnetic field is caused to change at the same frequency as the Thus the steel original sound waves. diaphragm {D) is caused to vibrate at this frequency and sound waves almost identical to those at the transmitter are reproduced at the receiver. If the transmitter and receiver circuits are directly linked, only weak signals This is because of the will be heard. very small changes in resistance in the carbon granules of the transmitter as compared to the total resistance of the circuit. The slight variations in current do not cause enough variation in the receiver electromagnet to give satisfactory reception. To overcome this, the transmitter is connected to the primary winding of a step-up transformer. The secondary coil is connected in the line circuit through which magnified effects are carried to both receivers as shown in Fig. 28:14. Refinements such as the bell signal, automatic dialing, relays, ultra high-frequency beam wireless, sheathed cables, coaxial cables, and others are employed in the modern system. However, the above still remains the basis for all telephone communication. V:66 SELF-INDUCTANCE It has been observed that when an is broken, as when a circuit electric in a house-lighting switch turned off, a spark or “electric arc” freThis is caused by selfquently occurs. inductance which is explained below. circuit is r RINGER CO IL j DIAL MECHANISM Bell Telephone Co. Canada The Dial Telephone. 326 When a coil of wire of many turns, wound about an iron core, is connected to a battery, the current that begins to flow causes a magnetic field about the wire. As an increasing number of lines of force cut the conductor, an E.M.F. ELECTROMAGNETIC INDUCTION Sec. V:67 will be induced in the coil itself causing a current to flow in a direction opposite to the flow of current in the coil (Lenz’s Law). Thus the current will build uo more slowly than might be expected in such a circuit (Chap. 31, Exp. 40). When the connection to the battery is broken, the decreasing current causes a decreasing magnetic field about the wires. An E.M.F. causing a current that flows in the same direction as the main current will result. Since this is of relatively high voltage, a definite spark (arc) will be observed. In a circuit, an induced current that opposes any change in the flow of current through it caused by self-inductance is called a self-induced current. It is so similar to the inertia of matter that it has been termed electrois an important magnetic inertia. consideration in all alternating-current equipment design. It V : 67 THE INDUCTION COIL The induction coil is very like the transformer in construction, as both consist of primary and secondary coils of different numbers of turns of wire wound about a common soft iron core. Whereas the transformer can utilize the changing magnetic field caused by alternating current to step up the voltage, the same operation is impossible with a direct current for lack of changing magnetic field. However, it is possible to obtain high voltage from a low voltage direct current by the use of an induction coil. is accomplished by continually This interrupting the primary current, with the aid of a circuit breaker similar to (Sec. V:50). that in the electric bell This results in a changing magnetic field around the primary which induces a high voltage current in the secondary coil. This device was first invented by Page in the United States in 1836 and later reinvented by Ruhmkorff in Europe in 1851. The of coil (Fig. thick induction 28:15) consists of a primary coil T of a few insulated copper wire turns wound about a soft iron core B which is not solid but built up from a bundle of soft iron wires to prevent self-induction effects. Around the primary coil is a secondary coil C consisting of many turns of very fine copper wire wound in sections with parafifined-paper insulation between the layers of each section. The primary circuit is completed through Secondary Terminals a make-and-break device D. This consists of a stiff piece of spring steel with a soft iron armature on one side and a platinum (or tungsten) contact on the other. It is normally in contact with the point on the platinum (or tungsten) As the end of the adjusting screw. primary current flows, the core becomes magnetized and attracts the armature, thereby breaking the circuit. The magnetism of the core now collapses and the armature thus released is drawn back by the spring to make contact with the The primary circuit again screw. complete for the action to be repeated. As the magnetic field of the primary circuit builds up and collapses at each make and break, an alternating induced is 327 Chap. 28 MAGNETISM AND ELECTRICITY E.M.F. is set up in the secondary coil. This E.M.F. may be many tens of thousands of volts and can cause a spark several inches long to pass between the terminals of the secondary coil. Oneof each alternation has a much half greater E.M.F. than the other (Sec. V:60). As a result this current usually is viewed as a pulsating direct current. In actual practice the weaker of the two is practically eliminated by the use of a condenser. Induction
coils can be used for operalthough modern ating X-ray tubes, tubes are mainly transformer operated. They are also employed in some forms of laboratory research where moderately high voltages are required, and in the ignition systems of automobiles. Note: Because of the very large selfinductance of the primary winding, the upbuilding and decay of the primary current are both delayed, and the slower changes in the magnetic field mean correspondingly reduced voltages the secondary coil. The self-induced forward E.M.F. in the primary circuit at the break causes arcing across the contact points which become considerably worn as a result. To prevent this a condenser E, formed from alternate sheets of tinfoil and paraffined-paper, is placed across the make and break, and the self-induced current at break surges into the con- in denser instead of arcing across the points, thereby producing a very rapid break. The condenser plates, however, are connected through the primary winding, and no sooner is the condenser charged than it discharges through the primary coil, producing a current in it in a direction opposite to the now upbuilding current from the battery. As was noted before (Sec. V:60), the induced current is weaker at the make than at the break. The condenser weakens it still further. Accordingly, the coil produces practically 328 a pulsating, E.M.F. than that of the primary. direct current of higher V:68 THE AUTOMOBILE IGNITION SYSTEM the into fitted plugs The flammable mixture of gasoline vapour and air used in internal combustion engines is ignited by a series of sparks passing across the electrodes of spark cylinder heads. These sparks, which are correctly timed for each of the cylinders, are produced by a type of induction coil, the primary of which is connected to the battery through a circuit breaker, while the secondary is connected through a “distributor” to the electrodes of the spark plugs. The scheme of the ignition system, is shown in Fig. 28:16. Spark Plugs Fig. 28:16 Simple Automobile Ignition System. The circuit breaker is operated by a cam revolving at half speed and driven from the engine crankshaft. When the platinum points are separated, the resulting change of magnetic field of force produces an induced voltage in the secondary coil. A condenser is connected to prevent across contact points the time. 3. 4. ELECTROMAGNETIC INDUCTION Sec. V: 69 arcing (and consequent damage to the points) and to ensure a rapid break as In described in the previous section. order to withstand the high voltages (of the order of 10,000 volts), the leads from the secondary coil to the distributor and spark plugs are covered with thick in- sulation. V : 69 QUESTIONS A 1. Read a more complete story of Faraday in a good encyclopedia and write a brief account of his accomplish- 5. ments. 2. Describe briefly Faraday's method of producing induced current for the first (a) What is the cause of an induced current? (b) What factors affect the magnitude of the E.M.F. induced in a circuit? (c) State Lenz's Law. (d) On the following series of diagrams label of the current and indicate the polarity of the upper end of the solenoid. direction the 6. the inductor with regard to direction and magnitude as the coil is revolved through 360° in the earth's field. (a) Make a diagram of a simple generator with a single coil of wire rotating between the permanent regard to current produced with magnitude and direction as the coil rotates through 360°. (b) How is the current taken from the rotating armature and fed into the poles of a magnet. Describe the external circuit? (c) Describe how a generator differs direct-current from an alter- nating-current generator. (a) Describe, using simple diagrams, an (ii) a current produced alternating-current generator by the (i) direct-current generator. (b) What is meant by (i) a single (ii) 60-cycle current (iii) pulcycle sating direct-current? 7. Explain: (i) the structure (ii) the action of 10. 8. (a) a step-up transformer (b) a step-down transformer. (a) Describe (i) a simple telephone (ii) a simple telephone transmitter receiver with regard to purpose, (a) Describe the structure of an earth inductor. (b) Describe the current produced in structure, action. (b) Describe how transmitter and receiver circuits are linked in order to keep up the strength of signals. 9. What is the effect of self-inductance when a circuit containing a coil of wire is (a) completed (b) broken? Describe a simple induction coil as to (a) purpose (b) structure (c) action (d) uses. 329 Chap. 28 MAGNETISM AND ELECTRICITY B 1. A transformer is required to provide 6 volts to operate an electric door-bell on a 120 volt circuit. If the primary coil has 1 600 turns, how many turns should be on the secondary? 2 . A step-down transformer has 2000 turns on the primary and 200 turns on the secondary. If the primary voltage Is 22,000 volts, what is the secondary voltage? 3. An electric door-bell transformer has 960 turns on the primary and 80 turns on the secondary. When it is connected to the 110 volt circuit, what voltage is provided for the bell? 4 . A generator produces at 2200 volts. This current alternating is supplied through a step-up transformer with a voltage ratio of 1:100 to the transmission lines. This is fed to a step-down transformer of voltage ratio 1000:1. (a) What is the final voltage? (b) Draw a simple diagram to show these stages. 5 . A generator with an E.M.F. of 550 volts supplies energy to a 2200-volt line. has The transformer primary winding. How many turns must there be on its secondary? 1 00 turns on its 6 . A transformer, used to operate a toy electric train on a 1 20-volt line, has taps on the secondary to give 1 2, 8, 6 and 4 volts. If the primary winding has 960 turns, how many turns must there be on the secondary at each tap? 330 CHAPTER 29 ELECTRICAL ENERGY V:70 THE PRODUCTION OF ELECTRICITY towards industrial Not many years ago Canada was known throughout the world as an agricultural country. A survey of our exports at the beginning of the twentieth century shows that farm products formed the basis of our economy. In the years preceding the Second World War a gradual swing greater exFrom those pansion could be noted. years to the present time the change has been almost phenomenal. Canada has become an industrial nation. Our exports now, while still including a large part of the world wheat supply and other agricultural commodities, have been expanded to take in countless manufactured farm articles machinery to paper, clothing, electrical devices, chemicals and products of mining and smelting. ranging heavy from Closely connected with these changes has been the marvellous development of Canada’s hydro-electric power resources. The energy of falling water has been turn the huge turbines harnessed which rotate the armatures of large genSteam-generating erators plants have become common in which (Fig. 29:1). to the energy from coal is utilized. In recent years heat from an atomic pile is These being used in the same way. latter methods provide the alternating current that is so vital to our modern way of life. Millions of horsepower have been made available to turn the wheels of industry across the country and to provide the luxuries of cheap electricity in our homes. Men of foresight have planned for new and greater feats in engineering as they divert the waterways, tunnel through mountains and improve the generators to provide Canada with power for the future. V:71 THE TRANSMISSION OF ELECTRICITY Every high-school student knows that electrical energy is readily transformed into heat energy. A current of electricity consists of a shunting of free electrons in the circuit. The energy of these electrons is converted into heat as they encounter opposition to their motion in the conIt is apparent that the greater ductor. the number of electrons moving and the greater the resistance of the conductor, the greater will be the amount of heat produced. In 1841, James Joule proved experimentally that the heat produced was proportional to the square of the current, to the resistance, and to the time the current flowed {HaPRt). It is evident then that the amount of current in a conductor is the most important single factor in heat losses dur- 331 Aerial View of Hydro-Electric Plant, Niagara Falls. Ontario Hydro 332 ELECTRICAL ENERGY Sec. V: 72 Ing the transmission of electricity. The power of an electric current is equal to the voltage times the current strength, i.e., P— VI (Sec. V;75). This is constant for a given current. Thus, if the voltage is made higher, the current becomes smaller and heat losses will be reduced. For this reason, alternating current produced at a potential difference of 2,200 to 12,000 volts is stepped up to between 50,000 and 220,000 volts for transmission. the the voltage is stepped down by transformers so that the pole voltage is usually about 2,200 volts. The familiar pole transformers step this down further to 220 and 110 volts for household circuits. This current is used to operate electric ranges, heaters, lights and various other appliances. sub-station At V:72 FUSES ^ In designing any electrical circuit care must be taken to ensure that no circuit will carry an overload of current for fear of overheating the conductor. Re- Ontario Hydro House Fuse Box. call that the heat developed is proporcurrent the square of tional the to the times resistance of the strength, conductor. The wiring, then, must be of such a type as to carry the intended current safely. To prevent fires or failure in the circuit through melting of the conductor, fuses are put in as safeIn homes, these are usually guards. placed in special boxes near where the current enters. Each fuse of a length of metal alloy with a low melting-point enclosed in a case to confine the heat (Fig. 29:2). In the plug-type fuse one end is transparent, to permit inspection. In the cartridge fuse no opportunity for inspection is prov
ided and larger ones sometimes contain a powder to extin- consists 333 Fig. 29:2 Fuses (a) Plug Type (b) Cartridge Type. Chap. 29 MAGNETISM AND ELECTRICITY frustrating research ending in success. In the years preceding 1879 Edison had tried almost every kind of fibre in his to find a filament that would efforts carry the current and emit light. The guish the arc when the current becomes interrupted by the melting of the metal strip. Each fuse is marked with a number denoting the maximum current in amperes that it is designed to carry. One should be certain that the rating of the fuse is not greater than that of the ciris used. Any overload cuit in which it should melt the fuse wire instead of harming the rest of the circuit. V:73 THE INCANDESCENT LAMP In 1879 Thomas Edison exhibited the first incandescent lamp (Fig. 29:3) at his laboratory in New Jersey. The dethis lamp is a story of velopment of Fig. 29:4 A Carbon Filament Bulb. Canadian General Electric filament had to have high electrical resistance and a high melting-point; it could not be too brittle even at high temperatures; it had to glow brightly when the current passed through it; it had to be enclosed in an finally, evacuated glass chamber so that it would not burn or unite with oxygen from the Eventually, Edison discovered that air. a carbon filament made by heating bamboo fibres met these conditions and the carbon filament lamp was born (Fig. 29:4). Other problems had to be solved before the lamps could be used extensively. Many improvements were found, so that now the modern electric-light Thomas A. Edison of Canada Ltd. Fig. 29:3 The Incandescent Lamp. 334 ELECTRICAL ENERGY Sec. V:74 \ \ \ ' resistance wire. This is an alloy of nickel (80%) and copper (20%) which can be heated to a high temperature without melting, and oxidizing very slowly even when red hot. In addition, it has a resistance of about sixty times that of copper and this reduces the costs. / f { \ /1 1 I- fp [1 ^ lp ^ 1 Canadian General Electric Fig. 29:5 A Modern Tungsten Filament Bulb. bulb (Fig. 29:5) bears little resemblance to the original. Modern filaments are made of tungsten, giving greater service and whiter light. Instead of evacuated glass bulbs, we now have bulbs filled with an inactive mixture of nitrogen and argon which allows the filament to be heated to higher temperatures. Platinum lead-in wires passing through the glass have been replaced by inexpensive nickeliron alloys. Costs have gone down while efficiency has been raised to heights not dreamed of a few decades ago. V:74 OTHER APPLICATIONS OF THE HEATING EFFECT (a) Domestic Heating Appliances Nowadays, in many homes, space heaters, hot-water heaters, ranges, etc., use electricity. The appliances designed for these various purposes contain heating-elements usually made of “nichrome” the heaters, In electric heating-elements are generally supported on fireclay forms, and in some of the smaller heaters concave polished metal reflectors are used to concentrate the radiant heat into a beam. The electric range has nichrome heating-elements embedded in some suitable insulating material for protection. The elements often consist of two parts. When the switch is set at “high” these are in parallel; when at WVVWWVW •-VvWVVVWWS^ fo) (b) (c) Fig. 29:6 Regulation of Heat in an Electric Stove Element. “medium” one element is cut out; when at “low” both elements are in series (Fig. 29:6). The walls and door of the oven are filled with heat-insulating material such as fibre glass to reduce the heat loss, the oven temperature being regulated by some form of thermostat. (Fig. 29:7) With electric irons the heating-elements are usually wound on insulating-sheets of mica situated near the “sole” of the iron. These heatingelements are covered by a layer of asbestos and above this is a heavy iron plate firmly screwed to Electric kettles are also provided with nichrome heating-elements wound on mica and these, sandwiched between mica strips, place below the are In other kettles base the heating-element is of the immerof sion type and consists of a coil the base. firmly kettle. held the of in 335 Chap. 29 MAGNETISM AND ELECTRICITY slowly apart. This phenomenon was discovered by Davy in 1806, using two thousand simple cells and charcoal rods. These rods were later replaced by gas carbon and the intensity of the light produced by this means led to the extensive use of electric “arc lamps” in street lighting towards the end of the nineteenth century. Arc lights are used as spotlights in theatres. nichrome wire contained in a metal tube from which it is insulated. (h) The Electric Arc If two carbon rods are joined across an electric supply of not less than 45 volts, and the ends of the carbons are touched together, an intensely brilliant “arc” of white light appears between the ends of the rods on drawing them The heavy current passing on striking the arc produces intense heat between the carbon tips which are thus to incandescence. The arc-like raised flame consists of white-hot particles of carbon torn from one of the rods which thus steadily wears away. If used with a D.C. supply the positive carbon is consumed twice as quickly as the negative rod, and it is found that a “crater” develops at the end of it. Almost all the light originates at this “positive crater,” about of which is temperature the 3500°C. Electric Arc Welding. Air Reduction C.anada Ltd. 336 ELECTRICAL ENERGY Sec. V:75 ) (c) The Electric Furnace Some types of electric furnaces make use of the high temperatures produced by the arc discharge. These arc furnaces the manufacture of are employed in carborundum and calcium carbide—car- bon rods being immersed in the constituent chemicals through which an arcdischarge is passed. Another use of the arc furnace is in the “fixation” of atmospheric nitrogen for the preparation of fertilizers. (d) Electric Welding The high temperature developed by the electric arc is also applied in electric ships’ plates, Boiler welding. plates, etc., can be welded by connecting the plates to the negative of a D.C. supply, applying the positive side to the weldingrod where the weld is to be made. V:75 BUYING ELECTRICAL ENERGY Since electricity is able to do work, it is a form of energy. The faster work is done, the faster energy is utilized. The rate of doing work, i.e., the rate of utilizing energy, is the power. Electrical power is measured in watts. One watt is the power provided when a current of one ampere flows with a potential difference of one volt. Thus: Power (watts) = P.D. (volts) X Current (amps. or P = VI watts P ={IR)ir.'V = IR,Ohm'sLaw) or P = PR watts If the power of one watt is provided for one hour, the consumer must pay for 1 watt-hour of energy. Similarly, if one kilowatt (1000 watts) is used for one hour, the consumer must pay for 1 kilowatt-hour (k.w.h.) of energy. Thus electrical energy (k.w.h.) == power (k.w.) X time (hours). Note: Since 746 watts = 1 horsepower (h.p.) it follows that P or P = = VI 746 PR 746 h.p. h.p. Example Find the cost of operating an electric toaster for two hours if it draws 8 amperes on a 110 volt circuit. The electrical energy costs 4 cents per kilowatt-hour. I =z 8 amp. no volts / P = VI P=\10 X 8 = 880 watts 880= —r—- = .880 k.w. 1000 Electric energy used = .88 X 2 = 1.76 k.w.h, 1 k.w.h. costs 4 cents 1.76 k.w.h. cost 4 X 1.76 = 7.04 cents. The cost is 7 cents. 337 0 5 (i) Chap. 29 V : 76 MAGNETISM AND ELECTRICITY QUESTIONS A 1. (a) Describe what happens to alternating current from the time it produced at the generator until it is used to operate the electric toaster in your home. is Explain why the (b) changed before electricity is transmitted across the country. voltage is 2. (a) What is the function of a fuse in an electric circuit? (b) Describe the structure and action of a simple fuse. 3. (a) Make a labelled diagram of an electric-light bulb. (b) Explain the function of each part labelled in (a). 4. List three appliances found in each of the following, in which the heating effect of your employed, current electric (a) is house (b) your school (c) industry. 5. Define: power, watt, kilowatt, kilo- watt-hour. B 1. At the power-house, electricity at a potential of 1 2,000 volts is generated. It is prepared for long-distance transmission by being applied to a step-up transformer in which the number of turns of wire in the primary and secondary are in the ratio of 1 to 2000. (a) What will be the potential difference in the secondary? (b) How will the current strength be altered? (c) How will this affect the loss of energy through heating of the con- ductor during transmission? 2. (a) A pole transformer steps down the line potential from 5500 volts to 1 1 0 volts. What is the ratio of the number of turns of wire in the primary and secondary of the transformer? 338 (b) What change will occur in the current strength (ii) the heating effect? 3. Determine the power of an electric kettle which amperes when plugged into a 110 volt requires a current of 1 electric outlet. 4. What current is drawn by a 550 watt toaster when it is plugged into the 1 1 volt line? 5. What voltage is applied to a 2970 watt heating-element that draws a current of 1 3.5 amperes? 6. What is the power (a) in horsepower (b) in watts (c) in kilowatts, of a motor which draws a current of 3.38 amperes on the 1 1 0 volt line? 7. What is the maximum power that can be used in a circuit containing a 15 ampere fuse in the 1 1 0 volt line? 8. If an electric toaster of 550 watts and an electric kettle of 1000 watts were plugged into the same wall outlet in your home, would the 15 ampere fuse in the circuit melt? Explain your answer. 9. Find the cost of operating an electric toaster for 3 hours if it draws 5 amperes on a 110 volt circuit. The electricity costs 3.5 cents per kilowatt-hour. 10. What is the cost to a storekeeper of leaving a 40 watt light bulb burning near his safe for 36 hours if electricity costs 3 cents per kilowatt-hour? 11
. An electric stove element draws 5 amperes on the 220 volt circuit. is turned on for 4 hours, what is the cost of operating the stove at 2.2 cents per kilo- If it watt-hour? 12. A Vi h.p. electric motor In an oil furnace comes on for a period of 5 minutes 48 times each day on the average. What is the cost of the electricity for 30 days If it costs 2.5 cents per kilowatt-hour? 13 A boy on returning home from school. ELECTRICAL ENERGY Sec. V:76 at 4:30 p.m. turns on lights in the kitchen, basement, hall and bedroom. The kitchen has one 100 watt bulb, the basement two 60 watt bulbs, the hall one 40 watt bulb and the bedroom one 40 watt bulb and one 60 watt bulb. If these lights are left burning 1 0:00 p.m. what is the cost when electricity costs 3 cents per kilowatt-hour? until 339 CHAPTER 30 ELECTRONICS points of which are to the secondary of an induction coil, the discharge close together. The induction coil is turned on while the tube is being evacuated (Chap. 31, Exp, 42). At first a discharge discharge occurs between the V : 77 INTRODUCTION (a) When a lighted match, or the flame of a gas burner, is brought near the knob of a positively charged electroscope, the leaves will be seen to fall (Chap. 31, Exp. 41). The flame liberates electrons from the atoms of the gases causing positively charged ions. The electrons so set free are attracted to the electroscope and neutralize its charge. Had the charge on the instrument been negative, charge would have been lost positively charged gaseous ions. become them the to to its (b) An electrical discharge tube with an open side-arm (Fig. 30:1) is attached Fig. 30:1 Apparatus to Show Dis- charge of Electricity at Low Pressures. 340 points but, as the tube is partially evacuthe current ceases between the ated, points and begins between the tertube even though minals the they are farther apart. This discharge, though varying in appearance with the even degree of though the tube is evacuated as com- evacuation, continues inside pletely as possible. The current flows through the gas at reduced pressure because the gas becomes ionized. The electrons expelled from the atoms of gas are attracted to the anode where they re-enter the metallic part of the circuit. The positively charged gaseous ions are attracted to the cathode where they remove electrons from it. Where the electricity is passing through a vacuum, electrons are expelled from the cathode and proceed in straight lines directly to the anode. The behaviour of electricity as it passes through a gas or a vacuum is known as electronics. The vacuum-tube and the low pressure gas-filled tube have opened the way for a seemingly endless number of new The most applications familiar application is the radio, but this is only one of the many being used in our modern way of living. Electronics electricity. of ELECTRONICS Sec. V:78 ticularly near the anode end, shine with a bright green fluorescent light. The phenomena occurring at this stage of the discharge proved to be most important. If an object is placed some distance in front of the cathode a sharp shadow of the object is cast on the end of the tube Experiments suggest near the anode. is used to guide our ships and aircraft, to perforin calculations in minutes that would take a mathematician days, to time delicate processes, to operate radio, television, sound movies, and to perform amazing feats of mechanical control. Its future is likely to be as remarkable as its past—a challenging field for keen young scientists. V:78 CATHODE RAYS The main characteristics of the discharge of electricity through gases may be shown by the same method as was used in section V: 77(b). At normal pressures the insulating properties of the is of the the gas are too great for a spark to occur, but as the pressure is reduced a current passed and irregular streamers of light traverse the tube between the electrodes each of which is surrounded by a luminous glow. With continued lowdischarge ering pressure broadens out to a steady luminous column which extends from the anode to This is almost as far as the cathode. known as the positive column and is separated from the cathode glow by a the Faraday dark dark space called space. The pressure in the gas at this stage is about 5 mm. of mercury. Further reduction in the pressure of the gas to about positive column to shrink and to break up into alternate light and dark patches known as striations. The Faraday dark space increases in size at this low pressure, and the cathode glow moves away from the cathode, leaving a dark space between it and the cathode called the Crookes’ dark space (Fig. 30:2). When the pressure is about 0.1 mm. the positive column disappears altogether and negative glow becomes considerthe ably extended. At still lower pressures the 1 mm. causes negative disappears, glow the the Crookes’ dark space filling tube, and the glass walls of the tube, par- the Gases, (a) At Reduced Pressures— P, Positive Column; F, Faraday Dark Space; N, Cathode Glow; C, Crookes' Dark Space, (b) At Very Low Pressures— Cathode Rays in Crookes' Tube. that the fluorescence of the glass is due to something being emitted from the in Germany, cathode, and Goldstein, gave the name of cathode rays to this emanation. Crookes, in England, showed that the rays were shot out at right angles from the cathode, caused a rise of temperature of bodies interposed in their path, and that the rays were also capable of producing a mechanical force. 341 . Chap. 30 MAGNETISM AND ELECTRICITY He hazarded a guess that the rays consisted of a stream of particles, and it was later shown that the rays could be deflected by a magnetic field and that they carried a negative charge (Chap. 31, Exp. 43) It was Sir J. J. Thomson, in England, of however, who showed in a series masterly experiments, carried out at the close of the nineteenth century, that the rays consisted of tiny, identical, negatively electrified particles which we now call electrons. In one of his experiments, in which the rays were deflected by both magnetic and electric fields, he was able to calculate their velocity and also the ratio of the charge to the mass of the the particles. The magnitude of electronic charge had been found by other means, and so he was able to obtain a value for the mass of the electron which he showed to be about 1/1840 part of the mass of the hydrogen atom. Various forms of the cathode ray tube are in use. One of these, the X-ray tube will be described in the next section. The other, the television picture-tube will be referred to in the next chapter. V : 79 X-RAYS While experimenting with discharge tubes at very low pressure, Roentgen, in Germany, in a fluorescent screen some distance away glowed brightly even when the discharge tube was covered with black paper, and 1895, noted that A Chest X-Ray Machine. General Electric 342 ELECTRONICS Sec. V:79 objects placed between the tube and screen produced shadows. Roentgen attributed these effects to a new form of tube, radiation its unknown which, of nature, was called X-rays. emanating account from the on cyanide, which is the substance frequently used on X-ray viewing-screens. They affect photographic plates, produce chemical changes, and are capable of discharging electrified bodies since they air. the are X-rays ionize electromagnetic waves of the same nature as visible or ultra-violet rays but having a very much shorter wave-length. One of the most characteristic properties of the rays is their penetrating power. They through many substances can opaque to ordinary light waves, the degree of penetration varying inversely as the density of the substance. Roentgen himself noticed this power of penetration and observed that human flesh was more transparent to the rays than bone. The pass It X-rays, is now known that or Roentgen rays, are produced whenever cathode rays impinge on a solid object Fig. 30:3 shows (Chap. 31, Exp. 44). a modern form of X-ray tube in which the electrons, constituting the cathode stream, originate from a heated filament (Sec. V:80) and are focused by a concave cathode on to a tungsten or molybdenum target set in a block of copper called the anti-cathode. The face of the anti-cathode is set at an angle of 45° to the axis of the tube, the X-rays coming off in the direction indicated. Most of the energy of the cathode stream is dissipated as heat, only about 1 per cent of the energy is converted into X-rays. X-Ray Photograph of a Human Hand. St. Michael’s Hospital X-rays travel in straight lines, are not deflected by either magnetic or electric fields and produce fluorescence in suitable materials such as platinum-barium medical uses of X-rays in locating fractures and foreign bodies in the human system are their most important application. Another use of X-rays is the detec- 343 Chap. 30 MAGNETISM AND ELECTRICITY tion of flaws in metal castings, and research scientists find them a useful aid in many of their investigations. V : 80 THE ELECTRONIC TUBE Just as a tap acts as a valve to control the flow of water through the pipes of a house, so the electronic tube acts as a valve to control the flow of electrons in an electric circuit (Chap. 31, Exp. Electronic tubes are able to 45, 46). increase or decrease the amount of current flow, or even to start or stop it, and the action is immediate. They use praccircuit. No tically no power in attempt will be made in this text to do any more than to outline the structure and action of the simplest forms of these the tubes. (a) Structure Any electronic tube has four basic parts (Fig. 30:4) : 1. A glass metal envelope enclosing the tube serves to sustain the vacuum or low pressure within the tube; 2. A cathode, which is heated either directly by an electric current or indirectly by a heater element which carries the current, serves to give ofT the electrons when heated; 3. An anode serves to receive the electrons from the cathode; 4. A base which has four or eight terminals or pro
ngs for connecting the tube into its circuit. Such a tube is called a diode. A triode has an additional part called the grid (part d). (b) Action When the tube is connected into a circuit, the circuit is actually broken. When the cathode is heated electrons Hot Filament Fig. 30:5 Action of a Diode Tube. If a positive charge is leave its surface. applied to the anode, these electrons will be attracted across the intervening space and thus a flow of electrons will occur (Fig. 30:5). The anode is frequently referred to as the ‘‘plate” and the current 344 ELECTRONICS Sec. V:80 passing through the plate is called the “plate current”. (c) Rectification If the anode is connected to an alternating-current circuit (Fig. 30:6) it will be alternately positiv^ely and negatively While the anode is positive charged. electrons will flow from the cathode to the anode and through the plate circuit. During those half cycles when the anode the negative, electrons about the is Fig. 30:6 Electrons are Attracted Only When Plate Rectifier Circuit. is Positive. (a) (b) 345 Chap. 30 MAGNETISM AND ELECTRICITY flow. current Thus the cathode will be repelled and no current passes will through the tube in one direction only, from cathode to anode, and a direct current flows in the plate circuit. The tube has been used to obtain direct current from an alternating-current supply. Such a process is called rectification. The tube used for the purpose is designated a rectifier tube. Half-wave rectification by a diode is illustrated in Fig. 30:7. (d) Amplification A triode tube has a fifth part called the grid (Fig. 30:8) which controls the current passing through the tube. The grid is a special wire or mesh between the cathode and the anode. Any change in grid voltage causes a proportional change in the amount of current flowing through the tube. Thus, if we have a very weak alternating current, too weak to be used directly, it would be applied to the grid as a signal voltage. As the grid voltage varies it causes an identical amplified variation in the plate current. The grid is given a negative charge just large enough that it never actually becomes positive when the signal is applied. Thus, no electrons are attracted to the grid itself so the tube operates at maximum efficiency. Fig. 30:9 illustrates this amplification action as a radio signal is impressed on the grid of the triode. 346 Triode Tube as an Amplifier. ELECTRONICS Sec. V:82 Sound Waves Microphone Aerial - Aerial Vacuum Tube Amplifiers Vacuum ' \ Tube Generator of High Frequency Waves ' Modulated Carrier Waves Tuner / Detector / 00000(5 ^ // 7 ‘ ( Tube Used as an Amplifier Vacuum Tube Used as a Modulator (a) TRANSMISSION Audio-frequency Amplifier Radio Frequency Amplifier (b) RECEPTION Fig. 30:10 Radio Transmission and Reception. V:81 RADIO TRANSMISSION AND RECEPTION The triode tube can be used as an oscillator which is merely a type of alternating-current generator capable of providing frequencies in the order of 150,000 to 1,500,000 cycles or more per As these variations in current second. are impressed on a transmitting antenna, waves, electromagnetic carrier called In the waves, are sent out into space. transmitting-studio a microphone transforms sound waves into a fluctuating This current is used to direct current. modulate, or cause variations in, the amplitude of the carrier waves from the oscillator. differences At the receiving-station the varying electromagnetic waves induce minute antenna. potential Certain of these, selected by the tuning are impressed on the grid of circuit, an amplifier where they cause correspondingly larger variations in the plate These current of the amplifier tube. the in currents pass to a detector circuit in which another electronic tube separates the audio frequency component of the wave (i.e., the part that can be converted to audible sound) from the radio These frequency of the carrier wave. audio frequency components are further amplified and used to operate the diaphragm of earphones or a magnetic speaker. A simple diagram illustrating the above principles is shown in Fig. Students desiring further infor30:10. mation on the principles and structure of radio circuits are referred to the Radio Ajnateurs’ Handbook or specialized texts on radio. V:82 PHOTOELECTRICITY In 1888, investigation. Photoelectric cells or, as they are commonly called, ‘‘electric eyes”, are the result of almost one hundred years of Hallcareful wacks, a German scientist, found that negatively charged zinc plates lost their charges when exposed ultra-violet light coming from an arc-lamp (Fig. 30:11). Thus electrons were shown to be substances when light falls upon them. This effect may be demonstrated visually if a zinc to an electroscope plate (Chap. 31, Exp. 47). This is called photo-emission of electrons. attached released certain from to is A short time later it was discovered that the alkali metals, sodium, potassium and caesium are sensitive even to orlight. Modern emissive dinary visible type photoelectric cells use these metals and are similar in appearance to radio vacuum-tubes. A layer of light-sensitive metal coats the entire inner surface of the tube with the exception of one small area which admits light. A positively charged anode is located in the centre of the tube and serves to attract elec- 347 Chap. 30 MAGNETISM AND ELECTRICITY trons from all sides as they are released (Fig. 30:12). The rate of electron emission is proportional to the amount and nature of the light falling upon the metal. Such cells control circuits which are used to open and close store doors, operate cesses with the exception of sound reproIt consists of a metal supportduction. ing disc upon which is deposited a layer of light-sensitive material. Over this is placed a transparent metal grid which acts as an electrode and collector for the electrons set free by light. Students Positively Charged Anode Serves as Plate to Attract Emitted Electrons Light-Sensitive Cathode Gives Off Electrons when Light Hits It Electrons Escape from Surface Electric Bell Fig, 30:12 Photoelectric Cell. burglar alarms, control dangerous machinery, reproduce sound for movies, count articles on production lines, and for hundreds of other purposes contributing to the efficiency and safety of industrial processes (Fig. 30:13). The photronic cell is used almost universally to-day for all the above pro- 348 Fig. 30:13 Photoelectric to Operate an Electric Bell. Cell Used ELECTRONICS Sec. V:83 The Component Parts of a Light-Meter. Canadian General Electric interested in photography will recognize the light-meter as a simple application of the photronic cell. In these devices light energy is con- verted into electrical energy. This, in part, provides the answer to the scientists’ dream of eventually tapping the huge resources of the sun itself. V : 83 QUESTIONS 1. (a) What is the meaning of the term electronics? (b) State two facts fundamental to electronics. thermionic (c) Distinguish emission and photo-emission of elec- between trons. 2. (a) What are cathode rays? (d) Describe cathode rays. (c) How are cathode rays produced? properties some of 3. (a) Explain how X-rays are produced. (b) Describe some properties of Xrays. (c) How are these properties associated with the uses of X-rays? 4. (a) Make a labelled diagram of a simple diode tube. State the purpose of each part. (b) How does a triode tube differ from a diode tube? (a) What is meant by rectification? (b) Explain why a diode tube may be used as a rectifier. (a) What is meant by amplification? (b) Explain why a triode tube may be used as an amplifier. 5. 6 . 7. Describe briefly how radio signals are sent from one station to another. 8 . (a) Describe the structure of a simple photoelectric cell (ii) (i) a photronic cell. (b) What transformation of energy takes place in each of the above cells? 349 CHAPTER 31 EXPERIMENTS ON MAGNETISM AND ELECTRICITY EXPERIMENT 1 To distinguish between magnetic and non-magnetic substances. (Ref. Sec. V:2) Apparatus Bar magnet, assortment of small articles made from iron, paper, glass, wood, copper, nickel, rubber, tin, silver, plastic, etc. Method Approach each of the articles in turn with the bar magnet. Tabulate your results. Observations State which objects were attracted and which were not attracted by the magnet. Conclusions 1. State which were magnetic substances and which were non-magnetic substances. 2. Define (a) magnetic substances, (b) non-magnetic substances. EXPERIMENT 2 To determine if the magnetic strength is equal in all parts of the bar magnet. (Ref. Sec. V;2) Apparatus Bar magnet, iron-filings. Method Sprinkle iron-filings liberally on a piece of paper. Roll the bar magnet in the filings and observe. Clean the magnet carefully with a cloth at the end of the experiment and return the filings to their container. Observations 1. Are the filings attracted equally to all parts of the bar magnet? 2. Where do most of the filings collect? 350 EXPERIMENTS ON MAGNETISM AND ELECTRICITY 3. Where does the least number of filings collect? 4. Compare the number of filings collected at each end of the magnet. Conclusions 1. Where is the magnetic strength of a bar magnet concentrated? What are the these areas called? 2. Where is the least magnetic strength in a bar magnet? EXPERIMENT 3 To determine the difference between the poles of a bar magnet, (Ref. Sec. V:2) Apparatus Bar magnet, chalk, non-magnetic stand, brass support. Method 1. Mark one end of the magnet with chalk so that it may be readily identified. 2. Suspend the magnet horizontally from the non-magnetic stand by means of thread supporting a brass support. 3. Allow the magnet to swing freely until it comes to rest. Note the direction in which the marked end is pointing. 4. Reverse the ends of the magnet in the support and again let it swing freely until it comes to rest. Observations 1. In what direction did the marked end of the magnet point when it came to rest t
he first time? 2. In what direction did the marked end of the magnet point when it came to rest the second time? Conclusions 1. Along what directional line will a freely suspended bar magnet always come to rest? 351 Chap. 31 2. MAGNETISM AND ELECTRICITY Why do we call one end of a bar magnet the north-seeking pole and the other end the south-seeking pole? What abbreviations for these are in common use? EXPERIMENT 4 To investigate the effect of bringing like and unlike magnetic poles together, (Ref. Sec. V:2) Apparatus Two bar magnets of known polarity, non-magnetic stand, brass support (Fig. 22:3). Method 1. Suspend one bar magnet from the non-magnetic stand by a thread and a brass support so that it is free to move in a horizontal plane, or balance a bar magnet horizontally on the convex surface of a watch glass. 2. Approach the N-pole of the suspended magnet with the N-pole of the other magnet. 3. Repeat, using the N-pole of the suspended magnet with the S-pole of of the other. 4. Now approach the S-pole of the suspended magnet with each end of the other. 5. Tabulate your results. Observations Pole of Suspended Magnet Pole of Other Magnet Observation N-pole N-pole S-pole S-pole N-pole S-pole N-pole S-pole Conclusions 1. What is the effect of like magnetic poles on each other? 2. How do unlike magnetic poles affect each other? 3. The above two conclusions combined into one statement give us the Law of Magnetism. State this important law. Questions 1. If the polarity of the magnets to be used is not known, how could you determine this information? 2. Why is it correct to say that the north magnetic pole of the earth contains S-pole magnetism? 352 EXPERIMENTS ON MAGNETISM AND ELECTRICITY EXPERIMENT 5 To study the magnetic field about a bar magnet, (Ref. Sec. V:4) Apparatus Two bar magnets, iron-filings, sheet of paper. Method 1. Lay a bar magnet on the desk and place a book of approximately the same thickness as the magnet on either side of it. Place the paper over the magnet so that the edges of the paper are supported by the books (a magnet board may be used to support the magnet Sprinkle iron-filings evenly over and paper instead of the books). the surface of the paper while tapping it gently. 2. Repeat the above procedure using the two bar magnets placed with their N-poles about 2 in. apart. 3. Repeat part 2 with an N-pole and an S-pole about 2 in. apart. Observations 1. Where are the lines of filings most concentrated? 2. Where are the lines least concentrated? 3. What is the general shape of the lines formed? 4. Do any of the lines appear to cross each other? 5. Where do the lines seem to begin and end in the case of the single bar magnet? 6. Describe the lines formed when like poles are adjacent to each other. 7. Describe the lines formed when unlike poles are adjacent to each other. Conclusions 1. What causes the iron-filings to take up their positions about the bar magnet? 2. What evidence have you found that supports the theory that lines of force repel each other? 3. What property of lines of force explains the attraction of unlike poles? 4. What evidence have you found that would indicate that the field is strongest close to the poles of the magnet? Questions 1. Make a diagram of each of the patterns formed by: (a) A single bar magnet. (b) Two bar magnets with like poles adjacent. (c) Two bar magnets with unlike poles adjacent. 2. Indicate by means of a diagram what you would expect the field of force about a horseshoe magnet to look like. Verify your prediction by experiment. 353 Chap. 31 MAGNETISM AND ELECTRICITY EXPERIMENT 6 To study induced magnetism. (Ref. Sec. V:6) Apparatus Bar magnet, long iron nail, iron tacks, magnetic compass. Fig. 31:2 Method 1, Place one end of the nail near the tacks to test for magnetic power. Now hold one end of the bar magnet near one end of the nail. Allow the other end of the nail to come in contact with the tacks. Lift the magnet and the nail together. Then move the magnet away from the nail. 2. With the head of the nail near the N-pole of the magnet, approach the N-pole of the compass-needle with the point of the nail. Repeat this with the head of the nail near the S-pole of the magnet. Observations 1. Did the nail by itself possess magnetic powers? 2. When a pole of the magnet was close to the nail, what was the effect on the tacks? 3. What occurred when the magnet was moved away from the nail? 4. With the N-pole of the magnet near the head of the nail, how was the N-pole of the compass-needle affected? 5. What occurred when the S-pole of the magnet was used? Conclusions 1. Why did the nail act as a magnet when under the influence of the bar magnet? 2. Why did it not remain a magnet when the bar magnet was removed? 3. What polarity was induced in the end of the nail (a) nearer to (b) farther from the pole of the magnet used? Questions 1. Why do we specify an iron nail and iron tacks? 2. Why should a long nail be used? 354 EXPERIMENTS ON MAGNETISM AND ELECTRICITY EXPERIMENT 7 To determine if the earth's magnetic field can cause induced magnetism. (Ref. Sec. V:6) Apparatus Dipping-needle (Fig. 22:8), soft iron bar about 12 in. long, hammer, magnetic compass. Method 1. Determine the direction of the magnetic lines of force at your location on the earth’s surface by means of the dipping-needle (Sec. V:5). 2. Test the iron bar for magnetic polarity by approaching the N-pole of the compass-needle with each end in turn. 3. Holding the iron bar at the angle indicated by the dipping-needle, strike the bar sharply several times with the hammer. 4. Again test the bar for magnetic polarity using the magnetic compass. Observations 1. What is the magnitude of the angle of inclination? 2. What was the result of testing each end of the iron bar with the compass in the first case? 3. After striking the bar what changes were observed when it was tested with the compass? Conclusions 1. Why was the test different in the second case than in the first? 2. From your observations what magnetic polarity must the north mag- netic pole of the earth possess? EXPERIMENT 8 To study magnetic shielding. (Ref. Sec. V:6) Apparatus Demonstration magnetic compass-needle, bar magnet, thin sheets of iron, glass, copper, wood. Method 1. Hold the bar magnet to one side of the compass-needle so that it will be attracted towards the magnet. 2. Move the point of the compass-needle to one side and release it, allowing it to begin swinging from side to side. Note the number of vibrations in 10 seconds and the time in which it finally comes to rest. 3. Insert the iron sheet between the compass-needle and the magnet and repeat the above procedures, taking care that all other conditions remain the same as before. 4. Repeat part 3 using other materials. 355 Chap. 31 MAGNETISM AND ELECTRICITY Observations 1. Number of v.p.s. without shield 2. Time to come to full stop without shield 3. Number of v.p.s. with iron shield 4. Time to come to full stop with iron shield. 5. Repeat the above observations when other materials are inserted. Conclusions What is magnetic shielding? Questions 1. Why is it necessary to shield a magnetic compass on board a ship? 2. How can such magnetic shielding be accomplished? EXPERIMENT 9 To magnetize a steel wire, (Ref. Sec. V:7) Apparatus Bar magnet, steel wire or thin knitting-needle, magnetic compass. Method 1. Test the ends of the wire for magnetism, using the compass. 2. Using the N-pole of the bar magnet stroke the wire from one end to the other several times, always in the same direction. Lift the magnet well away from the wire as you return it to the starting point for each stroke. Test each end of the wire for polarity, using the compass. 3. Repeat the above steps using the S-pole of the bar magnet to stroke the wire. Observations State observations for each part described above. 356 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Conclusions 1. What indication did you have that the wire was not magnetized before it was stroked with the magnet? 2. What makes you believe that the wire has become a bar magnet during the experiment? 3. What \vas the polarity of the end of the wire last touched by the pole of the bar magnet? Questions 1. Why were both ends of the wire tested for magnetic polarity? 2. Why was the bar magnet lifted away from the wire as it was returned at the end of each stroke? 3. Explain why the wire became magnetized. EXPERIMENT 10 To investigate some experimental evidence for the theory of magnetism. (Ref. Sec. V:7) Apparatus A magnetized steel wire, wire cutters, magnetic compass. Method 1. Determine the polarity of the steel wire. 2. Cut the wire in half and determine the polarity of each half. 3. Cut the pieces into smaller and smaller sections and determine the polarity of each piece. Observations By means of diagrams illustrate your observations. Conclusion From these observations what inference can be made about the basic structure of a magnet? EXPERIMENT 11 To investigate further experimental evidence for the theory of magnetism. (Ref. Sec. V:7) Apparatus Test-tube, iron-filings, stopper, bar magnet, magnetic compass. Method 1. Fill the test-tube with iron-filings and insert the stopper. Test the tube for magnetic polarity. 2. Stroke the test-tube from end to end several times using one pole of the bar magnet as was done in experiment 9. Test the tube again for magnetic polarity. 3. Shake the test-tube vigorously and again test it for polarity. 357 Chap. 31 MAGNETISM AND ELECTRICITY Observations 1. Did the tube of filings act as a magnet when first tested? 2. Was any movement of the filings noticed as the test-tube was stroked ? 3. What was observed in the second polarity test? 4. How did the shaking afifect the magnetic properties? Conclusion How does this experiment support the theory of magnetism? Question Would there be any limit to the magnetic strength produced? Explain. EXPERIMENT 12 To produce static electricity by friction, (Ref. Sec. V:10) Apparatus H
ard rubber or ebonite rod, glass rod, cat’s fur, piece of silk, small scraps of papers, sawdust and other light objects. Method 1. Rub the ebonite rod with the cat’s fur. Bring the rod near the objects. 2. Repeat the above procedure, rubbing the glass rod with the silk. Note Lucite rubbed with a sheet of rubber or polystyrene (the material used in transparent vegetable containers) is an excellent substitute for glass and silk. Observations Describe the behaviour of the objects as the rod approaches in each case. Conclusion What is the efTect of rubbing an ebonite rod with cat’s fur or a glass rod with silk? 358 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Questions 1. What transformation of energy is involved in producing static elec- tricity by friction? 2. What evidence is there that this new power of attraction is not magnetism? 3. Describe instances encountered static electricity. in your everyday life in which you have EXPERIMENT 13 To show that there are different kinds of electrical charges and to establish the law of electrical charges. (Ref. Sec. V:11) Apparatus Two ebonite rods, two glass cat’s fur, silk. rods, insulated stand, thread, stirrup, Method 1. Charge an ebonite rod by rubbing it with cat’s fur. Suspend the charged rod in the stirrup attached by a thread to the insulated stand. Approach the charged end of the ebonite rod with the other charged ebonite rod. 2. Approach the charged end of the ebonite rod with the charged end of a glass rod which has been rubbed with silk. 3. Replace the suspended ebonite rod by a charged glass rod and repeat the procedures outlined above. Observations What is observed in each of the above parts? Conclusions 1. Are there different kinds of electrical charges? Name each kind. 2. What simple law of electrical charges may now be stated? Question What standard method have we of producing each kind of charge in the laboratory? EXPERIMENT 14 To study the use of the pith-ball electroscope. (Ref. Sec. V : 13) Apparatus Pith-ball electroscope, ebonite rod, cat’s fur, glass rod, silk. Method 1. Charge the ebonite rod and slowly bring it near the pith balls. Allow them to touch the rod, and observe. 359 Chap. 31 MAGNETISM AND ELECTRICITY 2. After charging the pith balls approach them slowly with (a) a negatively charged rod and (b) a positively charged rod. 3. Repeat the above procedures using a charged glass rod in part 1. Observations Describe what is observed in each part above. Conclusions 1. Explain each observation. 2. How may the pith-ball electroscope be used to (a) detect (b) identify an electrical charge? Questions 1. What is meant by “charging by contact”? Explain this process. 2. Why is repulsion the only sure evidence that an object is charged? EXPERIMENT 15 To study the use of the gold-leaf electroscope. (Ref. Sec. V:13) Apparatus Gold-leaf electroscope, ebonite rod, cat’s fur, glass rod, silk, stick of sealing-wax, piece of flannel. Method 1. Charge the ebonite rod and slowly bring it near the knob of the electroscope until it touches. Withdraw the charged ebonite rod. 2. Approach but do not touch the knob of the charged electroscope with (a) another negatively charged rod (b) a positively charged glass rod (c) a stick of sealing-wax that has been rubbed with flannel. 3. Remove the charge from the electroscope by touching the knob with a finger. 4. Repeat parts 1 and 2, touching the knob with a charged glass rod. Observations Describe the observations for each part above. Conclusions 1. What charge is given to the electroscope when it is “charged by contact” with (a) a positively (b) a negatively charged rod? 2. What procedure can be used to identify an unknown charge on an object? EXPERIMENT 16 To distinguish between conductors and insulators. (Ref. Sec. V:14) 360 7 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Apparatus Gold-leaf electroscope, ebonite rod, cat’s room. fur, various objects in the Method 1. Place a negative charge on the electroscope by touching the knob with the charged ebonite rod. 2. Approach the knob of the charged electroscope with the charged ebonite rod. Note what happens. 3. Rub the charged rod on a water pipe and again approach the knob. Note what happens. 4. Recharge the rod after each test and repeat step 3, rubbing the charged rod on rubber, slate, wood, your hand, and dipping it in water. Observations Note the effect on the electroscope leaves in each case. Conclusions 1. List those materials which allowed the charge to escape. What are such materials called? 2. List those that did not permit the charge to escape. What are they called? EXPERIMENT 1 To produce induced electrical charges. (Ref. Sec. V:15) Apparatus Two metal spheres on insulated stands, proof-plane, ebonite rod, cat’s fur, glass rod, silk, gold-leaf electroscope. Method 1. Charge the electroscope negatively by the contact method. 2. Test each sphere for the presence of a charge by bringing it near the knob of the charged electroscope. 3. Place the two spheres A and B in contact as shown in Fig. 23:6. Approach, but do not touch, sphere A with the end of a negatively charged ebonite rod. Remove B from A and then remove the rod. Again test each sphere as in part 2. 4. Touch the spheres to each other and test each for a charge as in part 2. 5. Repeat parts 3 and 4, using a positively charged rod. Observations State what was observed. Explanation Explain your observations in this experiment with reference to the electron theory. 361 9 Chap. 31 MAGNETISM AND ELECTRICITY Conclusions 1. In each case what charge was induced on (a) the sphere near the inducing charge (sphere .4), (b) the sphere distant from the inducing charge (sphere B) ? 2. What do you conclude about the magnitude of these two induced charges ? EXPERIMENT 18 To charge a gold-leaf electroscope by induction. (Ref. Sec. V:16) Apparatus Gold-leaf electroscope, ebonite rod, cat’s fur, glass rod, silk. Method 1. Charge an ebonite rod negatively and approach, but do not touch, the knob of the electroscope with it. Keeping the charged rod near, touch the knob with a finger, i.e., “ground” the knob. Remove your finger. Then remove the charged rod. 2. Repeat the procedure, using a positively charged glass rod. Observations Record the behaviour of the leaves in each step of the method by means of a series of diagrams. Explanation Using the electron theory, indicate on the diagrams what occurred in each of the above steps. Conclusions What charge can be induced on the electroscope with (a) a negatively charged rod, (b) a positively charged rod? Question Describe how to charge an electrophorus (Sec. V:17) by induction. Explain fully. EXPERIMENT 1 To investigate the distribution of electric charges on a charged object. (Sec. V;19) Apparatus Electrophorus, proof-plane, gold-leaf electroscope, Biot’s spheres (Fig. 23:11), several insulated conductors of different shapes. Method A. Using Biot's Spheres 1. Charge the disc of the electrophorus and transfer the charge by con- tact to the insulated metal sphere. 362 EXPERIMENTS ON MAGNETISM AND ELECTRICITY 2. Test the insulated metal hemispheres for electric charge by means of a proof-plane and the electroscope. 3. Then place the hemispheres tightly around the charged sphere. Remove the hemispheres and using the proof-plane and electroscope again test them for charge. 4. Test the sphere for the presence of electric charge. Observations Describe clearly what occurred in the above steps. Explain fully. Conclusion Where do electric charges reside on a charged object? B. Using Conductors of Different Shapes 1. Using the disc of the electrophorus, charge the conductors by contact. 2. Touch various parts of each conductor with the proof-plane and bring it to a given distance from a charged electroscope each time. Note the amount of deflection in each case and compare the density of charge on the various parts of each conductor. (a) Fig. 31:5 Observations Describe clearly what occurred. Conclusion Where is there the greatest density of charge on the surface of an irregularly-shaped object? EXPERIMENT 20 To study the action of pointed conductors. (Ref. Sec. V:20) Apparatus Wimshurst machine, metal point, candle. Method Attach the metal point to one terminal of the Wimshurst machine. Turn the handle rapidly to generate an electric charge. Hold the lighted candle in front of the point when it has become highly charged. Observation What happens to the flame of the candle? 363 Chap. 31 MAGNETISM AND ELECTRICITY Explanation What causes this phenomenon? Conclusion What effect do points have on the charge residing on a conductor? Application Explain the action of lightning-rods. EXPERIMENT 2 1 To study the action of a voltaic cell. (Ref. Sec. V:23, 24) Apparatus Glass vessel, strips of copper and zinc metal, flash-light bulb, socket, water, potassium dichromate. Method 1. Slowly add 10 ml. of concentrated sulphuric acid to 200 ml. of water, with continuous stirring. Pour this solution (the electrolyte) into the glass vessel. Immerse the two plates in the electrolyte parallel to each other but not touching. Note what occurs. 2. Connect the plates by wires joined to the flash-light bulb. Note what happens at both plates and to the flash-light bulb. 3. Add 14 teaspoonful of potassium dichromate and stir until it dis- solves. Observations 1 . Is there any change noted at both plates in part 1 ? 2. (a) What changes occur at the plates in part 2? (b) What happens to the light bulb in part 2? 3. What is the result of adding potassium dichromate? 4. What eventually happens to the electrodes as the action continues? Explanations 1. What is the cause of the changes at the zinc plate? 2. What must have been produced to make the light bulb glow? 3. Account for the change in the glowing that is observed. 4. What is the action of the potassium dichromate? 5. Why will the action of the cell eventually stop? Conclusions 1. How is electric current produced in a voltaic cell? 2. What is (a) local action (b) polari
zation? How may they be prevented? Questions Why is the voltaic cell as used in this experiment not in common use to-day? 364 EXPERIMENTS ON MAGNETISM AND ELECTRICITY 2. What name, describing its action, can be given to the potassium dichromate in this experiment? EXPERIMENT 22 To determine the relationship between potential difference fV)^ current strength (I), and resistance (R) of an electric circuit (Ohm's Law). (Ref. Sec. V;33) Apparatus Resistance coil, a number of dry cells, galvanometer (Sec. V:53), knife switch. VVWWW^ ( G I I .X I Method Fig. 31:6 1. Connect one dry cell, the resistance coil, the galvanometer, and the knife switch in a series circuit as indicated in the diagram. Close the switch and note the deflection of the galvanometer. Open the switch. 2. Repeat the above procedure using two, three, four and then five dry cells in series. Record all your results in the table below. Observations Number of Cells Galvanometer Deflection (Divisions) No. OF Cells Galvanometer Deflection 1 2 3 4 5 Explanations 1. What is represented by number of cells? How could it be repre- sented using an electrical symbol? 2. What is represented by the galvanometer deflection? How could it be represented using an electrical symbol? 3. How do the ratios obtained by dividing the number of cells by the The instructor should ensure that the resistance used is of a correct value to give good results with the galvanometer and potentials employed. 365 Chap. 31 MAGNETISM AND ELECTRICITY galvanometer deflections compare? sistance R. This ratio is called the re- Conclusions 1. State Ohm’s Law. 2. Express it, using symbols. Questions 1. Name the units used for measuring potential difference, current strength, and resistance. 2. Define “ohm”. EXPERIMENT 23 To determine the resistance of an unknown resistance by the voltmeter-ammeter method. (Ref. Sec. V:38) Apparatus Voltmeter, ammeter, rheostat, dry cells, unknown resistance. Method 1. Connect the voltmeter, ammeter, rheostat and dry cells in a circuit with the unknown resistance X as shown in the diagram. 2. Adjust the rheostat, R, until any suitable value of current flows. 3. Determine the value of the current in amperes by reading the Determine the potential difference in volts by reading ammeter. the voltmeter. 4. Calculate the value of X in ohms. Observations V=^ I = Calculations Determine the value of X using Ohm’s Law. Conclusion What is the value of the unknown resistance? 366 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Questions 1. Why may the setting of the rheostat be varied without affecting the final result? 2. Why is the rheostat used in this circuit? 3. Why is the voltmeter connected in resistance? parallel with the unknown EXPERIMENT 24 To determine the resistance of an unknown resistance by the substitution method. (Ref. Sec. V:38) Apparatus Ammeter, resistance box, rheostat, dry cell, unknown resistance. Method 1. Connect the ammeter, rheostat and dry cell in series with the unknown resistance. Adjust the rheostat to get a large deflection of the ammeter. 2. Remove the unknown resistance and substitute the resistance box with all plugs removed. Replace the plugs singly and in groups of two or more until you get the same ammeter reading as before. Observation What is the total resistance indicated by the unreplaced plugs? Conclusion What is the value of the unknown resistance? EXPERIMENT 25 To study electrolysis of water. (Ref. Sec. V.41) Apparatus Simple electrolytic cell, pair of platinum electrodes, electrolyte consisting of one part sulphuric acid slowly added to approximately ten parts of water with constant stirring, source of direct current, test-tubes, wood splints. Fig. 31:8 367 . Chap. 31 MAGNETISM AND ELECTRICITY Method 1. Fill the electrolytic cell almost to the top with electrolyte. Completely fill the two test-tubes with the same liquid and invert one over each electrode as shown in the diagram. Connect the electrodes to the current source noting which is the cathode ( — ), and which is the anode ( + ) 2. Allow the current to flow until one test-tube is almost filled with gas. Measure and compare the lengths of the gas column in the two test-tubes. 3. Remove the test-tubes and immediately lower a burning splint into each in turn. Note the results. Observations 1. What is observed at the anode and at the cathode as the current flows? 2. Compare the volume of gas produced at the anode with that at the cathode in a given time. 3. What was observed when the splint was lowered into each of the gases in part 3? Conclusions 1. What gas is produced at (a) the anode (b) the cathode? 2. What are the relative volumes of the two gases? 3. Define cathode, anode, electrolyte, electrolysis. Explanation Write a brief explanation of electrolysis of water (Sec. V:41). Include in your explanation a description of what the current consists of in (a) the external circuit (b) the electrolyte. Questions 1. Why was sulphuric acid added to the water in making the electrolyte? 2. Why were the electrodes made of platinum? 3. What would be the eflfect on the quantities of gases produced of (a) increasing the current strength, (b) increasing the length of time that it flows. The Hoffman water voltameter (Fig. 26:2) may be used in place of the above apparatus. EXPERIMENT 26 To study electrolysis of copper sulphate solution. (Ref. Sec. V:42) Apparatus Electrolytic cell, pair of carbon electrodes, solution of copper sulphate, source of direct current, test-tube, wood splints. 368 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Method 1. Fill the electrolytic cell almost to the top with the copper sulphate solution. Connect the electrodes to the current source. Allow the current to flow for several minutes. Note what occurs at both electrodes. Note any change in the colour of the electrolyte. 2. Collect a test-tube full of the gas liberated at the anode as described in the previous experiment and test it with a glowing splint. examine the deposit at the cathode. Also, Observations Describe all observations. Conclusion What is obtained at (a) the anode (b) the cathode? Explanation Write a brief e.xplanation of the electrolysis of copper sulphate solution. EXPERIMENT 27 To discover the factors that affect the amounts of materials liberated at the electrodes during electrolysis. (Ref. Sec. V:43) Apparatus As in experiment 25, with ammeter and rheostat in series with the electrodes. Fig. 31:9 Method 1. (a) Set up the electrolytic cell as in experiment 25, and allow a current of known strength to flow for one minute. Measure the length of the oxygen column collected over the anode. (b) Repeat part 1 (a) allowing the current to flow for two minutes, three minutes, etc. 3G9 — Chap. 31 2. MAGNETISM AND ELECTRICITY Repeat part 1 the strength etc., of that previously used. (a) using a current of one-half the strength, twice Observations 1. Current Strength Constant Time Current Flows (min.) Length of Oxygen Column (cm.) 2. Time Constant Current Strength (amp.) Length of Oxygen Column (cm.) Conclusion What factors affect the amounts of materials liberated at the electrodes during electrolysis? EXPERIMENT 28 To determine the strength of an electric current using a copper voltameter. (Ref. Sec. V:44) Apparatus Copper voltameter, copper sulphate solution, ammeter, rheostat, switch, source of direct current, stop-watch, balance, alcohol, emery-paper. Method 1. Clean the cathode with the emery-paper. Wash it in water. Dip 4. it in alcohol and let it dry by evaporation. Weigh it accurately. 2. Arrange apparatus as in Fig. 26:3. Simultaneously close the switch and start the stop-watch. Take ammeter readings every minute and let the current flow for exactly 15 minutes. 3. Remove the cathode, being careful not to dislodge any of the deposit from it. Rinse in water, dip in alcohol, and allow it to dry as before. Weigh it accurately. Observations Ammeter Readings 1. Initial weight of cathode 1 = 2. Final weight of cathode Weight of copper deposited = = 3. Time of current flow gm. gm. gm. sec. Current strength (average of ammeter readings) = amp. Calculations 1. Determine the weight of copper deposited in 1 second. 370 EXPERIMENTS ON MAGNETISM AND ELECTRICITY 2. Knowing that the electrochemical equivalent of copper is 0.000329, calculate the current flowing in the circuit. Conclusions 1. What was the calculated current? 2. How docs it compare with the ammeter reading? 3. Define ampere, coulomb. Questions 1. What possible sources of error were there in this experiment? 2. W'hy would a silver voltameter be more accurate than a copper voltameter? EXPERIMENT 29 To study electroplating. (Ref. Sec. V:45) Apparatus Electrolytic cell, copper sulphate solution, copper anode, object to be plated (a key) as cathode, source of direct current, rheostat, emerypaper, carbon tetrachloride. Method Carefully clean the object with emery-paper, wash in carbon tetrachloride and wash in water. Arrange apparatus as in the diagram. Close the circuit. Adjust the rheostat so that only a small current flows. After several minutes, remove the electrodes and examine them. Observations What is observed? Conclusion What is meant by electroplating? 371 Chap. 31 MAGNETISM AND ELECTRICITY Explanation Write a brief explanation for copper plating. Questions 1. In order to electroplate with any metal, what must (a) the cathode (b) the anode (c) the electrolyte consist of? 2. (a) Why should the cathode be cleaned? (b) Why should a small current be used? 3. Why was there no change in the colour of the copper sulphate solution in the above experiment? EXPERIMENT 30 To illustrate the principle of the lead-acid storage cell, (Ref. Sec. V:46) Apparatus Two strips of lead metal, emery-paper, two dry cells, glass tumbler, solution of sulphuric acid (one part concentrated sulphuric acid added to ten parts water), flash-light bulb, socket, galvanometer with fuse. Sulphuric Acid Solution Method 1. Thoroughly clean the lead strips, using the emery-paper. F
ill the tumbler about three-quarters full of sulphuric acid solution. Immerse the lead strips in the solution so that they do not touch each other, connect them in series with the dry cells and let the current flow for about 5 minutes. Momentarily insert the galvanometer into the circuit to determine the direction of the current. Note any changes observed at each electrode. 372 2. EXPERIMENTS ON MAGNETISM AND ELECTRICITY Remove the diy cells and insert a flash-light bulb in their place. Note what happens. Momentarily insert the galvanometer to determine the direction of the current again. Let the current continue to flow until the flash-light bulb goes out and note any changes in the plates. Observations Describe all observations. Explanation 1. (a) What type of cell do you have in part 1 ? (b) What energy transformation takes place in it? 2. (a) What type of cell do you have in part 2? (b) What energy transformation takes place in it? 3. Describe briefly the chemical changes involved during the above energy transformations. Conclusions 1. What does a lead-acid storage cell consist of? 2. Describe what takes place during discharging and recharging. Questions 1. How may a hydrometer be used to determine the condition of a storage battery? 2. Why must water be added to a storage battery periodically? Why should it be distilled water? EXPERIMENT 31 To investigate the magnetic field surrounding a conductor of electricity. (Ref. Sec. VMS) Apparatus Several dry cells, conductor, iron-filings, small magnetic compass, piece of cardboard, switch. Method 1. Pass the conductor vertically through the piece of cardboard and connect it with the dry cell and switch so that the electrons will flow upward in the conductor when the switch is closed. Sprinkle ironfilings on the cardboard and gently tap it. Note the results. 2. Explore the magnetic field around the conductor with the compass- needle and plot the direction of the needle on a diagram. 3. Grasp the wire with your left hand so that the Angers point in the direction in which the N-poles are deflected. Compare the direction of the electron flow and that in which your thumb points. 4. Reverse the direction of electron flow and repeat the procedure in 3 and 4. 373 Chap. 31 MAGNETISM AND ELECTRICITY Observations Describe all observations made above. Conclusions 1. Describe the magnetic field around a wire carrying electric current. 2. State the Left-Hand Rule. Question Describe the magnetic field obtained if the above conductor were coiled into a single loop. EXPERIMENT 32 To investigate the magnetic field surrounding a helix carrying electric current, (Ref. Sec. V:49) Apparatus Dry cell, helix, iron-filings, magnetic compass, piece of cardboard. Method 1. Set the helix into a slit in the cardboard and support it horizontally. Join it to the dry cell and switch. Close the switch and sprinkle ironfilings on the cardboard, tapping it gently as you do so. Describe what occurs. 2. Determine the polarity of the magnetic field about the helix, using the N-pole of the compass. 3. Grasp the helix with your left hand so that the fingers point in the direction of the electron flow. Toward which pole does your thumb point? 4. Reverse the direction of electron flow and repeat parts 2 and 3 above. 374 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Observations Describe all the observations made above. Conclusions 1. Compare the magnetic field about a helix carrying a current, with that about a bar magnet. 2. State the Helix Rule. EXPERIMENT 33 To discover the factors affecting the strength of an electromagnet, (Ref. Sec. V:50) Apparatus Two di-y cells, about 5 ft. of copper wire, cylindrical piece of soft iron about 4 in. long and in. in diameter, compass-needle, small iron weights (tacks, etc.) switch. Method 1. Wind about 10 turns of wire into a loose coil around the piece of soft “iron”. Remove the iron core and join the coil to a dry cell and switch. Close the switch and test the coil for magnetism. Determine the number of iron weights that can be lifted. 2. Insert the soft iron core, close the switch and test as before, noting any difference in the strength of the magnetism. 3. Using twenty turns of magnet wire around the iron core repeat the observations made in parts 1 and 2. 4. Using the two dry cells in series, repeat part 3. Observations State the number of iron weights lifted in each of the above cases. Conclusion What factors affect the strength of an electromagnet? Explanation State why each of these factors increases the strength of an electromagnet. Questions The galvanoscope (Fig. 27:14) consists of several coils of wire with different numbers of turns, e.g., 1, 25, 100, in each coil. 1. Describe how this instrument may be used (a) to determine the to compare the strengths of different direction of a current (b) currents. 2. What is the purpose of the coils with different numbers of turns? 375 Chap. 31 MAGNETISM AND ELECTRICITY EXPERIMENT 34 To determine the effect of a magnetic field upon a conductor carrying electric current, (Ref. Sec. V:52) Apparatus Battery of dry cells, strong horseshoe magnet, switch. Method 1. Assemble the apparatus as in the diagram with the conductor suspended between the poles of the magnet at right angles to the Close the switch and note what direction of the lines of force. happens. 2. Reverse the direction of the current and repeat part 1. Observations Describe what is observed. Conclusion What is the effect of a magnetic field upon a conductor carrying an electric current? This is the motor principle. Explanation What causes the conductor to move in each case? 376 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Questions 1. If the above conductor were coiled into a helix and freely suspended in a magnetic field, what would be the effect (a) When a small current is passed through it? (b) When a larger current is passed through it? (c) When the current is reversed? (d) When the field is reversed? Make labelled sketches to indicate polarities and direction of movement. 2. What energy transformation occurs? Application Describe the construction of the D’Arsonval galvanometer. EXPERIMENT 35 To investigate the principle of operation of a simple directcurrent motor, (Ref. Sec. V:55) Apparatus St. Louis motor (Fig. 27:21), dry cells, rheostat, magnetic compass, switch. Method 1. Examine the motor carefully. Identify the field magnets, armature, Connect the brushes in series with the brushes and commutator. battery, rheostat and switch. Close the switch and give the armature a gentle push to start it rotating. 2. Remove the magnets. Close the switch and adjust the rheostat so that a small current passes through the armature winding. Determine the polarity of the armature. Note the position of the commutator segments and brushes. Slowly rotate the armature through 180°. Note the changes in the position of the commutator segments and brushes as you do so. Again determine the polarity of the armature. Replace the magnets. 3. (a) Vary the current by changing the rheostat setting. (b) Vary the magnetic field by changing the distance of the poles from the ends of the armature. What effect do each of these have on the speed of rotation of the armature? 4. (a) Reverse the direction of the current. (b) Reverse the direction of the magnetic field. What effect do each of these have on the direction of rotation of the armature? Observations Describe all the observations. 377 Chap. 31 MAGNETISM AND ELECTRICITY Conclusions 1. Why does the armature of a direct current motor rotate? What energy transformation takes place? 2. What governs the speed of rotation of the armature? 3. What governs the direction of rotation of the armature? Explanation 1. What is the purpose of the commutator in a direct-current motor? Explain its action. 2. Explain why the factors mentioned in conclusions 2 and 3 affect the rotation of the armature. Question How does this motor differ from commercial D.G. motors? EXPERIMENT 36 To determine the cause of an induced current, (Ref. Sec. V:58) Apparatus Galvanometer, two solenoids wound on hollow spools so that one with few turns of wire will slip entirely inside one with many turns, bar magnet, dry cell, switch. 1 mm T (a) Fig. 31:14 Method Connect the larger solenoid to the galvanometer. 1. Thrust the magnet into the centre of the solenoid. Keep it stationary for a few moments and then withdraw it quickly. 2. Connect the terminals of the smaller solenoid through a switch to a dry cell. Close the circuit. Thrust smaller solenoid into the centre 378 EXPERIMENTS ON MAGNETISM AND ELECTRICITY of the larger. withdraw it. Leave it stationary for a few moments and then 3. Place the smaller solenoid (the primary coil) inside the larger one After a few (the secondary coil), and close the primary circuit. moments open the circuit again. Observations Observe the galvanometer needle at each stage of the experiment. Conclusion State the cause of an induced current. EXPERIMENT 37 To determine what factors affect the magnitude of an induced electromotive force (E.M.FJ. (Ref. Sec. V:59) Apparatus Same as for experiment 36, iron core to fit in hollow core of primary coil. Method 1. With the larger solenoid attached to the galvanometer, insert the bar magnet into the hollow core first slowly and then rapidly. Compare the strengths of the induced E.M.F.’s. 2. With the larger solenoid attached to the galvanometer plunge the bar magnet into its hollow core. Then attach the galvanometer to the solenoid with fewer turns of wire and plunge the bar magnet into it at the same speed as before. 3. Attach the larger solenoid to the galvanometer. Connect the smaller to the dry cell to make an electromagnet. solenoid (primary coil) Plunge the electromagnet into the hollow core of the secondary coil. Now insert the iron core into the centre of the primary coil and plunge the primary into the secondary. Observations State what is observed in each part above. Conclusion What factors
affect the magnitude of an induced E.M.F.? Question Why were the induced currents produced at the “make” and “break” of the primary circuit in experiment 36 much greater than those produced by other means? EXPERIMENT 38 To investigate the direction of an induced E.M.F. (Ref. Sec. V:61) (Lenz's Law), 379 Chap. 31 MAGNETISM AND ELECTRICITY Apparatus Solenoid with many turns, bar magnet, galvanometer, dry cell, high resistance. Method 1. Connect the dry cell through the high resistance to the galvanometer. Note in what direction the galvanometer needle is deflected when a current of known direction is passing through it. 2. Connect the solenoid to the galvanometer and note the direction of the needle deflection and thus determine the direction of the current when: (a) The N-pole of the magnet is thrust into the coil. (b) The N-pole is withdrawn. (c) The S-pole is inserted. (d) The S-pole is withdrawn. 3. Using the Helix Law, determine the polarity of the upper end of the solenoid for each part of 2. Observations Direction of Current Across Front of Solenoid Polarity of Upper End of Solenoid Operation N-pole inserted N-pole withdrawn S-pole inserted S-pole withdrawn Conclusions 1. What effect does the magnetic field produced have on the motion that is inducing the current? 2. State Lenz’s Law. Questions 1. Make a series of diagrams to illustrate the above observations. Show the direction of motion of the magnet and the polarity produced. 2. Use Lenz’s Law to explain the production of an induced E.M.F. at the make and break of a primary circuit (Experiment 36, part 3). EXPERIMENT 39 To demonstrate the production of an alternating current and to study the principle of the A.C» and D.C. generator. (Ref. Sec. V:61 to 63) Apparatus Earth inductor (Fig. 28:4), galvanometer, connecting wires, St. Louis motor with both A.C. and D.C. armatures ( Fig. 27:21). 380 EXPERIMENTS ON MAGNETISM AND ELECTRICITY A. Using the Earth Inductor Method Connect the two leads of the earth inductor to the terminals of the galvanometer. Rotate the coil rapidly through 360° and note the action of the galvanometer needle. Observations State what is observed. Explanation 1. What is the cause of the induced E.M.F.? 2. Why does it vary in direction throughout the 360° rotation? Illus- trate your answer by a series of diagrams. Conclusion What is produced by each complete rotation of a coil in a magnetic held? B. Using the St. Louis Motor Model Method 1. Connect the leads from the A.C. armature to the galvanometer. Spin the armature with the huger and note the action of the galvanometer needle. 2. Replace the A.C. armature with the D.C. armature. Spin the armature as before and again note the action of the galvanometer needle. Observations State what occurred. Explanation 1. Answer the questions found in the explanation in part A. 2. How is the induced alternating current carried to an external circuit? 3. How may this induced alternating current be changed to direct current? Name and describe the device used to do this. Conclusions 1. What is a generator? 2. What energy transformation occurs in the operation of a generator? 3. What is the difference between an A.C. and a D.C. generator? Questions 1. How could you produce 60 cycle A.C. with the model used in this experiment? 2. What modihcation in the structure of the apparatus would permit a reduction in the speed of rotation of the armature in question 1? 381 Chap. 31 MAGNETISM AND ELECTRICITY EXPERIMENT 40 To study self-inductance. (Ref. Sec. V:67) Apparatus Coil with many turns wound around a soft iron core, three dry cells, switch, 6-volt lamp, neon lamp and socket. Fig. 31:15 Method 1. Connect the coil in series with the dry cells and switch as shown in the diagram. Join the lamp in parallel with the coil. 2. Close the switch. Note the effect on the lamp. 3. Open the switch. Note the effect on the lamp. Observations Make note of the observations. Explanation With the aid of Lenz’s Law account for the phenomena noted above. Conclusion What is meant by self-inductance? Question What causes the arc when a switch is opened? EXPERIMENT 41 To demonstrate the conductivity of a gas (An introduction to electronics). (Ref. Sec. V:77) Apparatus Gold-leaf electroscope, ebonite rod, cat’s fur, glass rod and silk, Bunsen burner. Method 1. Charge the electroscope positively. Hold a flame near the knob of the electroscope. Note what happens. 2. Charge the electroscope negatively and repeat step 1. 382 EXPERIMENTS ON MAGNETISM AND ELECTRICITY Observations Describe what is observed in each part above. Explanation What effect did the flame have on the particles of gas causing it to become a conductor of electricity? Why were both the positively and negatively charged electroscopes discharged? Conclusion What is electronics? EXPERIMENT 42 To study the changes in the conductivity of a gas as its pressure is reduced. (Ref. Sec. V:77) Apparatus Electrical discharge tube with open side-arm (Fig. 30; 1), vacuum-pump, induction coil, battery. Method Attach the electrodes of the discharge tube to the terminals of the secondaiy of the induction coil. The discharge points of the induction coil should be close together. Turn on the induction coil and evacuate the tube as completely as possible with the vacuum-pump. Note what happens. Observations 1. Where did the electrical discharge occur before and after evacuation of the tube? 2. Describe the discharge in the tube at various stages of evacuation. Explanation Why does the discharge occur in the evacuated tube rather than between the discharge points? Conclusion What effect does reducing the pressure of a gas have on its electrical conductivity? Question. According to the electron theory, what constitutes the current of electricity (a) through a gas at ordinary or reduced pressures (b) through a vacuum? EXPERIMENT 43 To study some properties of cathode rays. (Ref. Sec. V:78) Apparatus Crookes’ tube (Fig. 30:2b), induction coil, dry cells, bar magnet. 383 . Chap. 31 MAGNETISM AND ELECTRICITY Method 1. Set up apparatus as in diagram. Darken the room. Complete the circuit and observe the appearance of the end of the tube remote from the cathode. 2. Erect the target or metal cross in the path of the rays and again observe as in part 1 3. Approach the sides of the tube with first the N-pole and then the S-pole of the bar magnet and observe as above. Observations Describe what is observed. Conclusion List three properties of cathode rays. Question What evidence is provided in part 3 above to prove that cathode rays are streams of electrons? EXPERIMENT 44 To study the production and nature of X-rays. (Ref. Sec. V:79) Apparatus X-ray tube (Fig. 30:3), induction coil, gold-leaf electroscope. Method 1. Join the secondary terminals of the induction coil to the anode and cathode of the X-ray tube. Charge the electroscope and place it with its knob near the target of the tube. Start the induction coil and observe the leaf of the electroscope. 2. Repeat the above procedure placing (a) a piece of cardboard (b) a piece of wood (c) a piece of lead plate between the knob of the charged electroscope and the X-ray tube. Observations State what is observed in each case above. Conclusions 1. How are X-rays produced? 2. Account for the above observations. EXPERIMENT 45 To study thermionic emission of electrons in a vacuum-tube, (Ref. Sec. V:80) Apparatus Diode tube (B), two switches (Kj and Kg), “A” battery, “5” battery, galvanometer (G). 384 — EXPERIMENTS ON MAGNETISM AND ELECTRICITY Method 1. Connect the circuit as shown in the diagram using the plate and filament voltages recommended for the tube available. 2. With Ki open (a) close Kz making the plate positive and watch the galvanometer for any deflection. (b) reverse the terminals of the ‘‘5” battery so the plate is negative and repeat part (a). 3. With Kr closed— (a) close K 2 making the plate positive and note any galvanometer deflection. (b) reverse the terminals of the “B” battery so the plate is negative and repeat part (a). Observations Switch Plate Charge Galvanometer Deflection 2(a) 2(b) 3(a) 3(b) Explanation . 1. Explain the effect on the filament of closing switch Ki. 2. Account for the above observations. Conclusions What two conditions must be fulfilled so that electrons will flow through a vacuum-tube? 385 Chap. 31 MAGNETISM AND ELECTRICITY EXPERIMENT 46 To demonstrate rectification by a diode tube. (Ref. Sec. V:80) Apparatus Diode tube, two 45-volt “B” batteries, “A” battery, neon bulb. Method 1. Connect the two “B” batteries in series and join these to the neon bulb. Reverse the battery connections. 2. Connect the neon bulb to the 110-volt alternating-current outlet. 3. In the circuit shown in experiment 45 replace the “B” battery by the 110- volt alternating-current source, and the galvanometer by the neon bulb. Observations Describe the way in which the neon bulb glows in each step above. Explanation Account for each observation. Conclusions 1. What is meant by rectification? 2. Explain how rectification occurs in the diode tube. 3. Make a diagram to illustrate the direct current produced in the plate circuit on the diode tube in step 3. EXPERIMENT 47 To demonstrate photo-emission of electrons. (Ref. Sec. V;82) Apparatus Zinc plate, gold-leaf electroscope, arc-lamp. Method 1. Attach the knob of the electroscope to the zinc plate by a fine wire. Charge the zinc plate negatively. Allow the beam of light from the arc-lamp to fall upon the charged zinc plate. 2. Repeat part 1 with the zinc plate positively charged. Observations What is observed in each step above? Explanation Account for the observations. Conclusions What is meant by photo-emission of electrons? 386 MODERN DEVELOPMENTS IN PHYSICS This Electron Microscope Permits Viewing of Particles Smaller Than One 10-Millionth of an Inch in Any Diameter. It Provides Magnification 50 Times Greater Than Heretofore Possible— So Great That a Human Hair Would Take on the Dimens
ions of a Subway Tunnel. R.C.A. Victor Company, Ltd. CHAPTER 32 MODERN DEVELOPMENTS IN PHYSICS gets of research have become ordinary articles of commerce. This century has seen the “horseless carriage”, originally a luxury, become a necessity for business and pleasure. The airplane in one generation has undergone an amazing development for peace and war. Electronics has given birth in turn to radio, radar and television. In this chapter, a few modern developThese are ments are to be described. THIS TWENTIETH CENTURY This has been called the age of Physics —a time when apparent luxuries have become commonplace necessities or gad- 389 Chap. 32 MAGNETISM AND ELECTRICITY selected because the preceding chapters serve as a preparation for them and because they are topics that will be of considerable interest to many students. Cathode-Ray Oscilloscope Fig. The principle of the cathode-ray tube (Sec. V;78) is shown in 32:1. Electrons emitted by the hot cathode travel at high speed toward the (c) (a). Many pass through the anode opening in the anode and form the electron beam (b) which passes between the (di d 2 ) two sets of deflecting plates and impinge on the fluorescent screen (i). A spot of light appears on the electron beam the screen wherever strikes. The spot of light is made to travel by applying voltages to the deflecting plates, thus causing the electron beam to sweep from side to side, move up and down, or undergo both motions simultaneously. Each point on the screen struck by the beam continues to fluoresce for a short time after the beam has gone by and does not quite die out before the next passage of the beam revives it. This fact combined with persistence of vision RCA Victor Company, Ltd. Canadian Marconi Co. Fig. 32:2 Commercial Cathode Ray Oscilloscope and Tube. Fig. 32:3 The Image Orthicon. 390 MODERN DEVELOPMENTS IN PHYSICS of the human eye produces the illusion of a continuous line of light on the screen. Thus, it is easy to see that if one voltage is used to make the beam sweep from side to side at a known uniform rate, another voltage applied to the ver- Illus. Courtesy of Canadian Marconi Co. A Picture Tube deal deflection plates will make the spot write the autograph of the latter voltage on the screen, revealing in graphic form frequency, wave form and other its characteristics. The commercial cathode-ray oscilloscope and tube is pictured in Fig. 32:2. It combines in one circuit a cathode-ray tube, a sweep circuit, amplifiers, a power supply, all properly synchronized. This is probably the most widely used instrument in electronics, particularly in the testing and repair field. All kinds of sound—speech, noise, music—can be ana- lyzed by the use of this instrument. Television In radio, electromagnetic waves modulated by the original sound-effects are transmitted through space. In television, a picture controlled by the light from the scene is transmitted at the same time as the sound. The feeble energy from such waves controls electric currents and magnetic fields in such a way as to reproduce the original sound and scene. In the modern television camera, the essential component is an image orthicon. Fig. 32:3, not much bigger than a rolled-up magazine, but very complicated, sensitive and costly. It receives an optical image, /, of the scene on a thin metal coating, M, on the inside are surface tube. ejected from this screen in direct propor- Electrons the of Fluorescent Coating 391 Chap. 32 MAGNETISM AND ELECTRICITY Scene in Television Studio During a Broadcast. Canadian Broadcasting Corp. tion to the brightness of the light that falls on it. These collect on the target screen, T, nearby and produce an elec- Fig. 32:5 The Effect of the Earth's Cur- vature on Television Waves. tron image of the scene. This electron image is scanned by an electron beam, B, from the electron gun, G, in much 392 the same manner as you read a page in this book but much faster (covering 525 lines in 1/3 sec.). The beam is controlled as in the cathode-ray tube. The speed of the electrons in the beam is so regulated that some of the electrons are repelled in a return beam whose intensity varies with the concentration of electrons on the target screen and, consequently, with the brightness of light from the scene. This return beam constitutes a weak current which is amplified, sent to a transmitter where it modulates ultra high frequency carrier waves that are broadcast much as are radio waves. The main part of the receiving aptube, paratus called a kinescope (Fig. 32:4), that is another cathode-ray is MODERN DEVELOPMENTS IN PHYSICS essentially the same as in the cathoderay oscilloscope. The signal received by the antenna is amplified and then is Bell Telephone Company of Canada Fig. 32:6 A Relay Station. applied to the tube in a manner that controls the intensity electron beam shot out by its electron gun. The the of motion of this beam is synchronized with the image orthicon. the scanning of of the beam and Since the intensity hence the brightness of the spot on the screen varies with the signal received, the beam produces a reproduction of the original scene 30 times per second. The illusion of a continuous picture is explained by the rapidity of scanning and our persistence of vision. Television waves travel in a straight path. Each transmitter’s coverage is an area with a radius of about 130 miles on account of the earth’s curvature (Fig. 32:5). Obviously, this presents a problem in telecasting events of national coast. To importance from coast overcome the difficulty a series of relay stations (Fig. 32:6) has been developed which is used both for television and telephone conversations using short waves called microwaves to carry the energy. Each frequency carries several messages. Waves of this kind can be focused and aimed at the next station so that little energy is lost. to The sound system of television is of the type called frequency modulation (F.M.) unlike that of ordinary radio which is amplitude modulation (A.M.). In the latter, the signal modulates (alters (i) Carrier Frequency (ii) Sound Pattern (iii) Amplitude Modulated Carrier Fig. 32:7 (a) Amplitude Modulation. (b) Frequency Modulation. 393 Chap. 32 MAGNETISM AND ELECTRICITY the form of) the amplitude of the carrier wave as in Fig. 32:7a. Such a wave is affected by all kinds of outside elecIn tromagnetic disturbances the former, the signal modulates the frequency of in Fig. 32 : 7b. Such a wave is free from carrier wave as (static). the static. In one method for colour television, the camera has three separate orthicons operating behind red, green and blue filters respectively and giving rise to is the tube picture In the receiver three different signals. each signal puts its own corresponding component of into operation. The end of the picture tube dotted with “phosphors” that will glow red, green or blue, when properly stimulated. When all three phosphors are made to glow by their respective electron beams, the original scene is reproduced in full colour. Radar The word radar is a contraction of the words Radio Detection And Ranging which gives a good idea of what it means. Man has known for centuries that echoes are useful in detecting distant obstacles and determining their position. We know that the bat uses the echo of and icebergs. his high-pitched squeak to enable him to avoid objects as he flies back and forth Mariners have used the in the dark. echo of the ship’s whistle to locate shoreHigh frequency lines radio waves (300 to 3000 megacycles per second) are used for the same purpose in radar. At these frequencies they resemble light waves and can be reflected, refracted and focused. Like light they travel in straight lines. The radar principle, first used in 1924 in exploring the upper regions of the earth’s atmosphere was adapted by Sir Robert Watson Watt for the location of aircraft, ships and shorelines in World War II. It was this device which enabled those few valiant defenders of the British Isles (and of the freedom of the world) during the Battle those remarkable of words of Sir Winston Churchill when he . was so much owed by said, “Never . so many to so few”. to earn Britain . In the chapters on sound, we learned that the distance from an observer to an object may be computed if we know the time required for sound to travel to it and for the echo to return (Sec. 11:7). Similarly, the distance of an aircraft may be determined by sending out a pulse of high-frequency radio waves, P, Screen of Cathode Ray Tube Fig. 32:8 (a) Using Radar in Aircraft Detection, (b) View of Screen of Receiver. 394 MODERN DEVELOPMENTS IN PHYSICS Fig. 32:8a, from the transmitter, 7’, and detecting the reflected impulse, E, by a receiver, R. If the time interval required is determined and the velocity of radio is known, waves (186,000 mi. per sec.) the distance may be determined. spot The receiver contains a cathode-ray is made to tube in which the traverse the screen at a constant known speed. The outgoing and echo signals generate voltages which are applied to the vertical plates and produce upward “blips” on the (Fig. or The distance between these 32:8b). blips enables the operator to calculate trace jolts the time interval. To specify the position of an aircraft we must know the bearing and the elevation. The former is detected by comparthe echo signals ing the strength of from two sending aerials set at right angles. The elevation can be determined Canadian Aviation This Radar Screen Has Picked Up a Typhoon. with the aid of two aerials each above the other. To recognize friendly aircraft the radar pulse triggers a special transmitter and consequently the return pulse will be given a special mark which is distinguished on the cathode-ray tube. Radar has numerous peace-time apIt is used to follow the flight plications. of aircraft flying over lonely or thinlypopulated areas like the far north. Aerial and water transport employ it at night and in fog
to avoid striking obstacles. Highway traffic authorities find it valuable in tracking down speeding drivers. T he Electron Microscope It can be shown that the ability of a microscope to reveal the fine detail of an object (the resolving power) is limited by the wave-length of the light used. Consequently, the practical upper limit of magnifying power that can be obtained with optical microscopes—using visible light— is about 2000 diameters. Now it is known that an electron stream behaves in a manner very similar to a light beam but with an effective wave-length some 10,000 times smaller, permitting examination of bacteria, viruses and very large molecules. Accordingly scientists, especially Dr. F. Burton and his associates at the University of Toronto in the 1930’s explored the possibility of using electrons in microscopy. As a result of their researches the electron microscope, shown diagrammatically in Fig. 32:9, was developed. The electrons are produced by a heated cathode and are accelerated under a potential of some 50,000 volts applied to an anode with a central hole from which they emerge in a stream of swiftly moving electrons. The electron stream is converged and focused by magnetic and electric fields much as light is by glass lenses in an optical microscope. These “magnetic lenses” consist of short solenoids entirely encased in a steel shield except for a narrow gap in the centre through which the electrons pass and where they become deflected. The arrangement of lenses is similar to that in the optical microscope but the object to be studied is very much thinner (1/100,- 395 ) . Chap. 32 MAGNETISM AND ELECTRICITY ELECTRON MICROSCOPE OPTICAL MICROSCOPE ^SOURCE OF ILLUMINATION (Electrons) (Light) CONDENSER LENS (Magnetic) (Glass) SPECIMEN STAGE OBJEaiVE LENS (Magnetic) (Glass) PROJECTOR LENS (Magnetic) (Glass) IMAGE (Viewing Screen) (Eye Piece) — Fig. 32:9 A Comparison of Electron and Optical Microscopes. R.C.A. Victor Company Ltd. 000 cm. thick). The instrument must be evacuated to prevent deflection of the electron stream as it collides with air molecules. The varying penetrations of the electrons through the object produce a sort of shadow picture on the fluorescent viewing screen or photographic film. Magnifications of up to 100,000 diameters have been obtained. The electron microscope has proved to be a very valuable tool in the hands of the research worker. It has revealed new facts about particle shapes and sizes which affect such processes as the covering power of paint pigments, the wear- ing quality of rubber tires and the operation of chemical catalysts. It has Comparison of Some Magnifiers Object Diameter (Microns Magnification (Diameters) Fine Machine Work 25-100 Pond Life Fungi 10-25 Bacteria Structure of Bacteria Large Virus Colloidal Particles Small Viruses 1-2 0.25 0.10 0.05 0.01 Instrument Hand lens Low power optical Medium power optical 8 20 200 1000. High power optical 2000. Special optical microscope or electron microscope 4000. Electron microscope 20,000 Electron microscope Large Molecules 0.002 100,000 Electron microscope 396 MODERN DEVELOPMENTS IN PHYSICS salts themselves. erated by the This phenomenon was called radioactivity and was extensively investigated by Pierre and Marie Curie. Using pitch-blende, a natural ore rich in uranium compounds, they succeeded, after a long series of repeated crystallizations, in isolating two new elements, polonium and radium, which than radioactive more were y- Rays provided new information in such fields as metallurgy, chemistiy, ceramics, crystallography, plastics, textiles, biology and medicine. NUCLEAR PHYSICS defined. protons, orbits — were In Section V:12, the composition of the atom was presented without any There the comsupporting evidence. ponents — electrons, neutrons, This nucleus, knowledge of the atom came about by a process of gradual evolution. First was the discovery of the electron by J. Thomson as a result of his Sir J. (Sec. V;78). work on cathode rays studies on radioactivity contriLater, buted much to knowledge about the nucleus of the atom. Radioactivity Fig. 32:10 Radium Emanations. In 1896 the French physicist Becquerel noticed that salts of uranium affected a neighbouring photographic plate. He concluded that these salts must be emitting rays or particles spontaneously gen- uranium. Later work by the Curies and others has revealed, indeed, that quite a number of heavier elements are radioactive. the Fig. 32:11 The Uranium Disintegration Series. 397 Chap. 32 MAGNETISM AND ELECTRICITY in In the Fig. into that rays, 1903 three undertook the opposite Rutherford a careful examination of the nature of the radiation from radioactive substances. He showed, by applying a magnetic radiation field, (/?), and gamma (y) 32:10, could be resolved distinct groups of rays which he called alpha (a), beta the a-rays being deflected in one direction, the /?-rays direction, while the y-rays do not undergo deFurther experiments showed flection. that the a-rays were positively charged helium nuclei, the /?-rays, negative electrons, and that the y-rays were highly penetrating, short wave-length, energy waves. Rutherford came to the conclusion that the nuclei of radioactive substances are spontaneously disintegrating into the above rays and changing into new elements as they do so. Fig. 32:11 shows the sequence of changes through which uranium goes changes into lead. as it gradually The Structure of the Atom Rutherford found that a-particles a sheet positively would pass through a thin of metal, and concluded that there was considerable empty space in the atom. His concept of the atom was a miniature solar system consisting of electrons revolving about charged Later, while at McGill Uninucleus. versity, he was able to isolate the hydrogen ion or proton. Moseley’s experiments, in England, with X-ray phenomena on various elements, indicated the presence of these protons in every nucleus and a different number in the nuclei of different elements. The number of protons in the nucleus of the atom is the atomic number or nuclear charge. Since the atom is electrically neutral, the Fig. 32:12 Structure of Some Atoms. (MN = Mass Number; AN = Atomic Number) 398 MODERN DEVELOPMENTS IN PHYSICS Fig. 32:13 Isotopes of Hydrogen. Tritium MN = 3 AN = 1 number of electrons revolving about the nucleus must equal the number of protons in the nucleus, and hence must also equal the atomic number. of the This Rutherford’s atom served until the discovery by Sir James Chadwick in 1932 neutron. particle (which was earlier predicted by Rutherford) has the same mass as the proton It is now bebut possesses no charge. lieved that atomic nuclei are built up of protons and neutrons. The sum of the protons plus the neutrons make up what is called the mass number of the atom. In general the number of neutrons the mass in number minus the nuclear charge or atomic number. For example, the helium nucleus which has an atomic number of 2 and a mass number of 4 must contain 2 protons and 2 neutrons. the nucleus is equal to Finally, to complete the picture of the atom, Niels Bohr postulated that the electrons revolved in various orbits, or energy levels, about the nucleus. The Bohr Concept for a number of typical atoms is shown in Fig. 32:12. Isotopes About this time, it was discovered by Rutherford and Aston in England that although all atoms of an element have the same chemical properties some differed from others in mass. Such forms of any element are called isotopes. They postulated that all the isotopes of any one element must have the same number of protons and electrons and hence the same atomic number, but must differ in the number of neutrons in the nucleus and hence must differ in mass num- (Fig. 32:13). ber. Most elements consist of a number For example, hydrogen is of isotopes. thought to consist of a mixture of three Protium is the isotopes commonest isotope of hydrogen. Deuterium is the isotope of hydrogen that is found in heavy water. The nucleus of called is much used in experimental work in nuclear physics. Tritium is a very rare isotope of hydrogen that is of importance in the hydrogen bomb. deuterium, deuteron, a Artificial Transmutation The above interpretation of natural radioactivity suggested to Rutherford the possibility of bringing about the disinteThis was acgration of other nuclei. complished by Rutherford in 1919 while at McGill by using a-particles from a radioactive source to bombard nitrogen atoms. As a result of the collisions he found that some of the nitrogen atoms had a proton ejected from them, the a-particle remaining with the disrupted These atoms therefore had a nucleus. net increase of 1 proton and 2 neutrons and hence became oxygen atoms (actually a heavy isotope of oxygen) : 14 4 1 17 7 N 2 18 He-^P + O Thus was achieved the first transmutation of the elements—the old alchemists’ dream becoming a reality! artificial Rutherford next considered the use of artificially energized particles as a more effective means of bringing about nuclear changes, using protons accelerated under a potential difference of about half a million volts. These high-speed protons were directed on to a lithium target. 399 Chap. 32 MAGNETISM AND ELECTRICITY some of the atoms of which were split up to form helium atoms: l/+ P^-^He + He 3 12 2 Other light elements were disintegrated by this means, but it was realized that even more energetic particles were required to attack the nuclei of heavier elements on account of the large forces of repulsion between the bombarding particles and the larger nuclear positive Particles having the necessary charge. energy for this purpose can be produced by means of the cyclotron (Fig. 32:14). This latter machine, first invented by the American physicists Lawrence and Livingston in 1930, is capable of rapidly accelerating charged particles such as protons and deuterons. Nuclear Fission The discovery of the neutron provided physicists
with a most effective projectile with which to bombard atomic nuclei 400 because, having no charge, a neutron can approach the positive nucleus of an atom without any repulsive force. A common way of producing neutrons for to bombard beryllium this purpose is metal by a beam of fast-moving deu- Slow Neutron AN 92 MN 235 Beta Particles Uranium Excess Mass Changed into Energy Fig. 32:15 Nuclear Fission. terons from a cyclotron—neutrons being knocked out of the beryllium nuclei. It was while bombarding uranium 235 (a light unstable isotope of uranium) with neutrons in 1939, that Hahn and Strass- MODERN DEVELOPMENTS IN PHYSICS mann discovered an entirely new foiTn of nuclear transformation, Nuclear Fission, in which the uranium nucleus was split into two nearly equal parts with the release of a large amount of energy (Fig. 32:15). Another interesting feature of the fission of uranium is that during the process further neutrons are emitted in relatively larger numbers than those absorbed. Thus from each uranium nucleus undergoing fission two or more neutrons are released. The neutrons so released can then cause fission in other uranium nuclei and so on causIf allowed to ing a chain reaction. proceed in this way, provided that there is a suitable amount of fissionable matter present, the whole of the material is transformed in a very short time, with the release of vast quantities of energy. This is the principle of the atomic bomb. It has been proven that the combined mass of the products is a little less than the mass of original material. Dr. Albert Einstein as early as 1905 using his famous equation, E =; MC^, predicted what we now know—that the mass lost becomes transformed into the energy of the explosion. E represents the energy, M the loss in mass and C, a constant quantity equal to the speed of light (Sec. IV: 6). Thus we can see that a small amount of mass is transformed into a tremendous amount of energy. Nuclear Reactors If the neutrons from the fission of U 235 are absorbed or slowed down as they are produced, the chain reaction can be controlled and the energy released put to useful purposes. Graphite and certain other light substances which possess the property of absorbing neutrons, are used to control the rate of neutron emission in atomic fission and are called moderators. The assembly of uranium (or other fissile material) and graphite (in the form of blocks or rods) is known as an atomic pile or nuclear reactor (page 402), the rate of energy release being controlled by adjusting the length of the graphite rods inserted in the pile. The energy released is transfomied into heat which is used for various purposes. Energy is variously measured. The engineer may express it as B.T.U. per pound (Sec. 111:20). The nuclear physicist may express it in electron volts per atom. One B.T.U. equals 665 X 10^^ million electron volts. estimated 13,600 B.T.U. per pound, this is 4 electron volts per atom of carbon. By comparison one atom of uranium produces 200 million This means that one electron atom of uranium produces 50 million times as much energy as one atom of carbon or if you have equal masses of coal and uranium the amount of energy released from the uranium is 3 million times as great. There is little wonder, therefore, that the search for uranium goes on at a feverish pace. gives volts. that, coal It is if Atomic Fusion to In 1951 when man was congratulating himself on his conquest of the atom through turning fission his advana new and still more exciting tage, process was evolved by Dr. Edward Teller in the United States. You will recall Hans Bethe’s theory of the origin of the sun’s continuous heat by the union of four atoms of hydrogen to form an atom of helium accompanied by the conversion of a small amount of mass into energy (Sec. Dr. Teller employed this process using the isotopes of hydrogen—protium, deuterium and tritium—resulting in the production of a source of energy far greater than ever The hydrogen bomb realized was created which was fantastically devastating. A source of energy was born that was so amazing that it may well be the source of energy of the future. Ill: 3). before. 401 Chap. 32 MAGNETISM AND ELECTRICITY Atomic Knergy of Canada Ltd. Model of N.R.X. Reactor, Chalk River, Ontario. star Mewspaper Service U.S.S. Nautilus. The First Atomic Powered Submarine. 402 MODERN DEVELOPMENTS IN PHYSICS COOLING WATER TRAVELS Space does not permit dealing at any length with the many and varied uses of atomic energy. Suffice it to say that biological, medical, agricultural and industrial research are greatly benefiting from the developments made in this field as are some users of industrial and domestic power. are as yet beyond man's fondest dreams. The possibilities 403 ANSWERS Unit I—Mechanics A Chapter 2, Section I : 4—Page 16 4 (a) 1000 mm., 100 c.m., 10 dm., 1000 m. 10.000 dg., 1000 gm. 1, 3. 11.46 (ii) 8.3. 9 (b) 5, 5, 5, 5, 5 (b) 1,000,000 mg., 100,000 eg., 10 (b) 37., 34., 37. 11 (b) (i) B (a) A A B A 1520, 3.8, 605.4 cm.; (b) 1 2.3, 6,000,050.465 c.c. (c) 100 m. (i) 28.1 ft. (hi) 64.4 km. per hr. (ii) 22.5 cu. dm. (hi) 22.5 litres 225.000 dg., 22.5 kg.; (b) (i) 28.4 gm. (ii) 909. kg. (hi) 5.4 pints 102.70; (b) (ii) 0.35 (hi) 27 X 101; (j) (i) 14 x 102 (h) 0.50 (hi) 9.0 Chapter 3, Section I : 9—Page 22 30,000, 2.36, 60,005.44 sq. cm.; (c) 2,500,000, 2 (a) 37.4, 15.7, 39.4 in.; (b) (i) 30.5 cm. (ii) 1.61 km.; (ii) 856 cm.; (d) (i) 58.7 R. per sec. (ii) 17.9 m. per sec. 3 (a) (i) 1500 sq. cm. (ii) 15. sq. dm.; (b) (i) 22,500 c.c. 4 (a) 22,500 gm., 22,500,000 mg., 2,250,000 eg., 5 (i) 3.1 litres (ii) 3080 ml. 36. (i) 66.56 (ii) 71.25 (hi) 67.17; (c) 6 (a) (i) 1 (b) 2.70 gm. per c.c. 3 (b) 8.0. B 1 5 gm. per c.c. lb. 6 20.3 cu. ft. cu. ft. 16 72.8 gm. or 6. gm. per c.c. 24 1.18 gm. per c.c. 2 52.5 or 53. gm. 3 268 c.c. 4 62.5 or 63. lb. per cu. ft. 7 4.87 12 1781.3 or 18 X lO^ lb. 17 8.5 gm. per c.c. 8 54.1 c.c. 9 800 or 8 X 10^ gm. 13 7.64 gm. per c.c. 18 (a) 19.5% (b) 13.4% 5 64. 10 0.068 11 0.59 15 0.80 14 0.8722 20 5.9 19 0.42 c.c. 23 0.9 gm. per c.c. 21 9.9 gm. per c.c. 22 8.7 gm. per c.c. Chapter 4, Section I : 15—Page 32 1 (b) (i) 100 gm. (ii) 790 gm. 3 (c) (i) 2.6 gm. (ii) 4.3 c.c B 5 (i) 7.14 (ii) 70 c.c 2 (a) 68 gm.; (b) 625.0 lb. 4 10.5 1 (a) 30 gm.; (b) 162.5 lb. cu. ft. per cu. ft. 10 8.90, 0.840 or 1.6 X 102 lb. 20 18.8 or 19. cm. 26 14. cm. 25 1/3 11 32.0 gm. 7 (i) 105.9 gm. (ii) 8.30 gm. per c.c. 12 No 13 120 gm. 17 11.7 c.c. 22 100 gm. 16 36.6 or 37. gm. 21 20.4 or 20 cm. 3 (a) 25 ml. (b) 0.40 6 (i) 2.5 (ii) 0.096 cu. ft. (hi) 156 lb. 9 (i) 5.36 (ii) 0.73 15 156. 19 1.3 24 0.21 8 1.30 14 0.9 gm. per c.c. 18 0.90 gm. per c.c. 23 187.5 or 188. c.c. Unit II—Sound A Chapter 6, Section II : 9—Page 61 5(b) 1 100 ft. per sec. 9 1 150 ft. per sec. 10(a) 1 100 ft. per sec. 1 (a) 3.75 cm.; (b) 15 cm. (b) 288 m. per sec. 6 (a) 1099, 1055, 1135 ft. per sec.; (b) 335, 321.8, 345.8 m. per sec. (b) 6.7°C. 12 (a) 56.5 ft. 2 .67, .67 sec. 4 (a) 320 v.p.s.; (b) 667 v.p.s. 3 (a) 1170 or 11.7 X 10^ ft. per sec.; 5 (a) 1.50 ft.; (b) 56.0 cm. 7 (a) 15°C.; 11 (a) 357 m.persec.; (b) 42°C. 14 750 m.p.h.’ 10 259 v.p.s. 13 22.6 sec. 8 5595 ft. 9 1.33 m. Chapter 7, Section II : 18—Page 71 6 (b) (i) 750 v.p.s., (ii) 100 v.p.s.; (c) (i) 208.3 or 208 v.p.s. (ii) 50 v.p.s. 404 ANSWERS B (a) 1 4 150 teeth 8 25, 80, 35 cm. v.p.s. 13 160 v.p.s. 960, 1920, 7680 v.p.s.; (b) 240, 60, 15 v.p.s. 5 256, 320, 384, 512 v.p.s. 9 1080, 270 v.p.s. 2 240 v.p.s. 6 900, 225, 675 v.p.s. 10 312 v.p.s. 11 400, 2500 gm. 3 320 r.p.m. 7 587 v.p.s. 12 384 A Ch.\pter 8, Section II : 25— Page 79 2 (a) (i) 48 in. (ii) 1 200 ft. per sec.; (b) 24 in. B 1 (a) .302 m. (i) 5 ft. (ii) 5 15.5°C. A 7 1781 ft. Unit III—Heat A 4 4 X 10-3 lb. ft.; (b) 1100 ft. per sec. 1 6 33.3°C. 7 18.8 cm. 2 1:2 8 4 beats per sec. 3 320 v.p.s. 4 11.9 in., 9 486 or 474 v.p.s. Chapter 9, Section II : 34—Page 93 Chapter 11, Section III : 4—Page 114 Chapter 12, Section III : 11— Page 124 2 37.8, 176.7, —140, — 45.6°C. 3 (a) —40°; (b) 4 (a) (i) 330, 250° K. (ii) 25, -36° C.; (b) (i) 309.7, 255.2, °K. (ii) 212, B 1 59, 392, —76, — 459.4°F. 320° — 459.4°F. Chapter 14, Section III : 32—Page 160 2 1600, 8400 cal. (i) A (b) B 350. cal., (ii) 5328 B.T.U., (iii) 179.4 cal. 4 (b) 0.087. 7 4000, 9 (a) 27,000 cal; (b) 8100 cal.; (c) 21,700 cal. 1 100,000 or 10 X 104 4 6000 or 60 X 10“ B.T.U. cal. 8 40°C. gm. 9 0.104 14 204.3 or 204 gm. 18 187.5 or 188 gm. 24 27 gm. 28 534 cal. 1.2 X 102 25 46.6 or 47°C. 29 40°C. 30 53°C. 10 0.0933. 15 12.8 or 13°C. 19 208.3 gm. 2 39,000 or 39 X lO^ B.T.U. 5 1320 or 13.2 X 10^ cal. 11 0.21 12 0.031 16 0.44 20 80 cal. 26 530 or 5.3 X 10^ cal. 31 3640 cal. 6 101 cal. 3 91,000 or 91 X lO^ 7 56.3 13 292.6 or 293 gm, 17 1250 or 12.5 X 10^ gm. 22 82 cal. 23 0.50 27 541 or 5.4 X 10^ cal. 33 121 or 32 0.112 or 0.11 21 78 cal. Unit IV—Light A Chapter 16, Section IV : 8—Page 182 5 (d) 2 in. 7 (b) 8.3 min. B 1 30 ft. 2 0.8 in. 3 148 or 1.5 X 10^ m. 4 1.3 sec. 5 5.87 X lO^^ miles, 9.46 X 1042 km. 6 52.8 X 1042 7 4 3 yg^rs. Chapter 17, Section IV : 20—Page 194 A. 5 (a) 7. 405 A 2 (c) dioptres. B I 1.20 B (i) 10 o’clock (ii) 5.45 o’clock (v) — 60 cm. 5 — 9.23 in. 11 — 4.5 in. 12— 8.6 cm. 1 oc 13 20 in. 14 2.2 in. Chapter 18, Section IV ; 34— Page 211 ANSWERS 4 (a) 6 3 ft. (i) 50 cm. (ii) 60 cm. (iii) 90 cm. (iv) 7 4 in. 10 36 cm. 8 4.3 ft. 9 3 in. 124,000 miles per sec.; (d) 2.47 5 (b) glass 7 (b) water 9 (d) 6.7 2 479 X 102 3 (i) 35° (ii) 28° (iii) 20° per sec. ZD = 37° 5 One with critical angle of 30° 7 — 9.2 in. (v) — 60 cm. (iii) 90 cm. (iv) II 32 in. 13 6 in. (i) 0.9 in. (ii) 11 X 18 (a) 5.5 in.; (b) 36 in. x 60 in. cc 12 24 cm. 6 (a) 8 5 cm. 14 24 cm. 4 Zi = 48.5° (i) 50 cm. (ii) 60 cm. 10 80 X 17 9 9 metres 15 —20 cm. 16 140, 12.7 cm. Chapter 19, Section IV : 43— Page 225 1 (c) 7.3 X 1014 ^ p 3 (a) 3.1 in.; (b) 1.5 in. 7 (b) 160 X 9 (d) 192 cm. Unit V—Magnetism and Electricity Chapter 20, Section IV : 50—Page 235 6 (c) (i) 8.3 X (ii) 6.7 X (d)
(i) 25 mm. (ii) 20 mm. Chapter 25, Section V : 39— Page 293 A (b) 22 volts 4 (c) 1 ohms (ii) 5.5, 2.8 amp. (i) 90 ohms (ii) 6 (b) 14 ohms. 1.2 amp. (iii) 60, 48 volts; (d) (i) 13 B 2 212.5 or 21 X 10 volts 1 12 ohms 4 (a) 0.12 amp.; (b) 11.9 or 12 volts and 0.06 volts 5 (a) 22 ohms; (b) 5.5 ohms 6 (a) 41.3 ohms; (b) 0.333 amp.; (c) 2.66 amps. 7 (b) 6 volts; (c) 0.9 ohm 8 16.7 or 17 ohms. 9 (a) 3.4 ohms; (b) 3.3 amp.; (c) Chapter 26, Section V : 47—-Page 302 11 (a) 2.4, 1.6 amp.; (b) 18 volts. 1.7 amp. 10 0.2 amp. 3 3.3 amp. B (a) 3.55 gm.; (b) 12.1 gm. 4 0.003 amp. 8 0.002 cm. 5 1.26 amp. 9 3.95, 0.37 gm. Chapter 27, Section V : 56—^Page 315 6 0.02 amp. 10(a) 2 amp.; (b) 15 hr. 2 (a) 9.5 gm.; (b) 32 gm. 3 (a) 2.4 gm. (b) 0.30 7 (a) 0.18 amp.; (b) 0.0028 1 gm. gm B 1 (a) 0.08 ohm; (b) 0.008 ohm; (c) 0.0008 ohm, 20.002 ohms 3 (a) 115 ohms; (b) 1240 ohms; (c) 12490 ohms 4 9998 ohms. Chapter 28, Section V : 69—Page 329 B 4 220 volts 5 400 turns 6 (i) 96 turns 1 80 turns 2 2200 volts (ii) 64 turns (iii) 48 turns (iv) 32 turns. 3 9.2 volts B Chapter 29, Section V ; 76—Page 338 1 (a) 24 X 10® volts; (b) lZ2000th; (c) greatly diminished 50 times (ii) 2500 times (ii) 372 watts (iii) h.p. 4 5 amp. 7 1650 watts 3 1650 watts 0.372 kw. 8 No. cents 11 9.7 cents 12 $1.12 13 6 cents, 406 5 220 volts 2 (a) 50:1; (b) (i) 6 (i) 0.498 10 4.3 9 6 cents INDEX Aberration, chromatic, 210, 224; spherical, 193 Abnormal expansion of water, 124 Absolute, temperature, 123; zero, 123, 132 Absorption of radiant energy, 133, 142 {Exp. 168) _ Absorption spectra, 219 Accommodation, lens, 229 Accuracy, degree of, 11; limits of, 13 Achromatic lens, 224, 233 Acid, 295, 299 Acoustics of buildings, 87-89 Action of points, 276 {Exp. 363) Addition, 14 Additive theory of colour, 222 Air, lens, 205, 206; liquefaction of, 155; pres- sure effect on the boiling-point, 148 Airships, 31 Alcohol, 295 Alcoholometer, 29 Alkali, base, 299; metals, 347 Alloys, 291; densities of, 21 Alnico magnet, 266 Alpha rays, 398 Alternating current, 322; 25-cycle, 322; 60- cycle, 322 Alternating current generator, 321 {Exp. 380) Amalgamate with mercury, 281 Amber, 268 Ammeter, 311, 312 Ammonia gas, 152, 154 Amperes, 281, 288, 298 Ampere-hours, 302 Amplihcation, 346 Amplifier, 346 Amplitude, modulation, 393; of vibration, 50, 64, 76, 78; of wave, 52 Angle, critical, 200; of declination, 262; of deviation, 200; of incidence, 184; of inclination, 263; of reflection, 184; of refraction, 196, 198; visual, 231 Angstrom, 2l4 Anions, 295 Annular eclipse, 179 Anode, 295 Antenna, 285, 347, 393 Anthracene, 218 Anti-cathode, 343 Anti-freeze hydrometer, 29 Aperture, 227 Approximate numbers, 13; rules for using, 14 Aquastat, 119 Aqueous humour, 228 Arc, electric, 326, 336, 382; furnace, 337; lamp, 336, 386; welding, 336, 337 Archimedes, 1, 3, 24, 25 Archimedes’ principle, 25 {Exp. 39, 40); ap- plications of, 29-31 Area, measurement of, 10; of conductor and resistance, 288 Argon, 335 Aristotle, 1, 3, 49, 175 Armature, 321, 323, 377 Arrangement of cells, 283 Arrhenius, Svante, 295 Artificial, magnets, 260; transmutation of ele- ments, 399 Astigmatism, 230, 231 Astronomical telescope, 233, 234 {Exp. 254) Atmosphere, pressure of, 120, 148, 150 Atmospheric refraction, 202 Atom, 112, 132, 397; structure of, 269, 270, 398 ; Bohr concept, 399 Atomic, bomb, 401; energy, 3, 401; fusion, 401; number, 398; pile, 112, 331, 401 Atropine, 228 Attraction, electrical, 268 {Exp. 358) ; mag- netic, 260 {Exp. 352) Audibility, limits of, 65 Audio-frequency, 347 Auditoria, acoustics in, 59, 77, 88-89 Auditory nerve, 49, 82 Automobile, engine, 159; cooling system, 145; generator cut out relay, 307, 309; ignition system, 328 Axis, principal, 189, 204; secondary, 189 Bacon, Roger, 3, 226 Balance, 12, 34, 35 Balance wheel of watch, 116 Ball and ring apparatus, 115 {Exp. 164) Balloons, 30 Bar magnet, 260 {Exp. 350, 351) Base (Alkali), 295 Battery, 29, 279, 284; hydrometer, 29; rat- ing, 302 Beam, of light, 176; wireless, 326 Beats, 78, 79 {Exp. 106) Becquerel, Henri, 397 Bel, 64 Bell, Alexander Graham, 325 Bell-in-vacuo, 51 {Exp. 96) Bell-jar, 51 Bellows, camera, 226 Beryllium, 400 Beta ray, 398 Bethe, Dr. Hans, 113 Bifocal lenses, 231 Bimetallic strip, 115 Binary colours, 221 Binoculars, 201 Biot’s spheres, 276 {Exp. 362) Black, 221 Blind spot, 229 Blip, radar, 394 407 1 ; INDEX Bohr, Niels, 399; concept of the atom, 399 Boiling-point, 148; effect of air pressure, 148; table of, 152 Bomb calorimeter, 144 Break and make, 379 Breaking the sound barrier, 60, 61 Breezes, land and sea, 130, 131 Bridges, expansion in, 116 Brilliancy, 202 British system of measurement, 9 British thermal unit, 139, 401 Brushes, electric, 314, 321, 323 Bucket and cylinder, 39 Bunsen burner, 163, 164 Buoyancy, 24 Buoyant force, 24 Burette, 12 Burton, Dr. F., 395 Caesium, 219, 347 Calculations with approximate numbers, 13, 14, 15 Calibrating a thermometer, 120, 124 Calipers, 12 Caloric theory, 109 Calorie, 138; food, 139 Calorihc values of fuels, 144 Calorimeter, 142; bomb, 142, 144 Camera, lens, 226, 227; pin-hole, 176-178 Camouflage, 223 Can, overflow, 40, 43 Canadian National Exhibition Band Shell, 58 Carbon dioxide, 156 Carbon resistors, 291; rods, 297; filament, 334 Carrier waves, 347, 393 Catch bucket, 40, 43 Cathode, 295 Cathode rays, 341, 397 {Exp. 383) Cathode ray oscilloscope, 390; tube, 342, 389- 393 Cations, 295 Cat’s fur, 268 Cell, arrangement of, 283; dry, 279; electrolytic, 295, 296, 367, 369; primary, 300; secondary, 300; storage, 300, 372; voltaic, 279, 364 Celsius, 122; scale, 121 Centigrade scale, 121 Centigram, 10; centimetre, 9, 10 Centre of curvature, mirrors, 189; lens, 204 C. G. S. system of measurement, 9 Chadwick, Sir James, 399 Chain reaction, 401 Changes of state, 146 Characteristics of musical sounds, 63 Charges, electric, kinds of, 268, 269; distri- bution of, 276 Charging, by induction, 273 {Exp. 362) ; by contact, 271 {Exp. 359, 360) Churchill, Sir Winston, 394 Ciliary muscles, 229 Circuit, 281 ; breaker, 284, 327 Climate, 146, 150 Clock, 50, 117 Closed tube, 73, 74 {Exp. 102) Clouds, screening effect of, 135 Coal, 112-113, 156, 401 Coaxial cable, 326 Cobalt, 259 Coefficient of expansion, of gases, 123; of liquids, 119; of solids, 116 Cohesion, force of, 146 Coil, 305 Coil spring, 54 Cold frames, 135 Cold light, 176, 218 Collimator tube, 219 Colour, blindness, 223; chart, 221; disc, 215; filters, 221; pigments, 221, 223 {Exp. 253); importance of, 213, 223; nature of, 220222; printing, 223, 224; television, 394; theories of, 220, 222; uses of, 223; vision, 223 Coloured lights, 222 Commutator, 314, 322, 323, 377 Compass, magnetic, 261 Compensating for expansion, 116 Complementary colours, 221 Compound bar, 115, 165 Compound microscope, 232, 396 Compression of gases, ll2 Concave lens, 203; images in, 208 {Exp. 250) focal length, 205 {Exp. 248) Concave mirror, 188; images in, 190 {Exp. 241, 242) Concave reflector, 58, 235 Condensation, 54, 74, 75, 79, 325 Condenser, 328 Condensing lens, 233, 235 Conductance, 289 Conduction of heat, 126, 156; in gases, 128; in liquids, 128; in solids, 126, 134 {Exp. 166) Conductometer, 126, 166 Conductors, electric, 272, 288 {Exp. 361) ; dif- ferent shapes, 276 {Exp. 363) Conductors, of electricity, 272; of heat, 127 Cones, for vision, 229 Conservation of energy, 1 1 Constantan, 291 Continuous spectra, 219 Convection currents, 128-131 ; in liquids, 128130 {Exp. 166); in gases, 130, 135, 142 {Exp. 167) Convention of signs, 192, 205, 208 Converging (Convex) lens, 203; images in, 207 {Exp. 250) ; focal length, 205 {Exp. 248) Chemical, change, 112; compounds, 295; effects of electric current, 295; energy, 300 Converging pencil, 176 Convex mirror, 188; images in, 191 {Exp. Chemistry and colour, 220 Chlorine, 154 Choroid coat, 228 Chromatic, aberration, 210, 224; scale, 84 Chromium plating, 299 243) Cooling system of automobile, 145 Copper, 288; ions, 297; plate, 279; plating, 299 Copper sulphate, electrolysis, 297 {Exp. 370) 408 INDEX Copper voltameter, 298 {Exp. 370) Cornea, 228 Corpuscular theory- of light, 181 Corrosion, 299 Cosmic rays, 132 Coulombs, 281 Crest, 52 Critical, angle, 200; pressure, 155; tempera- ture, 155 Crookes, Sir William, 341 Crookes’ tube, 383; dark space, 341 Crystalline lens, 228 Cubic measure, 10 Curie, Pierre and Marie, 397 Current, see electric alternating, direct, convection, Curved mirrors, 188 Cut glass, 202 Cut-out relay, automobile generator, 307-309 Cycle, 50; alternating current, 322 Cyclotron, 400 Cylinder, volume of, 36; automobile, 328 D’Arsonval galvanometer, 310, 377 Davy, Sir Humphrey, 110, 156, 295, 317, 336 Decibel, 64, 92 Decimetre, 9 Declination, angle of, 262 Decomposition of water, 295, 296 Defects, of vision, 230 Degrees, 120; absolute (Kelvin), 123; centi- grade (Celsius), 121; Fahrenheit, 121 Demagnetization, 265 Density, 15; definition, 18 {Exp. 35, 36, 37); effect of expansion on, 124; in relation to convection, 129; maximum density of water, 19, 20; table of, 21 Depolarizing agents, 280 Depth, apparent change in {Exp. 246) Detector circuit, 347 Deuterium, 399 Deuterons, 399 Deviation, angle of, 200 {Exp. 247) Dewar flask, 156 Dial, telephone, 326; thermometer, 117 Diamonds, 199, 202 Diaphragm, camera, 226, 227; telephone, 325 Diatonic scale, 82, 83 Diesel engine, 156, 159, 160 Difference of potential, 280-282 Differential thermometer, 167 Diffusion of light, 184, 185 Dilatometer, 19, 20 Diode tube, 344, 345; as a rectifier, 345 {Exp. 386) ; thermionic emission, 344 {Exp. 384) Dioptre, 205 Dip, magnetic, 263 Dipping needle, 263 Direct current, 346; generator 323 380); motor, 313, 314 {Exp. 377) {Exp. Direction of electron flow, 374 Discharge, electrical, 341 ; through gases, 340, 341; tubes, 219 Discontinuous spectra, 219, 220 Discord, 84 Dispersion, of light, 213 {Exp. 251) Displacement, 30 Dissociation, 295, 296 Distance formula, lens, 208; mirror, 192 Distilled water, 295, 373 Distinct vision, least distance of, 229, 231 Distribu
tion of charges, 276 {Exp. 362) Distributor, 328 Diverging lens, 203 Diverging pencil, 176 Division, 14, 15 Doppler’s principle, 66 Drums, 87 Dry, cell, 279-281; dock, 30; ice, 156; steam, 151 Dunlap Observatory, 234 Dynamo, alternating current, 317, 321 Ear, 81, 82; frequency limits of hearing, 65 Earth inductor, 320 {Exp. 380) Earth’s magnetic field, 260 Ebonite rod, 268 Echoes, 58, 59, 394 Eclipses, of moon, 179; of sun, 178, 179 Edison, 334 Efficiencies, of heat engines, 156; of electric heater, 137 Einstein, Dr. Albert, 2, 4, 132, 401 Elasticity, 54 Electric, arc, 326, 336; bell, 306, 307; conductors, 361; eye, 348; furnace, 337; generator, A.C., 321, D.C., 323; heating appliances, 335; insulators, 361; light bulb, 334; organ, 88; power, 337; symbols, 285; welding, 337 Electric charges, 268; induced, 272, 273, 277 {Exp. {Exp. distribution 362) ; kinds of, 269; law of, 269 {Exp. 359) 361); 276 of, Electric circuit, 281; types of, 283 Electric current, 279, 281; and electron flow, 282; alternating, 346, 347; direct, 346; chemical effects of, 295 {Exp. 367-373); energy, 331; induced, 318 {Exp. 378-380); magnetic effects of, 304 {Exp. 373, 374) ; measurement of, 281, 298 {Exp. 370) ; rectified, 345 {Exp. 386) Electric motor, 309, 310; St. Louis, 313, 377; {Exp. 376) Electrical discharge tube, 340 {Exp. 382, 383) Electrical energy, 300, 301, 337 Electricity, 1; heat from, 331, 333; production See Static of, 331; transmission of, 331. Electricity, Electric Current Electrochemical equivalent, 298 Electrodes, 295 Electrolysis, 295, 296 ; uses, 297 ; laws of, 298 {Exp. 369) ; of water, 295, 296 {Exp. 367) ; of copper sulphate solution, 297 {Exp. 368) Electrolyte, 279, 295 Electrolytic cell, 295, 296, 367 Electromagnet, 52, 263, 304, 306 {Exp. 375) Electromagnetic, induction, 317, 318; inertia, 327 Electromagnetic, waves, 132, 181, 219, 347; spectrum, 217 Electromagnetism, 317 Electromotive force, 282 {Exp. 379, 380) 409 1 INDEX Electron, flow, 282, 296, 297 ; microscope, 388, 395; theory, 269; volt, 401 Electrons, 269, 342, 397 ; thermionic emission of, 344 [Exp. 384) ; photo-emission of, 347 {Exp. 386) Electronics, 283, 340, 389 {Exp. 382) Electronic tube, 344 ; as amplifier, 346 ; as rectifier, 345 Electrophorus, 273, 274 {Exp. 362) Electroplating, 297, 299 {Exp. 371) Electrorefining, 297 Electroscopes, 270. See gold-leaf and pith-ball Electrostatic, induction, 272, 273 {Exp. 361) ; machine, 275 Electrostatics, 268 Electrotyping, 299 Elements, 270, 295; discovery of, 219; mean- ing of, 112; transmutation of, 399 Elementary (molecular) magnets, 265 Emergent ray, 200 Emission of radiant energy, 133 {Exp. 167) Emission theory of light, 181 E.M.F., 282; induced, 318 {Exp. 379, 380) Energy, 3, 4, 137; atomic, 401; chemical and electrical, 300, 331; conservation of. 111; equation, 401 ; from foods, 139, 144; light, 175; transformation of, 3 Energy levels (orbits), 181, 399 Engines, 156-160 Equilibrium, 147, 148 Equivalent weight, 298 Error, percentage, 13; possible, 13 Ether, 132 Evaporation, boiling and, 148 Exact numbers, 1 Expansion, 115, 119, 122-124 {Exp. 164-165); and density, 124; unusual for water, 19, 20 Eye, 227 Eyepiece, 232 Fahrenheit, D. G., 121 ; scale, 121 Faraday, Michael, 3, 262, 295, 298, 317 Faraday’s, Laws, 298; {Exp. 369) ; dark space, 341 Farsightedness, 230 Field coils, 314 Field glasses, 201 Field, magnetic, 262; around conductor, 304 {Exp. 373) ; around helix, 305, 306 {Exp. 374) Filament, 334, 343, 344, 346 Filters, colour, 221, 394 {Exp. 253) Fire extinguisher, 156 Fission, atomic, 400 Fixed points on thermometer, 120 Fixed resistors, 291 Fizeau, A. H., 179 Flame of Bunsen burner, 164; producing dis- charge, 340 {Exp. 383) Flannel, 268 Floating docks, 30 Flotation, principle of, 26 {Exp. 43) Fluid, 26 Fluorescein, 196 Fluorescence, 176, 218 410 Focal 189 lens, 204, 205; mirror, length, {Exp. 241) Focal plane, 204 Focus, principal, of lens, 204; of mirror, 189 Foods, 137 ; energy from, 139, 144 Foot, 7, 9 Force of cohesion, 146, 147 Forced-air heating, 131 Forced vibrations, 66 Formulae for: electrical energy conversion of centigrade degrees to absolute (Kelvin), 123, centigrade degrees to Fahrenheit, 122; distance from velocity and time, energy 92; (Kilowatt-hours), equation 337; (Einstein), 401; frequency of sound wave, 53; heat from electricity, 331, 333; images in inclined mirrors, 188; index of refraction, 198; lens formulae, distance, magnification, 208 ; magnifying power, compound microscope, 231, telescope, 233 ; mirror formulae, distance, magnification, 192; Ohm’s law, 287, 288; pin-hole camera, 178; power of a lens, 205; power of electricity, horsepower, 337, watts, 333, 337; resistances, in parallel, 289, in series, 289; specific gravity, 20; transformer, of wave, 53; volumes of 324; solids, 36; wave-length of sound, 75, 77 magnifying velocity glass, 232, Four-cycle engine, 159 F.P.S. system of measurement, 9 Franklin, Benjamin, 268, 269, 276 Fraunhofer lines, 220 Freezing, 146; mixture, 155; point, 120; of pond, 20 Frequency of vibration, 50, 64, 65, 92; determination of, 64, 65; of waves, 53; of various sources, 65; ultrasonic, 65, 92 Frequency modulation, 393 Friction, 156; electricity produced by, 268, 270 {Exp. 358) Fuels, 113, 137, 144 Fundamental, 69, 75 Furnace, electric, 337 Fuses, 333 Fusion, meaning of, 146; atomic, 401; heat of, 148 {Exp. 170) Future of sound, 92 Galileo, 2, 3, 120 Galvani, L., 279, 295 Galvanometer, D’Arsonval, 310, 311 Galvanoscope, 309, 375 Gamma ray, 219, 398 Gas, 113; thermometer, 124 Gaseous ions, 340 Gases, thermal conductivity of, 128; convection in, 130 {Exp. 167) ; density of, 21; expansion of, 122-124; liquefaction of, 154; molecular motion of, 110; sound transmission in, 54; spectra of, 219 Generator, 308; A.C., 321 {Exp. 380); D.C., 323 {Exp. 380) Geographic North Pole, 260 Geometric construction of images, 186, 190, 204, 207 Gilbert, Dr. William, 2, 3, 4, 259, 262, 268 INDEX Glare, light, 185 Glass, insulator, 288; plate, 199 {Exp. 245); rod, 268 Gold-leaf electroscope. 271; charging by contact, 272; charging by induction, 273, 274 {Exp. 362) ; identifying charge, 272 {Exp. 360) Graduated cylinder, 12, 34 Gram, 10 Graphite, 299, 401 Greenhouse, 134, 135 Grid, 301, 344, 346, 348 Ground, 273, 274, 285 of, 128-131 Half-wave rectification, 346 Hammond organ, 88 Harmonic, 69 Headlights, car, 194 Hearn, R. L., Generating Station, 158 Heat, 1, 3; absorption of, 133 {Exp. 168); conductivity of, 126-128 {Exp. 166) ; con166) ; expanvection sion caused by, 122-124 {Exp. 164. ’65) ; insulators, 127, 128, 142; measurement of, 137-162; nature of, 110; of fusion, 146-150 {Exp. 170); of vaporization, 150132-136 153 {Exp. 167); sources of. 111; specific heat, and 139-146 {Exp. temperature, 169) ; 119-121; thermometers, 119-122; transfer of, 126; theory of, 109, 110; and work, 156 Heat exchange, principle of, 137, 140; dur- radiation 172) ; {Exp. {Exp. 109, of, ing changes of state, 146 Heat from electric current, 331, 333 Heat from nuclear fission, 401 Heating appliances, electrical, 335 Heating systems, hot water, 145, 152; hot air, 130, 131 119, 129, 130, Heavy water, 399 Helium, 113, 124, 220, 398, 400 Helix, 305; rule, 306 {Exp. 375) Herschel, Sir W., 216 Herschel’s divided tube, 77, 78 {Exp. 105) Hoffman water voltameter, 296 Horse-power, 337 Horseshoe, magnet, 260 ; electromagnet, 306, 307 Hot-air heating, 130, 131 Hot-water, heating, 119, 129, 130, 145, 152; supply, 129-130 frequency limits of Hues, 221 Human, ear, 82; hearing, 65; eye, 227 81, Humours, aqueous, vitreous, 228 Huygens, C., 181 Hydro-electric power, 331 Hydrogen, 113, 154, 280, 295, 399; bomb, 401; ion, 296, 398; isotopes, 399, 401 Hydrometer, 27-29 {Exp. 44, 45) Hydroxyl ion, 296, 297 Hypo (sodium thiosulphate), 196 Image, in camera, 226; real, 178, 190, 192, 207; virtual, 185, 186, 191, 192, 207; in {Exp. 250), convex, lenses, concave, 207 185 207 {Exp. 239), concave, 190 {Exp. 242), con{Exp. 243), inclined, 187 {Exp. vex, 191 240), in pin-hole camera, 178 {Exp. 237) in mirrors, plane, {Exp. 250) : Image orthicon, 391 Incandescence, 175 Incandescent lamp, 334 Inch, 7, 9, 10 Incident rays, 183 Inclination, angle of, 263 Index of refraction, 198, 224 {Exp. 243) Induced charges, 272, 273, 277 {Exp. 361) Induced current (E.M.F.) 318; cause of, 318 {Exp. 378); direction of, 319 {Exp. 380); self, 327 magnitude of, 318 {Exp. 379) ; {Exp. 382) Induced magnetism, 263, 264 {Exp. 354, 355) Induction, 263; charging electroscope by, 273, 274 {Exp. 362) Induction coil, 327 ; uses, 328, 383 Infra-red radiations, 132, 216 In parallel, 284, 312 Input, 112 In series, 283, 312 Instruments, measuring, 12; musical, 85-88 Insulators, of electricity, 272, 288 {Exp. 361) ; of heat, 128, 142; of sound, 64, 88, 89 Intensity of sound, 63, 64, 77, 78 Interference, 54, 55; of sound waves, 76-79 Internal combustion engine, 159; ignition sys- tem, 328 International ampere, 298 Invar, 117 Invisible radiation, 216, 218 Ionization, in gases, 340; theory of, 295 Ions, 295; gaseous, 340 Ion-pairs, 295 Iris, 228; reflex, 228 Iron, 259; electric, 335, 336; filings, 259, 260; soft, 263, 264 Irregular reflection, 185 Isotopes, 399 ; of hydrogen, 399 Jupiter, moons of, 179 Kaleidoscope, 188 Keeper, magnetic, 265, 266 Kelvin, Lord, 123 Kelvin (absolute) temperature, 123 Kettle, electric, 335 Kilo, gram, 10; metre, 9; volt, 282; watt, 337 Kilogram calorie, 139 Kilowatt-hour, 113, 337 Kinescope, 391 Kinetic theory, 110 Knife switch, 284 Knowledge, seven steps to, 4 Ice, dry, 156; heat of fusion of, 150 {Exp. 170) Ignition system, 328 Lactometer, 29 Lamp, arc, 336, 386; electric 334 Land breezes, 130, 146 Larynx, 81 411 ) INDEX Laterally, displaced, 199; inverted, 186 Law of: conservation of energy 111; electrolysis, 298; electrostatics, 269 {Exp. 359) ; of sound, 64; Lenz, 319 {Exp. intensity 380); magnetism, 260, 352; reflection, of sound, 58, of light, 184; refraction of light, 197; vibrating strings, 67-69 {Exp. 99) Lead,
397, 398 ; spongy, 301 Lead-acid storage battery, 300 {Exp. 372) Lead peroxide, 301 Left-hand rule, 305 {Exp. 374) Length, measurement of, 7, 8, 9; of conduc- tor and resistance, 288 Lenses, accommodation, 229; action of, 203 {Exp. 250); applications of, 210; crystalline, 228; focal length of, 205 {Exp. 249); formulae, 208 ; power of, 205 Lenz’s Law, 319 {Exp. 380) Lifting electromagnet, 308 Light, 1, 175; colour of, 213-214; diffusion of, 185; dispersion of, 213; nature of, 175; reflection of, 183 {Exp. 238) ; refraction of, 196 {Exp. 243); sources of, 175; theories of, 181 ; transmission of, 96, 132, 176; velocity of, 179, 180 Light, energy, 349; meter, 349; microscope, 396; sensitive metals, 347; year, 180; waves, 394 Lightning, 276; rods, 276, 277 Limits of audibility, 65 Line spectra, 219 Lines of magnetic force, 262, 263 {Exp. 353) Linear measure, table of, 9, 10 Liquefaction, 146; of gases, 154 Liquid, buoyancy of, 24, 25, 26 ; heat conductivity of, 128, 166; convection in, 128-130 {Exp. 166); density of, 18, 21; expansion of, 19-20, 119; specific gravity of, 21 {Exp. 38, 42, 45) Liquid air, 155, 156 Litre, 10 Local action, 280 {Exp. 364) Lodestone, 259 Longitudinal vibrations, 51 {Exp. 95) Longitudinal waves, 54 {Exp. 97) Loop or coil, 305 Loops, 57, 69, 74, 86 {Exp. 99) Loudness of sound, 50, 63-64, 77, 78 Lower fixed point, 120 Lucite, 358 Luminous bodies, 175 Magnet, 260; kinds, 266 Magnetic, circuit breakers, 307; compass, 261; effects of electric current, 304, 305 {Exp. 373, 374); field, {Exp. 353); lenses, 395, 396; poles, 260 {Exp. 350-351) ; shield, 264 separator, 264; substances, 259 {Exp. 350) {Exp. 355) ; 262, 306 Magnetism, 1, 259; induced, 263 {Exp. 354, 355) laws of, 260, 352 {Exp. 352); terrestrial, 260-263; theory of, 264 {Exp. 357) Magnetite, 259 Magnetization {Exp. 356) Magnification formula, lens, 208; mirror, 192 Magnifying glass, 231 412 Magnifying power, 231, 232, 233, 395 Major triad, 84 Make and break, 379 Manganin, 288, 291 Mass, 10, 11, 19, 112, 114; number, 398, 399 Matter, states of, 146; and energy, 112, 114, 401 Maximum density of water, 20 Mean, position, 50; solar day, 11 Measurement, 7 ; accuracy of, 11; of heat, 137-162; of of resistance, 292; standard units of, 8; systems of, 7 of mass, length, 10; 9; Measuring devices, 7, 11, 12 Mechanics, 1 Media, for transmitting sound, 51, 52 {Exp. 96); light, 52, 96; radiant energy, 132 Megacycle, 394 Megohm, 282 Melting, definition, 146; point, 147 Meniscus, 34 Mercury, 118-120, 128, 281; switches, 284 Metals, heat conductivity of, 126 {Exp. 166) ; 21; expansion of, density of, {Exp. 165); plating of, 299 {Exp. 371); purification of, 297; specific heat of, 139, 144, 146 {Exp. 169) 115 Meteorological balloons, 31 Method, of science, 4; of mixtures, 144, 148 Metre, 9 Metric system, 9 Mho, 289 Mica, insulator, 288 Michelson, A. A., 180 Microampere, 282 Micrometer screw gauge, 12 Microscope, electron, 388, 395-397 ; optical, 232, 396 Microwaves, 393 Mile, 9 Mini, 10; litre, gram, ampere, 282; metre, 9; volt, 282 Minimum deviation, 200 Mirage, 202 Mirrors, 183; applications of, 193; curved, 184; formulae, {Exp. 240); parallel, 187; plane, 183 {Exp. 239) 192; inclined, 187 10; Mixtures, method of, 144, 148 Moderators, 401 Modes of vibration, in air columns, 74, 75; in strings, 69 {Exp. 100) Modulation, 347, 393 Molecules, 19, 112, 128, 132, 134, 152, 297; motion of, 110; kinetic theory of, 110 Molecular theory of magnetism, 265 {Exp. 357) Monochromatic light, 199; flame attachment, 253 Moseley, 398 Motor, principle, 309, 310 {Exp. 376, 377); St. Louis, 313; structure of, 313; direct current, 314 {Exp. 377 Movies, projector, 235; sound, 90, 91 Multiplication, 14, 15 Museum of Science and Industry, 58 Music, 49; analysis of, 391 INDEX Musical, interval, scales, 82-84; sounds. 63 82; instruments, 85-88; Nautilus (submarine), 402 Nearsightedness, 230 Negative, charge, 269-270; photographic, 227 Neon, tube, 176; lamp, 382, 386 Nerve cells, 229 Neutral body, 268, 270 Neutralization, 297 Neutron, 269, 397, 399 Newton, Sir Isaac, 2, 3, 181, 213 Newton’s disc, 215 {Exp. 252) Niagara, power, 332; region, 146 Nichrome, 288, 335 Nickel, 259, 300 Nitrogen, 124, 156, 335, 399; fixation, 337 Node, 57, 69, 74, 99 Noise, 49, 63, 391 ; noise level, 64 Non-electrolyte, 295 Non-luminous, 175 Non-magnetic substances, 259 {Exp. 350) Non-periodic vibrations, 63 Normal, 184, 189 North magnetic pole, 260 N-pole (north-seeking pole), 260 Nuclear, charge, 398; fission, 400; physics, 1, 397 ; reactors, 112, 401, 402 Nucleus, 269, 397 Numbers, exact, 11; approximate, 13 Objective, optical, 232 Octave, 65, 82 Oersted, H. C., 304, 317 Ohm, G. S., 287 Ohm, 3, 281, 288 Ohm’s Law, 287 {Exp. 365) Oil, 112, 113, 160 Oil-immersion microscope, 232 Opaque, 134, 176 Open tube, 74, 86 Optical, centre, 204; density, 197, 199; disc, 183, 184; instruments, 226-235; microscope, 396 Optics, 175 Orbit (or energy level), 269, 399 Organ, electric, 88; pipes, 85-87 Origin of sound, 49 Orthicon, 391 Oscillator, 347 Oscilloscope, sound tracings, 63, 70 Ounce, 10 Output, 112 Overflow can, 40, 43 Overtone, 69, 75, 86 Oxidizing agent, 280 Oxygen, 156, 295, 399 Parabola, 194 Parabolic mirrors, 194, 234 Parallax, method of, 34 Parallel, circuit, 283, 284; mirrors, 187 Partial eclipse, 178 Pencils, converging and diverging, 176 Pendulum, 50, 117 {Exp. 94); clock, 12 Penumbra, 178 Percentage error, 13 Percussion, 112 Period of vibration, 50; of wave, 53 Periodic vibrations, 63 Periscope, 201 Perm.alloy, 263 Permanent magnets, 266 Permeability (magnetic), 263 Persistence of vision, 215, 390, 393 Perspex, 201 Phase, 52, 76-78 Phonograph, 90 Phosphors, 394 Phosphorus, 176 Photoelectric cell, 347 Photoelectricity, 347 Photoemission of electrons, 347 {Exp. 386) Photography, 178, 216, 223, 227, 349 Photon, 181 Photosynthesis, 113, 114 Photronic cell, 348 Physical states of matter, 146 Physics, 1 ; definition, 3, 8 Piano, 85 Pictures, motion, 90, 91, 235 Pigments, 221, 223 Pin-hole camera, 176, 177; image in, 226 {Exp. 237) 178, Pint, 10 Pipe organ, 85-87 Pipeless furnace, 131 Pitch, 50, 63-65, 214 {Exp. 99) Pitchblende, 397 Pith-ball electroscope, 270; charging by contact, 271; identifying charge, 271 {Exp. 359) Planck, Max, 181 Plane mirrors, 185; images in, 185, 186 {Exp. 239) Plate, 344; current, 345 Platinum, 296 Plimsoll line, 29, 30 Point of incidence, 184 Points, action of, 276 {Exp. 363) Polarization, 280, 364 Poles, magnetic, 260 Polonium, 397 Polystyrene, 358 Pond, freezing of, 20 Positive charge, 269, 270 Possible error, 13 Potassium, 347; dichromate, 280, 364; nitrate, {Exp. 350-352) 130; permanganate, 128, 166 Potential, difference, 280-282; energy, 112 Pound, 10 Power, meaning, 337; of electric current, 324, 337; of lens, 205 Prefixes (metric), 9 Presbyopia, 231 Pressure, atmospheric, 120; cooker, 147; critical, 155; effect on boiling point, 147, 148; in water, 282 Primary, cells, 300; coil, 323, 378; currents, 318, 319, 327 ; colours, 220-222 Principal axis, of lens, 204; of mirror, 189 Principal focus, of lens, 204; of mirror, 189 Principle, {Exp. 39, 40) ; Doppler’s, 66; of flotation, 26 {Exp. 43); Archimedes’, 25 413 7 INDEX of generator, 321, 323 {Exp. 380) ; of heat exchange, 137, 140 {Exp. 169) ; of hydrometer, 28 {Exp. 44) ; of lead-acid storage direct current cell, 300 {Exp. 372) ; motor, 309, 310 {Exp. 376, 377) of Printing, colour, 223, 224 Prisms, 199, 213; reversed, 215, 224 {Exp. 247) ; total reflection, 202 Projector, slide, 235 Proof plane, 272 Protium, 399 Proton, 269, 397, 398 Protractor, 263 Psychology and colour, 220 Pulsating current, 323 Pupil of eye, 228 Purification of metals, 297 Push button switch, 284 Pyrex, 1 1 Pyrite (fool’s gold), 109 Quality of sound, 63, 70, 81 {Exp. 101) Quantity, of heat, 137-139; of electricity, 281, 298 Quanta, 132, 181 Quantum theory, 181 Quartz, prism, 218; lenses, 218 Queen Elizabeth, the, 30 Quick-freeze units, 154 Radar, 394; screen, 395 Radiant, energy, 132; heating, 136 Radiation, 142, 135, 131, 126, 156 {Exp. 167, 168) Radioactivity, 397 Radio, frequency, 347 ; transmission and recep- tion, 347; waves, 219 Radiometer, 134 Radium, 112, 397 Rainbow, 199, 216 Range, electric, 335; finder, 227 Rarefaction, 54, 77, 79, 325 Rays, alpha, 398; beta, 398; cathode, 341, 397; cosmic, 132; gamma, 219, 398; infrared, 152, 216; light, 175, 176; ultra-violet, 132, 347 Reactors, nuclear, frontispiece, 401-403 Real image, 178, 207 Rear-vision mirror, 193 Receiver, telephone, 325 Receiving station, radio, 347 Reciprocating steam engine, 157 Recomposition of white light, 215 {Exp. 252) Recorder, sunlight, 206 ; tape, 90 Recordings, 90 Rectification, 345 {Exp. 386) Rectifier, tube, 302, 346 ; circuit, 345 Rectilinear propagation, 176, 177 Reducing heat loss, 127 Refining of metals, 297 Reflected rays, 183 Reflecting telescope, 233, 234 Reflection of light, 183, diffuse, 184; laws of, 184 {Exp. 238) ; regular, 184; total, 200 Reflection of sound, 57, 73; laws of, 57 Reflex time, 59 414 Refracting angle, 199 Refracting telescope, 233 Refraction, of light, 196; atmospheric, 202; 197, 198 {Exp. 243) ; laws of, index of, 197 {Exp. 243); through glass plate, 199 {Exp. {Exp. IM) 245) ; through prisms, 199 Refractive index, 197, 198 {Exp. 243) Refractometer, 198 Refrigerator, electric, 153; gas, 154 Regular reflection, 184 Relativity, Theory of, 132 Relay station, T.V., 393 Reproduction of sound, 90-92 Resistance, electrical, 112, 281, 288; factors affecting, 288; measurement of, 292 {Exp. 365-367) ; table of, 288; unit of, 281, 288 Resistance box, 291 Resistors, 288; in series, 288; in parallel, 289; types of, 291, 292 Resolving power, 395 Resonance, 73; and interference, 73-80; and velocity of sound, 60, 75 {Exp. 102) Retina, 229 Reverberation, 59, 88 Rheostat, 292 Rod, unit of measurement, 9 Rods, for vision, 229 Roentgen, 342 ; rays, 343 Rotor of steam turbine, 158 Rounding-off numbers, 14 Rubidium, 219 Ruhmkorff, 327 Ruler, 12, 34 Rumford, Count
, 109, 156 Rutherford, 398 Salts, 295 Saturation effect (magnetism), 265 Savart’s toothed wheel, 64 {Exp. 99) Scales, musical, 82-84; diatonic, 82; of equal temperament, 84, chromatic, 84 Scales, temperature, 121-123; absolute (Kelvin), 123; centigrade, 121; Fahrenheit, 121 Science, aim of, 8 Scientific method, 4, 259 Sclerotic coat, 227 Sea breeze, 131, 146 Searchlight, 194 Second, 11 Secondary, axis, 189; cells, 300; coil, 323, 327; currents, 318, 319 Segments, commutator, 314 Self-inductance, 326 {Exp. 382) Self-induced current, 327 Semitone, 83 Series circuit, 283 Shadows, 178 Shaving mirror, 193 Shorting plugs, 292 Shunt, 312; connection, 283 Shutter, camera, 226, 227 Significant digits, 13, 14 Silent points about tuning fork, 76, 77 {Exp. 104), Silk, 268 ; 1 INDEX Silver, 298; plating, 299; voltameter, 370 Simple hydrometer, 28 Sine of angle, 198 Slide projector, 235 Slip rings, 321 Smoke paper, 130 Sodium, 347; thiosulphate (hypo), 196 Soft iron, 263, 264 Solar, battery, 113; energy, 113; spectrum, 220 Solenoid, 305, 378 Solidihcation, 146 Solids, conductivity of heat, 126 {Exp. 166); density of, 21 {Exp. 35, 36); expansion of, 115 {Exp. 164, 165) ; specific gravity of, 21 [Exp. 41) 92; future, insulation, Sonar, 92 Sonometer, 66 Sound, 1, 49; characteristics, 63-72; dehnition, 89; 49; 54-57 intensity, {Exp. 104, 105, 106); media for transmisorigin, 49; pitch, 63-65 sion, 51 {Exp. 96) ; {Exp. 99); quality, 63 [Exp. 101); reflection, 57; reproduction, 90-92; resonance, 60, 73 {Exp. 102); transmission, 54; velocity of, 53 {Exp. 102) interference 63 ; 88, 50, 64, of, Sound, barrier, 60; ranging, 91; track, 90; waves, 54, 325 Sounding, 92; balloons, 31 Sources, of heat. 111, 114; of sound, 49; of light, 175 South magnetic pole, 261 Spark, 326; discharge, 275; plugs, 328 Specific gravity, 20, 301 ; determination of, 27, 28 {Exp. 38, 41, 42) Specific-gravity bottle, 38 Specific heat, definition of, 139 {Exp. 169) Spectacles, 230 Spectra, 213, 214; kinds, 219 {Exp. 251) Spectroscope, 218, 219 Spectrum analysis, 219 {Exp. 253) Sphere, volume of, 36 Spherical, aberration, 193; mirrors, 188 S-pole, 260 Spongy lead, 301 Spot lights, 336 Square measure, 10 Standard pressure, 120 Standing waves, 55-57 {Exp. 98) States of matter, 146 Static, 394; electricity, 268 {Exp. 358) Steam, engine, 157; generating plants, coloured, 223 158, 331; heating, 152; turbine, 157; trap, 151 Step, -down transformer, 324; -up, 324 Steps to knowledge, 4 St. Louis Motor, 313 {Exp. 377, 380) Storage battery, 279; hydrometer, 29 Storage cell, 300; action in, 301, 302 {Exp. 372); structure, 300, 301; uses, 302 Stringed instruments, 85 Strings, vibrating, modes of, 69-70 {Exp. 100); laws of, 67-69 {Exp. 99) Sublimation, 146, 156 Submarine, 30, 402 Subtraction, 14, 15 Subtractive theory of colour, 221 Substitution method for resistance, 292, 293 {Exp. 367) Sugar, 295 Sulphate ions, 296 Sulphuric acid, 279, 301 Sun, 1 12, 175, 349 Sunlight recorder, 206 Sunset, 202 Superposition, of light waves, 181; of .sound waves, 54-57 {Exp. 101) Switches, 284 Symbols, electric, 285 Sympathetic vibrations, 76 {Exp. 103) of liquids, 119; coefficient Tables of; boiling-points, 152; calorific values of fuels, 144; coefficient of cubical expansion of linear expansion of solids, 117; critical pressures and temperatures, 155; efficiencies of heat engines, 156; electrical conductors and insulators, 272; electrical symbols, 285; electrochemical electromagnetic waves, 217; heat conductivities, 127; heats of fusion, 150; heats of vaporization, 152; indices of refraction, 199; linear measure, 9; mass, 10; prefixes, 9; resistances, 288; specific gravities, 21; specific heats, 139; units of electricity, 282; velocity of sound and temperature, of sound in various media, 61; volume, 10; wave-lengths of coloured lights, 214 equivalents, velocity 298; 60; Tape recorder, 90 Telephone, 325, 326 Telescopes, 233, 234 {Exp. 254) Television, 391 ; colour, 394; picture tube, 391 Temperature, meaning of, 119; absolute, 123; critical, 155; scales, 121; and quantity of heat, 137; and resistance, 288 Temporary magnets, 266 Terrestrial telescope, 234 Theories, atomic, 270; colour (additive and subtractive), 220-223; electron, 269; heat, 109; light (corpuscular, 181; magnetism, 264-265; molecular (kinetic), (Einstein), 132; wave, of 110; relativity heat, 132, of light, 181, 197, of sound, 54 Thermionic emission of electrons, 344 {Exp. 295; quantum), (Arrhenius), ionization emission, 384) Thermodynamics, 109 Thermograph, 117, 118 Thermometers, 1 19-122 ; calibrating, 120, 124; construction of, 120; dial, 117; how to use, 164 1 Thermos bottle, 134, 156 Thermoscope, 134 Thermostat, 118 Thomson, Sir J. J., 342, 397 Time, 8, Tonic, 82 Total eclipse, 178 Total reflection, 200; prisms, 201, 202 Traffic signals, 223 Transfer of heat, 126 Transformers, 323, 324 415 INDEX Translucent, 176 Transmission of, 126, 128, 132; light, 176; radiant energy, 132134; radio waves, 347; sound, 49-62 electricity, 331 ; heat, Transmitter, telephone, 325 Transmitting studio, 347 Transmutation, artificial, 399 Transparent, 134, 176 Transverse vibrations, 50, 78, 132 {Exp. 94, 95) Transverse waves, 52, 78, 132 {Exp. 97) Triad, major, 84 Triode tube, 344, 346; amplifier, 346; oscil- lator, 347 Tritium, 399 Troughs, 52 Tubes, closed and open, 73, 85-87 Tuning-fork, 55 {Exp. 96) ; interference of sound around, 76 {Exp. l04) Tuning musical instruments, 79 Tungsten, 335 Turbines, 331; steam, 157 U, see Uranium below Ultrasonic frequencies, 65, 92 Ultra-violet, lamps, 218; radiation, 218; light, 132, 347 Umbra, 178 Unison, 79 Units, electrical, 282; heat, 138; measure- ment, 8-11; power, 337; sound, 64 Universal hydrometer, 28 Upper fixed point, 120 Uranium, U, 112; U-235, 400; disintegration series, 397; salts, 397; and coal, 401 Vacuum, 52, 132, 154; bottle, 134; evaporator, 147; tube, 132, 340 {Exp. 384, 386); tunnel (light), 180 Valves, 344 Vaporization, 146; heat of, 150 {Exp. 172) Variable resistors, 292 Velocity, definition, 53; of light, 179, 198; of radio waves, 132; of sound, 59, 75 {Exp. 102 ) Ventilation, 130 Vernier calipers, 12 Vertex, 189 Vibrating strings, laws of, 67-69 {Exp. 99) ; mode of vibrations 69 {Exp. 100) Vibration, 325; definition, 49; forced, 66; frequency for light, 214; kinds of, 50, 51; air columns, 74; of longitudinal, 51; of strings, 67-69; sympathetic, 76; transverse, 50 View finder, camera, 227 Virtual image, 185, 186, 191, 207 Vision, defects of, 230; persistence of, 215, 390, 393 Visual, angle, 231; cells, 229 Vita glass, 218 Vitreous humour, 228 Vocal cords, 81 Voice, 81 416 Volt, 282, 288 Volta, A., 3, 273 Voltage-drop, 282, 283, 288 Voltaic cell, 279 {Exp. 364) Voltameter, water, 296, 368; copper, 298 {Exp. 370) Voltmeter, 311-313 Voltmeter-ammeter method for resistance, 292 {Exp. 366) Volume, 10, 36 Watch, stop, 12; balance wheel, 116-117 Water, 295; boiling point of, 120; conductivity of, 127, 128; convection in, 129; density of, 18-20; displacement, 30; electrolysis of, 295, 296; expansion of, 124; heating {Exp. system, 170) ; heat of vaporization, 150-153 {Exp. 172); refractive index, 199; specific heat, 145; sound transmitted by, 52; supply (hot water), 129-130; total reflection in, 200, 201; turbine, 157, 331; voltameter, 296 129; heat of fusion, 148 Water, heavy, 399 Watt, Sir R. W., 394 Watt, definition, 337 ; hour, 337 Wave, 52; audio-frequency, 347; carrier, 347, 394; electromagnetic, 132, 181, 219, 347; formula, 53, 60, 75, 77, 86; front, 54, 197; heat, 132; length, 52, 70, 77, 214; light, 181; modulated, 347, 391, 393; motion, 3, 52-54 pulsating, 345; radio, 217, 219; sound, 54; train, 53 {Exp. 97) ; Wave length, of electromagnetic waves, 217; of sound waves, 75 Waves, electromagnetic, 132; longitudinal, 54; standing, 55-57 {Exp. 98) ; superposition of, 54-57 {Exp. 101); transverse, 52; television, 391, 392, 393 Weather balloons, 31 Weight, 7, 11 Welding, electric arc, 337 Wheel, balance, of watch, 116 Whirl, electric, 276 White, the colour, 221 White light, composition, 213, 215 Wimshurst machine, 275 Wind instruments, 85-86 Winds, 130 Wire-wound resistors, 291 Work from electricity, 337; and heat, 156; rate of doing, 337 X-rays, 132, 219, 342; nature of, 343 {Exp. 384) ; photograph, 343 X-ray tube, 328, 343 ; chest unit, 342 Yard, definition, 7, 9 Yellow spot, 229 Young-Helmholtz theory, 223 Zero, absolute, 123, 132 Zinc, photo-emission of electrons, 347, 348 {Exp. 386) ; plate in voltaic cell, 279 THB 85 Date Due 1 i 1 s QC 23 E8G 1958 C-3 EUBANK HOWARD LAWRENCE 1910- BASIC PHYSICS FOR SECONDARY SCHOOLS 399032G3 CURR HIST -00002252B7M3- OC 23 ESS 1958 C. Eubank , Howard Lawrence, 1910- 3 Basic physics for secondary schoo 1 39903263 CURR HIST ,X- CJ STUDENT ohool BASIC PHYSICS THE MACMILLAN COMPANY OF CANADA LIMITED
checks (NOTE TO SELF: Add to this table as we go along with examples from each section.) Now you don’t have to memorise this table but you should read it. The best thing to do is to refer to it every time you do a calculation. 1.9 Temperature We need to make a special mention of the units used to describe temperature. The unit of temperature listed in Table 1.1 is not the everyday unit we see and use. Normally the Celsius scale is used to describe temperature. As we all know, Celsius temperatures can be negative. This might suggest that any number is a valid temperature. In fact, the temperature of a gas is a measure of the average kinetic energy of the particles that make up the gas. As we lower the temperature so the motion of the particles is reduced until a point is reached 8 where all motion ceases. The temperature at which this occurs is called absolute zero. There is no physically possible temperature colder than this. In Celsius, absolute zero is at 273oC. Physicists have deﬂned a new temperature scale called the Kelvin scale. According to this scale absolute zero is at 0K and negative temperatures are not allowed. The size of one unit kelvin is exactly the same as that of one unit Celsius. This means that a change in temperature of 1 degree kelvin is equal to a change in temperature of 1 degree Celsius\| the scales just start in diﬁerent places. Think of two ladders with steps that are the same size but the bottom most step on the Celsius ladder is labelled -273, while the ﬂrst step on the Kelvin ladder is labelled 0. There are still 100 steps between the points where water freezes and boils. ¡ water boils ---> \|----\| \|----\| \|----\| \|----\| \|----\| 102 Celsius 101 Celsius 100 Celsius Celsius 99 Celsius 98 \|----\| \|----\| \|----\| \|----\| \|----\| 375 Kelvin 374 Kelvin 373 Kelvin 372 Kelvin 371 Kelvin ice melts ---> \|----\| \|----\| \|----\| \|----\| \|----\| 2 1 0 -1 -2 Celsius Celsius Celsius Celsius Celsius . . . . . . \|----\| \|----\| \|----\| \|----\| \|----\| 275 Kelvin 274 Kelvin 273 Kelvin 272 Kelvin 271 Kelvin \|----\| \|----\| \|----\| \|----\| \|----\| -269 Celsius -270 Celsius -271 Celsius -272 Celsius -273 Celsius \|----\| \|----\| \|----\| \|----\| \|----\| 4 Kelvin 3 Kelvin 2 Kelvin 1 Kelvin 0 Kelvin absolute zero ---> (NOTE TO SELF: Come up with a decent picture of two ladders with the labels \|water boiling and freezing\|in the same place but with diﬁerent labelling on the steps!) This makes the conversion from kelvin to Celsius and back very easy. To convert from Celsius to kelvin add 273. To convert from kelvin to Celsius subtract 273. Representing the Kelvin temperature by TK and the Celsius temperature by ToC, TK = ToC + 273: (1.1) It is because this conversion is additive that a diﬁerence in temperature of 1 degree Celsius is equal to a diﬁerence of 1 kelvin. The majority of conversions between units are multiplicative. For example, to convert from metres to millimetres we multiply by 1000. Therefore a change of 1m is equal to a change of 1000mm. 1.10 Scientiﬂc Notation, Signiﬂcant Figures and Rounding (NOTE TO SELF: still to be written) 9 1.11 Conclusion In this chapter we have discussed the importance of units. We have discovered that there are many diﬁerent units to describe the same thing, although you should stick to SI units in your calculations. We have also discussed how to convert between diﬁerent units. This is a skill you must acquire. 10 Chapter 2 Waves and Wavelike Motion Waves occur frequently in nature. The most obvious examples are waves in water, on a dam, in the ocean, or in a bucket. We are most interested in the properties that waves have. All waves have the same properties so if we study waves in water then we can transfer our knowledge to predict how other examples of waves will behave. 2.1 What are waves? Waves are disturbances which propagate (move) through a medium 1. Waves can be viewed as a transfer energy rather than the movement of a particle. Particles form the medium through which waves propagate but they are not the wave. This will become clearer later. Lets consider one case of waves: water waves. Waves in water consist of moving peaks and troughs. A peak is a place where the water rises higher than when the water is still and a trough is a place where the water sinks lower than when the water is still. A single peak or trough we call a pulse. A wave consists of a train of pulses. So waves have peaks and troughs. This could be our ﬂrst property for waves. The following diagram shows the peaks and troughs on a wave. Peaks Troughs In physics we try to be as quantitative as possible. If we look very carefully we notice that the height of the peaks above the level of the still water is the same as the depth of the troughs below the level of the still water. The size of the peaks and troughs is the same. 2.1.1 Characteristics of Waves : Amplitude The characteristic height of a peak and depth of a trough is called the amplitude of the wave. The vertical distance between the bottom of the trough and the top of the peak is twice the amplitude. We use symbols agreed upon by convention to label the characteristic quantities of 1Light is a special case, it exhibits wave-like properties but does not require a medium through which to propagate. 11 the waves. Normally the letter A is used for the amplitude of a wave. The units of amplitude are metres (m). 2 x Amplitude Worked Example 1 Amplitude Amplitude Question: (NOTE TO SELF: Make this a more exciting question) If the peak of a wave measures 2m above the still water mark in the harbour what is the amplitude of the wave? Answer: The deﬂnition of the amplitude is the height that the water rises to above when it is still. This is exactly what we were told, so the answer is that the amplitude is 2m. 2.1.2 Characteristics of Waves : Wavelength Look a little closer at the peaks and the troughs. The distance between two adjacent (next to each other) peaks is the same no matter which two adjacent peaks you choose. So there is a ﬂxed distance between the peaks. Looking closer you’ll notice that the distance between two adjacent troughs is the same no matter which two troughs you look at. But, more importantly, its is the same as the distance between the peaks. This distance which is a characteristic of the wave is called the wavelength. Waves have a characteristic wavelength. The symbol for the wavelength is ‚. The units are metres (m). ‚ ‚ ‚ The wavelength is the distance between any two adjacent points which are in phase. Two points in phase are separate by an integer (0,1,2,3,...) number of complete wave cycles. They don’t have to be peaks or trough but they must be separated by a complete number of waves. 2.1.3 Characteristics of Waves : Period Now imagine you are sitting next to a pond and you watch the waves going past you. First one peak, then a trough and then another peak. If you measure the time between two adjacent peaks you’ll ﬂnd that it is the same. Now if you measure the time between two adjacent troughs you’ll 12 ﬂnd that its always the same, no matter which two adjacent troughs you pick. The time you have been measuring is the time for one wavelength to pass by. We call this time the period and it is a characteristic of the wave. Waves have a characteristic time interval which we call the period of the wave and denote with the symbol T. It is the time it takes for any two adjacent points which are in phase to pass a ﬂxed point. The units are seconds (s). 2.1.4 Characteristics of Waves : Frequency There is another way of characterising the time interval of a wave. We timed how long it takes for one wavelength to pass a ﬂxed point to get the period. We could also turn this around and say how many waves go by in 1 second. We can easily determine this number, which we call the frequency and denote f. To determine the frequency, how many waves go by in 1s, we work out what fraction of a waves goes by in 1 second by dividing 1 second by the time it takes T. If a wave takes 1 2 a second to go by then in 1 second two waves must go by. 1 = 2. The unit of frequency is the Hz or s¡1. 1 2 Waves have a characteristic frequency. f = 1 T f T : frequency (Hz or s¡1) : period (s) 2.1.5 Characteristics of Waves : Speed Now if you are watching a wave go by you will notice that they move at a constant velocity. The speed is the distance you travel divided by the time you take to travel that distance. This is excellent because we know that the waves travel a distance ‚ in a time T. This means that we can determine the speed. v = ‚ T v ‚ T : speed (m:s¡1) : wavelength (m) : period (s) There are a number of relationships involving the various characteristic quantities of waves. A simple example of how this would be useful is how to determine the velocity when you have the frequency and the wavelength. We can take the above equation and substitute the relationship between frequency and period to produce an equation for speed of the form v = f ‚ v ‚ f : speed (m:s¡1) : wavelength (m) : frequency (Hz or s¡1) Is this correct? Remember a simple ﬂrst check is to check the units! On the right hand side we have velocity which has units ms¡1. On the left hand side we have frequency which is 13 measured in s¡1 multiplied by wavelength which is measure in m. On the left hand side we have ms¡1 which is exactly what we want. 2.2 Two Types of Waves We agreed that a wave was a moving set of peaks and troughs and we used water as an example. Moving peaks and troughs, with all the characteristics we described, in any medium constitute a wave. It is possible to have waves where the peaks and troughs are perpendicular to the direction of motion, like in the case of water waves. These waves are called transverse waves. There is another type of wave. Called a longitudinal wave and it has the peaks and troughs in the same direction as the wave is moving. The question is how do we construct such a wave? An example of a longitudinal wave is a pressure wave moving through a gas. The peaks in thi
s wave are places where the pressure reaches a peak and the troughs are places where the pressure is a minimum. In the picture below we show the random placement of the gas molecules in a tube. The piston at the end moves into the tube with a repetitive motion. Before the ﬂrst piston stroke the pressure is the same throughout the tube. JLK When the piston moves in it compresses the gas molecules together at the end of the tube. If the piston stopped moving the gas molecules would all bang into each other and the pressure would increase in the tube but if it moves out again fast enough then pressure waves can be set up. UWV 465 When the piston moves out again before the molecules have time to bang around then the increase in pressure moves down the tube like a pulse (single peak). The piston moves out so fast that a pressure trough is created behind the peak. ØŁ æ Æ As this repeats we get waves of increased and decreased pressure moving down the tubes. We can describe these pulses of increased pressure (peaks in the pressure) and decreased pressure (troughs of pressure) by a sine or cosine graph. ´˜ˆ £¥⁄ ƒ¤§ ˇ— ¢¡ 14 ! " # $ % & ’ ( ) * + , . / Incident ray There are a number of examples of each type of wave. Not all can be seen with the naked eye but all can be detected. 2.3 Properties of Waves We have discussed some of the simple characteristics of waves that we need to know. Now we can progress onto some more interesting and, perhaps, less intuitive properties of waves. 2.3.1 Properties of Waves : Reection When waves strike a barrier they are reected. This means that waves bounce oﬁ things. Sound waves bounce oﬁ walls, light waves bounce oﬁ mirrors, radar waves bounce oﬁ planes and it can explain how bats can y at night and avoid things as small as telephone wires. The property of reection is a very important and useful one. (NOTE TO SELF: Get an essay by an air tra–c controller on radar) (NOTE TO SELF: Get an essay by on sonar usage for ﬂshing or for submarines) When waves are reected, the process of reection has certain properties. If a wave hits an obstacle at a right angle to the surface (NOTE TO SELF: diagrams needed) then the wave is reected directly backwards. If the wave strikes the obstacle at some other angle then it is not reected directly backwards. The angle that the waves arrives at is the same as the angle that the reected waves leaves at. The angle that waves arrives at or is incident at equals the angle the waves leaves at or is reected at. Angle of incidence equals angle of reection i = r 15 (2.1) Incident ray i = r i r : angle of incidence : angle of reection i r In the optics chapter you will learn that light is a wave. This means that all the properties we have just learnt apply to light as well. Its very easy to demonstrate reection of light with a mirror. You can also easily show that angle of incidence equals angle of reection. If you look directly into a mirror your see yourself reected directly back but if you tilt the mirror slightly you can experiment with diﬁerent incident angles. Phase shift of reected wave When a wave is reected from a more dense medium it undergoes a phase shift. That means that the peaks and troughs are swapped around. The easiest way to demonstrate this is to tie a piece of string to something. Stretch the string out at and then ick the string once so a pulse moves down the string. When the pulse (a single peak in a wave) hits the barrier that the string is tide to it will be reected. The reected wave will look like a trough instead of a peak. This is because the pulse had undergone a phase change. The ﬂxed end acts like an extremely dense medium. If the end of the string was not ﬂxed, i.e. it could move up and down then the wave would still be reected but it would not undergo a phase shift. To draw a free end we draw it as a ring around a line. This signiﬂes that the end is free to move. 16 2.3.2 Properties of Waves : Refraction Sometimes waves move from one medium to another. The medium is the substance that is carrying the waves. In our ﬂrst example this was the water. When the medium properties change it can aﬁect the wave. Let us start with the simple case of a water wave moving from one depth to another. The speed of the wave depends on the depth 2. If the wave moves directly from the one medium to the other then we should look closely at the boundary. When a peak arrives at the boundary and moves across it must remain a peak on the other side of the boundary. This means that the peaks pass by at the same time intervals on either side of the boundary. The period and frequency remain the same! But we said the speed of the wave changes, which means that the distance it travels in one time interval is diﬁerent i.e. the wavelength has changed. Going from one medium to another the period or frequency does not change only the wave- length can change. Now if we consider a water wave moving at an angle of incidence not 90 degrees towards a change in medium then we immediately know that not the whole wavefront will arrive at once. So if a part of the wave arrives and slows down while the rest is still moving faster before it arrives the angle of the wavefront is going to change. This is known as refraction. When a wave bends or changes its direction when it goes from one medium to the next. If it slows down it turns towards the perpendicular. 2 17 Air Water If the wave speeds up in the new medium it turns away from the perpendicular to the medium surface. Air Water When you look at a stick that emerges from water it looks like it is bent. This is because the light from below the surface of the water bends when it leaves the water. Your eyes project the light back in a straight line and so the object looks like it is a diﬁerent place. 18 Air Water 2.3.3 Properties of Waves : Interference If two waves meet interesting things can happen. Waves are basically collective motion of particles. So when two waves meet they both try to impose their collective motion on the particles. This can have quite diﬁerent results. If two identical (same wavelength, amplitude and frequency) waves are both trying to form a peak then they are able to achieve the sum of their eﬁorts. The resulting motion will be a peak which has a height which is the sum of the heights of the two waves. If two waves are both trying to form a trough in the same place then a deeper trough is formed, the depth of which is the sum of the depths of the two waves. Now in this case the two waves have been trying to do the same thing and so add together constructively. This is called constructive interference. A=0.5 B=1 A+B=1.5 If one wave is trying to form a peak and the other is trying to form a trough then they are competing to do diﬁerent things. In this case they can cancel out. The amplitude of the resulting 19 wave will depend on the amplitudes of the two waves that are interfering. If the depth of the trough is the same as the height of the peak nothing will happen. If the height of the peak is bigger than the depth of the trough a smaller peak will appear and if the trough is deeper then a less deep trough will appear. This is destructive interference. A=0.5 B=1 B-A=0.5 2.3.4 Properties of Waves : Standing Waves When two waves move in opposite directions, through each other, interference takes place. If the two waves have the same frequency and wavelength then a speciﬂc type of constructive interference can occur: standing waves can form. Standing waves are disturbances which don’t appear to move, they look like they stay in the same place even though the waves that from them are moving. Lets demonstrate exactly how this comes about. Imagine a long string with waves being sent down it from either end. The waves from both ends have the same amplitude, wavelength and frequency as you can see in the picture below: 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 To stop from getting confused between the two waves we’ll draw the wave from the left with a dashed line and the one from the right with a solid line. As the waves move closer together when they touch both waves have an amplitude of zero: 20 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 If we wait for a short time the ends of the two waves move past each other and the waves overlap. Now we know what happens when two waves overlap, we add them together to get the resulting wave. 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 Now we know what happens when two waves overlap, we add them together to get the resulting wave. In this picture we show the two waves as dotted lines and the sum of the two in the overlap region is shown as a solid line: 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 The important thing to note in this case is that there are some points where the two waves always destructively interfere to zero. If we let the two waves move a little further we get the picture below: 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 Again we have to add the two waves together in the overlap region to see what the sum of the waves looks like. 1 0 -1 -5 -4 -3 -2 -1 0 1 2 3 4 5 In this case the two waves have moved half a cycle past each other but because they are out of phase they cancel out completely. The point at 0 will always be zero as the two waves move past each other. 21 When the waves have moved past each other so that they are overlapping for a large region the situation looks like a wave oscillating in place. If we focus on the range -4, 4 once the waves have moved over the whole region. To make it clearer the arrows at the top of the picture show peaks where maximum positive constructive interference is taking place. The arrows at the bottom of the picture show places where maximum negative interference is taking place. 1 0 -1 0 As time goes by the peaks become smaller and the troughs become shallower but they do not -3 -1 -2 -4 1 3 2 4 move. 1 0 -1 For an instant the entire region will look completely at. -4 -3 -2 -1 0 1 2 3 4 1 0 -1 The various points continue their motion in the same manner. -4 -3
-2 -1 0 1 2 3 4 1 0 -1 0 Eventually the picture looks like the complete reection through the x-axis of what we started -2 -1 -4 -3 4 3 2 1 with: 1 0 -1 0 Then all the points begin to move back. Each point on the line is oscillating up and down -1 -3 -4 -2 2 1 4 3 with a diﬁerent amplitude. 22 1 0 -1 -3 -2 -4 0 If we superimpose the two cases where the peaks were at a maximum and the case where the same waves were at a minimum we can see the lines that the points oscillate between. We call this the envelope of the standing wave as it contains all the oscillations of the individual points. A node is a place where the two waves cancel out completely as two waves destructively interfere in the same place. An anti-node is a place where the two waves constructively interfere. -1 1 4 3 2 Important: The distance between two anti-nodes is only 1 2 ‚ because it is the distance from a peak to a trough in one of the waves forming the standing wave. It is the same as the distance between two adjacent nodes. This will be important when we workout the allowed wavelengths in tubes later. We can take this further because half-way between any two anti-nodes is a node. Then the distance from the node to the anti-node is half the distance between two anti-nodes. This is half of half a wavelength which is one quarter of a wavelength, 1 4 ‚. Anti-nodes To make the concept of the envelope clearer let us draw arrows describing the motion of points along the line. Nodes Every point in the medium containing a standing wave oscillates up and down and the amplitude of the oscillations depends on the location of the point. It is convenient to draw the envelope for the oscillations to describe the motion. We cannot draw the up and down arrows for every single point! Reection from a ﬂxed end If waves are reected from a ﬂxed end, for example tieing the end of a rope to a pole and then sending waves down it. The ﬂxed end will always be a node. Remember: Waves reected from a ﬂxed end undergo a phase shift. The wavelength, amplitude and speed of the wave cannot aﬁect this, the ﬂxed end is always a node. Reection from an open end If waves are reected from end, which is free to move, it is an anti-node. For example tieing the end of a rope to a ring, which can move up and down, around the pole. Remember: The waves sent down the string are reected but do not suﬁer a phase shift. 23 Wavelengths of standing waves with ﬂxed and open ends There are many applications which make use of the properties of waves and the use of ﬂxed and free ends. Most musical instruments rely on the basic picture that we have presented to create speciﬂc sounds, either through standing pressure waves or standing vibratory waves in strings. The key is to understand that a standing wave must be created in the medium that is oscillating. There are constraints as to what wavelengths can form standing waves in a medium. For example, if we consider a tube of gas it can have both ends open (Case 1) one end open and one end closed (Case 2) both ends closed (Case 3). † † † Each of these cases is slightly diﬁerent because the open or closed end determines whether a node or anti-node will form when a standing wave is created in the tube. These are the primary constraints when we determine the wavelengths of potential standing waves. These constraints must be met. In the diagram below you can see the three cases diﬁerent cases. It is possible to create standing wave with diﬁerent frequencies and wavelengths as long as the end criteria are met. Case 1 L Case 2 L Case 3 L The longer the wavelength the less the number of anti-nodes in the standing waves. We cannot have a standing wave with 0 no anti-nodes because then there would be no oscillations. We use n to number to anti-nodes. If all of the tubes have a length L and we know the end constraints we can workout the wavelenth, ‚, for a speciﬂc number of anti-nodes. Lets workout the longest wavelength we can have in each tube, i.e. the case for n = 1. ‚ = 2L ‚ = 4L n = 1 Case 1: In the ﬂrst tube both ends must be nodes so we can place one anti-node in the 2 ‚ and we also know this middle of the tube. We know the distance from one node to another is 1 distance is L. So we can equate the two and solve for the wavelength: 1 2 ‚ = L ‚ = 2L Case 2: In the second tube one ends must be a node and the other must be an anti-node. We are looking at the case with one anti-node we we are forced to have it at the end. We know the distance from one node to another is 1 2 ‚ but we only have half this distance contained in the tube. So : 1 2 ( 1 2 ‚) = L ‚ = 4L 24 NB: If you ever calculate a longer wavelength for more nodes you have made a mistake. Remember to check if your answers make sense! Case 3: Here both ends are closed and so we must have two nodes so it is impossible to construct a case with only one node. Next we determine which wavelengths could be formed if we had two nodes. Remember that we are dividing the tube up into smaller and smaller segments by having more nodes so we expect the wavelengths to get shorter. ‚ = L ‚ = 4 3 L ‚ = 2L n = 2 Case 1: Both ends are open and so they must be anti-nodes. We can have two nodes inside the tube only if we have one anti-node contained inside the tube and one on each end. This means we have 3 anti-nodes in the tube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength between the left side and the middle and another half wavelength between the middle and the right side so there must be one wavelength inside the tube. The safest thing to do is workout how many half wavelengths there are and equate this to the length of the tube L and then solve for ‚. Even though its very simple in this case we should practice our technique: 2( 1 2 ‚) = L ‚ = L Case 2: We want to have two nodes inside the tube. The left end must be a node and the right end must be an anti-node. We can have one node inside the tube as drawn above. Again we can count the number of distances between adjacent nodes or anti-nodes. If we start from the left end we have one half wavelength between the end and the node inside the tube. The distance from the node inside the tube to the right end which is an anti-node is half of the distance to another node. So it is half of half a wavelength. Together these add up to the length of the tube ‚) = Case 3: In this case both ends have to be nodes. This means that the length of the tube is one half wavelength: So we can equate the two and solve for the wavelength: 1 2 ‚ = L ‚ = 2L To see the complete pattern for all cases we need to check what the next step for case 3 is when we have an additional node. Below is the diagram for the case where n=3 25 Case 1: Both ends are open and so they must be anti-nodes. We can have three nodes inside the tube only if we have two anti-node contained inside the tube and one on each end. This means we have 4 anti-nodes in the tube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength between every adjacent pair of anti-nodes. We count how many gaps there are between adjacent anti-nodes to determine how many half wavelengths there are and equate this to the length of the tube L and then solve for ‚. 3( 1 2 ‚) = L ‚ = 2 3 L Case 2: We want to have three nodes inside the tube. The left end must be a node and the right end must be an anti-node, so there will be two nodes between the ends of the tube. Again we can count the number of distances between adjacent nodes or anti-nodes, together these add up to the length of the tube. Remember that the distance between the node and an adjacent anti-node is only half the distance between adjacent nodes. So starting from the left end we 3 nodes so 2 half wavelength intervals and then only a node to anti-node distance: 2( 1 2 ‚) + ‚) = Case 3: In this case both ends have to be nodes. With one node in between there are two sets of adjacent nodes. This means that the length of the tube consists of two half wavelength sections: 2( 1 2 ‚) = L ‚ = L 2.3.5 Beats If the waves that are interfering are not identical then the waves form a modulated pattern with a changing amplitude. The peaks in amplitude are called beats. If you consider two sound waves interfering then you hear sudden beats in loudness or intensity of the sound. The simplest illustration is two draw two diﬁerent waves and then add them together. You can do this mathematically and draw them yourself to see the pattern that occurs. Here is wave 1: 26 Now we add this to another wave, wave 2: When the two waves are added (drawn in coloured dashed lines) you can see the resulting wave pattern: To make things clearer the resulting wave without the dashed lines is drawn below. Notice that the peaks are the same distance apart but the amplitude changes. If you look at the peaks they are modulated i.e. the peak amplitudes seem to oscillate with another wave pattern. This is what we mean by modulation. 2Amax 2Amin The maximum amplitude that the new wave gets to is the sum of the two waves just like for constructive interference. Where the waves reach a maximum it is constructive interference. The smallest amplitude is just the diﬁerence between the amplitudes of the two waves, exactly like in destructive interference. The beats have a frequency which is the diﬁerence between the frequency of the two waves that were added. This means that the beat frequency is given by fB = f1 j f2 j ¡ (2.2) j ¡ f2 fB = f1 j : beat frequency (Hz or s¡1) : frequency of wave 1 (Hz or s¡1) : frequency of wave 2 (Hz or s¡1) fB f1 f2 27 2.3.6 Properties of Waves : Diﬁraction One of the most interesting, and also very useful, properties of waves is diﬁraction. When a wave strikes a barrier with a hole only part of the wave can move through the hole. If the hole is similar in size to the wavelength of the wave diﬁractions occurs. The waves that comes through the hole no longer looks like a straight wave front. It
bends around the edges of the hole. If the hole is small enough it acts like a point source of circular waves. This bending around the edges of the hole is called diﬁraction. To illustrate this behaviour we start by with Huygen’s principle. Huygen’s Principle Huygen’s principle states that each point on a wavefront acts like a point source or circular waves. The waves emitted from each point interfere to form another wavefront on which each point forms a point source. A long straight line of points emitting waves of the same frequency leads to a straight wave front moving away. To understand what this means lets think about a whole lot of peaks moving in the same direction. Each line represents a peak of a wave. If we choose three points on the next wave front in the direction of motion and make each of them emit waves isotropically (i.e. the same in all directions) we will get the sketch below: What we have drawn is the situation if those three points on the wave front were to emit waves of the same frequency as the moving wave fronts. Huygens principle says that every point on the wave front emits waves isotropically and that these waves interfere to form the next wave front. To see if this is possible we make more points emit waves isotropically to get the sketch below: 28 You can see that the lines from the circles (the peaks) start to overlap in straight lines. To make this clear we redraw the sketch with dashed lines showing the wavefronts which would form. Our wavefronts are not perfectly straight lines because we didn’t draw circles from every point. If we had it would be hard to see clearly what is going on. It Huygen’s principle is a method of analysis applied to problems of wave propagation. recognizes that each point of an advancing wave front is in fact the center of a fresh disturbance and the source of a new train of waves; and that the advancing wave as a whole may be regarded as the sum of all the secondary waves arising from points in the medium already traversed. This view of wave propagation helps better understand a variety of wave phenomena, such as diﬁraction. Wavefronts Moving Through an Opening Now if allow the wavefront to impinge on a barrier with a hole in it, then only the points on the wavefront that move into the hole can continue emitting forward moving waves - but because a lot of the wavefront have been removed the points on the edges of the hole emit waves that bend round the edges. 29 The wave front that impinges (strikes) the wall cannot continue moving forward. Only the points moving into the gap can. If you employ Huygens’ principle you can see the eﬁect is that the wavefronts are no longer straight lines. For example, if two rooms are connected by an open doorway and a sound is produced in a remote corner of one of them, a person in the other room will hear the sound as if it originated at the doorway. As far as the second room is concerned, the vibrating air in the doorway is the source of the sound. The same is true of light passing the edge of an obstacle, but this is not as easily observed because of the short wavelength of visible light. This means that when waves move through small holes they appear to bend around the sides because there aren’t enough points on the wavefront to form another straight wavefront. This is bending round the sides we call diﬁraction. 2.3.7 Properties of Waves : Dispersion Dispersion is a property of waves where the speed of the wave through a medium depends on the frequency. So if two waves enter the same dispersive medium and have diﬁerent frequencies they will have diﬁerent speeds in that medium even if they both entered with the same speed. We will come back to this topic in optics. 2.4 Practical Applications of Waves: Sound Waves 2.4.1 Doppler Shift The Doppler shift is an eﬁect which becomes apparent when the source of sound waves or the person hearing the sound waves is moving. In this case the frequency of the sounds waves can 30 be diﬁerent. This might seem strange but you have probably experienced the doppler eﬁect in every day life. When would you notice it. The eﬁect depends on whether the source of the sound is moving away from the listener or if it is moving towards the listener. If you stand at the side fo the road or train tracks then a car or train driving by will at ﬂrst be moving towards you and then away. This would mean that we would experience the biggest change in the eﬁect. We said that it eﬁects the frequency of the sound so the sounds from the car or train would sound diﬁerent, have a diﬁerent frequency, when the car is coming towards you and when it is moving away from you. Why does the frequency of the sound change when the car is moving towards or away from you? Lets convince ourselves that it must change! Imagine a source of sound waves with constant frequency and amplitude. Just like each of the points on the wave front from the Huygen’s principle section. Remember the sound waves are disturbances moving through the medium so if the source moves or stops after the sound has been emitted it can’t aﬁect the waves that have been emitted already. The Doppler shift happens when the source moves while emitting waves. So lets imagine we have the same source as above but now its moving to the right. It is emitting sound at a constant frequency and so the time between peaks of the sound waves will be constant but the position will have moved to the right. In the picture below we see that our sound source (the black circle) has emitted a peak which moves away at the same speed in all directions. The source is moving to the right so it catches up a little bit with the peak that is moving away to the right. 31 When the second peak is emitted it is from a point that has moved to the right. This means that the new or second peak is closer to the ﬂrst peak on the right but further away from the ﬂrst peak on the left. If the source continues moving at the same speed in the same direction (i.e. with the same velocity which you will learn more about later). then the distance between peaks on the right of the source is the constant. The distance between peaks on the left is also constant but they are diﬁerent on the left and right. This means that the time between peaks on the right is less so the frequency is higher. It is higher than on the left and higher than if the source were not moving at all. On the left hand side the peaks are further apart than on the right and further apart than if the source were at rest - this means the frequency is lower. So what happens when a car drives by on the freeway is that when it approaches you hear higher frequency sounds and then when it goes past you it is moving away so you hear a lower frequency. 2.4.2 Mach Cone Now we know that the waves move away from the source at the speed of sound. What happens if the source moves at the speed of sound? This means that the wave peaks on the right never get away from the source so each wave is emitted on top of the previous on on the right hand side like in the picture below. 32 If the source moves faster than the speed of sound a cone of wave fronts is created. This is called a Mach cone. Sometimes we use the speed of sound as a reference to describe the speed of the object. So if the object moves at exactly the speed of sound we can say that it moves at 1 times the speed of sound. If it moves at double the speed of sound then we can say that it moves at 2 times the speed of sound. A convention for this is to use Mach numbers, so moving at the speed of sound is Mach one (one times the speed of sound) and moving at twice the speed of sound is called Mach two (twice the speed of sound). 2.4.3 Ultra-Sound Ultrasound is sound with a frequency greater than the upper limit of human hearing, approximately 20 kilohertz. Some animals, such as dogs, dolphins, and bats, have an upper limit that is greater than that of the human ear and can hear ultrasound. Ultrasound has industrial and medical applications. Medical ultrasonography can visualise muscle and soft tissue, making them useful for scanning the organs, and obstetric ultrasonography is commonly used during pregnancy. Typical diagnostic ultrasound scanners operate in the frequency range of 2 to 13 megahertz. More powerful ultrasound sources may be used to generate local heating in biological tissue, with applications in physical therapy and cancer treatment. Focused ultrasound sources may be used to break up kidney stones. Ultrasonic cleaners, sometimes called supersonic cleaners, are used at frequencies from 20-40 kHz for jewellery, lenses and other optical parts, watches, dental instruments, surgical instruments and industrial parts. These cleaners consist of containers with a uid in which the object to be cleaned is placed. Ultrasonic waves are then sent into the uid. The main mechanism for cleaning action in an ultrasonic cleaner is actually the energy released from the collapse of millions of microscopic cavitation events occurring in the liquid of the cleaner. 33 Interesting Fact: Ultrasound generator/speaker systems are sold with claims that they frighten away rodents and insects, but there is no scientiﬂc evidence that the devices work; controlled tests have shown that rodents quickly learn that the speakers are harmless. Medical Ultrasonography Medical ultrasonography makes uses the fact that waves are partially reected when the medium in which they are moving changes density. If the density increases then the reected waves undergoes a phase shift exactly like the case where the waves in a string were reected from a ﬂxed end. If the density decreases then the reected waves has the same phase exactly like the case where the waves in a string were reected from a free end. Combining these properties of waves with modern computing technology has allowed medical professionals to develop an imaging technology to help with many aspects of diagnosis. Typical ultrasound units have a hand-held prob
e (often called a scan head) that is placed directly on and moved over the patient: a water-based gel ensures good contact between the patient and scan head. Ultrasonic waves are emitted from the scan head and sent into the body of the patient. The scan head also acts a receiver for reected waves. From detailed knowledge of interference and reection an image of the internal organs can be constructed on a screen by a computer programmed to process the reected signals. Interesting Fact: Medical ultrasonography was invented in 1953 at Lund University by cardiologist Inge Edler and Carl Hellmuth Hertz, the son of Gustav Ludwig Hertz, who was a graduate student at the department for nuclear physics. Uses Ultrasonography is widely utilized in medicine, primarily in gastroenterology, cardiology, gynaecology and obstetrics, urology and endocrinology. It is possible to perform diagnosis or therapeutic procedures with the guidance of ultrasonography (for instance biopsies or drainage of uid collections). Strengths of ultrasound imaging It images muscle and soft tissue very well and is particularly useful for ﬂnding the interfaces between solid and uid-ﬂlled spaces. It renders "live" images, where the operator can dynamically select the most useful section for diagnosing and documenting changes, often enabling rapid diagnoses. It shows the structure as well as some aspects of the function of organs. † † † 34 It has no known long-term side eﬁects and rarely causes any discomfort to the patient. Equipment is widely available and comparatively exible; examinations can be performed at the bedside. Small, easily carried scanners are available. † † † Much cheaper than many other medical imaging technology. † Weaknesses of ultrasound imaging † † † Ultrasound has trouble penetrating bone and performs very poorly when there is air between the scan head and the organ of interest. For example, overlying gas in the gastrointestinal tract often makes ultrasound scanning of the pancreas di–cult. Even in the absence of bone or air, the depth penetration of ultrasound is limited, making it di–cult to image structures that are far removed from the body surface, especially in obese patients. The method is operator-dependent. A high level of skill and experience is needed to acquire good-quality images and make accurate diagnoses. Doppler ultrasonography Ultrasonography can be enhanced with Doppler measurements, which employ the Doppler eﬁect to assess whether structures (usually blood) are moving towards or away from the probe. By calculating the frequency shift of a particular sample volume, e.g. within a jet of blood ow over a heart valve, its speed and direction can be determined and visualised. This is particularily useful in cardiovascular studies (ultrasonography of the vasculature and heart) and essential in many areas such as determining reverse blood ow in the liver vasculature in portal hypertension. The Doppler information is displayed graphically using spectral Doppler, or as an image using colour Doppler or power Doppler. It is often presented audibly using stereo speakers: this produces a very distinctive, despite synthetic, sound. 2.5 Important Equations and Quantities Frequency: f = 1 T : (2.3) Quantity Amplitude Period Wavelength Frequency Speed Symbol A T ‚ f v S.I. Units m s m or m:s¡1 Hz s¡1 Direction \| \| \| \| \| Table 2.1: Units used in Waves 35 Speed: v = f ‚ ‚ T = 36 Chapter 3 Geometrical Optics In this chapter we will study geometrical optics. Optics is the branch of physics that describes the behavior and properties of light and the interaction of light with matter. By geometrical optics we mean that we will study all the optics that we can treat using geometrical analysis (geometry), i.e. lines, angles and circles. As we have seen in the previous chapters, light propagates as a wave. The waves travel in straight lines from the source. So we will consider light as a set of rays. The wavelike nature will become apparent when the waves go from one medium to another. Using rays and Snell’s law to describe what happens when the light ray moves from one medium to another we can solve all the geometrical optics problems in this chapter. 3.1 Refraction re-looked We have seen that waves refract as they move from shallower to deeper water or vise versa, thus light also refracts as it moves between two mediums of diﬁerent densities. We may consider the parallel beams of light in Fig 1(a) as a set of wheels connected by a straight rod placed through their centres as in Fig1(b). As the wheels roll onto the grass (representing higher density), they begin to slow down, while those still on the tarmac move at a relatively faster speed. This shifts the direction of movtion towards the normal of the grasstarmac barrier. Likewise light moving from a medium of low density to a medium of high density (Fig 2) moves towards the normal, hence the angle of incidence (i) is greater than the angle of Refraction (r): i > r for d1 < d2 (NOTE TO SELF: Diagram of ray moving into more dense medium) (NOTE TO SELF: Discussion of Snell’s Law) (NOTE TO SELF: Follow by discussion of how we can reverse rays) (NOTE TO SELF: Show its all the same) 37 where both angles are measured from the normal to the ray. I is the incident ray and R is the refracted ray. Rays are reversible 3.1.1 Apparent and Real Depth: If you submerge a straight stick in water and look at the stick where it passes through the surface of the water you will notice that it looks like the stick bends. If you remove you will see that the stick did not bend. The bending of the stick is a very common example of refraction. How can we explain this? We can start with a simple object under water. We can see things because light travels from the object into our eyes where it is absorbed and sent to the brain for processing. The human brain interprets the information it receives using the fact that light travels in straight lines. This is why the stick looks bent. The light did not travel in a straight line from the stick underwater to your eye. It was refracted at the surface. Your brain assumes the light travelled in a straight line and so it intrepets the information so that it thinks the stick is in a diﬁerent place. This phenomenon is easily explained using ray diagrams as in Fig??. The real light rays are represented with a solid line while dashed lines depict the virtual rays. The real light rays undergo refraction at the surface of the water hence move away from the normal. However the eye assumes that light rays travel in straight lines, thus we extend the refracted rays until they converge to a point. These are virtual rays as in reality the light was refracted and did not originate from that point. 38 air water (NOTE TO SELF: Insert an example with water - can be worked example!) We note that the image of is seen slightly higher and ahead of the object. Where would we see the object if it was submerged in a uid denser than water? (NOTE TO SELF: Insert an example with denser medium show change - can be worked example!) 3.1.2 Splitting of White Light How is a rainbow formed? White is a combination of all other colours of light. Each colour has a diﬁerent wavelength and is thus diﬁracted through diﬁerent angles. red yellow blue The splitting of white light into its component colours may be demonstrated using a triangular prism (Fig4). White light is incident on the prism. As the white light enters the glass its component colours are diﬁracted through diﬁerent angles. This separation is further expanded as the light rays leave the prism. Why? What colour is diﬁracted the most? If red has the longest wavelength and violet the shortest, what is the relation between refraction and wavelength? As the sun appears after a rainstorm, droplets of water still remain in the atmosphere. These act as tiny prisms that split the suns light up into its component colours forming a rainbow. 39 white light red yellow blue 3.1.3 Total Internal Reection Another useful application of refraction is the periscope. We know that as light move from higher to lower density mediums, light rays tend to be diﬁracted towards the normal. We also know that the angle of refraction is greater than the angle of incidence and increases as we increase the angle of incidence (Fig 5(a),(b)). At a certain angle of incidence, c, the refracted angle equals 90o (Fig 5(c)), this angle is called the critical angle. For any angle of incidence greater than c the light will be refracted back into the incident medium, such refraction is called Total Internal reection (Fig 5(d)). A periscope uses two 90o triangular prisms as total internal reectors. Light enters the periscope and is reected by the ﬂrst prism down the chamber, where again the light is reected to the observer. This may be illustrated using a ray diagram as in (Fig 6). 3.2 Lenses Lenses are used in many aspects of technology ranging from contact lenses to projectors. We shall again use light rays to explain the properties on lenses. There are two types of lenses, namely: Converging: Lenses that cause parallel light rays to converge to a point (Fig 7). Diverging: Lenses that cause parallel light rays to diverge (Fig 8). Such deviation of light rays are caused by refraction. It is now a good time to introduce a few new deﬂnitions: Optical Centre: The centre of the lens Principal Axis: The line joining two centres of curvature of the surfaces of the lens. Focal Point: The point at which light rays, parallel to the principal axis, converge, or appear to converge, after passing through the lens. Focal Length: The distance between the optical centre and the focal point. 40 As seen in Fig 7 and 8, the focal point of a converging lens is real, while the focal point of a diverging lens is vitual. 3.2.1 Convex lenses Convex lenses are in general converging lenses. Hence they possess real focal points (with one exception, discussed later). Such focal points allow for real images to b
e formed. One may verify the above by placing an illuminated object on one side of a convex lens, ensuring that the distance between them is greater than the focal length (explained later). By placing a screen on the other side of the lens and varing the distance, one will acquire a sharp image of the object on the screen. The ray diagrams below illustrate the images formed when an object is placed a distance d from the optical centre of a convex lens with focal length F. d Image type Magniﬂcation Orientation Image Position (I) Figure >2F Real <1 Inverted F<I<2F 9(a) 2F Real 1 Inverted 2F 9(b) F<d<2F Real >1 Inverted >2F 9(c) F No Image - - - 9(d) <F Virtual >1 Original >d 9(e) 3.2.2 Concave Lenses Concave lenses disperse parallel rays of light and are hence diverging lenses. All images formed by concave lenses are virtual and placed between the lens and the object. Furthermore, the image retains its original orientation while it is smaller in size (Fig 10). B0 A0 B A 41 B0 A0 B A 3.2.3 Magniﬂcation By Deﬂnition the magniﬂcation, m, is: m = (Height of Image) / (Height of Object) however, using similar triangles, we may prove that: m = (Distance of Image) / (Distance of Object) where both distances are measured from the optical centre. The above method allows us to accurately estimate the magniﬂcation of an image. This is commonly used in a compound microscope as discussed in the next section. Worked Example 2 A n object is placed 0.5m in front of a convex a lens. † † At what distance should a screen be placed in order to create an image with a magniﬂcation of 1.5? If the height of the image and object are equal, what is the focal length of the lens? Solutions: a)m = (Distance of Image) / (Distance of Object) therefore Distance of Image = (Distance of Object) x m Distance of Image = 0.5m x 1.5 = 0.75m b)This implies that the magniﬂcation m = 1. Therefore from the above table it is seen that d = I = 2F Therefore F = d/2 = 0.25m 42 3.2.4 Compound Microscope This type of microscope uses two convex lenses. The ﬂrst creates a real magniﬂed image of the object that is in turn used by the second lens to create the ﬂnal image. This image is virtual and again enlarged. The ﬂnal image, as seen in Fig 11, is virtual, enlarged and inverted. The lens, L1 that forms the real image is called the objective lens, while L2 is referred to as the eyepiece. B A A1 B1 3.2.5 The Human Eye The eye also contains a biconvex lens that is used to focus objects onto the retina of the eye. However, in some cases the lens maybe abnormal and cause defects in ones vision. Hyperopia (Long-Sightedness) This occurs when the image is focussed beyond the retina. Hyperopia is due to the eyeball being too short or the lens not being convex enough. A convex lens is used to correct this defect (Fig12). Miopia (Short-Sightedness) Images are focussed before the retina. The lens being too convex or the eyeball being too long causes this. A concave lens corrects this defect (Fig 13). Astigmatism When this occurs, one is able to focus in one plane (e.g. vertical) but not another. This again is due to a distortion in the lens and may be cured by using relevant lenses. 43 3.3 Introduction Light is at ﬂrst, something we feel incredibly familiar with. It can make us feel warm, it allows us to see, allows mirrors and lenses to ‘work’, allows for ... Under more careful study light exhibits many fascinating and wonderful properties. The study of light has led to many important and amazing scientiﬂc discoveries and much progress. For example ﬂbre optics, lasers and CCD’s play a huge role in modern technology. † † † † Light is a form of energy. This is demonstrated by Crookes Radiometer or Solar Cells Light travels in straight lines. This is demonstrated by an experiment involving a card with an hole looking at light source e.g. candle. Also the simple camera: candle black box one side pin hole other side grease proof paper. We see the candle upside down on the paper Light travels at a constant speed. The speed depends on the medium it is in. (substance like air or water)1. Nothing travels faster than the speed of light in a vacuum c = 2:99790 of the fundamental constants in physics. £ 108. This is one Probably the most important use of light by the majority of living things on earth is that it allows them to see. If the light from an object enters our optical detectors/sensors i.e. our eyes!, we can see that object. In some cases the light originates at the object itself. Objects which give out light are said to be luminous objects e.g. a lighted candle/torch/bulb, the Sun, stars. The moon is not a luminous object! Why? Most objects however are not luminous, objects which do not give out their own light. We can see them because they reect light into our eyes. 3.4 Reection 3.4.1 Diﬁuse reection (NOTE TO SELF: diag of light rays hitting a rough surface) (NOTE TO SELF: diag of light rays hitting a polished surface) Most objects do not have perfectly smooth surfaces. Because of this diﬁerent parts of the surface reect light rays in diﬁerent directions (angles). 3.4.2 Regular reection Mirrors and highly polished surfaces give regular reection. A mirror is a piece of glass with a thin layer of silver (aluminium is commonly used) on the rear surface Light is reected according to the laws of reection. 3.4.3 Laws of reection (NOTE TO SELF: diag of i N r rays striking a mirror deﬂne i r N) Laws of reection: The angle of incidence is always equal to the angle of reection The incident ray, the normal and the reected ray are all in the same plane 1There are some substances where light moves so slowly you can walk faster than it moves - more on this later! 44 Note that the angle of incidence i is between the incident ray and the normal2 to the surface, not between the incident ray and the surface of the mirror. There are two forms of an image formed by reection: real and virtual. A real image is formed by the actual intersection of light rays. It is always inverted (upside down) and can be formed on a screen A virtual image is formed by the apparent intersection of light rays. It is always erect and cannot be formed on a screen. Images formed in plane mirrors are virtual images. Real images are formed by lenses (e.g. the image on a cinema screen) or by curved mirrors. 3.4.4 Lateral inversion Where some thing is back to front e.g. AMBULANCE (NOTE TO SELF: How a periscope works, why images are right!) 3.5 Curved Mirrors 3.5.1 Concave Mirrors (Converging Mirrors) (NOTE TO SELF: how to remember a con cave from a convex mirror) (NOTE TO SELF: diag of light rays striking a concave mirror base ray diag! deﬂne things) P = pole of mirror F = focal point (focus) C = centre of curvature \|CP\| = radius of curvature (r) \|FP\| = focal length (f) r = 2f Rules for ray tracing: Rays of light that arrive parallel to principal axis leave the surface of the mirror through the focal point Rays of light that arrive through the focal point leave the surface of the mirror parallel to principal axis Rays of light that arrive through the centre of curvature leave the surface back through the centre of curvature (NOTE TO SELF: Maybe diag for each of these above!) In order to obtain the position, nature (real or virtual) and size of the image we need just apply the rules above. The details depend on the distance of the object from the mirror surface.. Object outside C diag Image - located between c and f - inverted (upside down) - diminished (reduced in size, smaller) - real Object at C diag Image - located at c - inverted (upside down) - same size - real Object between C and f diag Image - located outside c - inverted (upside down) - magniﬂed (increased in size, larger) - real Object at f diag 2The normal is a line that is perpendicular to the surface i.e. the angle between the line and the surface is 90o 45 Image - located at inﬂnity Object between f and P diag Image - located behind the mirror - erect (right side up) - magniﬂed (increased in size, larger) - virtual 3.5.2 Convex Mirrors diag base ray diag! deﬂne things Image - located behind the mirror - erect (right side up) - diminished (reduced in size, smaller) - virtual We can also arrive at the position and nature of the image by calculation, using the following formula 1/u + 1/v = 1/f AND Magniﬂcation (=M) m = v/u (We will deal with the ‘proﬁ’ for these later) N.B. distance between f and P = f (focal length negative for convex mirror) distance between o and P = u (object distance) distance between i and P = v (image distance negative for virtual image) concave mirror f is positive convex mirror f is negative real image v is positive virtual image v is negative Lots of examples with numbers e.g. u = 10cm v = 20cm m=2 things are bigger if m > 1 (and -1) things are smaller if 0 > m > 1 (and -1) Uses of Convex mirrors image is always erect wide range of view † † They often used in shops, double decker busses, dangerous bends in roads, wing mirrors of cars Disadvantage False sense of distance (objects seem closer than they actually are) uses of concave mirrors They are usually used as make-up mirrors or shaving mirrors and in reecting telescopes Advantages if object is inside f the image is magniﬂed † MAYBE MENTION ‘Virtual object’ in convex mirror for completeness 3.5.3 Refraction When light travels from one medium to another it changes direction, except when it is incident normally on the separating surface. The change of direction is caused by the change in the velocity of the light as it passes from one medium to the other. 46 4 3 2 1 0 -1 -2 -3 -4 D A C B -4 -5 -3 diag deﬂning angles i, r and N Light is refracted according to the laws of refraction: -1 -2 2 1 0 3 4 5 3.5.4 Laws of Refraction Laws of Refraction: sin i /sin r is a constant for two given media (Snell’s Law) The incident ray, the normal and the refracted ray are all in the same plane Sin i /sin r is known as the refractive index (n) EXP verify Snell’s law EXP coin in an empty cup, move
head till it disappears. Then ﬂll with water. Can you see it? diag real apparent depth Due to refraction a body which is at O beneath the water appears to be at I when seen from above. As a result of refraction the pool appears to be 1.5m instead of 2m. The relationship between the real depth and the apparent depth is determined by the refractive index of water, which is 4/3 diag of water depth in pool n = sin i /sin r =\|AP\|/\|PI\| / \|AP\|/\|PO\| \|PO\|/\|PI\| However the above diagram is greatly exaggerated in size. In practise \|PO\| \|AO\| and \|PI\| \|AI\| So n =\|AO\|/\|AI\| == real/apparent depth! More than two media .g. air glass water The refractive index of glass is 3/2 and the refractive index of water is 4/3. The refractive £ index from water to glass a n w= 4/3 => w n a =3/=4 £ 3=2 = 9=8 £ 47 3.5.5 Total Internal Reection This can only happen when light is travelling into a less dense medium Refrectionnumbers are n1 = 2 and n2 = 11 -2 -3 6 4 3 0 5 1 2 -2 -3 -7 -1 -4 -5 -8 -6 diag air glass critical angle i r When light travels from glass to air it is refracted away from the normal as in a above. As the angle of incidence is increased the angle of refraction eventually reaches 90o as in b If the angle of incidence is increased further beyond this value total internal reection occurs as in c The critical angle c is the angle of incidence corresponding to the angle of refraction of 90o From the above the refractive index from from air to glass n=sin90/sin c = 1/sin c Also sin c = 1 / n = 2/3 in the case of glass. Sin c = 0.6667 => critical angle for glass is 41o490 Total internal reection has some very useful properties eg diag prism for turning light 180o daig prism for turning light 90o The size of the ﬂnal image in a pair of binoculars depends on the distance travelled by the light within the binoculars. By using two prisms this distance can be increased without increasing the length of the binoculars. Fibre Optics In the same Water can be directed from one place to another by conﬂning it within a pipe. way light can be directed from one place to another by conﬂning it within a single glass ﬂbre. The light is kept within the ﬂbre by total internal reection. The amount of light which can be carried by a single ﬂbre is very small so it is usual to form a light tube tapping a few thousand ﬂbres together. On great advantage of such a light tube is exibility; it can be ties in knots 48 B 7 8 and still function. However since total internal reection only occurs when light is going from a medium to a less dense medium, it is necessary ti coat each ﬂbre with glass of a lower refractive index. Otherwise light would leak from one ﬂbre at their points of contact. Light tube can be used to bring light from a lamp to an object, thus illuminating the object. A second light tube can then used to carry light from the illuminated object to an observer, thus enabling the object to be seen and photographed. The procedure has been used to photograph the digestive system the reproductive system and many other parts of the human body. In the case of the light tube carrying light from the object to the observer, it is vital that the individual ﬂbres in the tube do not cross each other, otherwise the image will become garbled. Like radio waves, light waves are electromagnetic. (cf section em) However, their shorter wavelength and higher frequency means that a single light beam can carry far more telephone conservations at one time compared with a radio wave. In the case of long ﬂbre cables it would be necessary to incorporate a device to boost the intensity of the light to make up for losses due to absorption. Nevertheless the system has great potential for the communication industry, including the possibility of transmitting pictures over long distances. As we said earlier, the reason why light bends when going from one medium to another is because of the change of velocity. This will be dealt with in more detail later. For the moment we consider only the implication for the refractive index. 1 n 2 = velocity medium 1 / velocity medium 2 3.5.6 Mirage Yet another eﬁect of refraction is the mirage. The most common mirage occurs in warm weather when motorists see what appears to be a pool of water on the road close to the horizon. The explanation is this: when air is heated it expands and becomes less dense and when it cools it contracts and becomes more dense. In summer the ground is hot and the layer of air nearest the ground is hot. The layer above that is cooler, etc etc. A of light coming from the blue sky passes down through the layers of air which are getting progressively less dense. As it does so it is progressively bent, as shown in the diagram. As a result an image of the blue sky is seen on the road and is taken to be a pool of water! 3.6 The Electromagnetic Spectrum They are transverse waves. They can travel through empty space. They travel at the speed of light. Short wavelength Long wavelength Dangers Microwaves kill living cells Ultraviolet light causes skin cancer and can kill living cells Uses Radio waves are used in TV’s and radios Microwaves are used to heat foods, they are also used to transmit signals to satellites and back to earth. Infrared light is used to asses heat loss from buildings, weather forecasting, locating survivors in earthquakes and in TV remote controls. Visible light is used in ﬂbre optics. Ultraviolet light produces vitamins in the skin if it is in small doses. Suntan. X-Rays are used to look at bones inside your body, my body and everybody’s body. g-rays are used to treat cancer. 49 3.7 Important Equations and Quantities Quantity Symbol Unit S.I. Units Direction Units or Table 3.1: Units used in Optics 50 Chapter 4 Vectors (NOTE TO SELF: SW: initially this chapter had a very mathematical approach. I have toned this down and tried to present in a logical way the techniques of vector manipulation after ﬂrst exploring the mathematical properties of vectors. Most of the PGCE comments revolved around the ommision of the graphical techniques of vector addition (i.e. scale diagrams), incline questions and equilibrium of forces. I have addressed the ﬂrst two; equilibrium of forces and the triangle law of three forces in equilibrium I think would be better oﬁ in the Forces Chapter. Also the Forces chapter should include some examples of incline planes. Inclines are introduced here merely as an example of components in action!) 4.1 PGCE Comments † † † Use a+b = c to illustrate that a = c-b under subtraction. Summarise properties of vectors at end. Mention that vector multiplication exists in two forms- neither introduced at school, but used implicitly in work and motor rules. 4.2 ‘TO DO’ LIST † † † † † Mention that the direction of a vector is measured from its tail. Also include explanations of the ways in which to specify direction (e.g. bearing, compass points, angle from vertical, etc.) Include exam-type questions involving owing rivers or aeroplanes in wind. These questions link displacement, velocity and time and also test the learners understanding of vectors combining to give a combined result (i.e. resultant). Mention that displacement at a point is the directed line from the start to that point. Rewrite section on velocity to make clear the distinction between average and instantaneous rates of change (velocity and speed). The ¢’s in the equations imply we are calculating average quantities. Mention that we take the limit of a small time interval to give instantaneous quantities. Perhaps the example of a parabola with average gradient and gradient of 51 tangent can be used as an illustration. Else defer until chapter on Graphs and Equations of Motion. Instantaneous velocity: reading on the speedometer in a direction tangent to the path. Instantaneous speed is magnitude of instantaneous velocity but average speed is not equal to magnitude of average velocity. Average speed and average velocity are the total distance and resultant displacement over the time interval related to that part of the path. The example of the circular track uses these deﬂnitions and is an important illustration of the diﬁerences. Instantaneous calculated at a certain instant in time while average is calculated over an interval. Include relative velocity. Address PGCE comments above and comments in the text. † † 4.3 Introduction \A vector is ‘something’ that has both magnitude and direction." \‘Things’ ? What sorts of ‘things’ ?" Any piece of information which contains a magnitude and a related direction can be a vector. A vector should tell you how much and which way. Consider a man driving his car east along a highway at 100 km=h. What we have given here is a vector{ the car’s velocity. The car is moving at 100 km=h (this is the magnitude) and we know where it is going{ east (this is the direction). Thus, we know the speed and direction of the car. These two quantities, a magnitude and a direction, form a vector we call velocity. Deﬂnition: A vector is a measurement which has both magnitude and direction. In physics magnitudes often have directions associated with them. If you push something it is not very useful knowing just how hard you pushed. A direction is needed too. Directions are extremely important, especially when dealing with situations more complicated than simple pushes and pulls. Diﬁerent people like to write vectors in diﬁerent ways. Anyway of writing a vector so that it has both magnitude and direction is valid. Are vectors physics? No, vectors themselves are not physics. Physics is just a description of the world around us. To describe something we need to use a language. The most common language used to describe physics is mathematics. Vectors form a very important part of the mathematical description of physics, so much so that it is absolutely essential to master the use of vectors. 4.3.1 Mathematical representation Numerous notations are commonly used to denote vectors. In this text, vectors will be denote
d by symbols capped with an arrow. As an example, ¡!s , ¡!v and ¡!F are all vectors (they have both magnitude and direction). Sometimes just the magnitude of a vector is required. In this case, is another the arrow is ommitted. In other words, F denotes the magnitude of vector ¡!F . way of representing the size of a vector. ¡!F j j 52 4.3.2 Graphical representation Graphically vectors are drawn as arrows. An arrow has both a magnitude (how long it is) and a direction (the direction in which it points). For this reason, arrows are vectors. In order to draw a vector accurately we must specify a scale and include a reference direction in the diagram. A scale allows us to translate the length of the arrow into the vector’s magnitude. For instance if one chose a scale of 1cm = 2N (1cm represents 2N ), a force of magnitude 20N would be represented as an arrow 10cm long. A reference direction may be a line representing a horizontal surface or the points of a compass. Worked Example 3 Drawing vectors Question: Using a scale of 1cm = 2m:s¡1 represent the following velocities: a) 6m:s¡1 north b) 16m:s¡1 east Answer: scale 1cm = 2m:s¡1 m c 3 a) b) W N S E 8cm 4.4 Some Examples of Vectors 4.4.1 Displacement Imagine you walked from your house to the shops along a winding path through the veld. Your route is shown in blue in Figure 12.3. Your sister also walked from the house to the shops, but she decided to walk along the pavements. Her path is shown in red and consisted of two straight stretches, one after the other. Although you took very diﬁerent routes, both you and your sister walked from the house to the shops. The overall eﬁect was the same! Clearly the shortest path from your house to the shops is along the straight line between these two points. The length of this line and the direction from the start point (the house) to the end point (the shops) forms a very special vector known as displacement. Displacement is assigned the symbol ¡!s . Deﬂnition: Displacement is deﬂned as the magnitude and direction of the straight line joining one’s starting point to one’s ﬂnal point. 53 Finish (Shop) D isplace m ent Start (House) Figure 4.1: Illustration of Displacement OR Deﬂnition: Displacement is a vector with direction pointing from some initial (starting) point to some ﬂnal (end) point and whose magnitude is the straight-line distance from the starting point to the end point. (NOTE TO SELF: choose one of the above) In this example both you and your sister had the same displacement. This is shown as the black arrow in Figure 12.3. Remember displacement is not concerned with the actual path taken. It is only concerned with your start and end points. It tells you the length of the straight-line path between your start and end points and the direction from start to ﬂnish. The distance travelled is the length of the path followed and is a scalar (just a number). Note that the magnitude of the displacement need not be the same as the distance travelled. In this case the magnitude of your displacement would be considerably less than the actual length of the path you followed through the veld! 4.4.2 Velocity Deﬂnition: Velocity is the rate of change of displacement with respect to time. The terms rate of change and with respect to are ones we will use often and it is important that you understand what they mean. Velocity describes how much displacement changes for a certain change in time. We usually denote a change in something with the symbol ¢ (the Greek letter Delta). You have probably seen this before in maths{ the gradient of a straight line is ¢y ¢x . The gradient is just how much y changes for a certain change in x. In other words it is just the rate of change of y with respect to x. This means that velocity must be ¡!v = ¢¡!s ¢t = ¡!s f inal ¡ ¡!s initial tinitial tf inal ¡ : (4.1) 54 (NOTE TO SELF: This is actually average velocity. For instantaneous ¢’s change to differentials. Explain that if ¢ is large then we have average velocity else for inﬂnitesimal time interval instantaneous!) What then is speed? Speed is how quickly something is moving. How is it diﬁerent from velocity? Speed is not a vector. It does not tell you which direction something is moving, only how fast. Speed is the magnitude of the velocity vector (NOTE TO SELF: instantaneous speed is the magnitude of the instantaneous velocity.... not true of averages!). Consider the following example to test your understanding of the diﬁerences between velocity and speed. Worked Example 4 Speed and Velocity Question: A man runs around a circular track of radius 100m. It takes him 120s to complete a revolution of the track. If he runs at constant speed calculate: 1. his speed, 2. his instantaneous velocity at point A, 3. his instantaneous velocity at point B, 4. his average velocity between points A and B, 5. his average velocity during a revolution. A 100m B W N S E Direction the man runs Answer: 1. To determine the man’s speed we need to know the distance he travels and how long it takes. We know it takes 120s to complete one revolution of the track. Step 1 : First we ﬂnd the distance the man travels What distance is one revolution of the track? We know the track is a circle and we know its radius, so we can determine the perimeter or distance around the circle. We start with the equation for the circumference of a circle C = 2…r = 2…(100m) = 628:3 m: So we know the distance the man covers in one revolution is 628:3 m. 55 Step 2 : Now we determine speed from the distance and time. We know that speed is distance covered per unit time. So if we divide the distance covered by the time it took we will know how much distance was covered for every unit of time. v = Distance travelled time taken = 628:3m 120s = 5:23 m:s¡1 2. Step 3 : Determine his instantaneous velocity at A Consider the point A in the diagram. A Direction the man runs We know which way the man is running around the track and we know his speed. His velocity at point A will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of his velocity). The instant that he arrives at A he is moving as indicated in the diagram below. A So his velocity vector will be 5:23 m:s¡1 West. 3. Step 4 : Determine his instantaneous velocity at B Consider the point B in the diagram. B Direction the man runs 56 We know which way the man is running around the track and we know his speed. His velocity at point B will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of his velocity). The instant that he arrives at B he is moving as indicated in the diagram below. B So his velocity vector will be 5:23 m:s¡1 South. 4. Step 5 : Now we determine his average velocity between A and B (NOTE TO SELF: add this here to further stress the diﬁerence between average and instantaneous velocities, as well as the diﬁerence between magnitude of average velocity and average speed!) 5. Step 6 : Now we calculate his average velocity over a complete revolution. The deﬂnition of average velocity is given earlier and requires that you know the total displacement and the total time. The total displacement for a revolution is given by the vector from the initial point to the ﬂnal point. If the man runs in a circle then he ends where he started. This means the vector from his initial point to his ﬂnal point has zero length. So a calculation of his average velocity follows: ¡!v = ¢¡!s ¢t 0m 120s = 0 m:s¡1 = Remember displacement can be zero when even is distance not! 4.4.3 Acceleration Deﬂnition: Acceleration is the rate of change of velocity with respect to time. Acceleration is also a vector. Remember that velocity was the rate of change of displacement with respect to time so we expect the velocity and acceleration equations to look very similar. In fact, ¡!a = ¢¡!v ¢t = ¡!v f inal ¡ ¡!v initial tinitial tf inal ¡ (4.2) (NOTE TO SELF: average and instantaneous distiction again! expand further- what does it mean.) Acceleration will become very important later when we consider forces. 57 Using vectors is an important skill you MUST master! 4.5 Mathematical Properties of Vectors Vectors are mathematical objects and we will use them to describe physics in the language of mathematics. However, ﬂrst we need to understand the mathematical properties of vectors (e.g. how they add and subtract). We will now use arrows representing displacements to illustrate the properties of vectors. Remember that displacement is just one example of a vector. We could just as well have decided to use forces to illustrate the properties of vectors. 4.5.1 Addition of Vectors If we deﬂne a displacement vector as 2 steps in the forward direction and another as 3 steps in the forward direction then adding them together would mean moving a total of 5 steps in the forward direction. Graphically this can be seen by ﬂrst following the ﬂrst vector two steps forward and then following the second one three steps forward: 2 steps + 3 steps = = 5 steps We add the second vector at the end of the ﬂrst vector, since this is where we now are after the ﬂrst vector has acted. The vector from the tail of the ﬂrst vector (the starting point) to the head of the last (the end point) is then the sum of the vectors. This is the tail-to-head method of vector addition. As you can convince yourself, the order in which you add vectors does not matter. In the example above, if you decided to ﬂrst go 3 steps forward and then another 2 steps forward, the end result would still be 5 steps forward. The ﬂnal answer when adding vectors is called the resultant. Deﬂnition: The resultant of a number of vectors is the single vector whose eﬁect is the same as the individual vectors acting together. In other words, the individual vectors can be replaced by the resultant{ the overall eﬁect is the same. If vectors ¡!a and ¡!b have a resultant ¡!R , this can be represented mathematically as, ¡!R = ¡!a + ¡!b : Let us consider some more examples o
f vector addition using displacements. The arrows tell you how far to move and in what direction. Arrows to the right correspond to steps forward, while arrows to the left correspond to steps backward. Look at all of the examples below and check them. 1 step 1 step 1 step 1 step + + = = 2 steps 2 steps = = 2 steps 2 steps 58 Let us test the ﬂrst one. It says one step forward and then another step forward is the same as an arrow twice as long{ two steps forward. It is possible that you end up back where you started. In this case the net result of what you have done is that you have gone nowhere (your start and end points are at the same place). In this case, your resultant displacement is a vector with length zero units. We use the symbol ¡!0 to denote such a vector: 1 step 1 step 1 step 1 step + + = = 1 step 1 step 1 step 1 step = ¡!0 = ¡!0 Check the following examples in the same way. Arrows up the page can be seen as steps left and arrows down the page as steps right. Try a couple to convince yourself! + = = + = = + = = ¡!0 + = = ¡!0 It is important to realise that the directions aren’t special{ ‘forward and backwards’ or ‘left and right’ are treated in the same way. The same is true of any set of parallel directions: + = = + = = + = = ¡!0 + = = ¡!0 In the above examples the separate displacements were parallel to one another. However the same ‘tail-to-head’ technique of vector addition can be applied to vectors in any direction. + = = + = = 59 + = = Now you have discovered one use for vectors; describing resultant displacement{ how far and in what direction you have travelled after a series of movements. Although vector addition here has been demonstrated with displacements, all vectors behave in exactly the same way. Thus, if given a number of forces acting on a body you can use the same method to determine the resultant force acting on the body. We will return to vector addition in more detail later. 4.5.2 Subtraction of Vectors What does it mean to subtract a vector? Well this is really simple; if we have 5 apples and we subtract 3 apples, we have only 2 apples left. Now lets work in steps; if we take 5 steps forward and then subtract 3 steps forward we are left with only two steps forward: 5 steps - 3 steps 2 steps = What have we done? You originally took 5 steps forward but then you took 3 steps back. That backward displacement would be represented by an arrow pointing to the left (backwards) with length 3. The net result of adding these two vectors is 2 steps forward: 5 steps + 3 steps 2 steps = Thus, subtracting a vector from another is the same as adding a vector in the opposite direction (i.e. subtracting 3 steps forwards is the same as adding 3 steps backwards). This suggests that in this problem arrows to the right are positive and arrows to the left are negative. More generally, vectors in opposite directions diﬁer in sign (i.e. if we deﬂne up as positive, then vectors acting down are negative). Thus, changing the sign of a vector simply reverses its direction: - = - = - = - = - = - = 60 In mathematical form, subtracting ¡!a from ¡!b gives a new vector ¡!c : ¡!c = ¡!b ¡ ¡!a = ¡!b + ( ¡¡!a ) This clearly shows that subtracting vector ¡!a from ¡!b is the same as adding ( Look at the following examples of vector subtraction. ¡¡!a ) to ¡!b . - = + = ¡!0 - = + = 4.5.3 Scalar Multiplication What happens when you multiply a vector by a scalar (an ordinary number)? Going back to normal multiplication we know that 2 2 is just 2 groups of 2 added together to give 4. We can adopt a similar approach to understand how vector multiplication works. £ 2 x = + = 4.6 Techniques of Vector Addition Now that you have been acquainted with the mathematical properties of vectors, we return to vector addition in more detail. There are a number of techniques of vector addition. These techniques fall into two main categories- graphical and algebraic techniques. 4.6.1 Graphical Techniques Graphical techniques involve drawing accurate scale diagrams to denote individual vectors and their resultants. We next discuss the two primary graphical techniques, the tail-to-head technique and the paralelogram method. The Tail-to-head Method In describing the mathematical properties of vectors we used displacements and the tail-tohead graphical method of vector addition as an illustration. In the tail-to-head method of vector addition the following strategy is followed: † † Choose a scale and include a reference direction. Choose any of the vectors to be summed and draw it as an arrow in the correct direction and of the correct length{ remember to put an arrowhead on the end to denote its direction. 61 † † † Take the next vector and draw it as an arrow starting from the arrowhead of the ﬂrst vector in the correct direction and of the correct length. Continue until you have drawn each vector{ each time starting from the head of the previous vector. In this way, the vectors to be added are drawn one after the other tail-to-head. The resultant is then the vector drawn from the tail of the ﬂrst vector to the head of the last. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram. Worked Example 5 Tail-to-Head Graphical Addition I Question: A ship leaves harbour H and sails 6km north to port A. From here the ship travels 12km east to port B, before sailing 5:5km south-west to port C. Determine the ship’s resultant displacement using the tail-to-head technique of vector addition. Answer: Now, we are faced with a practical issue: in this problem the displacements are too large to draw them their actual length! Drawing a 2km long arrow would require a very big book. Just like cartographers (people who draw maps), we have to choose a scale. The choice of scale depends on the actual question{ you should choose a scale such that your vector diagram ﬂts the page. Before choosing a scale one should always draw a rough sketch of the problem. In a rough sketch one is interested in the approximate shape of the vector diagram. Step 1 : Draw a rough sketch of the situation Its easy to understand the problem if we ﬂrst draw a quick sketch. N A 6km H W E S B 45o 5:5km 12km 62 C In a rough sketch one should include all of the information given in the problem. All of the magnitudes of the displacements are shown and a compass has been included as a reference direction. Step 2 : Next we choose a scale for our vector diagram It is clear from the rough sketch that choosing a scale where 1cm represents 1km (scale: 1cm = 1km) would be a good choice in this problem ){ the diagram will then take up a good fraction of an A4 page. We now start the accurate construction. Step 3 : Now we construct our scaled vector diagram Contruction Step 1: Starting at the harbour H we draw the ﬂrst vector 6cm long in the direction north (remember in the diagram 1cm represents 1km): m c 6 A H W N S E 1cm = 1km Construction Step 2: Since the ship is now at port A we draw the second vector 12cm long starting from this point in the direction east: 63 A m c 6 H 12cm B W N S E 1cm = 1km 64 Construction Step 3: Since the ship is now at port B we draw the third vector 5:5cm long starting from this point in the direction south-west. A protractor is required to measure the angle of 45o. A m c 6 H 12cm B 45o C 5.5c m W N S 1cm = 1km E Construction Step 4: As a ﬂnal step we draw the resultant displacement from the starting point (the harbour H) to the end point (port C). We use a ruler to measure the length of this arrow and a protractor to determine its direction A m c 6 H 12cm B 45o C 5.5c m W N S 1cm = 1km E 75:4o 8 . 3 8 c m 65 Step 4 : Apply the scale conversion We now use the scale to convert the length of the resultant in the scale diagram to the actual displacement in the problem. Since we have chosen a scale of 1cm = 1km in this problem the resultant has a magnitude of 8:38 km. The direction can be speciﬂed in terms of the angle measured either as 75:4o east of north or on a bearing of 75:4o. Step 5 : Quote the ﬂnal answer The resultant displacement of the ship is 8:38 km on a bearing of 75:4o! Worked Example 6 Tail-to-Head Graphical Addition II Question: A man walks 40 m East, then 30 m North. a) What was the total distance he walked? b) What is his resultant displacement? N S E W m 0 3 40m Answer: Step 1 : a) Determine the distance that the man traveled In the ﬂrst part of his journey he traveled 40 m and in the second part he traveled 30 m. This gives us a total distance traveled of 40 + 30 = 70 m. Step 2 : b) Determine his resultant displacement - start by drawing a rough sketch The man’s resultant displacement is the vector from where he started to where he ended. It is the sum of his two separate displacements. We will use the tail-to-head method of accurate construction to ﬂnd this vector. Here is our rough sketch: 66 N S E W m 0 3 R es ulta n t 40m Step 3 : Choose a suitable scale A scale of 1cm represents 5m (1cm = 5m) is a good choice here. Now we can begin the process of construction. Step 4 : Draw the ﬂrst vector according to scale We draw the ﬂrst displacement as an arrow 8cm long (according to the scale 8cm = 8 5m = 40m) in the direction east: £ W N S E 1cm = 5m 8cm Step 5 : Draw the second vector according to scale Starting from the head of the ﬂrst vector we draw the second displacement as an 5m = 30m) in the direction north: arrow 6cm long (according to the scale 6cm = 6 £ 67 W N S E 1cm = 5m m c 6 8cm Step 6 : Determine the resultant vector Now we connect the starting point to the end point and measure the length and direction of this arrow (the resultant) W N S E 1cm = 5m 36:9o m c 6 1 0 c m 8cm Step 7 : Apply the scale conversion Finally we use the scale to convert the length of the resultant in the scale diagram to the actual magnitude of the resultant displacement. According to the chosen scale 1cm = 5m. Therefore 10cm represents 50m. The resultant displacement is then 50m 36:9o
north of east. 68 The Parallelogram Method When needing to ﬂnd the resultant of two vectors another graphical technique can be applied- the parallelogram method. The following strategy is employed: † † † † † Choose a scale and a reference direction. Choose either of the vectors to be added and draw it as an arrow of the correct length in the correct direction. Draw the second vector as an arrow of the correct length in the correct direction from the tail of the ﬂrst vector. Complete the parallelogram formed by these two vectors. The resultant is then the diagonal of the parallelogram. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram. Worked Example 7 Parallelogram Method of Graphical Addition I Question: A force of F1 = 5N is applied to a block in a horizontal direction. A second force F2 = 4N is applied to the object at an angle of 30o above the horizontal. N 4 = F 2 30o F1 = 5N Determine the resultant force acting on the block using the parallelogram method of accurate construction. Answer: Step 1 : Firstly make a rough sketch of the vector diagram N 4 30o 5N 69 Step 2 : Choose a suitable scale In this problem a scale of 1cm = 0:5N would be appropriate, since then the vector diagram would take up a reasonable fraction of the page. We can now begin the accurate scale diagram. Step 3 : Draw the ﬂrst scaled vector Let us draw F1 ﬂrst. According to the scale it has length 10cm: 10cm 1cm = 0:5N Step 4 : Draw the second scaled vector Next we draw F2. According to the scale it has length 8cm. We make use of a protractor to draw this vector at 30o to the horizontal: c m 8 30o 10cm Step 5 : Determine the resultant vector Next we complete the parallelogram and draw the diagonal: 1cm = 0:5N c m 8 1 7 : 4 c m 13:3o 10cm 1cm = 0:5N Step 6 : Apply the scale conversion Finally we use the scale to convert the measured length into the actual magnitude. Since 1cm = 0:5N , 17:4cm represents 8:7N . Therefore the resultant force is 8:7N at 13:3o above the horizontal. 70 The parallelogram method is restricted to the addition of just two vectors. However, it is arguably the most intuitive way of adding two forces acting at a point. 4.6.2 Algebraic Addition and Subtraction of Vectors Vectors in a Straight Line Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or some acting up and others down) you can use a very simple algebraic technique: † † † Choose a positive direction. As an example, for situations involving displacements in the directions west and east, you might choose west as your positive direction. In that case, displacements east are negative. Next simply add (or subtract) the vectors with the appropriate signs. As a ﬂnal step the direction of the resultant should be included in words (positive answers are in the positive direction, while negative resultants are in the negative direction). Let us consider a couple of examples. Worked Example 8 Adding vectors algebraically I Question: A tennis ball is rolled towards a wall which is 10m away to the right. If after striking the wall the ball rolls a further 2:5m along the ground to the left, calculate algebraically the ball’s resultant displacement. (NOTE TO SELF: PGCE suggest a ‘more real looking’ diagram, followed by a diagram one would draw to solve the problem (like our existing one with the positive direction shown as an arrow)) Answer: Step 1 : Draw a rough sketch of the situation 10 m 2.5 m Wall Start Step 2 : Decide which method to use to calculate the resultant We know that the resultant displacement of the ball (¡!s resultant) is equal to the sum of the ball’s separate displacements (¡!s 1 and ¡!s 2): ¡!s resultant = ¡!s 1 + ¡!s 2 Since the motion of the ball is in a straight line (i.e. the ball moves left and right), we can use the method of algebraic addition just explained. 71 Step 3 : Choose a positive direction Let’s make to the right the positive direction. This means that to the left becomes the negative direction. Step 4 : Now deﬂne our vectors algebraically With right positive: ¡!s 1 = +10:0m and ¡!s 2 = 2:5m ¡ Step 5 : Add the vectors Next we simply add the two displacements to give the resultant: ¡!s resultant = (+10m) + ( 2:5m) ¡ = (+7:5)m Step 6 : Quote the resultant Finally, in this case right means positive so: ¡!s resultant = 7:5m to the right Let us consider an example of vector subtraction. Worked Example 9 Subtracting vectors algebraically I Question: Suppose that a tennis ball is thrown horizontally towards a wall at 3m:s¡1 to the right. After striking the wall, the ball returns to the thrower at 2m:s¡1. Determine the change in velocity of the ball. Answer: Step 1 : Draw a sketch A quick sketch will help us understand the problem (NOTE TO SELF: Maybe a sketch here?) Step 2 : Decide which method to use to calculate the resultant Remember that velocity is a vector. The change in the velocity of the ball is equal to the diﬁerence between the ball’s initial and ﬂnal velocities: ¢¡!v = ¡!v f inal ¡ ¡!v initial Since the ball moves along a straight line (i.e. left and right), we can use the algebraic technique of vector subtraction just discussed. Step 3 : Choose a positive direction Let’s make to the right the positive direction. This means that to the left becomes the negative direction. 72 Step 4 : Now deﬂne our vectors algebraically With right positive: ¡!v initial = +3m:s¡1 and ¡!v f inal = 2m:s¡1 ¡ Step 5 : Subtract the vectors Thus, the change in velocity of the ball is: ¢¡!v = ( = ( 2m:s¡1) 5)m:s¡1 ¡ ¡ (+3m:s¡1) ¡ Step 6 : Quote the resultant Remember that in this case right means positive so: ¢¡!v = 5m:s¡1 to the left Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line. A More General Algebraic technique In worked example 3 the tail to head method of accurate construction was used to determine the resultant displacement of a man who travelled ﬂrst east and then north. However, the man’s resultant can be calculated without drawing an accurate scale diagram. Let us revisit this example. Worked Example 10 An Algebraic solution to Worked Example 3 Question: A man walks 40 m East, then 30 m North. Calculate the man’s resultant displacement. Answer: Step 1 : Draw a rough sketch As before, the rough sketch looks as follows: 73 N S E W m 0 3 R es ulta n t ﬁ 40m Step 2 : Determine the length of the resultant Note that the triangle formed by his separate displacement vectors and his resultant displacement vector is a right-angle triangle. We can thus use Pythogoras’ theorem to determine the length of the resultant. If the length of the resultant vector is called s then: s2 = (40m)2 + (30m)2 s2 = 2500m2 s = 50m Step 3 : Determine the direction of the resultant Now we have the length of the resultant displacement vector but not yet its direction. To determine its direction we calculate the angle ﬁ between the resultant displacement vector and East. We can do this using simple trigonometry: tan ﬁ = tan ﬁ = opposite adjacent 30 40 ﬁ = arctan(0:75) ﬁ = 36:9o Step 4 : Quote the resultant Our ﬂnal answer is then: Resultant Displacement: 50 m at 36:9o North of East This is exactly the same answer we arrived at after drawing a scale diagram! In the previous example we were able to use simple trigonometry to calculate a man’s resultant displacement. This was possible since the man’s directions of motion were perpendicular (north 74 and east). Algebraic techniques, however, are not limited to cases where the vectors to be combined are along the same straight line or at right angles to one another. The following example illustrates this. Worked Example 11 Further example of vector addition by calculation Question: A man walks from point A to point B which is 12km away on a bearing of 45o. From point B the man walks a further 8km east to point C. Calculate the man’s resultant displacement. Answer: Step 1 : Draw a rough sketch of the situation 8km C B 45o G 12k m F A 45o B ^AF = 45o since the man walks initially on a bearing of 45o. Then, A ^BG = B ^AF = 45o (alternate angles parallel lines). Both of these angles are included in the rough sketch. Step 2 : Calculate the length of the resultant The resultant is the vector AC. Since we know both the lengths of AB and BC and the included angle A ^BC, we can use the cosine rule: AC 2 = AB2 + BC 2 = (12)2 + (8)2 = 343:8 2 2 ¢ ¢ ¡ ¡ AC = 18:5 km BC cos(A ^BC) AB (12)(8) cos(135o) ¢ Step 3 : Determine the direction of the resultant Next we use the sine rule to determine the angle : 75 sin 8 = sin = sin 1350 18:5 8 £ sin 135o 18:5 = arcsin(0:3058) = 17:8o Thus, F ^AC = 62:8o. Step 4 : Quote the resultant Our ﬂnal answer is then: Resultant Displacement: 18:5km on a bearing of 62:8o 4.7 Components of Vectors In the discussion of vector addition we saw that a number of vectors acting together can be combined to give a single vector (the resultant). In much the same way a single vector can be broken down into a number of vectors which when added give that original vector. These vectors which sum to the original are called components of the original vector. The process of breaking a vector into its components is called resolving into components. While summing a given set of vectors gives just one answer (the resultant), a single vector can be resolved into inﬂnitely many sets of components. In the diagrams below the same black vector is resolved into diﬁerent pairs of components. These components are shown in red. When added together the red vectors give the original black vector (i.e. the original vector is the resultant of its components). In practice it is most useful to resolve a vector into components which are at right angles to one another. Worked Example 12 Resolving a vector into components 76 Question: A motorist undergoes a displacement of 250km in a direction 30o north of east. Resolve this di
splacement into components in the directions north (¡!s N ) and east (¡!s E). Answer: Step 1 : Draw a rough sketch of the original vector W N S 30o E k m 0 5 2 Step 2 : Determine the vector component Next we resolve the displacement into its components north and east. Since these directions are orthogonal to one another, the components form a right-angled triangle with the original displacement as its hypotenuse 30o ¡!s E N ¡!s Notice how the two components acting together give the orginal vector as their resultant. Step 3 : Determine the lengths of the component vectors Now we can use trigonometry to calculate the magnitudes of the components of the original displacement: sN = 250 sin 30o = 125 km 77 and sE = 250 cos 30o = 216:5 km Remember sN and sE are the magnitudes of the components{ they are in the directions north and east respectively. (NOTE TO SELF: SW: alternatively these results can be arrived at by construction. Include?) 4.7.1 Block on an incline As a further example of components let us consider a block of mass m placed on a frictionless surface inclined at some angle to the horizontal. The block will obviously slide down the incline, but what causes this motion? The forces acting on the block are its weight mg and the normal force N exerted by the surface on the object. These two forces are shown in the diagram below. N W k ? W mg Now the object’s weight can be resolved into components parallel and perpendicular to the inclined surface. These components are shown as red arrows in the diagram above and are at right angles to each other. The components have been drawn acting from the same point. Applying the parallelogram method, the two components of the block’s weight sum to the weight vector. To ﬂnd the components in terms of the weight we can use trigonometry: Wk = mg sin W? = mg cos The component of the weight perpendicular to the slope W? exactly balances the normal force N exerted by the surface. The parallel component, however, Wk is unbalanced and causes the block to slide down the slope. 78 Figure 4.2: An example of two vectors being added to give a resultant 4.7.2 Vector addition using components In Fig 4.3 two vectors are added in a slightly diﬁerent way to the methods discussed so far. It might look a little like we are making more work for ourselves, but in the long run things will be easier and we will be less likely to go wrong. In Fig 4.3 the primary vectors we are adding are represented by solid lines and are the same vectors as those added in Fig 4.2 using the less complicated looking method. Each vector can be broken down into a component in the x-direction and one in the ydirection. These components are two vectors which when added give you the original vector as the resultant. Look at the red vector in ﬂgure 4.3. If you add up the two red dotted ones in the x-direction and y-direction you get the same vector. For all three vectors we have shown their respective components as dotted lines in the same colour. But if we look carefully, addition of the x components of the two original vectors gives the x component of the resultant. The same applies to the y components. So if we just added all the components together we would get the same answer! This is another important property of vectors. Worked Example 13 Adding Vectors Using Components Question: Lets work through the example shown in Fig. 4.3 to determine the resultant. Answer: Step 1 : Decide how to tackle the problem The ﬂrst thing we must realise is that the order that we add the vectors does not matter. Therefore, we can work through the vectors to be added in any order. 79 Figure 4.3: Components of vectors can be added as well as the vectors themselves Step 2 : Resolve the red vector into components Let us start with the bottom vector. If you are told that this vector has a length of 5:385 units and an angle of 21:8o to the horizontal then we can ﬂnd its components. We do this by using known trigonometric ratios. First we ﬂnd the vertical or y component: sin = y hypotenuse sin(21:8) = y 5:385 y = 5:385 sin(21:8 Secondly we ﬂnd the horizontal or x component: x hypotenuse cos = cos(21:8) = x 5:385 x = 5:385 cos(21:8) x = 5 80 We now know the lengths of the sides of the triangle for which our vector is the hypotenuse. If you look at these sides we can assign them directions given by the dotted arrows. Then our original red vector is just the sum of the two dotted vectors (its components). When we try to ﬂnd the ﬂnal answer we can just add all the dotted vectors because they would add up to the two vectors we want to add. Step 3 : Now resolve the second vector into components . The green vector has a length of 5 units and a direction of 53.13 degrees to the horizontal so we can ﬂnd its components. sin(53:13) = sin = y hypotenuse y 5 y = 5 sin(53:13) y = 4 5 4 3 cos(53:13) = cos = x hypotenuse x 5 x = 5 cos(53:13) x = 3 Step 4 : Determine the components of the resultant vector Now we have all the components. If we add all the x-components then we will have the x-component of the resultant vector. Similarly if we add all the y-components then we will have the y-component of the resultant vector. The x-components of the two vectors are 5 units right and then 3 units right. This gives us a ﬂnal x-component of 8 units right. The y-components of the two vectors are 2 units up and then 4 units up. This gives us a ﬂnal y-component of 6 units up. Step 5 : Determine the magnitude and direction of the resultant vector Now that we have the components of the resultant, we can use Pythagoras’ theorem to determine the length of the resultant. Let us call the length of the hypotenuse l and we can calculate its value; 81 l2 = (6)2 + (8)2 l2 = 100 l = 10: 8 6 1 0 The resultant has length of 10 units so all we have to do is calculate its direction. We can specify the direction as the angle the vectors makes with a known direction. To do this you only need to visualise the vector as starting at the origin of a coordinate system. We have drawn this explicity below and the angle we will calculate is labeled ﬁ. 82 Using our known trigonometric ratios we can calculate the value of ﬁ; tan ﬁ = 6 8 ﬁ = arctan ﬁ = 36:8o: 6 8 Step 6 : Quote the ﬂnal answer Our ﬂnal answer is a resultant of 10 units at 36:8o to the positive x-axis. 4.8 Do I really need to learn about vectors? Are they really useful? Vectors are essential to do physics. Absolutely essential. This is an important warning. If something is essential we had better stop for a moment and make sure we understand it properly. 4.9 Summary of Important Quantities, Equations and Con- cepts Vector A vector is a measurement which has both magnitude and direction. Displacement Displacement is a vector with direction pointing from some initial (starting) point to some ﬂnal (end) point and whose magnitude is the straight-line distance from the starting point to the ﬂnal point. 83 Distance The distance travelled is the length of your actual path. Velocity Velocity is the rate of change of displacement with respect to time. Acceleration Acceleration is the rate of change of velocity with respect to time. Resultant The resultant of a number of vectors is the single vector whose eﬁect is the same as the individual vectors acting together. Quantity Displacement Velocity Distance Speed Acceleration Symbol ¡!s ¡!u ,¡!v d v ¡!a S.I. Units Direction m m:s¡1 m m:s¡1 m:s¡2 X X { { X Table 4.1: Summary of the symbols and units of the quantities used in Vectors 84 Chapter 5 Forces 5.1 ‘TO DO’ LIST introduce concept of a system for use in momentum- in NIII for instance, also concept of isolated system and external forces add incline plane examples generally ‘bulk-up’ the Newton’s Laws sections † † † 5.2 What is a force? The simplest answer is to say a ‘push’ or a ‘pull’. If the force is great enough to overcome friction the object being pushed or pulled will move. We could say a force is something that makes objects move. Actually forces give rise to accelerations! In fact, the acceleration of a body is directly proportional to the net force acting on it. The word net is important{ forces are vectors and what matters in any situation is the vector sum of all the forces acting on an object. The unit of force is the newton (symbol N ). It is named after Sir Isaac Newton, whose three laws you will learn about shortly. Interesting Fact: Force was ﬂrst described by Archimedes. Archimedes of Syracuse (circa 287 BC - 212 BC), was a Greek mathematician, astronomer, philosopher, physicist and engineer. He was killed by a Roman soldier during the sack of the city, despite orders from the Roman general, Marcellus, that he was not to be harmed. 5.3 Force diagrams The resultant force acting on an object is the vector sum of the set of forces acting on that one object. It is very important to remember that all the forces must be acting on the same object. 85 The easiest way to determine this resultant force is to construct what we call a force diagram. In a force diagram we represent the object by a point and draw all the force vectors connected to that point as arrows. Remember from Chapter ?? that we use the length of the arrow to indicate the vector’s magnitude and the direction of the arrow to show which direction it acts in. The second step is to rearrange the force vectors so that it is easy to add them together and ﬂnd the resultant force. Let us consider an example to get started: Two people push on a box from opposite sides with a force of 5N . 5N 5N When we draw the force diagram we represent the box by a dot. The two forces are represented by arrows, with their tails on the dot. Force Diagram: 5N 5N See how the arrows point in opposite directions and have the same magnitude (length). This means that they cancel out and there is no net force acting on the object. This result can be obtained algebraically too, since the two forces act along the same line. Firstly we choose a positive direction and then add the two vector
s taking their directions into account. Consider to the right as the positive direction Fres = (+5N ) + ( = 0N ¡ 5N ) As you work with more complex force diagrams, in which the forces do not exactly balance, 2N ). What does this mean? you may notice that sometimes you get a negative answer (e.g. Does it mean that we have something the opposite of force? No, all it means is that the force acts in the opposite direction to the one that you chose to be positive. You can choose the positive direction to be any way you want, but once you have chosen it you must keep it. ¡ Once a force diagram has been drawn the techniques of vector addition introduced in the previous chapter can be implemented. Depending on the situation you might choose to use a graphical technique such as the tail-to-head method or the parallelogram method, or else an algebraic approach to determine the resultant. Since force is a vector all of these methods apply! Always remember to check your signs Worked Example 14 Single Force on a block Question: A block on a frictionless at surface weighs 100N . A 75N force is applied to the block towards the right. What is the net force (or resultant force) on the block? Answer: 86 Step 1 : Firstly, draw a force diagram for the block Fnormal = 100N Fapplied = 75N Fweight = 100N Be careful not to forget the two forces perpendicular to the surface. Every object with mass is attracted to the centre of the earth with a force (the object’s weight). However, if this were the only force acting on the block in the vertical direction then the block would fall through the table to the ground. This does not happen because the table exerts an upward force (the normal force) which exactly balances the object’s weight. Step 2 : Answer Thus, the only unbalanced force is the applied force. This applied force is then the resultant force acting on the block. 5.4 Equilibrium of Forces At the beginning of this chapter it was mentioned that resultant forces cause objects to accelerate. If an object is stationary or moving at constant velocity then either, no forces are acting on the object, or † the forces acting on that object are exactly balanced. † A resultant force would cause a stationary object to start moving or an object moving at constant velocity to speed up or slow down. In other words, for stationary objects or objects moving with constant velocity, the resultant force acting on the object is zero. The object is said to be in equilibrium. If a resultant force acts on an object then that object can be brought into equilibrium by applying an additional force that exactly balances this resultant. Such a force is called the equilibrant and is equal in magnitude but opposite in direction to the original resultant force acting on the object. Deﬂnition: The equilibrant of any number of forces is the single force required to produce equilibrium. 87 Objects at rest or moving with constant velocity are in equilibrium and have a zero resultant force Resultant of ¡!F1 and ¡!F2 ¡!F1 ¡!F2 ¡!F3 Equilibrant of ¡!F1 and ¡!F2 In the ﬂgure the resultant of ¡!F1 and ¡!F2 is shown in red. The equilibrant of ¡!F1 and ¡!F2 is then the vector opposite in direction to this resultant with the same magnitude (i.e. ¡!F3). ¡!F1, ¡!F2 and ¡!F3 are in equilibrium ¡!F3 is the equilibrant of ¡!F1 and ¡!F2 † † ¡!F1 and ¡!F2 are kept in equilibrium by ¡!F3 † As an example of an object in equilibrium, consider an object held stationary by two ropes in the arrangement below: 50o 40o Rope 1 Rope 2 Let us draw a force diagram for the object. In the force diagram the object is drawn as a dot and all forces acting on the object are drawn in the correct directions starting from that dot. In this case, three forces are acting on the object. 88 50o ¡!T1 40o ¡!T2 ¡!W Each rope exerts a force on the object in the direction of the rope away from the object. These tension forces are represented by ¡!T1 and ¡!T2. Since the object has mass, it is attracted towards the centre of the earth. This weight is represented in the force diagram as ¡!W . Since the object is stationary, the resultant force acting on the object is zero. In other words the three force vectors drawn tail-to-head form a closed triangle: 40o ¡!T2 50o ¡!T1 ¡!W In general, when drawn tail-to-head the forces acting on an object in equilibrium form a closed ﬂgure with the head of the last vector joining up with the tail of the ﬂrst vector. When only three forces act on an object this closed ﬂgure is a triangle. This leads to the triangle law for three forces in equilibrium: Triangle Law for Three Forces in Equilibrium: Three forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in order. Worked Example 15 89 Equilibrium Question: A car engine of weight 2000N is lifted by means of a chain and pulley system. In sketch A below, the engine is suspended by the chain, hanging stationary. In sketch B, the engine is pulled sideways by a mechanic, using a rope. The engine is held in such a position that the chain makes an angle of 30o with the vertical. In the questions that follow, the masses of the chain and the rope can be ignored. 30o CHAIN CHAIN ENGINE ENGINE ROPE Sketch A Sketch B i) Draw a force diagram representing the forces acting on the engine in sketch A. ii) Determine the tension in the chain in sketch A. iii) Draw a force diagram representing the forces acting on the engine in sketch B. ii) In sketch B determine the magnitude of the applied force and the tension in the chain. Answer: Step 1 : Force diagram for sketch A i) Just two forces are acting on the engine in sketch A: ¡!T chain ¡!W 90 Step 2 : Determine the tension in the chain ii) Since the engine in sketch A is stationary, the resultant force on the engine is zero. Thus the tension in the chain exactly balances the weight of the engine, Tchain = W = 2000N Step 3 : Force diagram for sketch B iii) Three forces are acting on the engine in sketch B: ¡!F applied 30o ¡!W ¡!T chain 30o Since the engine is at equilibrium (it is held stationary) the three forces drawn tailto-head form a closed triangle. Step 4 : Calculate the magnitude of the forces in sketch B iv) Since no method was speciﬂed let us calculate the magnitudes algebraically. Since the triangle formed by the three forces is a right-angle triangle this is easily done: Fapplied W = tan 30o Fapplied = (2000) tan 30o and = 1155 N Tchain W Tchain = = 1 cos 30o 2000 cos 30o = 2309 N 5.5 Newton’s Laws of Motion Our current laws of motion were discovered by Sir Isaac Newton. It is said that Sir Isaac Newton started to think about the nature of motion and gravitation after being struck on the head by a falling apple. Newton discovered 3 laws describing motion: 91 5.5.1 First Law Deﬂnition: Every object will remain at rest or in uniform motion in a straight line unless it is made to change its state by the action of an external force. For example, a cement block isn’t going to move unless you push it. A rocket in space is not going to speed up, slow down nor change direction unless the engines are switched on. Newton’s First Law may seem rather surprising to you when you ﬂrst meet it. If you roll a ball along a surface it always stops, but Newton’s First Law says that an item will remain in uniform motion in a straight line unless a force acts upon it. It doesn’t seem like there is any force acting on the ball once you let it go, but there is! In real life (unlike many physics problems) there is usually friction acting. Friction is the external force acting on the ball causing it to stop. Notice that Newton’s First Law says that the force must be external. For example, you can’t grab your belt and pull yourself up to the ceiling. Of course, you could get someone else to pull you up, but then that person would be applying an external force. Worked Example 16 Newton’s First Law in action Question: Why do passengers get thrown to the side when the car they are driving in goes around a corner? Answer: Newton’s First Law Step 1 : What happens before the car turns Before the car starts turning both you and the car are travelling at the same velocity. (picture A) Step 2 : What happens while the car turns The driver turns the wheels of the car, which then exert a force on the car and the car turns. This force acts on the car but not you, hence (by Newton’s First Law) you continue moving with the same velocity. (picture B) Step 3 : Why passengers get thrown to the side If the passenger is wearing a seatbelt it will exert a force on the passenger until the passenger’s velocity is the same as that of the car (picture C). Without a seatbelt the passenger may hit the side of the car or even the windscreen! 92 A: Both the car and the person travelling at the same velocity B: The cars turns but not the person C: Both the car and the person are travelling at the same velocity again 5.5.2 Second Law Deﬂnition: The resultant force acting on a body results in an acceleration which is in the same direction as the resultant force and is directly proportional to the magnitude of this force and inversely proportional to the mass of the object. In mathematical form this law states, ¡!F Res = m¡!a (5.1) So if a resultant force ¡!F Res acts on an object with mass m it will result in an acceleration ¡!a . It makes sense that the direction of the acceleration is in the direction of the resultant force. If you push something away from you it doesn’t move toward you unless of course there is another force acting on the object towards you! Worked Example 17 Newton’s Second Law Question: A block of mass 10kg is accelerating at 2m:s¡2. What is the magnitude of the resultant force acting on the block? Answer: Step 1 : Decide what information has been supplied We are given the block’s mass the block’s acceleration † † all in the correct units. 93 Step 2 : Determine the force on the block We are asked to ﬂnd the magnitude of the force applied to the block. Newton’s Second L
aw tells us the relationship between acceleration and force for an object. Since we are only asked for the magnitude we do not need to worry about the directions of the vectors: FRes = ma 2m:s¡2 = 10kg = 20N £ Thus, there must be a resultant force of 20N acting on the box. Worked Example 18 Newton’s Second Law 2 Question: A 12N force is applied in the positive x-direction to a block of mass 100mg resting on a frictionless at surface. What is the resulting acceleration of the block? Answer: Step 1 : Decide what information has been supplied We are given the block’s mass the applied force † † but the mass is not in the correct units. Step 2 : Convert the mass into the correct units 100mg = 100 1000g = 1kg £ 10¡3g = 0:1g 1 1000g 1 = 1kg £ 1kg 1000g = 0:1g = 0:1g = 0:1g 1 1kg 1000g £ £ = 0:0001 kg Step 3 : Determine the direction of the acceleration We know that net force results in acceleration. Since there is no friction the applied force is the resultant or net force on the block (refer to the earlier example of the block pushed on the surface of the table). The block will then accelerate in the direction of this force according to Newton’s Second Law. 94 Step 4 : Determine the magnitude of the acceleration FRes = ma 12N = (0:0001kg)a a = = 120000 12N 0:0001kg N kg kg:m s2:kg m s2 = 120000 = 120000 = 1:2 £ 105 m:s¡2 From Newton’s Second Law the direction of the acceleration is the same as that of the 105 m:s¡2 resultant force. The ﬂnal result is then that the block accelerates at 1:2 in the positive x-direction. £ Weight and Mass You must have heard people saying \My weight is 60kg". This is actually incorrect because it is mass that is measured in kilograms. Weight is the force of gravity exerted by the earth on an object with mass: Fweight = mg (5.2) As such, weight is measured in newtons. If you compare this equation to Newton’s Second Law you will see that it looks exactly the same with the a replaced by g. Thus, when weight is the only force acting on an object (i.e. when Fweight is the resultant force acting on the object) the object has an acceleration g. Such an object is said to be in free fall. The value for g is the same for all objects (i.e. it is independent of the objects mass): g = 9:8ms¡2 10ms¡2 (5.3) … You will learn how to calculate this value from the mass and radius of the earth in Chapter ??. Actually the value of g varies slightly from place to place on the earth’s surface. The reason that we often get confused between weight and mass, is that scales measure your weight (in newtons) and then display your mass using the equation above. (NOTE TO SELF: include example of skydiver- calculate the acceleration before and after opening parachute{ free-fall and then negative acceleration) Worked Example 19 Calculating the resultant and then the acceleration 95 Question: A block (mass 20kg) on a frictionless at surface has a 45N force applied to it in the positive x-direction. In addition a 25N force is applied in the negative x-direction. What is the resultant force acting on the block and the acceleration of the block? Answer: Step 1 : Decide what information has been supplied We are given the block’s mass Force F1 = 45N in the positive x-direction Force F2 = 25N in the negative x-direction † † † all in the correct units. Step 2 : Figure out how to tackle the problem We are asked to determine what happens to the block. We know that net force results in an acceleration. We need to determine the net force acting on the block. Step 3 : Determine the magnitude and direction of the resultant force Since F1 and F2 act along the same straight line, we can apply the algebraic technique of vector addition discussed in the Vectors chapter to determine the resultant force. Choosing the positive x-direction as our positive direction: Positive x-direction is the positive direction: FRes = (+45N ) + ( = +20N = 20N in the positive x 25N ) ¡ direction ¡ where we remembered in the last step to include the direction of the resultant force in words. By Newton’s Second Law the block will accelerate in the direction of this resultant force. Step 4 : Determine the magnitude of the acceleration FRes = ma 20N = (20kg)a a = = 1 20N 20kg N kg kg:m s2:kg = 1 m:s¡2 = 1 The ﬂnal result is then that the block accelerates at 1 m:s¡2 in the positive x-direction (the same direction as the resultant force). 96 Worked Example 20 Block on incline Question: insert question here maybe with friction! N ? W mg W k 5.5.3 Third Law Deﬂnition: For every force or action there is an equal but opposite force or reaction. Newton’s Third Law is easy to understand but it can get quite di–cult to apply it. An important thing to realise is that the action and reaction forces can never act on the same object and hence cannot contribute to the same resultant. Worked Example 21 Identifying action-reaction pairs Question: Consider pushing a box on the surface of a rough table. 1. Draw a force diagram indicating all of the forces acting on the box. 2. Identify the reaction force for each of the forces acting on the box. Answer: 1. The following force diagram shows all of the forces acting on the box 97 Remember to draw the object in your force diagram as a dot Fnormal Fpush Ff riction Fweight 2. The following table lists each of forces acting on the box (the actions) together with their corresponding reactions: Action Fpush : Person pushing on the box Reaction Box pushing on the person Fweight : The earth attracting The box attracting the box (the weight) Fnormal : The box pushing on the table the earth The table pushing on the box Ff riction : The table acting on The box acting on the box the table Notice that to ﬂnd the reaction force you need to switch around the object supplying the force and the object receiving the force. Be careful not to think that the normal force is the reaction partner to the weight of the box. The normal force balances the weight force but they are both forces acting on the box so they cannot possibly form an action-reaction pair. There is an important thing to realise which is related to Newton’s Third Law. Think about dropping a stone oﬁ a cliﬁ. It falls because the earth exerts a force on it (see Chapter ??) and it doesn’t seem like there are any other forces acting. So is Newton’s Third Law wrong? No, the reactionary force to the weight of the stone is the force exerted by the stone on the earth. This is illustrated in detail in the next worked example. Worked Example 22 Newton’s Third Law Question: A stone of mass 0:5kg is accelerating at 10 m:s¡2 towards the earth. 98 1. What is the force exerted by the earth on the stone? 2. What is the force exerted by the stone on the earth? 3. What is the acceleration of the earth, given that its mass is 5:97 Answer: 1. Step 1 : Decide what information has been supplied We are given 1027kg? £ the stone’s mass the stone’s acceleration (g) † † all in the correct units. Step 2 : The force applied by the earth on the stone This force is simply the weight of the stone. Step 3 : The magnitude of the force of the earth on the stone Applying Newton’s Second Law, we can ﬂnd the magnitude of this force, FRes = ma = 0:5kg = 5N £ 10 m:s¡2 Therefore the earth applies a force of 5N towards the earth on the stone. 2. By Newton’s Third Law the stone must exert an equal but opposite force on the earth. Hence the stone exerts a force of 5N towards the stone on the earth. 3. We have the force acting on the earth the earth’s mass † † in the correct units Step 4 : To ﬂnd the earth’s acceleration we must apply Newton’s Second Law to ﬂnd the magnitude FRes = ma 5N = (5:97 a = 5:97 £ = 8:37521 1027kg)a £ 5N 1027kg 10¡28 m:s¡2 £ The earth’s acceleration is directed towards the stone (i.e. in the same direction as the force on the earth). The earth’s acceleration is really tiny. This is why you don’t notice the earth moving towards the stone, even though it does. 99 Interesting Fact: Newton ﬂrst published these laws in Philosophiae Naturalis Principia Mathematica (1687) and used them to prove many results concerning the motion of physical objects. Only in 1916 were Newton’s Laws superceded by Einstein’s theory of relativity. The next two worked examples are quite long and involved but it is very important that you understand the discussion as they illustrate the importance of Newton’s Laws. Worked Example 23 Rockets Question: How do rockets accelerate in space? Answer: † † † † Gas explodes inside the rocket. This exploding gas exerts a force on each side of the rocket (as shown in the picture below of the explosion chamber inside the rocket). Note that the forces shown in this picture are representative. With an explosion there will be forces in all directions. Due to the symmetry of the situation, all the forces exerted on the rocket are balanced by forces on the opposite side, except for the force opposite the open side. This force on the upper surface is unbalanced. This is therefore the resultant force acting on the rocket and it makes the rocket accelerate forwards. Systems and External Forces The concepts of a system and an external forces are very important in physics. A system is any collection of objects. If one draws an imaginary box around such a system then an external force is one that is applied by an object or person outside the box. Imagine for example a car pulling a trailer. 100 5.6 Examples of Forces Studied Later Most of physics revolves around forces. Although there are many diﬁerent forces we deal with them all in the same way. The methods to ﬂnd resultants and acceleration do not depend on the type of force we are considering. At ﬂrst glance, the number of diﬁerent forces may seem overwhelming - gravity, drag, electrical forces, friction and many others. However, physicists have found that all these forces can be classiﬂed into four groups. These are gravitational forces, electromagnetic forces, strong nuclear force and weak nuclear force. Even better, all the forces that you will
come across at school are either gravitational or electromagnetic. Doesn’t that make life easy? 5.6.1 Newtonian Gravity Gravity is the attractive force between two objects due to the mass of the objects. When you throw a ball in the air, its mass and the earth’s mass attract each other, which leads to a force between them. The ball falls back towards the earth, and the earth accelerates towards the ball. The movement of the earth toward the ball is, however, so small that you couldn’t possibly measure it. 5.6.2 Electromagnetic Forces Almost all of the forces that we experience in everyday life are electromagnetic in origin. They have this unusual name because long ago people thought that electric forces and magnetic forces were diﬁerent things. After much work and experimentation, it has been realised that they are actually diﬁerent manifestations of the same underlying theory. The Electric Force If we have objects carrying electrical charge, which are not moving, then we are dealing with electrostatic forces (Coulomb’s Law). This force is actually much stronger than gravity. This may seem strange, since gravity is obviously very powerful, and holding a balloon to the wall seems to be the most impressive thing electrostatic forces have done, but think about it: for gravity to be detectable, we need to have a very large mass nearby. But a balloon rubbed in someone’s hair can stick to a wall with a force so strong that it overcomes the force of gravity\|with just the charges in the balloon and the wall! Magnetic force The magnetic force is a diﬁerent manifestation of the electromagnetic force. It stems from the interaction between moving charges as opposed to the ﬂxed charges involved in Coulomb’s Law. Examples of the magnetic force in action include magnets, compasses, car engines, computer data storage and your hair standing on end. Magnets are also used in the wrecking industry to pick up cars and move them around sites. 101 Friction We all know that Newton’s First Law states that an object moving without a force acting on it will keep moving. Then why does a box sliding on a table stop? The answer is friction. Friction arises from the interaction between the molecules on the bottom of a box with the molecules on a table. This interaction is electromagnetic in origin, hence friction is just another view of the electromagnetic force. The great part about school physics is that most of the time we are told to neglect friction but it is good to be aware that there is friction in the real world. Friction is also useful sometimes. If there was no friction and you tried to prop a ladder up against a wall, it would simply slide to the ground. Rock climbers use friction to maintain their grip on cliﬁs. Drag Force This is the force an object experiences while travelling through a medium. When something travels through the air it needs to displace air as it travels and because of this the air exerts a force on the object. This becomes an important force when you move fast and a lot of thought is taken to try and reduce the amount of drag force a sports car experiences. The drag force is very useful for parachutists. They jump from high altitudes and if there was no drag force, then they would continue accelerating all the way to the ground. Parachutes are wide because the more surface area you show, the greater the drag force and hence the slower you hit the ground. 5.7 Summary of Important Quantities, Equations and Con- cepts Equilibrium Objects at rest or moving with constant velocity are in equilibrium and have a zero resultant force. Equilibrant The equilibrant of any number of forces is the single force required to produce equilibrium. Triangle Law for Forces in Equilibrium Three forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in order. Newton’s First Law Every object will remain at rest or in uniform motion in a straight line unless it is made to change its state by the action of an external force. Newton’s Second Law The resultant force acting on a body results in an acceleration which is in the same direction as the resultant force and is directly proportional to the magnitude of this force and inversely proportional to the mass of the object. Quantity Mass Acceleration Force Symbol m ¡!a ¡!F S.I. Units kg m:s¡2 kg:m:s¡2 or N Direction \| X X Table 5.1: Summary of the symbols and units of the quantities used in Force 102 Newton’s Third Law For every force or action there is an equal but opposite force or reaction. 103 Chapter 6 Rectilinear Motion 6.1 What is rectilinear motion? Rectilinear motion means motion along a straight line. This is a useful topic to study for learning how to describe the movement of cars along a straight road or of trains along straight railway In this section you have only 2 directions to worry about: (1) along the direction of tracks. motion, and (2) opposite to the direction of motion. To illustrate this imagine a train heading east. Train W P N S E If it is accelerating away from the station platform (P), the direction of acceleration is the same as the direction of the train’s velocity - east. If it is braking the direction of acceleration is opposite to the direction of its motion, i.e. west. 6.2 Speed and Velocity Let’s take a moment to review our deﬂnitions of velocity and speed by looking at the worked example below: 104 Worked Example 24 Speed and Velocity C B m 0 4 A 30m Question: A cyclist moves from A through B to C in 10 seconds. Calculate both his speed and his velocity. Answer: Step 1 : Decide what information has been supplied The question explicitly gives the distance between A and B the distance between B and C the total time for the cyclist to go from A through B to C † † † all in the correct units! Step 2 : Determine the cyclist’s speed His speed - a scalar - will be v = = s t 30m + 40m 10s = 7 m s Step 3 : First determine the cyclist’s resultant displacement Since velocity is a vector we will ﬂrst need to ﬂnd the resultant displacement of the cyclist. His velocity will be ¡!v = ¡!s t The total displacement is the vector from A to C, and this is just the resultant of the two displacement vectors, ie. ¡!s = ¡!AC = ¡¡!AB + ¡¡!BC 105 Remember to check the units! Using the rule of Pythagoras: ¡!s = (30m)2 + (40m)2 = 50m in the direction f rom A to C q Step 4 : Now we can determine the average velocity from the displacement and the time ) ¡!v = 50m 10s m s = 5 in the direction f rom A to C For this cyclist, his velocity is not the same as his speed because there has been a change in the direction of his motion. If the cyclist traveled directly from A to C without passing through B his speed would be v = 50m 10s m s = 5 and his velocity would be ¡!v = 50m 10s m s = 5 in the direction f rom A to C In this case where the cyclist is not undergoing any change of direction (ie. he is traveling in a straight line) the magnitudes of the speed and the velocity are the same. This is the deﬂning principle of rectilinear motion. Important: For motion along a straight line the magnitudes of speed and velocity are the same, and the magnitudes of the distance and displacement are the same. 6.3 Graphs In physics we often use graphs as important tools for picturing certain concepts. Below are some graphs that help us picture the concepts of displacement, velocity and acceleration. 6.3.1 Displacement-Time Graphs Below is a graph showing the displacement of the cyclist from A to C: 106 50 ) ¢¡!s ¢t time (s) 10 This graphs shows us how, in 10 seconds time, the cyclist has moved from A to C. We know the gradient (slope) of a graph is deﬂned as the change in y divided by the change in x, i.e ¢y ¢x . In this graph the gradient of the graph is just ¢¡!s - and this is just the expression for velocity. ¢t Important: The slope of a displacement-time graph gives the velocity. The slope is the same all the way from A to C, so the cyclist’s velocity is constant over the entire displacement he travels. In ﬂgure 6.1 are examples of the displacement-time graphs you will encounter) time ) time ) time Figure 6.1: Some common displacement-time graphs: a) shows the graph for an object stationary over a period of time. The gradient is zero, so the object has zero velocity. b) shows the graph for an object moving at a constant velocity. You can see that the displacement is increasing as time goes on. The gradient, however, stays constant (remember: its the slope of a straight line), so the velocity is constant. Here the gradient is positive, so the object is moving in the direction we have deﬂned as positive. c) shows the graph for an object moving at a constant acceleration. You can see that both the displacement and the velocity (gradient of the graph) increase with time. The gradient is increasing with time, thus the velocity is increasing with time and the object is accelerating. 107 6.3.2 Velocity-Time Graphs Look at the velocity-time graph below ¢¡!v ¢t time This is the velocity-time graph of a cyclist traveling from A to B at a constant acceleration, i.e. with steadily increasing velocity. The gradient of this graph is just ¢¡!v - and this is just the expression for acceleration. Because the slope is the same at all points on this graph, the acceleration of the cyclist is constant. ¢t Important: The slope of a velocity-time graph gives the acceleration. Not only can we get the acceleration of an object from its velocity-time graph, but we can also get some idea of the displacement traveled. Look at the graph below: 10 ) time (s) 5 This graph shows an object moving at a constant velocity of 10m=s for a duration of 5s. The area between the graph and the time axis (the (NOTE TO SELF: SHADED) area) of the above plot will give us the displacement of the object during this time. In this case we just need to calculate the area of a rectangle with width 5s and height 10m/s area of rectangle = height width £ = ¡!v = 10 t £ m s £ 5s = 50m = ¡!s = displace
ment 108 So, here we’ve shown that an object traveling at 10m/s for 5s has undergone a displacement of 50m. Important: The area between a velocity-time graph and the ‘time’ axis gives the displacement of the object. Here are a couple more velocity-time graphs to get used to) time y t i c o l e v b) time Figure 6.2: Some common velocity-time graphs: In ﬂgure 6.2 are examples of the displacement-time graphs you may encounter. a) shows the graph for an object moving at a constant velocity over a period of time. The gradient is zero, so the object is not accelerating. b) shows the graph for an object which is decelerating. You can see that the velocity is decreasing with time. The gradient, however, stays constant (remember: its the slope of a straight line), so the acceleration is constant. Here the gradient is negative, so the object is accelerating in the opposite direction to its motion, hence it is decelerating. 6.3.3 Acceleration-Time Graphs In this chapter on rectilinear motion we will only deal with objects moving at a constant acceleration, thus all acceleration-time graphs will look like these two) time ) time Here is a description of the graphs below: a) shows the graph for an object which is either stationary or traveling at a constant velocity. Either way, the acceleration is zero over time. 109 b) shows the graph for an object moving at a constant acceleration. In this case the acceler- ation is positive - remember that it can also be negative. We can obtain the velocity of a particle at some given time from an acceleration time graph it is just given by the area between the graph and the time-axis. In the graph below, showing an object at a constant positive acceleration, the increase in velocity of the object after 2 seconds corresponds to the (NOTE TO SELF: shaded) portion time (s) t area of rectangle = ¡!a £ m = 5 s2 £ m s = 10 2s = ¡!v Its useful to remember the set of graphs below when working on problems. Figure 6.3 shows how displacement, velocity and time relate to each other. Given a displacement-time graph like the one on the left, we can plot the corresponding velocity-time graph by remembering that the slope of a displacement-time graph gives the velocity. Similarly, we can plot an acceleration-time graph from the gradient of the velocity-time graph. 110 Displacement Velocity Acceleration Figure 6.3: A Relationship Between Displacement, Velocity and Acceleration 6.3.4 Worked Examples Worked Example 25 Relating displacement-, velocity-, and acceleration-time graphs Question: Given the displacement-time graph below, draw the corresponding velocitytime and acceleration-time graphs, and then describe the motion of the object time (s) 4 6 Answer: Step 1 : Decide what information is supplied The question explicitly gives a displacement-time graph. Step 2 : Decide what is asked? 3 things are required: 1. Draw a velocity-time graph 2. Draw an acceleration-time graph 111 3. Describe the behaviour of the object Step 3 : Velocity-time graph - 0-2 seconds For the ﬂrst 2 seconds we can see that the displacement remains constant - so the object is not moving, thus it has zero velocity during this time. We can reach this conclusion by another path too: remember that the gradient of a displacement-time graph is the velocity. For the ﬂrst 2 seconds we can see that the displacement-time graph is a horizontal line, ie. it has a gradient of zero. Thus the velocity during this time is zero and the object is stationary. Step 4 : Velocity-time graph - 2-4 seconds For the next 2 seconds, displacement is increasing with time so the object is moving. Looking at the gradient of the displacement graph we can see that it is not constant. In fact, the slope is getting steeper (the gradient is increasing) as time goes on. Thus, remembering that the gradient of a displacement-time graph is the velocity, the velocity must be increasing with time during this phase. Step 5 : Velocity-time graph - 4-6 seconds For the ﬂnal 2 seconds we see that displacement is still increasing with time, but this time the gradient is constant, so we know that the object is now travelling at a constant velocity, thus the velocity-time graph will be a horizontal line during this stage. So our velocity-time graph looks like this one below. Because we haven’t been given any values on the vertical axis of the displacement-time graph, we cannot ﬂgure out what the exact gradients are and hence what the values of the velocity are. In this type of question it is just important to show whether velocities are positive or negative, increasing, decreasing or constant time (s) 4 6 Once we have the velocity-time graph its much easier to get the acceleration-time graph as we know that the gradient of a velocity-time graph is the just the acceleration. Step 6 : Acceleration-time graph - 0-2 seconds For the ﬂrst 2 seconds the velocity-time graph is horizontal at zero, thus it has a gradient of zero and there is no acceleration during this time. (This makes sense because we know from the displacement time graph that the object is stationary during this time, so it can’t be accelerating). Step 7 : Acceleration-time graph - 2-4 seconds For the next 2 seconds the velocity-time graph has a positive gradient. This gradient its constant) throughout these 2 seconds so there must be a is not changing (i.e. 112 constant positive acceleration. Step 8 : Accleration-time graph - 4-6 seconds For the ﬂnal 2 seconds the object is traveling with a constant velocity. During this time the gradient of the velocity-time graph is once again zero, and thus the object is not accelerating. The acceleration-time graph looks like this time (s) 4 6 Step 9 : A description of the object’s motion A brief description of the motion of the object could read something like this: At t = 0s and object is stationary at some position and remains stationary until t = 2s when it begins accelerating. It accelerates in a positive direction for 2 seconds until t = 4s and then travels at a constant velocity for a further 2 seconds. Worked Example 26 Calculating distance from a velocity-time graph Question: The velocity-time graph of a car is plotted below. Calculate the displacement of the car has after 15 seconds. 113 2 5 time (s) 12 15 Answer: Step 1 : Decide how to tackle the problem We are asked to calculate the displacement of the car. All we need to remember here is that the area between the velocity-time graph and the time axis gives us the displacement. Step 2 : Determine the area under the velocity-time graph For t = 0s to t = 5s this is the triangle on the left: Area 4 h £ b = 1 2 1 2 = 10m 5s = 4m=s £ For t = 5s to t = 12s the displacement is equal to the area of the rectangle h Area⁄ = w = 7s £ = 28m £ 4m=s For t = 12s to t = 14s the displacement is equal to the area of the triangle above the time axis on the right Area 4 = = 1 2 1 2 h b £ 2s £ 4m=s = 4m For t = 14s to t = 15s the displacement is equal to the area of the triangle below the 114 time axis Area 4 = = 1 2 1 2 h b £ 1s £ 2m=s = 1m Step 3 : Determine the total displacement of the car Now the total displacement of the car is just the sum of all of these areas. HOWEVER, because in the last second (from t = 14s to t = 15s) the velocity of the car is negative, it means that the car was going in the opposite direction, i.e. back where it came from! So, to get the total displacement, we have to add the ﬂrst 3 areas (those with positive displacements) and subtract the last one (because it signiﬂes a displacement in the opposite direction). ¡!s = 10 + 28 + 4 1 ¡ = 41m in the positive direction Worked Example 27 Velocity from a displacement-time graph Question: Given the diplacement-time graph below, 1. what is the velocity of the object during the ﬂrst 4 seconds? 2. what is the velocity of the object from t = 4s to t = 7s? 2 ) m ( s 4 t (s) 7 Answer: Step 1 : The velocity during the ﬂrst 4 seconds The velocity is given by the slope of a displacement-time graph. During the ﬂrst 4 seconds, this is ¡!v = ¢s ¢t 2m 4s = 0:5m=s = 115 Step 2 : The velocity during the last 3 seconds For the last 3 seconds we can see that the displacement stays constant, and that the gradient is zero. Thus ¡!v = 0m=s Worked Example 28 From an acceleration-time graph to a velocity-time graph Question: Given the acceleration-time graph below, assume that the object starts from rest and draw its velocity-time graph2 2 4 t (s) 6 Answer: Once again attempt to draw the graph in time sections, i.e. ﬂrst draw the velocity-time graph for the ﬂrst 2 seconds, then for the next 2 seconds and so ons) 6 116 6.4 Equations of Motion This section is about solving problems relating to uniformly accelerated motion. We’ll ﬂrst introduce the variables and the equations, then we’ll show you how to derive them, and after that we’ll do a couple of examples. u = starting velocity (m/s) at t = 0 v = ﬂnal velocity (m/s) at time t s = displacement (m) t = time (s) a = acceleration (m/s2) v = u + at s = (u + v) 2 t s = ut + at2 1 2 v2 = u2 + 2as (6.1) (6.2) (6.3) (6.4) Make sure you can rhyme these oﬁ, they are very important! There are so many diﬁerent types of questions for these equations. Basically when you are answering a question like this: 1. Find out what values you have and write them down. 2. Figure out which equation you need. 3. Write it down!!! 4. Fill in all the values you have and get the answer. Interesting Fact: Galileo Galilei of Pisa, Italy, was the ﬂrst to determined the correct mathematical law for acceleration: the total distance covered, starting from rest, is proportional to the square of the time. He also concluded that objects retain their velocity unless a force { often friction { acts upon them, refuting the accepted Aristotelian hypothesis that objects "naturally" slow down and stop unless a force acts upon them. This principle was incorporated into Newton’s laws of motion (1st law). Equation 6.1 By the deﬂnition of acceleration a = ¢v t 117 where ¢v is the chan
ge in velocity, i.e. ¢v = v a = ¡ v u. Thus we have u ¡ t v = u + at Equation 6.2 In the previous section we saw that displacement can be calculated from the area between a velocity-time graph and the time-axis. For uniformly accelerated motion the most complicated velocity-time graph we can have is a straight line. Look at the graph below - it represents an object with a starting velocity of u, accelerating to a ﬂnal velocity v over a total time t time (s) To calculate the ﬂnal displacement we must calculate the area under the graph - this is just the area of the rectangle added to the area of the triangle. (NOTE TO SELF: SHADING) Area £ £ vt ¡ (v u) ¡ 1 2 ut h Area⁄ = w = t £ = ut £ u Displacement = Area⁄ + Area 4 1 2 ¡ ut s = ut + vt 1 2 (u + v) 2 t = Equation 6.3 This equation is simply derived by eliminating the ﬂnal velocity v in equation 6.2. Remembering from equation 6.1 that v = u + at 118 then equation 6.2 becomes s = = t u + u + at 2 2ut + at2 2 1 2 at2 = ut + Equation 6.4 This equation is just derived by eliminating the time variable in the above equation. From Equation 6.1 we know Substituting this into Equation 6.3 gives t = v u ¡ a s = u( u ) + v ¡ a u )2 ¡ a v2 v a( 1 2 1 2 v2 2a ¡ a( + u2 uv a ¡ a u2 uv a ¡ a 2u2 + v2 + u2 + = = 2as = ¡ v2 = u2 + 2as ¡ 2uv + u2 a2 ) uv a + u2 2a (6.5) This gives us the ﬂnal velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable. Worked Example 29 Question: A racing car has an initial velocity of 100m=s and it covers a displacement of 725m in 10s. Find its acceleration. Answer: Step 1 : Decide what information has been supplied We are given the quantities u, s and t - all in the correct units. We need to ﬂnd a. Step 2 : Find an equation of motion relating the given information to the acceleration We can use equation 6.3 s = ut + at2 1 2 Step 3 : Rearrange the equation if needed We want to determine the acceleration so we rearrane equation 6.3 to put acceleration on the left of the equals sign: 2(s a = ut) ¡ t2 119 Step 4 : Do the calculation Substituting in the values of the known quantities this becomes 2(725m a = 100 m 10s) ¡ 102s2 s ¢ = = 2( 275m) ¡ 100s2 m 5:5 s2 ¡ Step 5 : Quote the ﬂnal answer The racing car is accelerating at -5.5 m s2 , or we could say it is decelerating at 5.5 m s2 . Worked Example 30 Question: An object starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64m in 4s. Calculate its acceleration its ﬂnal velocity at what time the object had covered half the total distance what distance the object had covered in half the total time. † † † † Answer: Step 1 : Decide what information is supplied We are given the quantities u, s and t in the correct units. Step 2 : Acceleration: Find an equation to calculate the acceleration and rearrange To calculate the acceleration we can use equation 6.3. Step 3 : Rearrange to make a subject of the formula 2(s a = ut) ¡ t2 Step 4 : Do the calculation Substituting in the values of the known quantities this becomes 2(64m 0 m s 4s ¡ 42s2 a = = 128m 16s2 m s2 = 8 Step 5 : Final velocity: Find an equation to calculate the ﬂnal velocity We can use equation 6.1 - remember we now also know the acceleration of the object. v = u + at 120 Step 6 : Do the calculation v = 0 + (8 m s2 )(4s) m s m s = 32 Step 7 : Time at half the distance: Find an equation to relate the unknown and known quantities Here we have the quantities s, u and a so we do this in 2 parts, ﬂrst using equation 6.4 to calculate the velocity at half the distance, i.e. 32m: v2 = u2 + 2as = (0m)2 + 2(8m=s2)(32m) = 512m2=s2 v = 22:6m=s Now we can use equation 6.2 to calculate the time at this distance: t = 2s u + v = (2)(32m) 0m=s + 22:6m=s = 2:8s Step 8 : Distance at half the time: Find an equation to relate the distance and time To calculate the distance the object has covered in half the time. Half the time is 2s. Thus we have u, a and t - all in the correct units. We can use equation 6.3 to get the distance: s = ut + at2 1 2 = (0m=s)(2s) + = 16m 1 2 (8 m s2 )(2s)2 Worked Example 31 Question: A ball is thrown vertically upwards with a velocity of 10m=s from the balcony of a tall building. The balcony is 15m above the ground and gravitational accleration is 10m=s2. Find a) the time required for the ball to hit the ground, and b) the velocity with which it hits the ground. Answer: Step 1 : Draw a rough sketch of the problem In most cases helps to make the problem easier to understand if we draw ourselves a picture like the one below: 121 balcony s1 v1 u1 u2 a1; a2 s2 v2 ground where the subscript 1 refers to the upward part of the ball’s motion and the subscript 2 refers to the downward part of the ball’s motion. Step 2 : Decide how to tackle the problem First the ball goes upwards with gravitational acceleration slowing it until it reaches its highest point - here its speed is 0m=s - then it begins descending with gravitational acceleration causing it to increase its speed on the way down. We can separate the motion into 2 stages: Stage 1 - the upward motion of the ball Stage 2 - the downward motion of the ball. We’ll choose the upward direction as positive - this means that gravitation acceleraton is negative. We have used this and we’ll begin by solving for all the variables of Stage 1. Step 3 : For Stage 1, decide what information is given We have these quantities: u1 = 10m=s v1 = 0m=s a1 = ¡ t1 = ? s1 = ? 10m=s2 Step 4 : Find the time for stage 1 Using equation 6.1 to ﬂnd t1: v1 = u1 + a1t1 v1 u1 t1 = ¡ a1 0m=s = ¡ = 1s 10m=s ¡ 10m=s2 Step 5 : Find the distance travelled during stage 1 122 We can ﬂnd s1 by using equation 6.4 s1 = 1 = u2 v2 1 + 2a1s1 u2 v2 1 ¡ 1 2a (0m=s)2 2( = = 5m ¡ (10m=s)2 ¡ 10m=s2) Step 6 : For Stage 2, decide what information is supplied For Stage 2 we have the following quantities: u2 = 0m=s v2 = ? a2 = ¡ t2 = ? s2 = 15m ¡ 10m=s2 5m = 20m ¡ Step 7 : Determine the velocity at the end of stage 2 We can determine the ﬂnal velocity v2 using equation 6.4: 2 = u2 v2 2 + 2a2s2 = (0m=s)2 + 2( = 400(m=s)2 ¡ 10m=s2)( 20m) ¡ v2 = 20m=s downwards Step 8 : Determind the time for stage 2 Now we can determine the time for Stage 2, t2, from equation 6.1: v2 = u2 + a2t2 v2 u2 t2 = ¡ a2 20m=s 0m=s = ¡ = 2s ¡ 10m=s2 ¡ Step 9 : Quote the answers to the problem Finally, a) the time required for the stone to hit the ground is t = t1 + t2 = 1s + 2s = 3s b) the velocity with which it hits the ground is just v2 = 20m=s ¡ These questions do not have the working out in them, but they are all done in the manner described on the previous page. Question: A car starts oﬁ at 10 m/s and accelerates at 1 m/s2 for 10 seconds. What is it’s ﬂnal velocity? Answer: 20 m/s 123 Question: A car starts from rest, and accelerates at 1 m/s2 for 10 seconds. How far does it move? Answer: 50 m Question: A car is going 30 m/s and stops in 2 seconds. What is it’s stopping distance for this speed? Answer: 30 m Question: A car going at 20 m/s stops in a distance of 20 m/s. 1. What is it’s deceleration? 2. If the car is 1 Tonne (1000 Kg, or 1 Mg) how much force do the brakes exert? 124 6.5 Important Equations and Quantities Quantity Displacement Velocity Distance Speed Acceleration Units Symbol Unit ¡!s ¡!u ,¡!v s v ¡!a - Base S.I. Units m + direction m:s¡1 + direction m m:s¡1 m:s¡1 + direction Table 6.1: Units used in Rectilinear Motion 125 Chapter 7 Momentum 7.1 What is Momentum? Momentum is a physical quantity which is closely related to forces. We will learn about this connection a little later. Remarkably momentum is a conserved quantity. This makes momentum extremely useful in solving a great variety of real-world problems. Firstly we must consider the deﬂnition of momentum. Deﬂnition: The momentum of an object is deﬂned as its mass multiplied by its velocity. Mathematically, ¡!p = m¡!v : momentum (kg:m:s¡1 + direction) ¡!p m : mass (kg) ¡!v : velocity (m:s¡1 + direction) Thus, momentum is a property of a moving object and is determined by its velocity and mass. A large truck travelling slowly can have the same momentum as a much smaller car travelling relatively fast. Note the arrows in the equation deﬂning momentum{ momentum is a vector with the same direction as the velocity of the object. Since the direction of an object’s momentum is given by the direction of its motion, one can calculate an object’s momentum in two steps: Momentum is a vector with the same direction as the velocity. calculate the magnitude of the object’s momentum using, † p = mv : magnitude of momentum (kg:m:s¡1) p m : mass (kg) v : magnitude of velocity (m:s¡1) 126 include in the ﬂnal answer the direction of the object’s motion † Worked Example 32 Calculating Momentum 1 Question: A ball of mass 3kg moves at 2m:s¡1 to the right. Calculate the ball’s momentum. Answer: Step 1 : Decide what information has been supplied The question explicitly gives the ball’s mass, and the ball’s velocity † † in the correct units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the ball’s momentum. From the deﬂnition of momentum, ¡!p = m¡!v ; we see that we need the mass and velocity of the ball, which we are given. Step 3 : Do the calculation We calculate the magnitude of the ball’s momentum, p = mv = (3kg)(2m:s¡1) = 6 kg:m:s¡1: Step 4 : Quote the ﬂnal answer We quote the answer with the direction of the ball’s motion included, ¡!p = 6 kg:m:s¡1 to the right Worked Example 33 Calculating Momentum 2 Question: A ball of mass 500g is thrown at 2m:s¡1. Calculate the ball’s momentum. Answer: Step 1 : Decide what information is supplied The question explicitly gives the ball’s mass, and the magnitude of the ball’s velocity † † but with the ball’s mass in the incorrect units! 127 Remember to check the units! Remember to check the units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the momentum which is deﬂned as ¡!p = m¡!v : Thus, we need
the mass and velocity of the ball but we have only its mass and the magnitude of its velocity. In order to determine the velocity of the ball we need the direction of the ball’s motion. If the problem does not give an explicit direction we are forced to be general. In a case like this we could say that the direction of the velocity is in the direction of motion of the ball. This might sound silly but the lack of information in the question has forced us to be, and we are certainly not wrong! The ball’s velocity is then 2m:s¡1 in the direction of motion. Step 3 : Convert the mass to the correct units 1000g = 1kg 1 = 1kg 1000g 500g £ 1 = 500g £ = 0:500kg 1kg 1000g Step 4 : Do the calculation Now, let us ﬂnd the magnitude of the ball’s momentum, p = mv = (0:500kg)(2m:s¡1) = 1 kg:m:s¡1 Step 5 : Quote the ﬂnal answer Remember to include the direction of the momentum: ¡!p = 1 kg:m:s¡1 in the direction of motion of the ball Worked Example 34 Calculating the Momentum of the Moon Question: The moon is 384 400km away from the earth and orbits the earth in 1022kg1 what is the magnitude of its 27.3 days. If the moon has a mass of 7:35 momentum if we assume a circular orbit? Answer: Step 1 : Decide what information has been supplied The question explicitly gives £ 1This is 1 81 of the mass of the earth 128 with mass in the correct units but all other quantities in the incorrect units. The units we require are Remember to check the units! the moon’s mass, the distance to the moon, and the time for one orbit of the moon seconds (s) for time, and metres (m) for distance † † † † † Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate only the magnitude of the moon’s momentum (i.e. we do not need to specify a direction). In order to do this we require the moon’s mass and the magnitude of its velocity, since Step 3 : Find the speed or magnitude of the moon’s velocity Speed is deﬂned as, p = mv: speed = Distance time We are given the time the moon takes for one orbit but not how far it travels in that time. However, we can work this out from the distance to the moon and the fact that the moon’s orbit is circular. Firstly let us convert the distance to the moon to the correct units, 1km = 1000m 1000m 1km 1 = 384 400km £ 1000m 1km 1 = 384 400km £ = 384 400 000m 108 m = 3:844 £ Using the equation for the circumference, C, of a circle in terms of its radius, we can determine the distance travelled by the moon in one orbit: C = 2…r = 2…(3:844 = 2:42 £ 109 m: 108 m) £ Next we must convert the orbit time, T , into the correct units. Using the fact that a day contains 24 hours, an hour consists of 60 minutes, and a minute is 60 seconds long, 1day = (24)(60)(60)seconds 1 = (24)(60)(60)s 1day 27:3days £ 1 = 27:3days (24)(60)(60)s 1day = 2:36 106s £ 129 Therefore, T = 2:36 106s: £ Combining the distance travelled by the moon in an orbit and the time taken by the moon to complete one orbit, we can determine the magnitude of the moon’s velocity or speed, v = Distance time = C T = 1:02 103 m:s¡1: £ Step 4 : Finally calculate the momentum and quote the answer The magnitude of the moon’s momentum is: p = mv = (7:35 = 7:50 £ £ 1022kg)(1:02 £ 1025 kg:m:s¡1: 103 m:s¡1) 7.2 The Momentum of a System In Chapter ?? the concept of a system was introduced. The bodies that make up a system can have diﬁerent masses and can be moving with diﬁerent velocities. In other words they can have diﬁerent momenta. Deﬂnition: The total momentum of a system is the sum of the momenta of each of the objects in the system. Since momentum is a vector, the techniques of vector addition discussed in Chapter ?? must be used to calculate the total momentum of a system. Let us consider an example. Worked Example 35 Calculating the Total Momentum of a System Question: Two billiard balls roll towards each other. They each have a mass of 0:3kg. Ball 1 is moving at v1 = 1 m:s¡1 to the right, while ball 2 is moving at v2 = 0:8 m:s¡1 to the left. Calculate the total momentum of the system. Answer: Step 1 : Decide what information is supplied The question explicitly gives the mass of each ball, the velocity of ball 1, ¡!v1, and † † 130 the velocity of ball 2, ¡!v2, † all in the correct units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the total momentum of the system. In this example our system consists of two balls. To ﬂnd the total momentum we must sum the momenta of the balls, Remember to check the units! ¡!p total = ¡!p1 + ¡!p2 Since ball 1 is moving to the right, its momentum is in this direction, while the second ball’s momentum is directed towards the left. System m1 ¡!p 1 m2 ¡!p 2 Thus, we are required to ﬂnd the sum of two vectors acting along the same straight line. The algebraic method of vector addition introduced in Chapter ?? can thus be used. Step 3 : Choose a positive direction Let us choose right as the positive direction, then obviously left is negative. Step 4 : Calculate the momentum The total momentum of the system is then the sum of the two momenta taking the directions of the velocities into account. Ball 1 is travelling at 1 m:s¡1 to the right or +1 m:s¡1. Ball 2 is travelling at 0:8 m:s¡1 to the lef t or 0:8 m:s¡1. Thus, ¡ Right is the positive direction ¡!p total = m1¡!v1 + m2¡!v2 = (0:3kg)(+1 m:s¡1) + (0:3kg)( = (+0:3 kg:m:s¡1) + ( = +0:06 kg:m:s¡1 = 0:06 kg:m:s¡1 to the right ¡ ¡ 0:24 kg:m:s¡1) 0:8 m:s¡1) In the last step the direction was added in words. Since the result in the second last line is positive, the total momentum of the system is in the positive direction (i.e. to the right). 7.3 Change in Momentum If either an object’s mass or velocity changes then its momentum too will change. If an object has an initial velocity ¡!u and a ﬂnal velocity ¡!v , then its change in momentum, ¢¡!p , is 131 ¢¡!p = ¡!p f inal ¡ ¡!p initial = m¡!v ¡ m¡!u Worked Example 36 Change in Momemtum Question: A rubber ball of mass 0:8kg is dropped and strikes the oor at a velocity of 6 m:s¡1. It bounces back with an initial velocity of 4 m:s¡1. Calculate the change in momentum of the rubber ball caused by the oor. Answer: Step 1 : Decide what information has been supplied The question explicitly gives the ball’s mass, the ball’s initial velocity, and the ball’s ﬂnal velocity † † † all in the correct units. Do not be confused by the question referring to the ball bouncing back with an \initial velocity of 4 m:s¡1". The word \initial" is included here since the ball will obviously slow down with time and 4 m:s¡1 is the speed immediately after bouncing from the oor. Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the change in momentum of the ball, ¢¡!p = m¡!v m¡!u : ¡ We have everything we need to ﬂnd ¢¡!p . Since the initial momentum is directed downwards and the ﬂnal momentum is in the upward direction, we can use the algebraic method of subtraction discussed in the vectors chapter. Step 3 : Choose a positive direction Let us choose down as the positive direction. Then substituting, Down is the positive direction Step 4 : Do the calculation and quote the answer Remember to check the units! ¢¡!p = m¡!v m¡!u ¡ = (0:8kg)( = (0:8kg)( = = 8 kg:m:s¡1 up 8 kg:m:s¡1 ¡ ¡ ¡ 4 m:s¡1) ¡ 10 m:s¡1) (0:8kg)(+6 m:s¡1) where we remembered in the last step to include the direction of the change in momentum in words. 132 ¡!u 1 m1 ¡!u 2 m2 Figure 7.1: Before the collision. 7.4 What properties does momentum have? You may at this stage be wondering why there is a need for introducing momentum. Remarkably momentum is a conserved quantity. Within an isolated system the total momentum is constant. No matter what happens to the individual bodies within an isolated system, the total momentum of the system never changes! Since momentum is a vector, its conservation implies that both its magnitude and its direction remains the same. This Principle of Conservation of Linear Momentum is one of the most fundamental principles of physics and it alone justiﬂes the deﬂnition of momentum. Since momentum is related to the motion of objects, we can use its conservation to make predictions about what happens in collisions and explosions. If we bang two objects together, by conservation of momentum, the total momentum of the objects before the collision is equal to their total momentum after the collision. Momentum is conserved in isolated systems! Principle of Conservation of Linear Momentum: The total linear momentum of an isolated system is constant. or In an isolated system the total momentum before a collision (or explosion) is equal to the total momentum after the collision (or explosion). Let us consider a simple collision of two pool or billiard balls. Consider the ﬂrst ball (mass m1) to have an initial velocity (¡!u1). The second ball (mass m2) moves towards the ﬂrst ball with an initial velocity ¡!u2. This situation is shown in Figure 7.1. If we add the momenta of each ball we get a total momentum for the system. This total momentum is then ¡!p total bef ore = m1¡!u1 + m2¡!u2; After the two balls collide and move away they each have a diﬁerent momentum. If we call the ﬂnal velocity of ball 1 ¡!v1 and the ﬂnal velocity of ball 2 ¡!v2 (see Figure 7.2), then the total momentum of the system after the collision is ¡!p total af ter = m1¡!v1 + m2¡!v2; This system of two balls is isolated since there are no external forces acting on the balls. Therefore, by the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision. This gives the equation for the conservation of momentum in a collision of two objects, 133 ¡!v 1 m1 m2 ¡!v 2 Figure 7.2: After the collision. ¡!p total bef ore = ¡!p total af ter m1¡!u1 + m2¡!u2 = m1¡!v1 + m2¡!v2 m1 m2 : mass of object 1 (kg) : mass of object 2 (kg) ¡!u1 ¡!u2 ¡!v1 ¡!v2 : initial velocity of object 1 (m:s¡1 + direction) : initial velocity of object 2 (m:s¡1 + direction) : ﬂnal ve
locity of object 1 (m:s¡1 + direction) : ﬂnal velocity of object 2 (m:s¡1 + direction) This equation is always true- momentum is always conserved in collisions. The chapter ‘Collisions and Explosions’ (Chapter ??) deals with applications of momentum conservation. 7.5 Impulse At the beginning of this chapter it was mentioned that momentum is closely related to force. We will now explain the nature of this connection. Consider an object of mass m moving with constant acceleration ¡!a . During a time ¢t the object’s velocity changes from an initial velocity ¡!u to a ﬂnal velocity ¡!v (refer to Figure 7.3). We know from Newton’s First Law that there must be a resultant force ¡!F Res acting on the object. Starting from Newton’s Second Law, Momentum is always conserved in collisions! ¡!F Res = m¡!a = m( ¡!v ) ¡ ¡!u ¢t m¡!u m¡!v = ¡ ¢t = ¡!p f inal ¡ ¡!p initial ¢t since ¡!a = ¡!v ¡ ¡!u ¢t = ¢¡!p ¢t This alternative form of Newton’s Second Law is called the Law of Momentum. 134 ¡!F Res m ¡!u t = 0 t = ¢t ¡!v ¡!F Res m Figure 7.3: An object under the action of a resultant force. Law of Momentum: The applied resultant force acting on an object is equal to the rate of change of the object’s momentum and this force is in the direction of the change in momentum. Mathematically, ¡!F Res = ¢¡!p ¢t ¡!F Res ¢¡!p ¢t : resultant force (N + direction) : change in momentum (kg:m:s¡1 + direction) : time over which ¡!F Res acts (s) Rearranging the Law of Momentum, The product ¡!F Res¢t is called impulse, ¡!F Res¢t = ¢¡!p : Impulse · ¡!F Res¢t = ¢¡!p From this equation we see, that for a given change in momentum, ¡!F Res¢t is ﬂxed. Thus, if FRes is reduced, ¢t must be increased (i.e. the resultant force must be applied for longer). Alternatively if ¢t is reduced (i.e. the resultant force is applied for a shorter period) then the resultant force must be increased to bring about the same change in momentum. Worked Example 37 Impulse and Change in momentum 135 Question: A 150 N resultant force acts on a 300 kg object. Calculate how long it takes this force to change the object’s velocity from 2 m:s¡1 to the right to 6 m:s¡1 to the right. Answer: Step 1 : Decide what information is supplied The question explicitly gives the object’s mass, the object’s initial velocity, the object’s ﬂnal velocity, and the resultant force acting on the object † † † † all in the correct units! Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the time taken ¢t to accelerate the object from the given initial velocity to ﬂnal velocity. From the Law of Momentum, ¡!F Res¢t = ¢¡!p m¡!u = m¡!v ¡ ¡ ¡!u ): = m(¡!v Thus we have everything we need to ﬂnd ¢t! Step 3 : Choose a positive direction Although not explicitly stated, the resultant force acts to the right. This follows from the fact that the object’s velocity increases in this direction. Let us then choose right as the positive direction. Step 4 : Do the calculation and quote the ﬂnal answer Right is the positive direction Remember to check the units! ¡!F Res¢t = m(¡!v ¡ ¡!u ) (+150N )¢t = (300kg)((+6 (+150N )¢t = (300kg)(+4 m s m s (300kg)(+4 m s ) +150N ) ¢t = (+2 ) ¡ m s )) ¢t = 8s Worked Example 38 Calculating Impulse Question: A cricket ball weighing 156g is moving at 54 km:hr¡1 towards a batsman. It is hit by the batsman back towards the bowler at 36 km:hr¡1. Calculate i) the 136 Remember to check the units! ball’s impulse, and ii) the average force exerted by the bat if the ball is in contact with the bat for 0:13s. Answer: Step 1 : Decide what information is supplied The question explicitly gives the ball’s mass, the ball’s initial velocity, the ball’s ﬂnal velocity, and the time of contact between bat and ball † † † † all except the time in the wrong units! Answer to (i): Step 2 : Decide how to tackle the problem What is being asked? We are asked to calculate the impulse Impulse = ¢¡!p = ¡!F Res¢t: Since we do not have the force exerted by the bat on the ball (¡!F Res), we have to calculate the impulse from the change in momentum of the ball. Now, since ¢¡!p = ¡!p f inal ¡ ¡!p initial = m¡!v m¡!u ; ¡ we need the ball’s mass, initial velocity and ﬂnal velocity, which we are given. Step 3 : Convert to S.I. units Firstly let us change units for the mass 1000g = 1kg 1 = 1kg 1000g 156g £ 1 = 156g £ = 0:156kg 1kg 1000g Next we change units for the velocity 1km = 1000m 1000m 1km 1 = 3600s = 1hr 1 = 1hr 3600s 54 km hr £ 1 £ 1 = 54 = 15 km hr £ m s 1000m 1km £ 1hr 3600s 137 36 km hr £ 1 £ 1 = 36 = 10 km hr £ m s 1000m 1km £ 1hr 3600s Step 4 : Choose your convention Next we must choose a positive direction. Let us choose the direction from the batsman to the bowler as the postive direction. Then the initial velocity of the ball 15 m:s¡1, while the ﬂnal velocity of the ball is ¡!v = +10 m:s¡1 is ¡!u = Step 5 : Calculate the momentum Now we calculate the change in momentum, ¡ Direction from batsman to bowler is the positive direction ¢¡!p = ¡!p f inal ¡ ¡!p initial = m¡!v m¡!u ¡ ¡ ¡!u ) = m(¡!v = (0:156kg)((+10 m:s¡1) = +3:9 kg:m:s¡1 = 3:9 kg:m:s¡1 in the direction from batsman to bowler 15 m:s¡1)) ( ¡ ¡ where we remembered in the last step to include the direction of the change in momentum in words. Step 6 : Determine the impulse Finally since impulse is just the change in momentum of the ball, Impulse = ¢¡!p = 3:9 kg:m:s¡1 in the direction from batsman to bowler Answer to (ii): Step 7 : Determine what is being asked What is being asked? We are asked to calculate the average force exerted by the bat on the ball, ¡!F Res. Now, Impulse = ¡!F Res¢t = ¢¡!p : We are given ¢t and we have calculated the change in momentum or impulse of the ball in part (i)! Step 8 : Choose a convention Next we choose a positive direction. Let us choose the direction from the batsman to the bowler as the postive direction. Step 9 : Calculate the force Then substituting, Direction from batsman to bowler is the positive direction 138 ¡!F Res¢t = Impulse ¡!F Res(0:13s) = +3:9 kg:m s +3:9 kg:m s 0:13s ¡!F Res = = +30 kg:m s2 = 30N in the direction from batsman to bowler where we remembered in the ﬂnal step to include the direction of the force in words. 7.6 Summary of Important Quantities, Equations and Con- cepts Quantity Momentum Mass Velocity Change in momentum Force Impulse Units Symbol Unit ¡!p m ¡!u ,¡!v ¢¡!p ¡!F J N - S.I. Units Direction kg:m:s¡1 kg m:s¡1 kg:m:s¡1 kg:m:s¡2 kg:m:s¡1 X \| X X X X Table 7.1: Summary of the symbols and units of the quantities used in Momentum Momentum The momentum of an object is deﬂned as its mass multiplied by its velocity. Momentum of a System The total momentum of a system is the sum of the momenta of each of the objects in the system. Principle of Conservation of Linear Momentum: ‘The total linear momentum of an isolated system is constant’ or ‘In an isolated system the total momentum before a collision (or explosion) is equal to the total momentum after the collision (or explosion)’. Law of Momentum: The applied resultant force acting on an object is equal to the rate of change of the object’s momentum and this force is in the direction of the change in momentum. 139 Chapter 8 Work and Energy 8.1 What are Work and Energy? During this chapter you will discover that work and energy are very closely related: We consider the energy of an object as its capacity to do work and doing work as the process of transferring energy from one object or form to another. In other words, † † an object with lots of energy can do lots of work. when work is done, energy is lost by the object doing work and gained by the object on which the work is done. Lifting objects or throwing them requires that you do work on them. Even making electricity ow requires that something do work. Something must have energy and transfer it through doing work to make things happen. 8.2 Work To do work on an object, one must move the object by applying a force with at least a component in the direction of motion. The work done is given by W = Fks W : work done (N:m or J) Fk s : component of applied force parallel to motion (N ) : displacement of the object (m) It is very important to note that for work to be done there must be a component of the applied force in the direction of motion. Forces perpendicular to the direction of motion do no work. As with all physical quantities, work must have units. As follows from the deﬂnition, work is measured in N:m. The name given to this combination of S.I. units is the joule (J). Deﬂnition: 1 joule is the work done when an object is moved 1m under the application of a force of 1N in the direction of motion. 140 The work done by an object can be positive or negative. Since force (Fk) and displacement (s) are both vectors, the result of the above equation depends on their directions: † † If Fk acts in the same direction as the motion then positive work is being done. In this case the object on which the force is applied gains energy. If the direction of motion and Fk are opposite, then negative work is being done. This means that energy is transferred in the opposite direction. For example, if you try to push a car uphill by applying a force up the slope and instead the car rolls down the hill you are doing negative work on the car. Alternatively, the car is doing positive work on you! Worked Example 39 Calculating Work Done I Question: If you push a box 20m forward by applying a force of 15N in the forward direction, what is the work you have done on the box? Answer: Step 1 : Analyse the question to determine what information is provided † † † The force applied is F = 15N . The distance moved is s = 20m. The applied force and distance moved are in the same direction. Therefore, Fk = 15N . These quantities are all in the correct units, so no unit conversions are required. Step 2 : Analyse the question to determine what is being asked † We are asked to ﬂnd the work done on the box. We know from the deﬂnition that work done is W = Fks Step 3 : Next we substitute
the values and calculate the work done W = Fks = (15N )(20m) = 300 N = 300 J m ¢ Remember that the answer must be positive as the applied force and the motion are in the same direction (forwards). In this case, you (the pusher) lose energy, while the box gains energy. 141 Worked Example 40 Calculating Work Done II Question: What is the work done by you on a car, if you try to push the car up a hill by applying a force of 40N directed up the slope, but it slides downhill 30cm? Answer: Step 1 : Analyse the question to determine what information is provided † † † The force applied is F = 40N The distance moved is s = 30cm. This is expressed in the wrong units so we must convert to the proper S.I. units (meters): s = 30cm = 30cm 1m 100cm ¢ = 0:3m The applied force and distance moved are in opposite directions. Therefore, if we take s = 0:3m, then Fk = 40N . ¡ Step 2 : Analyse the question to determine what is being asked † We are asked to ﬂnd the work done on the car by you. We know that work done is W = Fks Step 3 : Substitute the values and calculate the work done Again we have the applied force and the distance moved so we can proceed with calculating the work done: W = Fks = ( 40N )(0:3m) = = ¡ 12N 12 J ¡ ¡ m ¢ Note that the answer must be negative as the applied force and the motion are in opposite directions. In this case the car does work on the person trying to push. What happens when the applied force and the motion are not parallel? If there is an angle between the direction of motion and the applied force then to determine the work done we have to calculate the component of the applied force parallel to the direction of motion. Note that this means a force perpendicular to the direction of motion can do no work. Worked Example 41 Calculating Work Done III Question: Calculate the work done on a box, if it is pulled 5m along the ground by applying a force of F = 10N at an angle of 60o to the horizontal. 142 F 60o Answer: Step 1 : Analyse the question to determine what information is provided The force applied is F = 10N The distance moved is s = 5m along the ground The angle between the applied force and the motion is 60o † † † These quantities are in the correct units so we do not need to perform any unit conversions. Step 2 : Analyse the question to determine what is being asked We are asked to ﬂnd the work done on the box. † Step 3 : Calculate the component of the applied force in the direction of motion Since the force and the motion are not in the same direction, we must ﬂrst calculate the component of the force in the direction of the motion. Fk F 60o Fjj From the force diagram we see that the component of the applied force parallel to the ground is Fjj = F cos(60o) ¢ = 10N = 5 N cos(60o) ¢ Step 4 : Substitute and calculate the work done Now we can calculate the work done on the box: W = Fks = (5N )(5m) = 25 J Note that the answer is positive as the component of the force Fk is in the same direction as the motion. We will now discuss energy in greater detail. 143 8.3 Energy As we mentioned earlier, energy is the capacity to do work. When positive work is done on an object, the system doing the work loses energy. In fact, the energy lost by a system is exactly equal to the work done by the system. Like work (W ) the unit of energy (E) is the joule (J). This follows as work is just the transfer of energy. A very important property of our universe which was discovered around 1890 is that energy is conserved. Energy is never created nor destroyed, but merely transformed from one form to another. Energy conservation and the conservation of matter are the principles on which classical me- chanics is built. Energy is conserved! IN THE ABSENCE OF FRICTION When work is done on an object by a system: -the object gains energy equal to the work done by the system Work Done = Energy Transferred IN THE PRESENCE OF FRICTION When work is done by a system: -only some of the energy lost by the system is transferred into useful energy -the rest of the energy transferred is lost to friction Total Work Done = Useful Work Done + Work Done Against Friction 8.3.1 Types of Energy So what diﬁerent types of energy exist? Kinetic, mechanical, thermal, chemical, electrical, radiant, and atomic energy are just some of the types that exist. By the principle of conservation of energy, when work is done energy is merely transferred from one object to another and from one type of energy to another. 144 Kinetic Energy Kinetic energy is the energy of motion that an object has. Objects moving in straight lines possess translational kinetic energy, which we often abbreviate as Ek. The translational kinetic energy of an object is given by Ek = 1 2 mv2 : kinetic energy (J) Ek m : mass of object (kg) v : speed of the object (m:s¡1) Note the dependence of the kinetic energy on the speed of the object{ kinetic energy is related to motion. The faster an object is moving the greater its kinetic energy. Worked Example 42 Calculation of Kinetic Energy Question: If a rock has a mass of 1kg and is thrown at 5m=s, what is its kinetic energy? Answer: Step 1 : Analyse the question to determine what information is provided The mass of the rock m = 1kg The speed of the rock v = 5m=s † † These are both in the correct units so we do not have to worry about unit conversions. Step 2 : Analyse the question to determine what is being asked † We are asked to ﬂnd the kinetic energy. From the deﬂnition we know that to work out Ek, we need to know the mass and the velocity of the object and we are given both of these values. Step 3 : Substitute and calculate the kinetic energy mv2 Ek = = 1 2 1 2 (1kg)(5 )2 m s m2 ¢ s2 kg = 12:5 = 12:5 J 145 To check that the units in the above example are in fact correct: kg m2 ¢ s2 m kg ¢ s2 ¢ ¶ = = J m = N m ¢ The units are indeed correct! Study hint: Checking units is an important cross-check and you should get into a habit of doing this. If you, for example, ﬂnish an exam early then checking the units in your calculations is a very good idea. Worked Example 43 Mixing Units and Kinetic Energy Calculations 1 Question: If a car has a mass of 900kg and is driving at 60km=hr, what is its kinetic energy? Answer: Step 1 : Analyse the question to determine what information is provided † † The mass of the car m = 900kg The speed of the car v = 60km=hr. These are not the units we want so before we continue we must convert to m=s. We do this by multiplying by one: 60 km hr £ 1 = 60 1000m 1km km hr £ m hr = 60000 Now we need to change from hours to seconds so we repeat our procedure: 60000 m hr £ 1 = 60000 = 16:67 1hr 3600s m hr £ m s and so the speed in the units we want is v = 16:67m=s. Step 2 : Analyse the question to determine what is being asked We are asked to ﬂnd the kinetic energy. † Step 3 : Substitute and calculate 146 We know we need the mass and the speed to work out Ek and we are given both of these quantities. We thus simply substitute them into the equation for Ek: Ek = = 1 2 1 2 mv2 (900kg)(16:67 m s )2 = 125 000 kgm2 s2 = 125 000 J Worked Example 44 Mixing Units and Kinetic Energy Calculations 2 Question: If a bullet has a mass of 150g and is shot at a muzzle velocity of 960m=s, what is its kinetic energy? Answer: Step 1 : Analyse the question to determine what information is provided † We are given the mass of the bullet m = 150g. This is not the unit we want mass to be in. We need to convert to kg. Again, we multiply by one: 150g ¢ 1 = 150g ¢ = 0:15kg 1kg 1000g † We are given the muzzle velocity which is just how fast the bullet leaves the barrel and it is v = 960m=s. Step 2 : Analyse the question to determine what is being asked We are asked to ﬂnd the kinetic energy. † Step 3 : Substitute and calculate We just substitute the mass and velocity (which are known) into the equation for Ek: Ek = = 1 2 1 2 mv2 (150kg)(960 m s )2 = 69 120 kgm2 s2 = 69 120 J 147 Potential Energy If you lift an object you have to do work on it. This means that energy is transferred to the object. But where is this energy? This energy is stored in the object and is called potential energy. The reason it is called potential energy is because if we let go of the object it would move. Deﬂnition: Potential energy is the energy an object has due to its position or state. As an object raised above the ground falls, its potential energy is released and transformed into kinetic energy. The further it falls the faster it moves as more of the stored potential energy is transferred into kinetic energy. Remember, energy is never created nor destroyed, but merely transformed from one type to another. In this case potential energy is lost but an equal amount of kinetic energy is gained. In the example of a falling mass the potential energy is known as gravitational potential energy as it is the gravitational force exerted by the earth which causes the mass to accelerate towards the ground. The gravitational ﬂeld of the earth is what does the work in this case. Another example is a rubber-band. In order to stretch a rubber-band we have to do work on it. This means we transfer energy to the rubber-band and it gains potential energy. This potential energy is called elastic potential energy. Once released, the rubber-band begins to move and elastic potential energy is transferred into kinetic energy. Gravitational Potential Energy As we have mentioned, when lifting an object it gains gravitational potential energy. One is free to deﬂne any level as corresponding to zero gravitational potential energy. Objects above this level then possess positive potential energy, while those below it have negative potential energy. To avoid negative numbers in a problem, always choose the lowest level as the zero potential mark. The change in gravitational potential energy of an object is given by: ¢EP = mg¢h ¢EP : Change in gravitational potential energy (J) m : mass of object (kg) g ¢h : acceleration due to gravity (m:s¡2) : change in height (m) When an object is lifted
it gains gravitational potential energy, while it loses gravitational potential energy as it falls. Worked Example 45 Gravitational potential energy 148 Question: How much potential energy does a brick with a mass of 1kg gain if it is lifted 4m. Answer: Step 1 : Analyse the question to determine what information is provided The mass of the brick is m = 1kg The height lifted is ¢h = 4m † † These are in the correct units so we do not have to worry about unit conversions. Step 2 : Analyse the question to determine what is being asked We are asked to ﬂnd the gain in potential energy of the object. † Step 3 : Identify the type of potential energy involved Since the block is being lifted we are dealing with gravitational potential energy. To work out ¢EP , we need to know the mass of the object and the height lifted. As both of these are given, we just substitute them into the equation for ¢EP . Step 4 : Substitute and calculate ¢EP = mg¢h = (1kg) 10 = 40 kg ‡ m2 ¢ s2 = 40 J (4m) m s2 · 8.4 Mechanical Energy and Energy Conservation Kinetic energy and potential energy are together referred to as mechanical energy. The total mechanical energy (U ) of an object is then the sum of its kinetic and potential energies: U = EP + EK 1 2 U = mgh + mv2 (8.1) Now, IN THE ABSENCE OF FRICTION Mechanical energy is conserved Ubef ore = Uaf ter This principle of conservation of mechanical energy can be a very powerful tool for solving physics problems. However, in the presence of friction some of the mechanical energy is lost: 149 IN THE PRESENCE OF FRICTION Mechanical energy is not conserved (The mechanical energy lost is equal to the work done against friction) ¢U = Ubef ore ¡ Uaf ter = Work Done Against Friction Worked Example 46 Using Mechanical Energy Conservation Question: A 2kg metal ball is suspended from a rope. If it is released from point A and swings down to the point B (the bottom of its arc) what is its velocity at point B? A 0.5m B Answer: Step 1 : Analyse the question to determine what information is provided The mass of the metal ball is m = 2kg The change in height going from point A to point B is h = 0:5m The ball is released from point A so the velocity at point A is zero (vA = 0m=s). † † † These are in the correct units so we do not have to worry about unit conversions. Step 2 : Analyse the question to determine what is being asked Find the velocity of the metal ball at point B. † Step 3 : Determine the Mechanical Energy at A and B To solve this problem we use conservation of mechanical energy as there is no friction. Since mechanical energy is conserved, UA = UB Therefore we need to know the mechanical energy of the ball at point A (UA) and at point B (UB). The mechanical energy at point A is UA = mghA + 1 2 m(vA)2 150 We already know m, g and vA, but what is hA? Note that if we let hB = 0 then hA = 0:5m as A is 0:5m above B. In problems you are always free to choose a line corresponding to h = 0. In this example the most obvious choice is to make point B correspond to h = 0. Now we have, UA = (2kg) 10 = 10 J ‡ m s2 · (0:5m) + (2kg)(0)2 1 2 As already stated UB = UA. Therefore UB = 10J, but using the deﬂnition of mechanical energy UB = mghB + 1 2 m(vB)2 = 1 2 m(vB)2 because hB = 0. This means that (2kg)(vB)2 10J = 1 2 (vB)2 = 10 J kg vB = p10 m s 8.5 Summary of Important Quantities, Equations and Con- cepts Quantity Work Kinetic Energy Potential Energy Mechanical Energy Units Symbol Unit W EK EP U J J J J S.I. Units N:m or kg:m2:s¡2 N:m or kg:m2:s¡2 N:m or kg:m2:s¡2 N:m or kg:m2:s¡2 Direction \| \| \| \| Table 8.1: Summary of the symbols and units of the quantities used in Energy Principle of Conservation of Energy: Energy is never created nor destroyed, but merely transformed from one form to another. Conservation of Mechanical Energy: In the absence of friction, the total mechanical energy of an object is conserved. 151 Essay 1 : Energy Author: Asogan Moodaly Asogan Moodaly received his Bachelor of Science degree (with honours) in Mechanical Engineering from the University of Natal, Durban in South Africa. For his ﬂnal year design project he worked on a 3-axis ﬂlament winding machine for composite (Glass re-enforced plastic in this case) piping. He worked in Vereeniging, Gauteng at Mine Support Products (a subsidiary of Dorbyl Heavy Engineering) as the design engineer once he graduated. He currently lives in the Vaal Triangle area and is working for Sasol Technology Engineering as a mechanical engineer, ensuring the safety and integrity of equipment installed during projects. Energy and electricity. Why the fuss? Disclaimer: The details furnished below are very basic and for illustration purposes only. Why do we need energy? Note that I use the word energy and not electricity. On a broad scale it stimulates economic growth, etc, etc but on a personal level it allows us to lead a comfortable lifestyle. e.g. Flick a switch and Heat for cooking Entertainment such as television and radio Heat for water and interior of house Ironing Electronic and electrical devices such as alarms, garage doors, etc. In a modern household this energy is provided in the form of electricity which is powered via fossil fuels or nuclear. How is electricity made? In a nutshell: By moving a magnet through or near a set of conducting coils. Linear Rotational Conducting wire typically copper Magnet Light bulb Magnet Iron core in coil increases electricity Most power stations produce steam through heat (nuclear reaction or burning fossil fuels), the steam drives a turbine which moves a magnet relative to a coil (the generator like the above but on a much larger scale i.e. bigger magnets, bigger coils, etc), which produces electricity that is transmitted via a power network to our homes. Gas ﬂred plants burn gas directly in a gas turbine to produce the same desired relative motion between permanent magnet and coil. 152 Steam rushes through Turbine, causing the blades to spin (attached to the shaft so shaft spins) Generator Boiler Shaft Steam Water Magnet attached to shaft Heat Turbine blades Coils Coal Electricity transmitted through power cable Power Station Pylons support cables transformers houses transmission cables Coal, oil and gas are fossil fuels. Fossil fuels were created by decomposing organic (plant and animal) matter a long, long time ago and are typically found underground. Diﬁerent temperatures and pressures resulted in the organic matter transforming into coal, oil or gas. Why the fuss about fossil fuels? 1. Fossil fuel power is bad news in the long run. It pollutes and contributes to the greenhouse eﬁect (global warming resulting in melting polar ice caps, oods, droughts, disease, etc). 2. Its not going to last forever. 3. Nuclear power is cleaner in terms of emissions but theres no proven way of disposing of the nuclear waste. Oh, and it wont last forever either! Renewable Energy As the name suggests renewable energy lasts forever. Solar (sun), wind, geothermal, wave, hydro and biomass (organic) are all sources of energy that will last until the sun eventually explodes many millions of years from now. Hopefully the human race will have moved from the earth by then! Generally the principal of renewable electricity generation is similar to fossil fuel electricity generation in that electricity is generated by moving a magnet relative to a conducting coil. What is diﬁerent is the way energy is supplied to cause that motion. The below are a few diﬁerent types of available renewable energy technologies. Solar There are diﬁerent types of solar electricity technologies, the main ones being solar thermal and photovoltaic. 153 Sun Heated oil stored in insulated tank for night use Transmission cables & pylon Steam pipes Steam Turbine Generator Mirrored trough reflector, reflects sunlight onto black pipes Black pipes carrying oil Heat exchanger transfers heat from hot oil to water Water becomes steam Water pipes Solar thermal uses the heat of the sun to produce electricity. Sun is concentrated using mirrors. This heat either creates steam which drives a turbine which in turn drives a generator (as per fossil fuel generation), or drives an air engine (engine that uses expanding air to obtain motion) that drives a generator. Photovoltaic panels convert sunlight directly into electricity. The beneﬂt of photovoltaic panels is that there are no moving parts, and is therefore relatively maintenance free. The downside is that its very expensive at this stage (17/06/2004). Solar Water Heaters could save up to 30% of the total electricity used in a house. Heater on roof Hot Water out Insulated geyser to store water Cold water in Wooden box with glass top provides insulation. Silicon gel should be used to make the box airtight. If air escapes, so does heat. copper Water (in pipes painted black) is heated by sunlight. Cold water sinks. Hot water rises into the geyser. Wind Wind turbines catch wind that spins the blades. The blades are connected to a shaft that spins because of the wind. This spinning shaft spins another shaft that turns a permanent magnet relative to conducting coils. Note that gears are used to convert the slow spinning of the 1st shaft to a faster spin on the 2nd shaft. The generator shaft needs to spin at the correct speed to produce the right amount and quality of electricity. Some generators are now being modiﬂed to run at slower speeds. This saves money as gears are not needed. 154 Vessel Nacelle containing all moving parts Blades Detail of Nacelle Vessel (top view) to blades Shaft 1 (slow spin) Shaft 2 (Fast spin) Gears Generator Biomass Biomass is anything organic i.e. plant or animal matter. It can be used in the place of coal as per a normal coal ﬂred plant and is renewable as long as the biomass e.g. wood; is handled in a sustainable manner. By sustainable I mean that suitable farming practices are used so that the land is not over farmed which will result in the soil becoming barren and nothing growing there again. Pipe
carrying biogas for heat or power Processed biomass fertilizer Biogas bubbles Upside down floating container to catch biogas Input biomass Water forms a seal to keep air out Input biomass being digested Biomass can also be processed using anaerobic digestion to produce a gas that can be burned for heat or electricity. This biogas is made up of a number of other gases that are similar to those found in fossil fuel natural gas Except the amount of the gases are diﬁerent. E.g. Natural gas has about 94 Anaerobic digestion: Anaerobic means No air. Therefore anaerobic digestion means to digest in the absence of air. Bacteria that naturally exist in organic matter will convert organic matter to biogas and fertilizer when all the air is removed. Thousands of anaerobic digesters have been installed in rural India, Nepal and China in rural areas where cow dung, human waste and chicken litter (faeces) are all processed using anaerobic digestion to produce gas that can be burned in the home for cooking and heating. The leftover is used as fertilizer. Geothermal Energy In some places on earth, the earths crust is thinner than others. As a result the heat from the earths core escapes. The heat can be captured by converting water to steam, and using the 155 steam to drive a steam generator as discussed above. Hydroelectric power Water from a river is diverted to turn a water turbine to create electricity similar to the principles of steam generation. The water is returned to the river after driving the turbine. Pipe diverting water Transmission cables Generator house River Wave Energy Some wave energy generators work similarly to wind turbines except that underwater ocean currents turns the blades instead of wind; and of course most of the structure is under water! Ocean current blades Ocean floor Underwater pipe to shore carrying transmission cable Another concept uses the rising and falling of the tides to suck air in using a one way valve. As a result air becomes compressed in a chamber and the compressed air is let out to drive a turbine which in turn drives a generator One way valve allows air to be sucked in then shuts when air tries to get out. Generator house with turbine and generator Chamber with compressed air Cliff face Ocean floor As the water level falls and rises, air is sucked in and compressed 156 These are relatively new technologies. Liquid Fuels Liquid fuels are used mainly for transportation. Petrol and diesel are the most common liquid fuels and are obtained from oil. Sasol is the only company in the world that makes liquid fuels from coal; and will be one of the leading companies in the world to make liquid fuels from natural gas! The Sasol petro-chemical plants are based in Sasolburg on the border of the Free State and in Secunda in Mpumalanga. However, as discussed above coal, gas and oil are fossil fuels and are not renewable. Petrol and diesel are obtained from fossil fuels and therefore pollute and contribute to the green house eﬁect (global warming). Alternatives Biodiesel Oil can be extracted from plants such as the soya bean, sunower and rapeseed by pressing it through a ﬂlter. This oil if mixed correctly with either methanol or dry ethanol and Sodium Hydroxide will separate the plant oil into biodiesel, glycerol and fertilizer. The biodiesel can be used as produced in a conventional diesel engine with little or no mod- iﬂcations required. The glycerol can be reﬂned a bit further for pharmaceutical companies to use, or can be used to make soap. Ethanol Corn, maize and sugar cane can be used to make ethanol as a fuel substitute for petrol. Its made by the same fermentation process used to make alcohol. Enzymes are often used to speed up the process. In ethanol from sugar cane production, the leftover bagasse (the ﬂbre part of the sugar cane) can be burned in a biomass power station to produce electricity. Hydrogen Through the process of electrolysis electricity (hopefully clean, renewable electricity!) can split water into hydrogen and oxygen. The stored hydrogen can be used in a fuel cell to create electricity in a process that is opposite to electrolysis; to drive electric motors in a car. The hydrogen can also be burned directly in a modiﬂed internal combustion engine. In both cases the waste product is water. 157 Essay 2 : Tiny, Violent Collisions Author: Thomas D. Gutierrez Tom Gutierrez received his Bachelor of Science and Master degrees in physics from San Jose State University in his home town of San Jose, California. As a Master’s student he helped work on a laser spectrometer at NASA Ames Research Centre. The instrument measured the ratio of diﬁerent isotopes of carbon in CO2 gas and could be used for such diverse applications as medical diagnostics and space exploration. Later, he received his Ph.D. in physics from the University of California, Davis where he performed calculations for various reactions in high energy physics collisions. He currently lives in Berkeley, California where he studies proton-proton collisions seen at the STAR experiment at Brookhaven National Laboratory on Long Island, New York. High Energy Collisions Take an orange and expanded it to the size of the earth. The atoms of the earth-sized orange would themselves be about the size of regular oranges and would ﬂll the entire \earth-orange". Now, take an atom and expand it to the size of a football ﬂeld. The nucleus of that atom would be about the size of a tiny seed in the middle of the ﬂeld. From this analogy, you can see that atomic nuclei are very small objects by human standards. They are roughly 10¡15 meters in diameter { one-hundred thousand times smaller than a typical atom. These nuclei cannot be seen or studied via any conventional means such as the naked eye or microscopes. So how do scientists study the structure of very small objects like atomic nuclei? The simplest nucleus, that of hydrogen, is called the proton. Faced with the inability to isolate a single proton, open it up, and directly examine what is inside, scientists must resort to a brute-force and somewhat indirect means of exploration: high energy collisions. By colliding protons with other particles (such as other protons or electrons) at very high energies, one hopes to learn about what they are made of and how they work. The American physicist Richard Feynman once compared this process to slamming delecate watches together and ﬂguring out how they work by only examining the broken debris. While this analogy may seem pessimistic, with su–cent mathematical models and experimental precision, considerable information can be extracted from the debris of such high energy subatomic collisions. One can learn about both the nature of the forces at work and also about the sub-structure of such systems. The experiments are in the category of \high energy physics" (also known as \subatomic" physics). The primary tool of scientiﬂc exploration in these experiments is an extremely violent collision between two very, very small subatomic objects such as nuclei. As a general rule, the higher the energy of the collisions, the more detail of the original system you are able to resolve. These experiments are operated at laboratories such as CERN, SLAC, BNL, and Fermilab, just to name a few. The giant machines that perform the collisions are roughly the size of towns. For example, the RHIC collider at BNL is a ring about 1 km in diameter and can be seen from space. The newest machine currently being built, the LHC at CERN, is a ring 9 km in diameter! Let’s examine the kinematics of such a collisions in some detail... 158 Chapter 9 Collisions and Explosions In most physics courses questions about collisions and explosions occur and to solve these we must use the ideas of momentum and energy; with a bit of mathematics of course! This section allows you to pull the momentum and energy ideas together easily with some speciﬂc problems. 9.1 Types of Collisions We will consider two types of collisions in this section Elastic collisions Inelastic collisions † † In both types of collision, total energy and total momentum is always conserved. Kinetic energy is conserved for elastic collisions, but not for inelastic collisions. 9.1.1 Elastic Collisions Deﬂnition: An elastic collision is a collision where total momentum and total kinetic energy are both conserved. (NOTE TO SELF: this should be in an environment for deﬂnitions!!) This means that the total momentum and the total kinetic energy before an elastic collision is the same as after the collision. For these kinds of collisions, the kinetic energy is not changed into another type of energy. Before the Collision In the following diagram, two balls are rolling toward each other, about to collide ¡!p 1, K1 ¡!p 2, K2 159 Before the balls collide, the total momentum of the system is equal to all the individual momenta added together. The ball on the left has a momentum which we call ¡!p 1 and the ball on the right has a momentum which we call ¡!p 2, it means the total momentum before the collision is ¡!p Before = ¡!p 1 + ¡!p 2 (9.1) We calculate the total kinetic energy of the system in the same way. The ball on the left has a kinetic energy which we call K1 and the ball on the right has a kinetic energy which we call K2, it means that the total kinetic energy before the collision is KBefore = K1 + K2 (9.2) After the Collision The following diagram shows the balls after they collide ¡!p 3, K3 ¡!p 4, K4 After the balls collide and bounce oﬁ each other, they have new momenta and new kinetic energies. Like before, the total momentum of the system is equal to all the individual momenta added together. The ball on the left now has a momentum which we call ¡!p 3 and the ball on the right now has a momentum which we call ¡!p 4, it means the total momentum after the collision is ¡!p After = ¡!p 3 + ¡!p 4 The ball on the left now has a kinetic energy which we call K3 and the ball on the right now has a kinetic energy which
we call K4, it means that the total kinetic energy after the collision is (9.3) KAfter = K3 + K4 (9.4) Since this is an elastic collision, the total momentum before the collision equals the total momentum after the collision and the total kinetic energy before the collision equals the total kinetic energy after the collision Before ¡!p Before ¡!p 1 + ¡!p 2 After = ¡!p After = ¡!p 3 + ¡!p 4 and KBefore K1 + K2 = KAfter = K3 + K4 (9.5) (9.6) Worked Example 47 An Elastic Collision 160 We will have a look at the collision between two pool balls. Ball 1 is at rest and ball 2 is moving towards it with a speed of 2 [m:s¡1]. The mass of each ball is 0.3 [Kg]. After the balls collide elastically, ball 2 comes to a stop and ball 1 moves oﬁ. What is the ﬂnal velocity of ball 1? Step 1 : Draw the \before" diagram Before the collision, ball 2 is moving; we will call it’s momentum P2 and it’s kinetic energy K2. Ball 1 is at rest, so it has zero kinetic energy and momentum. 2 1 ¡!p 2, K2 ¡!p 1 = 0, K1 = 0 Step 2 : Draw the \after" diagram After the collision, ball 2 is at rest but ball 1 has a momentum which we call P3 and a kinetic energy which we call K3. ¡!p 4 = 0, K4 = 0 ¡!p 3, K3 2 1 Because the collision is elastic, we can solve the problem using momentum conservation or kinetic energy conservation. We will do it both ways to show that the answer is the same, whichever method you use. Step 3 : Show the conservation of momentum We start by writing down that the momentum before the collision ¡!p Before is equal to the momentum after the collision ¡!p After Before After ¡!p Before = ¡!p After ¡!p 1 + ¡!p 2 = ¡!p 3 + ¡!p 4 0 + ¡!p 2 = ¡!p 3 + 0 ¡!p 2 = ¡!p 3 We know that momentum is just P = mv, and we know the masses of the balls, so we can rewrite the conservation of momentum in terms of the velocities of the balls (9.7) ¡!p 2 = ¡!p 3 m2v2 = m3v3 0:3v2 = 0:3v3 v2 = v3 So ball 1 exits with the velocity that ball 2 started with! v3 = 2[m:s¡1] 161 (9.8) (9.9) Step 4 : Show the conservation of kinetic energy We start by writing down that the kinetic energy before the collision KBefore is equal to the kinetic energy after the collision KAfter Before After KBefore = KAfter K1 + K2 = K3 + K4 0 + K2 = K3 + 0 K2 = K3 (9.10) We know that kinetic energy is just K = mv 2 , and we know the masses of the balls, so we can rewrite the conservation of kinetic energy in terms of the velocities of the balls 2 K2 = K3 (9.11) = m3v2 3 2 m2v2 2 2 0:15v2 2 = 0:15v2 3 2 = v2 v2 3 v2 = v3 So ball 1 exits with the velocity that ball 2 started with, which agrees with the answer we got when we used the conservation of momentum. v3 = 2[m:s¡1] (9.12) Worked Example 48 Elastic Collision 2 Question: Now for a slightly more di–cult example. We have 2 marbles. Marble 1 has mass 50 g and marble 2 has mass 100 g. I roll marble 2 along the ground towards marble 1 in the positive x-direction. Marble 1 is initially at rest and marble 2 has a velocity of 3 m:s¡1 in the positive x-direction. After they collide elastically, both marbles are moving. What is the ﬂnal velocity of each marble? Answer: Step 1 : Put all the quantities into S.I. units So: m1 = 0:05kg and m2 = 0:1kg Step 2 : Draw a rough sketch of the situation Before the collision: 2 1 ¡!p2, Ek2 ¡!p1 = 0, Ek1 = 0 162 After the collision: 2 1 ¡!p3, Ek3 ¡!p4, Ek4 Step 3 : Decide which equations to use in the problem Since the collision is elastic, both momentum and kinetic energy are conserved in the collision. So: EkBef ore = EkAf ter and ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter There are two unknowns (¡!v1 and ¡!v2) so we will need two equations to solve for them. We need to use both kinetic energy conservation and momentum conservation in this problem. Step 4 : Solve the ﬂrst equation Let’s start with energy conservation. Then: 1 2 2 m1¡!u1 + 2 m1¡!u1 2 EkBef ore = EkAf ter 1 2 + m2¡!u2 1 2 = m1¡!v1 m1¡!v1 2 m2¡!u2 = 2 2 + 1 2 + m2¡!v2 2 m2¡!v2 2 But ¡!u1=0, and solving for ¡!v2 2 : 2 2 ¡!v2 ¡!v2 2 ¡!v2 2 = ¡!u2 ¡ = (3)2 = 9 ¡ ¡ 1 2 ¡!v1 2 m1 m2 ¡!v1 (0:05) (0:10) ¡!v1 2 2 (A) Step 5 : Solve the second equation Now we have simpliﬂed as far as we can, we move onto momentum conservation: ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter m1¡!u1 + m2¡!u2 = m1¡!v1 + m2¡!v2 But ¡!u1=0, and solving for ¡!v1: m2¡!u2 = m1¡!v1 + m2¡!v2 m2¡!v2 m1¡!v1 = m2¡!u2 m2 ¡!v2 ¡!u2 m1 ¡!v1 = ¡ m2 m1 ¡!v1 = 2(3) ¡!v1 = 6 ¡ ¡ 2¡!v2 ¡ 2¡!v2 (B) 163 Step 6 : Substitute one equation into the other Now we can substitute (B) into (A) to solve for ¡!v2: 2 2 2 2 2 2 ¡!v2 ¡!v2 ¡!v2 ¡!v2 3¡!v2 ¡!v2 = 9 = 9 = 9 ¡ ¡ ¡ 2¡!v2)2 2 1 2 ¡!v1 1 (6 2 1 2 ¡ 18 + 12¡!v2 (36 ¡ = 9 = ¡ = 4¡!v2 ¡ 9 + 12¡!v2 3 ¡ 24¡!v2 + 4¡!v2 2 ) 2 2¡!v2 ¡ 2 ¡!v2 (¡!v2 ¡ ¡ 3)(¡!v2 4¡!v2 + 3 = 0 1) = 0 ¡!v2 = 3 or ¡!v2 = 1 ¡ We were lucky in this question because we could factorise. If you can’t factorise, then you can always solve using the formula for solving quadratic equations. Remember: b x = ¡ § 4ac pb2 2a ¡ So, just to check: ¡!v2 = 4 § 42 ¡ 2(1) 4(1)(3) 4 ¡ § 12 ¡!v2 = p p16 2 p4 § 2 1 ¡!v2 = 3 or ¡!v2 = 1 same as before ¡!v2 = ¡!v2 = 2 § 4 Step 7 : Solve for and quote the ﬂnal answers So ﬂnally, substituting into equation (B) to get ¡!v1: ¡!v1 = 6 2¡!v2 ¡ If ¡!v2 = 3 m:s¡1 then ¡!v1 = 6 ¡ 2(3) = 0 m:s¡1 But, according to the question, marble 1 is moving after the collision. So ¡!v1 ¡!v2 = 3. Therefore: = 0 and ¡!v2 = 1 m:s¡1 in the positive x and ¡!v1 = 4 m:s¡1 in the positive x direction direction ¡ ¡ 164 6 6 9.1.2 Inelastic Collisions Deﬂnition: An inelastic collision is a collision in which total momentum is conserved but total kinetic energy is not conserved; the kinetic energy is transformed into other kinds of energy. So the total momentum before an inelastic collisions is the same as after the collision. But the total kinetic energy before and after the inelastic collision is diﬁerent. Of course this does not mean that total energy has not been conserved, rather the energy has been transformed into another type of energy. As a rule of thumb, inelastic collisions happen when the colliding objects are distorted in some way. Usually they change their shape. To modify the shape of an object requires energy and this is where the \missing" kinetic energy goes. A classic example of an inelastic collision is a car crash. The cars change shape and there is a noticeable change in the kinetic energy of the cars before and after the collision. This energy was used to bend the metal and deform the cars. Another example of an inelastic collision is shown in the following picture. Here an asteroid (the small circle) is moving through space towards the moon (big circle). Before the moon and the asteroid collide, the total momentum of the system is: ¡¡¡¡!pBef ore = ¡!pm + ¡!pa (¡!pm stands for ¡¡¡!pmoon and ¡!pa stands for ¡¡¡¡¡! pasteroid) and the total kinetic energy of the system is: EBef ore = Ekm + Eka ¡!pm, Ekm ¡!pa, Eka When the asteroid collides inelastically with the moon, its kinetic energy is transformed mostly into heat energy. If this heat energy is large enough, it can cause the asteroid and the area of the moon’s surface that it hit, to melt into liquid rock! From the force of impact of the asteroid, the molten rock ows outwards to form a moon crater. After the collision, the total momentum of the system will be the same as before. But since this collision is inelastic, (and you can see that a change in the shape of objects has taken place!), 165 total kinetic energy is not the same as before the collision. ¡¡¡¡!pAf ter, EkAf ter ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter ¡!pm + ¡!pa = ¡¡¡¡!pAf ter but = EkAf ter = EkAf ter EkBef ore Ekm + Eka So: Worked Example 49 Inelastic Collision Question: Let’s consider the collision of two cars. Car 1 is at rest and Car 2 is moving at a speed of 2m:s¡1 in the negative x-direction. Both cars each have a mass of 500kg. The cars collide inelastically and stick together. What is the resulting velocity of the resulting mass of metal? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: Car 1 ¡!p1 = 0 Car 2 ¡!p2 After the collision: ¡¡¡¡!pAf ter 166 6 6 Step 2 : Decide which equations to use in the problem We know the collision is inelastic and there was a deﬂnite change in shape of the objects involved in the collision - there were two objects to start and after the collision there was one big mass of metal! Therefore, we know that kinetic energy is not conserved in the collision but total momentum is conserved. So: EkBef ore = EkAf ter but pT Bef ore = ¡¡¡¡!pAf ter ¡¡¡¡¡¡! Step 3 : Solve for and quote the ﬂnal velocity So we must use conservation of momentum to solve this problem. Take the negative x-direction to have a negative sign: pT Bef ore = ¡¡¡¡!pAf ter ¡¡¡¡¡¡! ¡!p1 + ¡!p2 = ¡¡¡¡!pAf ter m1¡!u1 + m2¡!u2 = (m1 + m2)¡!v 2) = (500 + 500)¡!v 0 + 500( ¡ ¡ 1000 = 1000¡!v ¡!v = 1 m:s¡1 ¡ Therefore, ¡!v = 1 m:s¡1 in the negative x direction: ¡ 9.2 Explosions When an object explodes, it breaks up into more than one piece and it therefore changes its shape. Explosions occur when energy is transformed from one kind e.g. chemical potential energy to another e.g. heat energy or kinetic energy extremely quickly. So, like in inelastic collisions, total kinetic energy is not conserved in explosions. But total momentum is always conserved. Thus if the momenta of some of the parts of the exploding object are measured, we can use momentum conservation to solve the problem! Interesting Fact: The Tunguska event was an aerial explosion that occurred near the Podkamennaya (Stony) Tunguska River in what is now Evenkia, Siberia, at 7:17 AM on June 30, 1908. The size of the blast was later estimated to be equivalent to between 10 and 15 million tons of regular explosive. It felled an estimated 60 million trees over 2,150 square kilometers. At around 7:15 AM, Tungus natives and Russian settlers in the hills northwest of Lake Baikal observed a huge ﬂreball moving across the sky, nearly as bright as the Sun. A few minutes later, there was a ash that lit up h
alf of the sky, followed by a shock wave that 167 6 knocked people oﬁ their feet and broke windows up to 650 km away (the same as the distance from Bloemfontein to Durban!). The explosion registered on seismic stations across Europe and Asia, and produced uctuations in atmospheric pressure strong enough to be detected in Britain. Over the next few weeks, night skies over Europe and western Russia glowed brightly enough for people to read by. Had the object responsible for the explosion hit the Earth a few hours later, it would have exploded over Europe instead of the sparsely-populated Tunguska region, producing massive loss of human life. In the following picture, a closed can of baked beans is put on a stove or ﬂre: BAKED BEANS ¡¡!pcan; Ekcan Before the can heats up and explodes, the total momentum of the system is: ¡¡¡¡!pBef ore = ¡¡!pcan = 0 and the total kinetic energy of the system is: EkBef ore = Ekcan = 0 since the can isn’t moving. Once the mixture of beans, juice and air inside the can reach a certain temperature, the pressure inside the can becomes so great that the can explodes! Beans and sharp pieces of metal can y out in all directions. Energy in the system has been transformed from heat energy into kinetic energy. ¡!p4; Ek4 ¡!p3; Ek3 ¡!p1; Ek1 ¡!p2; Ek2 168 After the explosion, the can is completely destroyed. But momentum is always conserved, so: ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter ¡¡¡¡!pBef ore = ¡!p1 + ¡!p2 + ¡!p3 + ¡!p4 0 = ¡!p1 + ¡!p2 + ¡!p3 + ¡!p4 However, the kinetic energy of the system is not conserved. The can’s shape was changed in the explosion. Before the explosion the can was not moving, but after the explosion, the pieces of metal and baked beans were moving when they were ying out in all directions! So: EkB 6 = EkA Safety tip: Never heat a closed can on a stove or ﬂre! Always open the can or make a hole in the lid to allow the pressure inside and outside the can to remain equal. This will prevent the can from exploding! Worked Example 50 Explosions 1 Question: An object with mass mT ot = 10 kg is sitting at rest. Suddenly it explodes into two pieces. One piece has a mass of m1 = 5 kg and moves oﬁ in the negative x-direction at ¡!v1 = 3ms¡1. What is the velocity of the other piece? Answer: Step 1 : Draw a rough sketch of the situation Before the explosion, the object is at rest: ¡¡!pT ot = 0, EkT ot = 0 After the explosion, the two pieces move oﬁ: 1 2 ¡!p1, Ek1 ¡!p2, Ek2 Step 2 : Decide which equations to use in the problem Now we know that in an explosion, total kinetic energy is not conserved. There is a deﬂnite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So: ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter ¡¡¡¡!pBef ore = ¡!p1 + ¡!p2 169 But the object was initially at rest so: 0 = ¡!p1 + ¡!p2 0 = m1¡!v1 + m2¡!v2 (A) Step 3 : Find the mass of the second piece Now we know that m1= 5 kg but we do not know what the mass of m2 is. However, we do know that: mT ot = m1 + m2 m2 = mT ot ¡ = 10 kg ¡ m1 5 kg = 5 kg Step 4 : Solve for and quote the velocity of the other piece Now we can substitute all the values we know into equation (A) and solve for ¡!v2. Let’s choose the positive x-direction to have a postive sign and the negative x-direction to have a negative sign: (A) 0 = m1¡!v1 + m2¡!v2 0 = 5( 3) + 5¡!v2 0 = ¡ 5¡!v2 = 15 ¡!v2 = +3 m:s¡1 ¡ 15 + 5¡!v2 Therefore, ¡!v2 = 3 m:s¡1 in the positive x direction: ¡ Worked Example 51 Explosions 2 Question: An object with mass mT ot =15 kg is sitting at rest. Suddenly it explodes into two pieces. One piece has a mass of m1 = 5000 g and moves oﬁ in the positive x-direction at v1 = 7ms¡1. What is the ﬂnal velocity of the other piece? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: ¡¡!pT ot = 0, EkT ot = 0 170 After the collision: 2 1 ¡!p2, Ek2 ¡!p1, Ek1 Step 2 : Convert all units into S.I. units m1 = 5000 g m1 = 5 kg Step 3 : Decide which equations to use in the problem Now we know that in an explosion, total kinetic energy is not conserved. There is a deﬂnite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So: ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter ¡¡¡¡!pBef ore = ¡!p1 + ¡!p2 But the object was initially at rest so: 0 = ¡!p1 + ¡!p2 0 = m1¡!v1 + m2¡!v2 (A) Step 4 : Determine the mass of the other piece Now we know that m1= 5 kg but we do not know what the mass of m2 is. However, we do know that: mT ot = m1 + m2 m2 = mT ot ¡ = 15 kg ¡ m1 5 kg = 10 kg Step 5 : Solve for and quote the ﬂnal velocity of the other piece Now we can substitute all the values we know into equation (A) and solve for ¡!v2. Let’s choose the positive x-direction to have a postive sign and the negative x-direction to have a negative sign: (A) 0 = m1¡!v1 + m2¡!v2 0 = 5(7) + 10(¡!v2) 0 = 35 + 10(¡!v2) 35 3:5 m:s¡1 ¡ ¡ 10(¡!v2) = ¡!v2 = Therefore, ¡!v2 = 3:5 m:s¡1 in the negative x direction: ¡ 171 9.3 Explosions: Energy and Heat In explosions, you have seen that kinetic energy is not conserved. But remember that total energy is always conserved. Let’s look at what happens to the energy in some more detail. If a given amount of energy is released in an explosion it is not necessarily all transformed into kinetic energy. Due to the deformation of the exploding object, often a large amount of the energy is used to break chemical bonds and heat up the pieces. Energy is conserved but some of it is transferred through non-conservative processes like It will be radiated into the heating. This just means that we cannot get the energy back. environment as heat energy but it is all still accounted for. Now we can start to mix the ideas of momentum conservation with energy transfer to make longer problems. These problems are not more complicated just longer. We will start oﬁ short and them combine the diﬁerent ideas later on. Long problems should be treated like a number of smaller problems. Focus on them one at a time. Worked Example 52 Energy Accounting 1 Question: An object with a mass of mt = 17 kg explodes into two pieces of mass m1 = 7 kg and m2 = 10 kg. m1 has a velocity of 9ms¡1 in the negative x-direction and m2 has a velocity of 6:3ms¡1 in the positive x-direction. If the explosion released a total energy of 2000 J, how much was used in a non-conservative way? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: After the collision: ¡¡!pT ot = 0, EkT ot = 0 E = 2000 J 2 1 ¡!p2, Ek2 ¡!p1, Ek1 Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Determine what is being asked We are asked how much energy was used in a nonconservative fashion. This is the diﬁerence between how much energy was used in a conservative fashion and how much was used in total. We are lucky because we have everything we need to determine the kinetic energy of both pieces. The kinetic energy of the pieces is energy that was used in a conservative way. 172 Step 3 : Determine the total kinetic energy The sum of the kinetic energy for the two blocks is the total kinetic energy of the pieces. So: EkT ot = Ek1 + Ek2 2 + = m1¡!v1 1 2 1 2 2 1 m2¡!v2 2 1 2 = (7)(9)2 + (10)(6:3)2 = 283:5 + 198:45 EkT ot = 481:95 J Step 4 : Solve for and quote the ﬂnal answer The total energy that was transformed into kinetic energy is 481:95 J. We know that 2000 J of energy were released in total. the question makes no statements about other types of energy so we can assume that the diﬁerence was lost in a non-conservative way. Thus the total energy lost in non-conservative work is: EkT ot = 2000 481:95 ¡ = 1518:05 J E ¡ Worked Example 53 Energy Accounting 2 Question: An object at rest, with mass mT ot = 4 kg, explodes into two pieces (m1, m2) with m1 = 2:3 kg. m1 has a velocity of 17ms¡1 in the negative x-direction. If the explosion released a total energy of 800 J, 1. What is the velocity of m2? 2. How much energy does it carry? 3. And how much energy was used in a non-conservative way? Answer: Step 1 : Draw a rough sketch of the situation Before the collision: After the collision: ¡¡!pT ot = 0, EkT ot = 0 173 E = 800 J 1 ¡!p1; Ek1 2 ¡!p2; Ek2 Step 2 : N ow we know that in an explosion, total kinetic energy is not conserved. There is a deﬂnite change in shape of the exploding object! But we can always use momentum conservation to solve the problem. So: ¡¡¡¡!pBef ore = ¡¡¡¡!pAf ter ¡¡¡¡!pBef ore = ¡!p1 + ¡!p2 But the object was initially at rest so: 0 = ¡!p1 + ¡!p2 0 = m1¡!v1 + m2¡!v2 (A) Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) Now we know that m1= 2.3 kg but we do not know what the mass of m2 is. However, we do know that: mT ot = m1 + m2 m2 = mT ot ¡ = 4 kg ¡ = 1:7 kg m1 2:3 kg Step 4 : (NOTE TO SELF: step is deprecated, use westep instead.) Now we can substitute all the values we know into equation (A) and solve for ¡!v2. Let’s choose the positive x-direction to have a postive sign and the negative x-direction to have a negative sign: 0 = m1¡!v1 + m2¡!v2 0 = (2:3)( 17) + 1:7¡!v2 (A) ¡ 39:1 + 1:7¡!v2 0 = 1:7¡!v2 = 39:1 ¡ ¡!v2 = 23 m:s¡1 † ¡!v2 = 23 m:s¡1 in the positive x -direction Step 5 : (NOTE TO SELF: step is deprecated, use westep instead.) Now we need to calculate the energy that the second piece carries: Ek2 = = 2 m2¡!v2 (1:7)(23)2 1 2 1 2 = 449:65 J 174 The kinetic energy of the second piece is Ek2 = 449:65 J † Step 6 : (NOTE TO SELF: step is deprecated, use westep instead.) Now the amount of energy used in a non-conservative way in the explosion, is the diﬁerence between the amount of energy released in the explosion and the total kinetic energy of the exploded pieces: We know that: E ¡ EkT ot = 800 EkT ot ¡ EkT ot = Ek1 + Ek2 = = 1 2 1 2 m1¡!v1 2 + 449:65 (2:3)(17)2 + 449:65 = 332:35 + 449:65 = 782 J Step 7 : (NOTE TO SELF: step is deprecated, use westep instead.) So going back to: E ¡ EkT ot = 800 = 800 EkT ot 782 ¡ ¡ = 18 J 18 J of energy was used in a non-conservative way in the explosion † 9.4 Important Equations and Quantities Quantity velocity
momentum energy Symbol Unit \| \| J ¡!v ¡!p E Units m s kg:m s kg:m s2 2 S.I. Units or m:s¡1 or kg:m:s¡1 or kg:m2s¡2 Direction X X \| Table 9.1: Units commonly used in Collisions and Explosions Momentum: Kinetic energy: ¡!p = m¡!v Ek = 1 2 m¡!v 2 175 (9.13) (9.14) Chapter 10 Newtonian Gravitation 10.1 Properties Gravity is a force and therefore must be described by a vector - so remember magnitude and direction. Gravity is a force that acts between any two objects with mass. To determine the magnitude of the force we use the following equation: F = Gm1m2 r2 This equation describes the force between two bodies, one of mass m1, the other of mass m2 (both have units of Kilogrammes, or Kg for short). The G is Newton’s ‘Gravitational Constant’ 10¡11 [Nm2kg¡2]) and r is the straight line distance between the two bodies in meters. (6:673 This means the bigger the masses, the greater the force between them. Simply put, big things matter big with gravity. The 1=r2 factor (or you may prefer to say r¡2) tells us that the distance between the two bodies plays a role as well. The closer two bodies are, the stronger the gravitational force between them is. We feel the gravitational attraction of the Earth most at the surface since that is the closest we can get to it, but if we were in outer-space, we would barely even know the Earth’s gravity existed! (10.1) £ Remember that F = ma (10.2) which means that every object on the earth feels the same gravitational acceleration! That means whether you drop a pen or a book (from the same height), they will both take the same length of time to hit the ground... in fact they will be head to head for the entire fall if you drop them at the same time. We can show this easily by using the two equations above (10.1 and 10.2). The force between the Earth (which has the mass me) and an object of mass mo is F = Gmome r2 and the acceleration of an object of mass mo (in terms of the force acting on it) is So we substitute equation (10.3) into equation (10.4), and we ﬂnd that ao = F mo ao = Gme r2 176 (10.3) (10.4) (10.5) Since it doesn’t matter what mo is, this tells us that the acceleration on a body (due to the Earth’s gravity) does not depend on the mass of the body. Thus all objects feel the same gravitational acceleration. The force on diﬁerent bodies will be diﬁerent but the acceleration will be the same. Due to the fact that this acceleration caused by gravity is the same on all objects we label it diﬁerently, instead of using a we use g which we call the gravitational acceleration. 10.2 Mass and Weight Weight is a force which is measured in Newtons, it is the force of gravity on an object. People are always asking other people \What is your weight?" when in fact they should be asking \What is your mass?". Mass is measured in Kilograms (Kg) and is the amount of matter in an object, it doesn’t change unless you add or remove matter from the object (if you continue to study physics through to university level, you will ﬂnd that Einstein’s theory of relativity means that mass can change when you travel as fast as light does, but you don’t need to worry about that right now). There are 1000g in 1Kg and 1000Kg in a Tonne. To change mass into weight we use Newton’s 2nd Law which is F = Ma. The weight is the force and gravity the acceleration, it can be rewritten as: W is the Weight, measured in Newtons. M is the Mass, measured in Kg and g is the acceleration due to gravity, measured in m=s2 it is equal to 10 on the Earth. W = mg (10.6) 10.2.1 Examples 1. A bag of sugar has a mass of 1Kg, what is it’s weight? (Acceleration due to gravity = 10m=s2) - Step 1: Always write out the equation, it helps you to understand the question, and you will get marks as well. W = M g (10.7) - Step 2: Fill in all the values you know. (remember to make sure the mass is in Kg and NOT in grams or Tonnes!) W = 1 10 £ W = 10 (10.8) - Step 3: Write out the answer remembering to include the units! You will lose marks if you don’t W = 10Newtons (10.9) 2. A space-man has a mass of 90Kg, what is his weight (a) on the earth? (b) on the moon? (c) in outer space? (The acceleration due to gravity on the earth is 10m=s2, on the moon gravity is 1/6 of the gravity on earth). (a) W = M g 10 = 900 W = 90 £ W = 900Newtons 177 (10.10) (b) W = 90 W = M g 1=6 = 150 10 £ £ W = 150Newtons (10.11) (c) Weightless in outer space because g = 0. So now when somebody asks you your weight, you know to reply \Anything!! But my mass is a diﬁerent matter!" 10.3 Normal Forces If you put a book on a table it does not accelerate it just lies on the table. We know that gravity is acting on it with a force F = G mEmbook r2 (10.12) but if there is a net force there MUST be an acceleration and there isn’t. This means that the gravitational force is being balanced by another force1. This force we call the normal force. It is the reaction force between the book and the table. It is equal to the force of gravity on the book. This is also the force we measure when we measure the weight of something. The most interesting and illustrative normal force question, that is often asked, has to do with a scale in a lift. Using Newton’s third law we can solve these problems quite easily. When you stand on a scale to measure your weight you are pulled down by gravity. There is no acceleration downwards because there is a reaction force we call the normal force acting upwards on you. This is the force that the scale would measure. If the gravitational force were less then the reading on the scale would be less. Worked Example 54 Normal Forces 1 Question: A man weighing 100kg stands on a scale (measuring newtons). What is the reading on the scale? Answer: Step 1 : Decide what information is supplied We are given the mass of the man. We know the gravitational acceleration that acts on him - g = 10m=s2. Step 2 : Decide what equation to use to solve the problem The scale measures the normal force on the man. This is the force that balances gravity. We can use Newton’s laws to solve the problem: Fr = Fg + FN (10.13) where Fr is the resultant force on the man. 1Newton’s third law! 178 Step 3 : Firstly we determine the net force acting downwards on the man due to gravity Fg = mg m s2 = 100kg 9:8 = 980 £ kgm s2 = 980N downwards Step 4 : Now determine the normal force acting upwards on the man We now know the gravitational force downwards. We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is 0 - so Fr = 0. Fr = Fg + FN 0 = FN = 980N upwards 980N + FN ¡ Step 5 : Quote the ﬂnal answer The normal force is then 980N upwards. It exactly balances the gravitational force downwards so there is no net force and no acceleration on the man. Now we are going to add things to exactly the same problem to show how things change slightly. We will now move to a lift moving at constant velocity. Remember if velocity is constant then acceleration is zero. Worked Example 55 Normal Forces 2 s . What is the reading on the scale? Question: A man weighing 100kg stands on a scale (measuring newtons) inside a lift moving downwards at 2 m Answer: Step 1 : Decide what information is supplied We are given the mass of the man and the acceleration of the lift. We know the gravitational acceleration that acts on him. Step 2 : Decide which equation to use to solve the problem Once again we can use Newton’s laws. We know that the sum of all the forces must equal the resultant acceleration times the mass (This is the resultant force, Fr). Fr = Fg + FN (10.14) 179 Step 3 : Firstly we determine the net force acting downwards on the man due to gravity Fg = mg m s2 = 100kg 9:8 = 980 £ kgm s2 = 980N downwards Step 4 : Now determine the normal force acting upwards on the man The scale measures this normal force, so once we’ve determined it we will know the reading on the scale. Because the lift is moving at constant velocity the overall resultant acceleration of the man on the scale is 0. If we write out the equation: Fr = Fg + FN 0 = FN = 980N upwards 980N + FN ¡ Step 5 : Quote the ﬂnal answer The normal force is then 980N upwards. It exactly balances the gravitational force downwards so there is no net force and no acceleration on the man. In this second example we get exactly the same result because the net acceleration on the man was zero! If the lift is accelerating downwards things are slightly diﬁerent and now we will get a more interesting answer! Worked Example 56 Normal Forces 3 s2 . What is the reading on the scale? Question: A man weighing 100kg stands on a scale (measuring newtons) inside a lift accelerating downwards at 2 m Answer: Step 1 : Decide what information is supplied We are given the mass of the man and his resultant acceleration - this is just the acceleration of the lift. We know the gravitational acceleration that acts on him. Step 2 : Decide which equation to use to solve the problem Once again we can use Newton’s laws. We know that the sum of all the forces must equal the resultant acceleration times the mass (This is the resultant force, Fr). Fr = Fg + FN (10.15) 180 Step 3 : Firstly we determine the net force acting downwards on the man due to gravity, Fg Fg = mg m s2 = 100kg 9:8 = 980 £ kgm s2 = 980N downwards Step 4 : Now determine the normal force acting upwards on the man, FN We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is 2 m s2 downwards. If we write out the equation: 100kg 2) ( ¡ £ 200 ¡ kgm s2 Fr = Fg + FN m s2 = = ¡ 980N + FN 980N + FN ¡ 200N = ¡ FN = 780N upwards 980N + FN ¡ Step 5 : Quote the ﬂnal answer The normal force is then 780N upwards. It balances the gravitational force downwards just enough so that the man only accelerates downwards at 2 m s2 . Worked Example 57 Normal Forces 4 s2 . What is the reading on the scale? Question: A man weighing 100kg stands on a sc
ale (measuring newtons) inside a lift accelerating upwards at 4 m Answer: Step 1 : Decide what information is supplied We are given the mass of the man and his resultant acceleration - this is just the acceleration of the lift. We know the gravitational acceleration that acts on him. Step 2 : Decide which equation to use to solve the problem Once again we can use Newton’s laws. We know that the sum of all the forces must equal the resultant acceleration times the mass (This is the resultant force, Fr). Fr = Fg + FN (10.16) 181 Step 3 : Firstly we determine the net force acting downwards on the man due to gravity, Fg Fg = mg m s2 = 100kg 9:8 = 980 £ kgm s2 = 980N downwards Step 4 : Now determine the normal force upwards, FN We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is 2 m s2 downwards. if we write out the equation: 100kg (4) £ Fr = Fg + FN m s2 = = ¡ 980N + FN 980N + FN ¡ 400 kgm s2 400N = FN = 1380N upwards 980N + FN ¡ Step 5 : Quote the ﬂnal answer The normal force is then 1380N upwards. then in addition applies su–cient force to accelerate the man upwards at 4 m s2 . It balances the gravitational force and 10.4 Comparative problems Here always work with multiplicative factors to ﬂnd something new in terms of something old. Worked Example 58 Comparative Problem 1 Question:On Earth a man weighs 70kg. Now if the same man was instantaneously beamed to the planet Zirgon, which has the same size as the Earth but twice the mass, what would he weigh? (NOTE TO SELF: Vanessa: isn’t this confusing weight and mass?) Answer: 182 Step 1 : We start with the situation on Earth mEm r2 Step 2 : Now we consider the situation on Zirgon W = mg = G WZ = mgZ = G mZm r2 Z (10.17) (10.18) Step 3 : Relation between conditions on Earth and Zirgon but we know that mZ = 2mE and we know that rZ = r so we could write the equation again and substitute these relationships in: Step 4 : Substitute WZ = mgZ = G (2mE)m (r)2 WZ = 2(G (mE)m (r)2 ) Step 5 : Relation between weight on Zirgon and Earth Step 6 : Quote the ﬂnal answer so on Zirgon he weighs 140kg. WZ = 2(W ) (10.19) (10.20) (10.21) 10.4.1 Principles Write out ﬂrst case Write out all relationships between variable from ﬂrst and second case Write out second case Substitute all ﬂrst case variables into second case Write second case in terms of ﬂrst case † † † † † Interesting Fact: The acceleration due to gravity at the Earth’s surface is, by convention, equal to 9.80665 ms¡2. (The actual value varies slightly over the surface of the Earth). This quantity is known as g. The following is a list of the gravitational accelerations (in multiples of g) at the surfaces of each of the planets in our solar system: Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto 0.376 0.903 1 0.38 2.34 1.16 1.15 1.19 0.066 183 Note: The "surface" is taken to mean the cloud tops of the gas giants (Jupiter, Saturn, Uranus and Neptune) in the above table. Worked Example 59 Comparative Problem 2 Question: On Earth a man weighs 70kg. On the planet Beeble how much will he weigh if Beeble has mass half of that of the Earth and a radius one quarter that of the earth. Answer: Step 1 : Start with the situation on Earth Step 2 : Now consider the situation on Beeble W = mg = G mEm r2 WB = mgB = G mBm r2 B (10.22) (10.23) Step 3 : Relation between conditions on Earth and on Beeble We know that mB = 1 again and substitute these relationships in: Step 4 : Substitute 2 mE and we know that rB = 1 4 r so we could write the equation WB = mgB = G = mgB = G (mB)m (rB)2 ( 1 2 mE)m ( 1 4 r)2 = 8(G (mE)m (r)2 ) Step 5 : Relation between weight on Earth and weight on Beeble WB = 8(W ) (10.24) Step 6 : Quote the ﬂnal answer So the man weighs 560kg on Beeble! 184 Interesting Fact: Did you know that the largest telescope in the Southern Hemisphere is the South African Large Telescope (SALT) which came online in 2005 outside Sutherland in the Karoo. 10.5 Falling bodies Objects on the earth fall because there is a gravitation force between them and the earth - which results in an acceleration - as we saw above. So if you hold something in front of you and let it go - it will fall. It falls because of an acceleration toward the centre of the earth which results from the gravitational force between the two. These bodies move in a straight line from the point where they start to the centre for the earth. This means we can reuse everything we learnt in rectilinear motion. the only thing that needs thinking about is the directions we are talking about. We need to choose either up or down as positive just like we had to choose a positive direction in standard rectilinear motion problems. this is the hardest part. If you can do rectilinear motion you can do falling body problems. Just remember the acceleration they feel is constant and because of gravity - but once you have chosen your directions you can forget that gravity has anything to do with the problem - all you have is a rectilinear motion problem with a constant acceleration!! 10.6 Terminal velocity Physics is all about being simple - all we do is look at the world around us and notice how it really works. It is the one thing everyone is qualiﬂed to do - we spend most of our time when we are really young experimenting to ﬂnd out how things work. Take a book - wave it in the air - change the angle and direction. what happens of course there is resistance. diﬁerent angles make it greater - the faster the book moves the greater it is. The bigger the area of the book moving in the direction of motion the greater the force. So we know that air resistance exists! it is a force. So what happens when an object falls? of course there is air resistance - or drag as it is normally called. There is an approximate formula for the drag force as well. The important thing to realise is that when the drag force and the gravitational force are equal for a falling body there is no net force acting on it - which means no net acceleration. That does not mean it does not move - but it means that its speed does not change. It falls at a constant velocity! This velocity is called terminal velocity. 10.7 Drag force The actual force of air resistance is quite complicated. Experiment by moving a book through the air with the face of the book and then the side of the book forward, you will agree that the area of the book makes a diﬁerence as to how much you must work in order to move the book at the same speed in both cases. This is why racing cars are slim-lined in design, and not shaped like a big box! 185 Get a plastic container lid (or anything waterproof) swing it around in air and then try to swing it around under water. The density of the water is much larger than the air, making you have to work harder at swinging the lid in water. This is why boats and submarines are a lot slower than aeroplanes! So we know that density, area and speed all play a role in the drag force. The expression we use for drag force is 1 2 where C is a constant which depends on the object and uid interactions, ‰ is the density, A is the area and v is the velocity. C‰Av2 (10.25) D = 10.8 Important Equations and Quantities Quantity mass velocity force energy Symbol Unit \| \| N J m ¡!v ¡!F E Units kg m s kg:m s2 kg:m s2 2 S.I. Units or \| or m:s¡1 or kg:m:s¡2 or kg:m2:s¡2 Direction \| X X \| Table 10.1: Units used in Newtonian Gravitation 186 Chapter 11 Pressure 11.1 Important Equations and Quantities Quantity Symbol Unit S.I. Units Direction Units or Table 11.1: Units used in Pressure 187 Essay 3 : Pressure and Forces Author: Asogan Moodaly Asogan Moodaly received his Bachelor of Science degree (with honours) in Mechanical Engineering from the University of Natal, Durban in South Africa. For his ﬂnal year design project he worked on a 3-axis ﬂlament winding machine for composite (Glass re-enforced plastic in this case) piping. He worked in Vereeniging, Gauteng at Mine Support Products (a subsidiary of Dorbyl Heavy Engineering) as the design engineer once he graduated. He currently lives in the Vaal Triangle area and is working for Sasol Technology Engineering as a mechanical engineer, ensuring the safety and integrity of equipment installed during projects. Pressure and Forces In the mining industry, the roof (hangingwall) tends to drop as the face of the tunnel (stope) is excavated for rock containing gold. As one can imagine, a roof falling on one’s head is not a nice prospect! Therefore the roof needs to be supported. Roof Face The roof is not one big uniform chunk of rock. Rather it is broken up into smaller chunks. It is assumed that the biggest chunk of rock in the roof has a mass of less than 20 000 kgs therefore each support has to be designed to resist a force related to that mass. The strength of the material (either wood or steel) making up the support is taken into account when working out the minimum required size and thickness of the parts to withstand the force of the roof. Roof Face Supports 188 Sometimes the design of the support is such that the support needs to withstand the rock mass without the force breaking the roof.. Therefore hydraulic supports (hydro = water) use the principles of force and pressure such that as a force is exerted on the support, the water pressure increases. A pressure relief valve then squirts out water when the pressure (and thus the force) gets too large. Imagine a very large, modiﬂed doctor’s syringe. Hydraulic support Doctors syringe Force Force Seal to trap water in tube Water filled tube Pressure relief valve Rubber seal to trap medicine in syringe Pressure Pressure In the petrochemical industry, there are many vessels and pipes that are under high pressures. A vessel is a containment unit (Imagine a pot without handles, that has the lid welded to the pot that would be a small vessel) where chemicals mix and react to form other chemicals, amongst other uses. End Product
Chemical The end product chemicals are sold to companies that use these chemicals to make shampoo, dishwashing liquid, plastic containers, fertilizer, etc. Anyway, some of these chemical reactions require high temperatures and pressures in order to work. These pressures result in forces being applied to the insides of the vessels and pipes. Therefore the minimum thickness of the pipe and vessels walls must be determined using calculations, to withstand these forces. These calculations take into account the strength of the material (typically steel, plastic or composite), the diameter and of course the pressure inside the equipment. Let examine the concepts of force and pressure in further detail. 189 Chapter 12 Heat and Properties of Matter 12.1 Phases of matter 12.1.1 Density Matter is a substance which has mass and occupies space. The density of matter refers to how much mass is in a given volume. Said diﬁerently, you can imagine the density to be the amount of mass packed into a given volume. density = M ass V olume If we consider a bar of soap and a bar of steel with the same volume, the steel will have more mass because it has a greater density. The density is greater in steal because more atoms are closely packed in comparison to the soap. Although they are both the same size, the bar of steel will be "heavier" because it has more mass. Worked Example 60 Density of objects A bar of aluminum (Al) has dimensions 2cm x 3cm x 5cm with a mass of 81g. A bar of lead (Pb) has dimensions 3cm x 3cm x 5cm and a mass of 510.3g. Calculate the density of the aluminum and lead. Solution: First we calculate the volume of Al and Pb: For Aluminum: volume = 2cm ⁄ 3cm For Lead: volume = 3cm ⁄ ⁄ volume = Length W idth Height ⁄ ⁄ 5cm = 30cm3 3cm 5cm = 45cm3 ⁄ We can now calculate the densities using the mass and volume of each material. 30cm3 = 2:7g=cm3 For Aluminum: density = 81g For Lead: density = 510:3g 45cm3 = 11:34g=cm3 190 2cm 5cm 3cm 5cm 3cm 3cm Now that you know the density of aluminum and lead, which object would be bigger (larger volume): 1kg of Lead or 1kg of Aluminum. Solution: 1kg of aluminum will be much larger in volume than 1kg of lead. Aluminum has a smaller density so it will take a lot more of it to have a weight of 1kg. Lead is much more dense, so it will take less for it to weigh 1kg. The density of liquids and gases can be calculated the same way as in solids. If the mass and volume of a liquid is known, the density can be calculated. We can often determine which liquid has a greater density by mixing two liquids and seeing how they settle. The more dense liquid will fall towards the bottom, or ’sink’. If you have ever added olive oil to water, you have seen it sits on the surface, or ’oats’. This is because olive oil is less dense than water. Fog occurs when water vapor becomes more dense than air("a cloud that sinks in air"). This principle can be used with solids and liquids. In fact, it is the density of an object that determines if it will oat or sink in water. Objects with densities greater than water will sink. Worked Example 61 Objects oating in water Ivory soap is famous for "soap that oats". If a 5cm x 3cm x 10cm bar of ivory soap weighs 1.35 Newtons, show that its density is less than water. Solution: First calculate the bars volume: volume = 3cm Now we must determine the mass of the bar based on its weight. We will use Newton’s Second law (F = ma): 10cm = 150cm3 5cm ⁄ ⁄ W eight = mass gravity = ) ⁄ W eight = 9:8m=s2 M ass ⁄ 191 M ass = 1:35N 9:8m=s2 = :138kg Using the mass and the volume we determine the density of the soap: density = 138g 150cm3 = :92g=cm3 Water has a density of 1g=cm3, therefore the soap is less dense than water, allowing it to oat. 12.2 Phases of matter Although phases are conceptually simple, they are hard to deﬂne precisely. A good deﬂnition of a phase of a system is a region in the parameter space of the system’s thermodynamic variables in which the free energy is analytic. Equivalently, two states of a system are in the same phase if they can be transformed into each other without abrupt changes in any of their thermodynamic properties. All the thermodynamic properties of a system { the entropy, heat capacity, magnetization, compressibility, and so forth { may be expressed in terms of the free energy and its derivatives. For example, the entropy is simply the ﬂrst derivative of the free energy with temperature. As long as the free energy remains analytic, all the thermodynamic properties will be well-behaved. When a system goes from one phase to another, there will generally be a stage where the free energy is non-analytic. This is known as a phase transition. Familiar examples of phase transitions are melting (solid to liquid), freezing (liquid to solid), boiling (liquid to gas), and condensation (gas to liquid). Due to this non-analyticity, the free energies on either side of the transition are two diﬁerent functions, so one or more thermodynamic properties will behave very diﬁerently after the transition. The property most commonly examined in this context is the heat capacity. During a transition, the heat capacity may become inﬂnite, jump abruptly to a diﬁerent value, or exhibit a "kink" or discontinuity in its derivative. In practice, each type of phase is distinguished by a handful of relevant thermodynamic properties. For example, the distinguishing feature of a solid is its rigidity; unlike a liquid or a gas, a solid does not easily change its shape. Liquids are distinct from gases because they have much lower compressibility: a gas in a large container ﬂlls the container, whereas a liquid forms a puddle in the bottom. Many of the properties of solids, liquids, and gases are not distinct; for instance, it is not useful to compare their magnetic properties. On the other hand, the ferromagnetic phase of a magnetic material is distinguished from the paramagnetic phase by the presence of bulk magnetization without an applied magnetic ﬂeld. To take another example, many substances can exist in a variety of solid phases each corresponding to a unique crystal structure. These varying crystal phases of the same substance are called polymorphs. Diamond and graphite are examples of polymorphs of carbon. Graphite is composed of layers of hexagonally arranged carbon atoms, in which each carbon atom is strongly bound to three neighboring atoms in the same layer and is weakly bound to atoms in the neighboring layers. By contrast in diamond each carbon atom is strongly bound to four neighboring carbon atoms in a cubic array. The unique crystal structures of graphite and diamond are responsible for the vastly diﬁerent properties of these two materials. Metastable phases 192 Metastable states may sometimes be considered as phases, although strictly speaking they aren’t because they are unstable. For example, each polymorph of a given substance is usually only stable over a speciﬂc range of conditions. For example, diamond is only stable at extremely high pressures. Graphite is the stable form of carbon at normal atmospheric pressures. Although diamond is not stable at atmospheric pressures and should transform to graphite, we know that diamonds exist at these pressures. This is because at normal temperatures the transformation If we were to heat the diamond, the rate of from diamond to graphite is extremely slow. transformation would increase and the diamond would become graphite. However, at normal temperatures the diamond can persist for a very long time. Another important example of metastable polymorphs occurs in the processing of steel. Steels are often subjected to a variety of thermal treatments designed to produce various combinations of stable and metastable iron phases. In this way the steel properties, such as hardness and strength can be adjusted by controlling the relative amounts and crystal sizes of the various phases that form. Phase diagrams The diﬁerent phases of a system may be represented using a phase diagram. The axes of the diagrams are the relevant thermodynamic variables. For simple mechanical systems, we generally use the pressure and temperature. The following ﬂgure shows a phase diagram for a typical material exhibiting solid, liquid and gaseous phases. The markings on the phase diagram show the points where the free energy is non-analytic. The open spaces, where the free energy is analytic, correspond to the phases. The phases are separated by lines of non-analyticity, where phase transitions occur, which are called phase boundaries. In the above diagram, the phase boundary between liquid and gas does not continue indefinitely. Instead, it terminates at a point on the phase diagram called the critical point. This reects the fact that, at extremely high temperatures and pressures, the liquid and gaseous phases become indistinguishable. In water, the critical point occurs at around 647 K (374 C or 705 F) and 22.064 MPa. The existence of the liquid-gas critical point reveals a slight ambiguity in our above deﬂnitions. When going from the liquid to the gaseous phase, one usually crosses the phase boundary, but it is possible to choose a path that never crosses the boundary by going to the right of the critical point. Thus, phases can sometimes blend continuously into each other. We should note, however, that this does not always happen. For example, it is impossible for the solid-liquid phase boundary to end in a critical point in the same way as the liquid-gas boundary, because the solid and liquid phases have diﬁerent symmetry. An interesting thing to note is that the solid-liquid phase boundary in the phase diagram of most substances, such as the one shown above, has a positive slope. This is due to the solid phase having a higher density than the liquid, so that increasing the pressure increases the melting temperature. However, in the phase diagram for water the solid-liquid phase boundary has a negative slope. This reects the fact that ice has
a lower density than water, which is an unusual property for a material. 193 12.2.1 Solids, liquids, gasses 12.2.2 Pressure in uids 12.2.3 change of phase 12.3 Deformation of solids 12.3.1 strain, stress Stress () and strain (†) is one of the most fundamental concepts used in the mechanics of materials. The concept can be easily illustrated by considering a solid, straight bar with a constant cross section throughout its length where a force is distributed evenly at the ends of the bar. This force puts a stress upon the bar. Like pressure, the stress is the force per unit area. In this case the area is the cross sectional area of the bar. stress = F orce Areacrosssection = ) = F A (A) Bar under compression (B) Bar under tension Figure 12.1: Illustration of Bar The bar in ﬂgure 1a is said to be under compression. If the direction of the force (¡!F ), were reversed, stretching the bar, it would be under tension (ﬂg. 1b). Using intuition, you can imagine how the bar might change in shape under compression and tension. Under a compressive load, the bar will shorten and thicken. In contrast, a tensile load will lengthen the bar and make it thinner. Figure 12.2: Bar changes length under tensile stress For a bar with an original length L, the addition of a stress will result in change of length L and L we can now deﬂne strain as the ratio between the two. That is, strain is L. With 4 deﬂned as the fractional change in length of the bar: 4 Strain L 4 L · 12.3.2 Elastic and plastic behavior Material properties are often characterized by a stress versus strain graph (ﬂgure x.xx). One way in which these graphs can be determined is by tensile testing. In this process, a machine 194 L L 4 Figure 12.3: Left end of bar is ﬂxed as length changes \| \| + Figure 12.4: dashed line represents plastic recovery incomplete stretches a the material by constant amounts and the corresponding stress is measured and plotted. Typical solid metal bars will show a result like that of ﬂgure x.xx. This is called a Type II response. Other materials may exibit diﬁerent responses. We will only concern ourself with Type II materials. The linear region of the graph is called the elastic region. By obtaining the slope of the linear region, it is easy to ﬂnd the strain for a given stress, or vice-versa. This slope shows itself to be very useful in characterizing materials, so it is called the Modulus of Elasticity, or Young’s Modulus: E = stress strain = F=A ¢L=L The elastic region has the unique property that allows the material to return to its original shape when the stress is removed. As the stress is removed it will follow line back to zero. One may think of stretching a spring and then letting it return to its original length. When a stress is applied in the linear region, the material is said to undergo elastic deformation. When a stress is applied that is in the non-linear region, the material will no longer return to its original shape. This is referred to as plastic deformation. If you have overstretched a spring you have seen that it no longer returns to its initial length; it has been plastically deformed. The stress where plastic behavior begins is called the yield strength (point A, ﬂg x). When a material has plastically deformed it will still recover some of its shape (like an overstretched spring). When a stress in the non-linear region is removed, the stress strain graph will follow a line with a slope equal to the modulus of elasticity (see the dashed line in ﬂgure x.xx). The plastically deformed material will now have a linear region that follows the dashed line. Greater stresses in the plastic region will eventually lead to fracture (the material breaks). The maximum stress the material can undergo before fracture is the ultimate strength. 195 \| \| + Figure 12.5: dashed line represents plastic recovery incomplete 12.4 Ideal gasses Author: G¶erald Wigger G¶erald Wigger started his Physics studies at ETH in Zuerich, Switzerland. He moved to Cape Town, South Africa, for his Bachelor of Science degree (with honours) in Physics from the University of Cape Town in 1998. Returned to Switzerland, he ﬂnished his Diploma at ETH in 2000 and followed up with a PhD in the Solid State Physics group of Prof. Hans-Ruedi Ott at ETH. He graduated in the year 2004. Being awarded a Swiss fellowship, he moved to Stanford University where he is currently continuing his Physics research in the ﬂeld of Materials with novel electronic properties. Any liquid or solid material, heated up above its boiling point, undergoes a transition into a gaseous state. For some materials such as aluminium, one has to heat up to three thousand degrees Celsius (–C), whereas Helium is a gas already at -269 –C. For more examples see Table 12.1. As we ﬂnd very strong bonding between the atoms in a solid material, a gas consists of molecules which do interact very poorly. If one forgets about any electrostatic or intermolecular attractive forces between the molecules, one can assume that all collisions are perfectly elastic. One can visualize the gas as a collection of perfectly hard spheres which collide but which otherwise do not interact with each other. In such a gas, all the internal energy is in the form of kinetic energy and any change in internal energy is accompanied by a change in temperature. Such a gas is called an ideal gas. In order for a gas to be described as an ideal gas, the temperature should be raised far enough above the melting point. A few examples of ideal gases at room temperature are Helium, Argon and hydrogen. Despite the fact that there are only a few gases which can be accurately described as an ideal gas, the underlying theory is widely used in Physics because of its beauty and simplicity. A thermodynamic system may have a certain substance or material whose quantity can be expressed in mass or mols in an overall volume. These are extensive properties of the system. In the following we will be considering often intensive versus extensive quantities. A material’s intensive property, is a quantity which does not depend on the size of the material, such as temperature, pressure or density. Extensive properties like volume, mass or number of atoms on 196 Material Aluminium Water Ethyl alcohol Methyl ether Nitrogen Helium Temperature in Celsius Temperature in Kelvin 2467 –C 100 –C 78.5 –C -25 –C -195.8 –C -268.9 –C 2740 K 373.15 K 351.6 K 248 K 77.3 K 4.2 K Table 12.1: Boiling points for various materials in degrees Celsius and in Kelvin quantity pressure p volume V unit Pa m3 molar volume vmol m3/mol temperature T mass M density ‰ internal energy E K kg kg/m3 J intensive or extensive intensive extensive intensive intensive extensive intensive extensive Table 12.2: Intensive versus extensive properties of matter the other hand gets bigger the bigger the material is (see Table 12.2 for various intensive/extensive properties). If the substance is evenly distributed throughout the volume in question, then a value of volume per amount of substance may be used as an intensive property. For an example, for an amount called a mol, volume per mol is typically called molar volume. Also, a volume per mass for a speciﬂc substance may be called speciﬂc volume. In the case of an ideas gas, a simple equation of state relates the three intensive properties, temperature, pressure, and molar or speciﬂc volume. Hence, for a closed system containing an ideal gas, the state can be speciﬂed by giving the values of any two of pressure, temperature, and molar volume. 12.4.1 Equation of state The ideal gas can be described with a single equation. However, in order to arrive there, we will be introducing three diﬁerent equations of state, which lead to the ideal gas law. The combination of these three laws leads to a complete picture of the ideal gas. 1661 - Robert Boyle used a U-tube and Mercury to develop a mathematical relationship between pressure and volume. To a good approximation, the pressure and volume of a ﬂxed amount of gas at a constant temperature were related by V = constant p ¢ p V : pressure (P a) : Volume (m3) In other words, if we compress a given quantity of gas, the pressure will increase. And if we put it under pressure, the volume of the gas will decrease proportionally. 197 Figure 12.6: Pressure-Volume diagram for the ideal gas at constant temperature. Worked Example 62 compressed Helium gas A sample of Helium gas at 25–C is compressed from 200 cm3 to 0.240 cm3. pressure is now 3.00 cm Hg. What was the original pressure of the Helium? Solution: It’s always a good idea to write down the values of all known variables, indicating whether the values are for initial or ﬂnal states. Boyle’s Law problems are essentially special cases of the Ideal Gas Law: Initial: p1 = ?; V1 = 200 cm3; Final: p2 = 3.00 cm Hg; V2 = 0.240 cm3; Since the number of molecules stays constant and the temperature is not changed along the process, so Its V1 = p2 p1 ¢ V2 ¢ hence p1 = p2 ¢ V2=V1 = 3:00cmHg 0:240cm3=200cm3 ¢ Setting in the values yields p1 = 3.60 Did you notice that the units for the pressure are in cm Hg? You may wish to convert this to a more common unit, such as millimeters of mercury, atmospheres, or pascals. 3.60 10mm/1 cm = 3.60 10¡2 mm Hg 10¡3 Hg ¢ 10¡3 cm Hg. ¢ ¢ ¢ 198 3.60 ¢ 10¡3 Hg ¢ 1 atm/76.0 cm Hg = 4.74 10¡5 atm ¢ One way to experience this is to dive under water. There is air in your middle ear, which is normally at one atmosphere of pressure to balance the air outside your ear drum. The water will put pressure on the ear drum, thereby compressing the air in your middle ear. Divers must push air into the ear through their Eustacean tubes to equalize this pressure. Worked Example 63 pressure in the ear of a diver How deep would you have to dive before the air in your middle ear would be compressed to 75% of its initial volume? Assume for the beginning that the temperature of the sea is constant as you dive. Solution: First we write down the pressure a
s a function of height h: where we take for p0 the atmospheric pressure at height h = 0, ‰ is the density of water at 20 degrees Celsius 998.23 kg/m3, g = 9.81 ms¡2. p = p0 + ‰ h g ¢ ¢ As the temperature is constant, it holds for both heights h Now solving for h using the fact that V0 = (p0 + ‰gh) p0 ¢ Ve ¢ yields Ve=V0 = 0:75 h = (0:75 p0 ⁄ ¡ p0)=(‰g) Now, how far can the diver dive down before the membranes of his ear brake. Solution: As the result is negative, h determines the way he can dive down. h is given as roughly 2.6 m. In 1809, the French chemist Joseph-Louis Gay-Lussac investigated the relationship between the Pressure of a gas and its temperature. Keeping a constant volume, the pressure of a gas sample is directly proportional to the temperature. Attention, the temperature is measured in Kelvin! The mathematical statement is as follows: 199 p1=T1 = p2=T2 = constant p1;2 T1;2 : pressures (P a) : Temperatures (K) That means, that pressure divided by temperature is a constant. On the other hand, if we plot pressure versus temperature, the graph crosses 0 pressure for T = 0 K = -273.15 –C as shown in the following ﬂgure. That point is called the absolute Zero. That is where any motion of molecules, electrons or other particles stops. Figure 12.7: Pressure-temperature diagram for the ideal gas at constant volume. Worked Example 64 Gay-Lussac Suppose we have the following problem: A gas cylinder containing explosive hydrogen gas has a pressure of 50 atm at a temperature of 300 K. The cylinder can withstand a pressure of 500 atm before it bursts, causing a building-attening explosion. What is the maximum temperature the cylinder can withstand before bursting? Solution: Let’s rewrite this, identifying the variables: A gas cylinder containing explosive hydrogen gas has a pressure of 50 atm (p1) at a temperature of 300 K (T1). The cylinder can withstand a pressure of 500 atm 200 (p2) before it bursts, causing a building-attening explosion. What is the maximum temperature the cylinder can withstand before bursting? Plugging in the known variables into the expression for the Gay-Lussac law yields we ﬂnd the answer to be 3000 K. T2 = p2=p1 T1 = 500atm=50atm ⁄ 300K = 3000K ⁄ The law of combining volumes was interpreted by the Italian chemist Amedeo Avogadro in 1811, using what was then known as the Avogadro hypothesis. We would now properly refer to it as Avogadro’s law: Equal volumes of gases under the same conditions of temperature and pressure contain equal numbers of molecules. This can be understood in the following. As in an ideal gas, all molecules are considered to be tiny particles with no spatial extension which collide elastically with each other. So, the kind of gas is irrelevant. Avogadro found that at room temperature, in atmospheric pressure the 1023 molecules or atoms, occupies the volume of 22.4 volume of a mol of a substance, i.e. 6.022 ¢ l. Figure 12.8: Two diﬁerent gases occupying the same volume under the same circumstances. Combination of the three empirical gas laws, described in the preceding three sections leads to the Ideal Gas Law which is usually written as: p ¢ V = n R T ¢ ¢ : pressure (P a) : Volume (m3) : number of mols (mol) p V n R : gas konstant (J=molK) : temperature (K) T 201 where p = pressure, V = volume, n = number of mols, T = kelvin temperature and R the ideal gas constant. The ideal gas constant R in this equation is known as the universal gas constant. It arises from a combination of the proportionality constants in the three empirical gas laws. The universal gas constant has a value which depends only upon the units in which the pressure and volume are measured. The best available value of the universal gas constant is: 8.3143510 Another value which is sometimes convenient is 0.08206 dm3 atm/mol K. R is related to the molK or 8.3143510 kP adm molK J 3 Boltzmann-constant as: R = N0 kB ¢ where N0 is the number of molecules in a mol of a substance, i.e. 6.022 ¢ 1.308 10¡23 J/K is valid for one single particle. ¢ This ideal gas equation is one of the most used equations in daily life, which we show in the (12.1) 1023 and kB is following problem set: Worked Example 65 ideal gas 1 A sample of 1.00 mol of oxygen at 50 –C and 98.6 kPa occupies what volume? Solution: We solve the ideal gas equation for the volume V = nRT p and plug in the values n = 1, T = 273.15 + 50 K = 323.15 K and p = 98.6 ¢ yielding for the volume V = 0.0272 m3 = 27.2 dm3. 103 Pa, This equation is often used to determine the molecular masses from gas data. Worked Example 66 ideal gas 2 A liquid can be decomposed by electricity into two gases. In one experiment, one of the gases was collected. The sample had a mass of 1.090 g, a volume of 850 ml, a pressure of 746 torr, and a temperature of 25 –C. Calculate its molecular mass. Solution: To calculate the molecular mass we need the number of grams and the number of mols. We can get the number of grams directly from the information in the question. We can calculate the mols from the rest of the information and the ideal gas equation. 202 V = 850mL = 0:850L = 0:850dm3 P = 746torr=760torr = 0:982atm T = 25:0–C + 273:15 = 298:15K pV = nRT (0:982atm)(0:850L) = (n)(0:0821Latmmol 1K ¡ ¡ 1)(298:15K) n = 0:0341mol molecular mass = g/mol = 1.090 g/ 0.0341 mol = 31.96 g/mol. The gas is oxygen. Or the equation can be comfortably used to design a gas temperature controller: Worked Example 67 ideal gas 3 In a gas thermometer, the pressure needed to ﬂx the volume of 0.20 g of Helium at 0.50 L is 113.3 kPa. What is the temperature? Solution: We transform ﬂrst need to ﬂnd the number of mols for Helium. Helium consists of 2 protons and 2 neutrons in the core (see later) and therefore has a molar volume of 4 g/mol. Therefore, we ﬂnd plugging this into the ideal gas equation and solving for the temperature T we ﬂnd: n = 0:20g=4g=mol = 0:05mol T = pV nR = 113:3 103P a 0:5 10¡3m3 ¢ 0:05mol ¢ ¢ 8:314J=molK = 136:3K ¢ The temperature is 136 Kelvin. 12.4.2 Kinetic theory of gasses The results of several experiments can lead to a scientiﬂc law, which describes then all experiments performed. This is an empirical, that is based on experience only, approach to Physics. A law, however, only describes results; it does not explain why they have been obtained. Significantly stronger, a theory is a formulation which explains the results of experiments. A theory usually bases on postulates, that is a proposition that is accepted as true in order to provide a basis for logical reasoning. The most famous postulate in Physics is probably the one formulated by Walter Nernst which states that if one could reach absolute zero, all bodies would have the same entropy. 203 The kinetic-molecular theory of gases is a theory of great explanatory power. We shall see how it explains the ideal gas law, which includes the laws of Boyle and of Charles; Dalton’s law of partial pressures; and the law of combining volumes. The kinetic-molecular theory of gases can be stated as four postulates: † † † † A gas consists of particles (atoms or molecules) in continuous, random motion. Gas molecules inuence each other only by collision; they exert no other forces on each other. All collisions between gas molecules are perfectly elastic; all kinetic energy is conserved. The average energy of translational motion of a gas particle is directly proportional to temperature. In addition to the postulates above, it is assumed that the volumes of the particles are negligible as compared to container volume. These postulates, which correspond to a physical model of a gas much like a group of billiard balls moving around on a billiard table, describe the behavior of an ideal gas. At room temperatures and pressures at or below normal atmospheric pressure, real gases seem to be accurately described by these postulates, and the consequences of this model correspond to the empirical gas laws in a quantitative way. We deﬂne the average kinetic energy of translation Et of a particle in a gas as Et = 1=2 mv2 (12.2) ¢ where m is the mass of the particle with average velocity v. The forth postulate states that the average kinetic energy is a constant deﬂning the temperature, i.e. we can formulate Et = 1=2 ¢ mv2 = c T ¢ (12.3) where the temperature T is given in Kelvin and c is a constant, which has the same value for all gases. As we have 3 diﬁerent directions of motion and each possible movement gives kBT , we ﬂnd for the energy of a particle in a gas as Et = 1=2 ¢ mv2 = 3=2kBT = 3=2 R NA T (12.4) Hence, we can ﬂnd an individual gas particle’s speed rms = root mean square, which is the average square root of the speed of the individual particles (ﬂnd u) r where Mmol is the molar mass, i.e. the mass of the particle m times the Avogadro number vrms = 3RT Mmol (12.5) NA. Worked Example 68 kinetic theory 1 204 Calculate the root-mean-square velocity of oxygen molecules at room temperature, 25 –C. Solution: Using vrms = 3RT =Mmol ; the molar mass of molecular oxygen is 31.9998 g/mol; the molar gas constant has the value 8.3143 J/mol K, and the temperature is 298.15 K. Since the joule is the s¡2, the molar mass must be expressed as 0.0319998 kg/mol. The root-meankg ¢ square velocity is then given by: m2 ¢ p vrms = 3(8:3143)(298:15)=(0:0319998) = 482:1m=s A speed of 482.1 m/s is 1726 km/h, much faster than a jetliner can y and faster than most rie bullets. p The very high speed of gas molecules under normal room conditions would indicate that a gas molecule would travel across a room almost instantly. In fact, gas molecules do not do so. If a small sample of the very odorous (and poisonous!) gas hydrogen sulﬂde is released in one corner of a room, our noses will not detect it in another corner of the room for several minutes unless the air is vigorously stirred by a mechanical fan. The slow diﬁusion of gas molecules which are moving very quickly occurs because the gas molecules travel only short distances in straight lines before they are de
ected in a new direction by collision with other gas molecules. The distance any single molecule travels between collisions will vary from very short to very long distances, but the average distance that a molecule travels between collisions in a gas can be calculated. This distance is called the mean free path l of the gas molecules. If the rootmean-square velocity is divided by the mean free path of the gas molecules, the result will be the number of collisions one molecule undergoes per second. This number is called the collision frequency Z1 of the gas molecules. The postulates of the kinetic-molecular theory of gases permit the calculation of the mean free path of gas molecules. The gas molecules are visualized as small hard spheres. A sphere (d=2)2 and length vrms each of diameter d sweeps through a cylinder of cross-sectional area … second, colliding with all molecules in the cylinder. ¢ The radius of the end of the cylinder is d because two molecules will collide if their diameters overlap at all. This description of collisions with stationary gas molecules is not quite accurate, however, because the gas molecules are all moving relative to each other. Those relative velocities range between zero for two molecules moving in the same direction and 2vrms for a head-on collision. The average relative velocity is that of a collision at right angles, which is p2vrms. The total number of collisions per second per unit volume, Z1, is Z1 = …d2p2vrms (12.6) This total number of collisions must now be divided by the number of molecules which are present per unit volume. The number of gas molecules present per unit volume is found by rearrangement of the ideal gas law to n=V = p=RT and use of Avogadro’s number, n = N=NA; thus N=V = pNA=RT . This gives the mean free path of the gas molecules, l, as (urms=Z1)=(N=V ) = l = RT =…d2pNAp2 (12.7) 205 According to this expression, the mean free path of the molecules should get longer as the temperature increases; as the pressure decreases; and as the size of the molecules decreases. Worked Example 69 mean free path Calculate the length of the mean free path of oxygen molecules at room temperature, 25 –C, taking the molecular diameter of an oxygen molecule as 370 pm. Solution: Using the formula for mean free path given above and the value of the root-meansquare velocity urms, l = …(370 ¢ (8:3143kgm2s¡2=Kmol)(298:15K) 10¡12m)2(101325kg=ms2)(6:0225 1023mol¡1)p2 ¢ ; 10¡8 m = 67 nm. so l = 6.7 ¢ The apparently slow diﬁusion of gas molecules takes place because the molecules travel only a very short distance before colliding. At room temperature and atmospheric pressure, oxygen 10¡12 m) = 180 molecular diameters between collisions. 10¡8 m)/(370 molecules travel only (6.7 ¢ ¢ The same thing can be pointed out using the collision frequency for a single molecule Z1, which is the root-mean-square velocity divided by the mean free path: Z1 = …d2pNAp2 =RT = vrms=l (12.8) For oxygen at room temperature, each gas molecule collides with another every 0.13 nanosec10+9 collisions per second 10¡9 s), since the collision frequency is 7.2 onds (one nanosecond is 1.0 ¢ ¢ per molecule. For an ideal gas, the number of molecules per unit volume is given using pV = nRT and n = N=NA as N=V = NAp=RT (12.9) which for oxygen at 25 –C would be (6.022 1023 mol¡1)(101325 kg/m s2) / (8.3143 kg m2/s2 ¢ 1025 molecules/m3. The number of collisions between two molecules K mol)(298.15 K) or 2.46 ¢ in a volume, Z11, would then be the product of the number of collisions each molecule makes times the number of molecules there are, Z1N=V , except that this would count each collision twice (since two molecules are involved in each one collision). The correct equation must be Z11 = …d2p2N 2 Ap2vrms 2R2T 2 (12.10) If the molecules present in the gas had diﬁerent masses they would also have diﬁerent speeds, so an average value of vrms would be using a weighted average of the molar masses; the partial pressures of the diﬁerent gases in the mixture would also be required. Although such calculations involve no new principles, they are beyond our scope. 206 12.4.3 Pressure of a gas In the kinetic-molecular theory of gases, pressure is the force exerted against the wall of a container by the continual collision of molecules against it. From Newton’s second law of motion, the force exerted on a wall by a single gas molecule of mass m and velocity v colliding with it is: F = m a = m ¢ ¢v ¢t (12.11) In the above equation, the change in a quantity is indicated by the symbol ¢, that means by changing the time t by a fraction, we change the velocity v by some other minimal amount. It is assumed that the molecule rebounds elastically and no kinetic energy is lost in a perpendicular collision, so ¢v = v - (-v) = 2v (see ﬂgure below). If the molecule is moving perpendicular to the wall it will strike the opposite parallel wall, rebound, and return to strike the original wall again. If the length of the container or distance between the two walls is the path length l, then the time between two successive collisions on the same wall is ¢t = 2l/v. The continuous force which the molecule moving perpendicular to the wall exerts is therefore Figure 12.9: Change in momentum as a particle hits a wall. F = m 2v 2l=v = mv2 l (12.12) The molecules in a sample of gas are not, of course, all moving perpendicularly to a wall, but the components of their actual movement can be considered to be along the three mutually perpendicular x, y, and z axes. If the number of molecules moving randomly, N, is large, then on the average one-third of them can be considered as exerting their force along each of the three perpendicular axes. The square of the average velocity along each axis, v2(x), v2(y), or v2(z), will be one-third of the square of the average total velocity v2: v2(x) = v2(y) = v2(z) = v2=3 (12.13) The average or mean of the square of the total velocity can replace the square of the perpen- dicular velocity, and so for a large number of molecules N , Since pressure is force per unit area, and the area of one side of a cubic container must be l2, the pressure p will be given by F=l2 as: F = (N=3) mv2 l (12.14) 207 This equation rearranges to p = (N=3) mv2 l3 (12.15) pV = N mv2=3 (12.16) ¢ because volume V is the cube of the length l. The form of the ideal gas law given above shows the pressure-volume product is directly proportional to the mean-square velocity of the gas molecules. If the velocity of the molecules is a function only of the temperature, and we shall see in the next section that this is so, the kinetic-molecular theory gives a quantitative explanation of Boyle’s law. Worked Example 70 gas pressure A square box contains He (Helium) at 25 –C. If the atoms are colliding with the walls 1022 times per second, calculate the force perpendicularly (at 90–) at the rate of 4.0 ¢ (in Newtons) and the pressure exerted on the wall per mol of He given that the area of the wall is 100 cm2 and the speed of the atoms is 600 ms¡1. Solution: We use the equation 12.14 to calculate the force. mv2 l The fraction v=l is the collision frequency Z1 = 0.6679 s¡1. The product of N Z1 is the number of molecules impinging on the wall per second. This induces for the force: = (N=3)mv F = (N=3) v l ¢ F = (N=3)mv¿ = 6:022 1023=3 ¢ ¢ 0:004g=mol 6:022 1023 ¢ ¢ 600m=s ¢ 0:6679s¡1 yielding for the force F = 0.534 N. The pressure is the force per area: p = F=A = 0:534N=0:01m2 = 53:4P a: The calculated force is 0.534 N and the resulting pressure is 53.4 Pa. 12.4.4 Kinetic energy of molecules In the following, we will make the connection between the kinetic theory and the ideal gas laws. We will ﬂnd that the temperature is an important quantity which is the only intrinsic parameter entering in the kinetic energy of a gas. We will consider an ensemble of molecules in a gas, where the molecules will be regarded as rigid large particles. We therefore neglect any vibrations or rotations in the molecule. Hence, making this assumption, Physics for a molecular gas is the same as for a single atom gas. 208 The square of the velocity is sometimes di–cult to conceive, but an alternative statement can be given in terms of kinetic energy. The kinetic energy Ek of a single particle of mass m moving at velocity v is mv2=2. For a large number of molecules N , the total kinetic energy Ek will depend on the mean-square velocity in the same way: mv2=2 = n The second form is on a molar basis, since n = N=NA and the molar mass M = mNA where M v2=2 Ek = N (12.17) ¢ ¢ NA is Avogadro’s number 6.022 1023. The ideal gas law then appears in the form: ¢ (12.18) Compare pV = nM v2=2. This statement that the pressure-volume product of an ideal gas is directly proportional to the total kinetic energy of the gas is also a statement of Boyle’s law, since the total kinetic energy of an ideal gas depends only upon the temperature. pV = 2Ek=3 Comparison of the ideal gas law, pV = nRT , with the kinetic-molecular theory expression pV = 2Ek=3 derived in the previous section shows that the total kinetic energy of a collection of gas molecules is directly proportional to the absolute temperature of the gas. Equating the pV term of both equations gives which rearranges to an explicit expression for temperature, Ek = 3=2nRT ; T = 2 3R Ek n = M v2 3R (12.19) (12.20) We see that temperature is a function only of the mean kinetic energy Ek, the mean molecular velocity v, and the mean molar mass M . Worked Example 71 mean velocity 1 Calculate the kinetic energy of 1 mol of nitrogen molecules at 300 K? Solution: Assume nitrogen behaves as an ideal gas, then Ek = 3=2 ¢ RT = (3=2)8:3145J=(molK) ¢ 300K = 3742J=mol(or3:74kJ=mol) At 300 K, any gas that behaves like an ideal gas has the same energy per mol. As the absolute temperature decreases, the kinetic energy must decrease and thus the mean velocity of the molecules must decrease also. At T = 0, the absolute zero of temperature, all motion of gas molecules wou
ld cease and the pressure would then also be zero. No molecules would be moving. Experimentally, the absolute zero of temperature has never been attained, although modern experiments have extended to temperatures as low as 1 „K. However, at low temperatures, the interactions between the particles becomes important and we enter a new regime of Quantum Mechanics, which considers molecules, single atoms or protons and electrons simultaneously as waves and as rigid particles. However, this would go too far. 209 Worked Example 72 mean velocity 2 If the translational rms. speed of the water vapor molecules (H2O) in air is 648 m/s, what is the translational rms speed of the carbon dioxide molecules (CO2) in the same air? Both gases are at the same temperature. And what is the temperature we measure? Solution: The molar mass of H2O is ¢ As the temperature is constant we can write ¢ MH2O = 2 1g=mol + 1 16g=mol = 18g=mol T = M v2 3R = 0:018kg=mol (648m=s)2 K ¢ ¢ 3 8:314J=mol ¢ Now we calculate the molar mass of CO2 = 303:0K = 29:9–C MCO2 = 2 ¢ 16g=mol + 1 ¢ 12g=mol = 44g=mol The rms velocity is again calculated with eq. 12.20 vCO2 = 3 R T ¢ MCO2 s 3 ¢ = s 8:314J=molK 303:0K 0:044kg=mol ¢ = 414:5m=s The experiment was performed at 29.9 –C and the speed of the CO2-molecules is 414.5 m/s, that is much slower than the water molecules as they are much heavier. 12.5 Temperature Let us look back to the equation for the temperature of an ideal gas, T = 2 3R We can see that temperature is proportional to the average kinetic energy of a molecule in the gas. In other words temperature is a measure of how much energy is contained in an object { in hot things the atoms have a lot of kinetic energy, in cold things they have less. It may be surprising that ‘hot’ and ‘cold’ are really just words for how fast molecules or atoms are moving around, but it is true. (12.21) Ek n Deﬂnition: Temperature is a measure of the average kinetic energy of the particles in a body. It should now be clear that heat is nothing more than energy on the move. It can be carried 210 ice water Figure 12.10: A heat ow diagram showing the heat owing from the warmer water into the cooler ice cube. by atoms, molecules or electromagnetic radiation but it is always just transport of energy. This is very important when we describe movement of heat as we will do in the following sections. ‘Cold’ is not a physical thing. It does not move from place to place, it is just the word for a lack of heat, just like dark is the word for an absence of light. 12.5.1 Thermal equilibrium Now that we have deﬂned the temperature of an isolated object (usually referred to as a body) we need to consider how heat will move between bodies at diﬁerent temperatures. Let us take two bodies; A which has a ﬂxed temperature and B whose temperature is allowed to change. If we allow heat to move between the two bodies we say they are in thermal contact. First let us consider what happens if B is cooler than A. Remember { we have ﬂxed the temperture of A so we need only worry about the temperature of B changing. An example of such a situation is an ice cube being dropped into a large pan of boiling water on a ﬂre. The water temperature is ﬂxed i.e. does not change, because the ﬂre keeps it constant. It should be obvious that the ice cube will heat up and melt. In physical terms we say that the heat is owing out of the (warmer) boiling water, into the (cooler) ice cube. This ow of heat into the ice cube causes it to warm up and melt. In fact the temperature of any cooler object in thermal contact with a warmer one will increase as heat from the warmer object ows into it. The reverse would be true if B were warmer than A. We can now picture putting a small amount of warm water in to a freezer. If we come back in an hour or so the water will have cooled down and possibly frozen. In physical terms we say that the heat is owing out of the (warmer) water, into the (cooler) air in the freezer. This ow of heat out of the ice cube in to the aircauses it to cool down and (eventually) freeze. Again, any warm object in thermal contact with a cooler one will cool down due to heat owing out of it. There is one special case which we have not yet discussed { what happens if A and B are at the same temperature? In this case B will neither warm up nor cool down, in fact, its temperature will remain constant. When two bodies are at the same temperature we say that they are in thermal equilibrium. Another way to express this is to say that two bodies are in thermal equilibrium if the particles within those bodies have the same average kinetic energies. 211 water air Figure 12.11: A heat ow diagram showing the heat owing from the warmer water into the cooler air in the fridge. Convince yourself that the last three paragraghs are correct before you continue. You should notice that heat always ows from the warmer object to the cooler object, never the other way around. Also, we never talk about coldness moving as it is not a real physical thing, only a lack of heat. Most importantly, it should be clear that the ow of heat between the two objects always attempts to bring them to the same temperature (or in other words, into thermal equilibrium). The logical conclusion of all this is that if two bodies are in thermal contact heat will ow from the hotter object to the cooler one until they are in thermal equilibrium (i.e. at the same temperature). We will see how to deal with this if the temperature of object A is not ﬂxed in the section on heat capacities. 12.5.2 Temperature scales Temperature scales are often confusing and even university level students can be tricked into using the wrong one. For most purposes in physics we do not use the familiar celcius (often innaccurately called centigrade) scale but the closely related absolute (or kelvin) scale { why? Let us think about the celcius scale now that we have deﬂned temperature as a measure of the average kinteic energy of the atoms or molecules in a body. A scale is a way of assigning a number to a physical quantity. Consider distance { using a ruler we can measure a distance and ﬂnd its legnth. This legnth could be measured in metres, inches, or miles. The same is true of temperatures in that many diﬁerent scales exist to measure them. Table 12.3 shows a few of these scales. Just like a ruler the scales have two deﬂned points which ﬂx the scale (consider the values at the beginning and end of the ruler e.g. 0cm and 15cm). This is usually achieved by deﬂning the temperature of some physical process, e.g. the freezing point of water. Armed with our knowledge of temperture we can see that Celcius’s scale has a big problem { it allows us to have a negative temperature. 212 Scale Fahrenheit Symbol Deﬂnition –F Temperature at which an equal mixture of ice and salt melts = 0–F Temperature of blood = 96–F Celcius –C Temperature at which water freezes = 0–C Temperature at which water boils = 100–C Kelvin K Absolute zero is 0 K Triple point of water is 273:16 K Table 12.3: The most important temperature scales. We found that temperature is a measure of the average kinetic energy of the particles in a body. Therefore, a negative temperature suggests that that the particles have negative kinetic energy. This can not be true as kinetic energy can only be positive. Kelvin addressed this problem by redeﬂning the zero of the scale. He realised that the coldest temperature you could achieve would be when the particles in a body were not moving at all. There is no way to cool something further than this as there is no more kinetic energy to remove from the body. This temperature is called absolute zero. Kelvin chose his scale so that 0K was the same as absolute zero and chose the size of his degree to be the same as one degree in the celcius scale. Interesting Fact: Rankine did a similar thing to Kelvin but set his degree to be the same size as one degree fahrenheit. Unfortunately for him, almost everyone preferred Kelvin’s absolute scale and the rankine scale is now hardly ever used! It turns out that the freezing point of water, 0–C, is equal to 273:15 K. So, in order to convert from celcius to kelvin need to subtact 273.15. Deﬂnition: T (K) = T (–C) 273:15 ¡ 12.5.3 Practical thermometers It is often important to be able to determine an object’s temperature precisely. This can be a challenge at very high or low tempertures or in inaccesible places. Consider a scientist who wishes to know how hot the magma in a volcano is. They are not going to be able to just lower a thermometer in to the magma as it will just melt as it reaches the superheated rock. We will now look at some less extreme situations and show how a variety of thermometry techniques can be developed. Consider ﬂrst the gas cylinder which we tried to explode in worked example 7 by heating it while sealed. We decided that we would need to heat it to around 3000K before it explodes. How can we check this experimentally? In a sealed gas cylinder the volume of the gas and the number of moles of gas remain constant as we heat, this is why we could use the Gay-Lussac law in example 7. The Gay-Lussac law tells us that pressure is directly proportional 213 to temperature for ﬂxed volume and amount of gas. Therefore by mesuring the pressure in the cylinder (which can be done by ﬂtting a pressure gauge to the top of it) we can indirectly work out the temperature. This is similar to familiar alcohol or mercury thermometers. In these we use the fact that expansion of a liquid as it is heated is approximately proportional to temperature so we can use this expansion to as a measure of temperature. In fact, any thermometer you can imagine uses some physical property which varies with temperature to measure it indirectly. 12.5.4 Speciﬂc heat capacity Conversion of macroscopic energy to microscopic kinetic energy thus tends to raise the temperature, while the reverse conversion lowers it. It is easy to show experimentally that the amo
unt of heating needed to change the temperature of a body by some amount is proportional to the amount of matter in the body. Thus, it is natural to write ¢Q = M C¢T (23.4) where M is the mass of material, ¢Q is the amount of energy transferred to the material, and ¢T is the change of the material’s temperature. The quantity C is called the speciﬂc heat of the material in question and is the amount of energy needed to raise the temperature of a unit mass of material one degree in temperature. C varies with the type of material. Values for common materials are given in table 22.2. Table 22.2: Speciﬂc heats of common materials. Material C (J kg¡1 K¡1) brass 385 glass 669 ice 2092 steel 448 methyl alcohol 2510 glycerine 2427 water 4184 12.5.5 Speciﬂc latent heat It can be seen that the speciﬂc heat as deﬂned above will be inﬂnitely large for a phase change, where heat is transferred without any change in temperature. Thus, it is much more useful to deﬂne a quantity called latent heat, which is the amount of energy required to change the phase of a unit mass of a substance at the phase change temperature. 12.5.6 Internal energy In thermodynamics, the internal energy is the energy of a system due to its temperature. The statement of ﬂrst law refers to thermodynamic cycles. Using the concept of internal energy it is possible to state the ﬂrst law for a non-cyclic process. Since the ﬂrst law is another way of stating the conservation of energy, the energy of the system is the sum of the heat and work input, i.e., E = Q + W. Here E represents the heat energy of the system along with the kinetic energy and the potential energy (E = U + K.E. + P.E.) and is called the total internal energy of the system. This is the statement of the ﬂrst law for non-cyclic processes. For gases, the value of K.E. and P.E. is quite small, so the important internal energy function is U. In particular, since for an ideal gas the state can be speciﬂed using two variables, the state variable u is given by , where v is the speciﬂc volume and t is the temperature. Thus, by deﬂnition, , where cv is the speciﬂc heat at constant volume. 214 Internal energy of an Ideal gas In the previous section, the internal energy of an ideal gas was shown to be a function of both the volume and temperature. Joule performed an experiment where a gas at high pressure inside a bath at the same temperature was allowed to expand into a larger volume. picture required In the above image, two vessels, labeled A and B, are immersed in an insulated tank containing water. A thermometer is used to measure the temperature of the water in the tank. The two vessels A and B are connected by a tube, the ow through which is controlled by a stop. Initially, A contains gas at high pressure, while B is nearly empty. The stop is removed so that the vessels are connected and the ﬂnal temperature of the bath is noted. The temperature of the bath was unchanged at the and of the process, showing that the internal energy of an ideal gas was the function of temperature alone. Thus Joule’s law is stated as = 0. 12.5.7 First law of thermodynamics We now address some questions of terminology. The use of the terms \heat" and \quantity of heat" to indicate the amount of microscopic kinetic energy inhabiting a body has long been out of favor due to their association with the discredited \caloric" theory of heat. Instead, we use the term internal energy to describe the amount of microscopic energy in a body. The word heat is most correctly used only as a verb, e. g., \to heat the house". Heat thus represents the transfer of internal energy from one body to another or conversion of some other form of energy to internal energy. Taking into account these deﬂnitions, we can express the idea of energy conservation in some material body by the equation ¢E = ¢Q ¡ ¢W (ﬂrst law of thermodynamics) where ¢E is the change in internal energy resulting from the addition of heat ¢Q to the body and the work ¢W done by the body on the outside world. This equation expresses the ﬂrst law of thermodynamics. Note that the sign conventions are inconsistent as to the direction of energy ow. However, these conventions result from thinking about heat engines, i. e., machines which take in heat and put out macroscopic work. Examples of heat engines are steam engines, coal and nuclear power plants, the engine in your automobile, and the engines on jet aircraft. 12.6 Important Equations and Quantities Quantity Symbol Unit S.I. Units Direction Units or Table 12.4: Units used in Electricity and Magnetism 215 Chapter 13 Electrostatics 13.1 What is Electrostatics? Electrostatics is the study of electric charge which is not moving i.e. is static. 13.2 Charge All objects surrounding us (including people!) contain large amounts of electric charge. Charge can be negative or positive and is measured in units called coulombs (C). Usually, objects contain the same amount of positive and negative charge so its eﬁect is not noticeable and the object is called electrically neutral. However, if a small imbalance is created (i.e. there is a little bit more of one type of charge than the other on the object) then the object is said to be electrically charged. Some rather amusing examples of what happens when a person becomes charged are for example when you charge your hair by combing it with a plastic comb and it stands right up on end! Another example is when you walk fast over a nylon carpet and then touch a metal doorknob and give yourself a small shock (alternatively you can touch your friend and shock them!) Charge has 3 further important properties: † † † Charge is always conserved. Charge, just like energy, cannot be created or destroyed. Charge comes in discrete packets. The smallest unit of charge is that carried by one electron called the elementary charge, e, and by convention, it has a negative sign (e = 10¡19C). 1:6 ¡ £ Charged objects exert electrostatic forces on each other. Like charges repel and unlike charges attract each other. Interesting Fact: The word ‘electron’ comes from the Greek word for amber! The ancient Greeks observed that if you rubbed a piece of amber, you could use it to pick up bits of straw. (Attractive electrostatic force!) 216 You can easily test the fact that like charges repel and unlike charges attract by doing a very simple experiment. Take a glass rod and rub it with a piece of silk, then hang it from its middle with a piece string so that it is free to move. If you then bring another glass rod which you have also charged in the same way next to it, you will see the rod on the string turn away from the rod in your hand i.e. it is repelled. If, however, you take a plastic rod, rub it with a piece of fur and then bring it close to the rod on the string, you will see the rod on the string turn towards the rod in your hand i.e. it is attracted! ////////// ////////// + + + + + + - - - - What actually happens is that when you rub the glass with silk, tiny amounts of negative charge are transferred from the glass onto the silk, which causes the glass to have less negative charge than positive charge, making it positively charged. When you rub the plastic rod with the fur, you transfer tiny amounts of negative charge onto the rod and so it has more negative charge than positive charge on it, making it negatively charged. Conductors and Insulators Some materials allow charge carriers to move relatively freely through them (e.g. most metals, tap water, the human body) and these materials are called conductors. Other materials, which do not allow the charge carriers to move through them (e.g. plastic, glass), are called nonconductors or insulators. Aside: As mentioned above, the basic unit of charge, namely the elementary charge, e, is carried by the electron. In a conducting material (e.g. copper), when the atoms bond to form the material, some of the outermost, loosely bound electrons become detached from the individual atoms and so become free to move around. The charge carried by these electrons can move around in the material! In insulators, there are very few, if any, free electrons and so the charge cannot move around in the material. 217 If an excess of charge is put onto an insulator, it will stay where it is put and there will be a concentration of charge in that area on the object. However, if an excess of charge is put onto a conductor, the charges of like sign will repel each other and spread out over the surface of the object. When two conductors are made to touch, the total charge on them is shared between the two. If the two conductors are identical, then each conductor will be left with half of the total charge. 13.3 Electrostatic Force As we now know, charged objects exert a force on one another. If the charges are at rest then this force between them is known as the electrostatic force. An interesting characteristic of the electrostatic force is that it can be either attractive or repulsive, unlike the gravitational force which is only ever attractive. The relative charges on the two objects is what determines whether the force between the charged objects is attractive or repulsive. If the objects have opposite charges they attract each other, while if their charges are similar they repel each other (e.g. two metal balls which are negatively charged will repel each other, while a positively charged ball and negatively charged ball will attract one another). F F - - F F - + It is this force that determines the arrangement of charge on the surface of conductors. When we place a charge on a spherical conductor the repulsive forces between the individual like charges cause them to spread uniformly over the surface of the sphere. However, for conductors with non-regular shape there is a concentration of charge near the point or points of the object. +++++ + + + + + + +++++ ----- - - - - - - - - ---- This collection of charge can actually allow charge to leak oﬁ the conductor if the point is sh
arp enough. It is for this reason that buildings often have a lightning rod on the roof to remove any charge the building has collected. This minimises the possibility of the building being struck by lightning. This \spreading out" of charge would not occur if we were to place the charge on an insulator since charge cannot move in insulators. 13.3.1 Coulomb’s Law The behaviour of the electrostatic force was studied in detail by Charles Coulomb around 1784. Through his observations he was able to show that the electrostatic force between two point-like 218 charges is inversely proportional to the square of the distance between the objects. He also discovered that the force is proportional to the product of the charges on the two objects. F / Q1Q2 r2 ; where Q1 is the charge on the one point-like object, Q2 is the charge on the second, and r is the distance between the two. The magnitude of the electrostatic force between two point-like charges is given by Coulomb’s Law: F = k Q1Q2 r2 (13.1) and the proportionality constant k is called the electrostatic constant. We will use the value k = 8:99 109N ¢ £ m2=C2: The value of the electrostatic constant is known to a very high precision (9 decimal places). Not many physical constants are known to as high a degree of accuracy as k. Aside: Notice how similar Coulomb’s Law is to the form of Newton’s Universal Law of Gravitation between two point-like particles: FG = G m1m2 r2 ; where m1 and m2 are the masses of the two particles, r is the distance between them, and G is the gravitational constant. It is very interesting that Coulomb’s Law has been shown to be correct no matter how small the distance, nor how large the charge: for example it still applies inside the atom (over distances smaller than 10¡10m). Let’s run through a simple example of electrostatic forces. Worked Example 73 Coulomb’s Law I 10¡9C Question: Two point-like charges carrying charges of +3 are 2m apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive. Answer: Step 1 : (NOTE TO SELF: step is deprecated, use westep instead.) First draw the situation: +3 10¡9C and 10¡9C 10¡9C 5 ¡ £ £ 5 ¡ £ £ ¢ ¢ 2m Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Is everything in the correct units? Yes, charges are in coulombs [C] and distances in meters [m]. 219 Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) Determine the magnitude of the force: Using Coulomb’s Law we have F = k Q1Q2 r2 = (8:99 £ = 3:37 ¡ £ 109N m2=C2) ¢ 10¡8N (+3 £ 10¡9C)( 5 ¡ (2m)2 £ 10¡9C) Thus the magnitude of the force is 3:37 two point charges having opposite signs. Step 4 : (NOTE TO SELF: step is deprecated, use westep instead.) Is the force attractive or repulsive? Well, since the two charges are oppositely charged, the force is attractive. We can also conclude this from the fact that Coulomb’s Law gives a negative value for the force. 10¡8N. The minus sign is a result of the £ Next is another example that demonstrates the diﬁerence in magnitude between the gravita- tional force and the electrostatic force. Worked Example 74 Coulomb’s Law II Question: Determine the electrostatic force and gravitational force between two electrons 1”Aapart (i.e. the forces felt inside an atom) Answer: Step 1 : (NOTE TO SELF: step is deprecated, use westep instead.) £ First draw the situation: e 10¡19C 10¡19C 1:60 1:60 £ e ¡ ¡ 1”A Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Get everything into S.I. units: The charge on an electron is 10¡31kg, and 1”A=1 of an electron is 9:11 Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) Calculate the electrostatic force using Coulomb’s Law: 10¡10m 1:60 £ £ ¡ £ 10¡19C, the mass FE = k e e ¢ 1”A2 Q1Q2 r2 = k 109N £ ¢ 10¡8N = (8:99 = 2:30 £ m2=C2) 1:60 ( ¡ £ 10¡19C)( 1:60 ¡ (10¡10m)2 10¡19C) £ 10¡8N. Hence the magnitude of the electrostatic force between the electrons is 2:30 (Note that the electrons carry like charge and from this we know the force must be repulsive. Another way to see this is that the force is positive and thus repulsive.) £ 220 Step 4 : (NOTE TO SELF: step is deprecated, use westep instead.) Calculate the gravitational force: FE = G m1m2 r2 = G me ¢ me (1”A)2 = (6:67 10¡11N ¢ £ m2=kg2) (9:11 £ 10¡31C)(9:11 (10¡10m)2 £ 10¡31kg) = 5:54 10¡51N £ The magnitude of the gravitational force between the electrons is 5:54 10¡51N £ Notice that the gravitational force between the electrons is much smaller than the electrostatic force. For this reason, the gravitational force is usually neglected when determining the force between two charged objects. We mentioned above that charge placed on a spherical conductor spreads evenly along the surface. As a result, if we are far enough from the charged sphere, electrostatically, it behaves as a point-like charge. Thus we can treat spherical conductors (e.g. metallic balls) as point-like charges, with all the charge acting at the centre. Worked Example 75 Coulomb’s Law: Challenge Question Question: In the picture below, X is a small negatively charged sphere with a mass of 10kg. It is suspended from the roof by an insulating rope which makes an angle of 60o with the roof. Y is a small positively charged sphere which has the same magnitude of charge as X. Y is ﬂxed to the wall by means of an insulating bracket. Assuming the system is in equilibrium, what is the magnitude of the charge on X? /////////// 60o 10kg X { Y + 50cm n n n n n Answer: How are we going to determine the charge on X? Well, if we know the force between X and Y we can use Coulomb’s Law to determine their charges as we know the distance between them. So, ﬂrstly, we need to determine the magnitude of the electrostatic force between X and Y. Step 1 : (NOTE TO SELF: step is deprecated, use westep instead.) Is everything in S.I. units? The distance between X and Y is 50cm = 0:5m, and the mass of X is 10kg. 221 Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Draw the forces on X (with directions) and label. T : tension from the thread 60o FE: electrostatic force X Fg: gravitational force Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) Determine the magnitude of the electrostatic force (FE). Since nothing is moving (system is in equlibrium) the vertical and horizonal components of the forces must cancel. Thus FE = T sin(60o); Fg = T sin(60o): The only force we know is the gravitational force Fg = mg. Now we can calculate the magnitude of T from above: T = Fg sin(60o) = (10kg)(10m=s2) sin(60o) = 1155N: Which means that FE is: FE = T cos(60o) = 1154N cos(60o) = 577:5N ¢ Step 4 : (NOTE TO SELF: step is deprecated, use westep instead.) Now that we know the magnitude of the electrostatic force between X and Y, we can calculate their charges using Coulomb’s Law. Don’t forget that the magnitudes of the charges on X and Y are the same: . The magnitude of the electrostatic = j force is QX j QY j j FE = k j QX j j = = j = k Q2 X r2 QXQY r2 FEr2 k r 8:99 s (577:5N)(0:5m)2 109N £ 10¡4C ¢ m2=C2 = 1:27 Thus the charge on X is 1:27 ¡ £ 10¡4C £ 222 13.4 Electric Fields We have learnt that objects that carry charge feel forces from all other charged objects. It is useful to determine what the eﬁect of a charge would be at every point surrounding it. To do this we need some sort of reference. We know that the force that one charge feels due to another depends on both charges (Q1 and Q2). How then can we talk about forces if we only have one charge? The solution to this dilemna is to introduce a test charge. We then determine the force that would be exerted on it if we placed it at a certain location. If we do this for every point surrounding a charge we know what would happen if we put a test charge at any location. This map of what would happen at any point we call a ﬂeld map. It is a map of the electric It tells us how large the force on a test charge would be and in what ﬂeld due to a charge. direction the force would be. Our map consists of the lines that tell us how the test charge would move if it were placed there. 13.4.1 Test Charge This is the key to mapping out an electric ﬂeld. The equation for the force between two electric charges has been shown earlier and is: F = k Q1Q2 r2 : (13.2) If we want to map the ﬂeld for Q1 then we need to know exactly what would happen if we put Q2 at every point around Q1. But this obviously depends on the value of Q2. This is a time when we need to agree on a convention. What should Q2 be when we make the map? By convention we choose Q2 = +1C. This means that if we want to work out the eﬁects on any other charge we only have to multiply the result for the test charge by the magnitude of the new charge. The electric ﬂeld strength is then just the force per unit of charge and has the same magnitude and direction as the force on our test charge but has diﬁerent units: E = k Q1 r2 (13.3) The electric ﬂeld is the force per unit of charge and hence has units of newtons per coulomb [N/C]. So to get the force the electric ﬂeld exerts we use: F = EQ (13.4) Notice we are just multiplying the electric ﬂeld magnitude by the magnitude of the charge it is acting on. 13.4.2 What do ﬂeld maps look like? The maps depend very much on the charge or charges that the map is being made for. We will start oﬁ with the simplest possible case. Take a single positive charge with no other charges around it. First, we will look at what eﬁects it would have on a test charge at a number of points. 223 Positive Charge Acting on Test Charge At each point we calculate the force on a test charge, q, and represent this force by a vector. +Q We can see that at every point the positive test charge, q, would experience a force pushing it away from the charge, Q. This is because both charges are positive and so they repel. Also notice that at points further away the vectors are shorter. That is because the force is smaller if you are further away. If the charge were negative we w
ould have the following result. Negative Charge Acting on Test Charge -Q Notice that it is almost identical to the positive charge case. This is important { the arrows are the same length because the magnitude of the charge is the same and so is the magnitude of the test charge. Thus the magnitude of the force is the same. The arrows point in the opposite direction because the charges now have opposite sign and so the test charge is attracted to the charge. Now, to make things simpler, we draw continuous lines showing the path that the test charge would travel. This means we don’t have to work out the magnitude of the force at many diﬁerent points. 224 Electric Field Map due to a Positive Charge +Q Some important points to remember about electric ﬂelds: There is an electric ﬂeld at every point in space surrounding a charge. Field lines are merely a representation { they are not real. When we draw them, we just pick convenient places to indicate the ﬂeld in space. Field lines always start at a right-angle (90o) to the charged object causing the ﬂeld. Field lines never cross! † † † † 13.4.3 Combined Charge Distributions We look at the ﬂeld of a positive charge and a negative charge placed next to each other. The net resulting ﬂeld would be the addition of the ﬂelds from each of the charges. To start oﬁ with let us sketch the ﬂeld maps for each of the charges as though it were in isolation. Electric Field of Negative and Positive Charge in Isolation +Q -Q 225 Notice that a test charge starting oﬁ directly between the two would be pushed away from the positive charge and pulled towards the negative charge in a straight line. The path it would follow would be a straight line between the charges. +Q -Q Now let’s consider a test charge starting oﬁ a bit higher than directly between the charges. If it starts closer to the positive charge the force it feels from the positive charge is greater, but the negative charge does attract it, so it would move away from the positive charge with a tiny force attracting it towards the negative charge. As it gets further from the positive charge the force from the negative and positive charges change and they are equal in magnitude at equal distances from the charges. After that point the negative charge starts to exert a stronger force on the test charge. This means that the test charge moves towards the negative charge with only a small force away from the positive charge. +Q -Q Now we can ﬂll in the other lines quite easily using the same ideas. The resulting ﬂeld map is: +Q -Q Two Like Charges I: The Positive Case For the case of two positive charges things look a little diﬁerent. We can’t just turn the arrows around the way we did before. In this case the test charge is repelled by both charges. This tells us that a test charge will never cross half way because the force of repulsion from both charges will be equal in magnitude. 226 +Q +Q The ﬂeld directly between the charges cancels out in the middle. The force has equal magnitude and opposite direction. Interesting things happen when we look at test charges that are not on a line directly between the two. +Q +Q We know that a charge the same distance below the middle will experience a force along a reected line, because the problem is symmetric (i.e. if we ipped vertically it would look the same). This is also true in the horizontal direction. So we use this fact to easily draw in the next four lines. +Q +Q 227 Working through a number of possible starting points for the test charge we can show the electric ﬂeld map to be: +Q +Q Two Like Charges II: The Negative Case We can use the fact that the direction of the force is reversed for a test charge if you change the sign of the charge that is inuencing it. If we change to the case where both charges are negative we get the following result: -Q -Q 13.4.4 Parallel plates One very important example of electric ﬂelds which is used extensively is the electric ﬂeld between two charged parallel plates. In this situation the electric ﬂeld is constant. This is used for many practical purposes and later we will explain how Millikan used it to measure the charge on the electron. 228 Field Map for Oppositely Charged Parallel Plates + + + + + + + + + - - - - - - - - - This means that the force that a test charge would feel at any point between the plates would be identical in magnitude and direction. The ﬂelds on the edges exhibit fringe eﬁects, i.e. they bulge outwards. This is because a test charge placed here would feel the eﬁects of charges only on one side (either left or right depending on which side it is placed). Test charges placed in the middle experience the eﬁects of charges on both sides so they balance the components in the horizontal direction. This isn’t the case on the edges. The Force on a Test Charge between Oppositely Charged Parallel Plates + + + + + + + + + - - - - - - - - - 13.4.5 What about the Strength of the Electric Field? When we started making ﬂeld maps we drew arrows to indicate the strength of the ﬂeld and the direction. When we moved to lines you might have asked \Did we forget about the ﬂeld strength?". We did not. Consider the case for a single positive charge again: 229 A m mg h B Figure 13.1: A mass under the inuence of a gravitational ﬂeld. +Q Notice that as you move further away from the charge the ﬂeld lines become more spread out. In ﬂeld map diagrams the closer ﬂeld lines are together the stronger the ﬂeld. This brings us to an interesting case. What is the electric ﬂeld like if the object that is charged has an irregular shape. 13.5 Electrical Potential 13.5.1 Work Done and Energy Transfer in a Field When a charged particle moves in an electric ﬂeld work is done and energy transfers take place. This is exactly analogous to the case when a mass moves in a gravitational ﬂeld such as that set up by any massive object. Work done by a ﬂeld Gravitational Case A mass held at a height h above the ground has gravitational potential energy since, if released, it will fall under the action of the gravitational ﬂeld. Once released, in the absence of friction, only the force of gravity acts on the mass and the mass accelerates in the direction of the force (towards the earth’s centre). In this way, work is done by the ﬂeld. When the mass 230 + + + + + + + A +Q QE s B - - - - - - - Figure 13.2: A charged particle under the inuence of an electric ﬂeld. falls a distance h (from point A to B), the work done is, W = F s = mgh In falling, the mass loses gravitational potential energy and gains kinetic energy. The work done by the ﬂeld is equal to the energy transferred, Energy is conserved! W = Gain in Ek = Loss in Ep (a falling mass) Electrical Case A charge in an electric ﬂeld has electrical potential energy since, if released, it will move under the action of the electric ﬂeld. When released, in the absence of friction, only the electric force acts on the charge and the charge accelerates in the direction of the force (for positive charges the force and acceleration are in the direction of the electric ﬂeld, while negative charges experience a force and acceleration in the opposite direction to the electric ﬂeld.) Consider a positive charge +Q placed in the uniform electric ﬂeld between oppositely charged parallel plates. The positive charge will be repelled by the positive plate and attracted by the negative plate (i.e. it will move in the direction of the electric ﬂeld lines). In this way, work is done by the ﬂeld. In moving the charge a distance s in the electric ﬂeld, the work done is, W = F s = QEs since E = F Q : In the process of moving, the charge loses electrical potential energy and gains kinetic energy. The work done by the ﬂeld is equal to the energy transferred, W = Gain in Ek = Loss in Ep (charge moving under the inuence of an electric ﬂeld) Work done by us Gravitational Case In order to return the mass m in Fig.13.1 to its original position (i.e. lift it a distance h from B back to A) we have to apply a force mg to balance the force of gravity. An amount of work mgh is done by the lifter. In the process, the mass gains gravitational potential energy, mgh = Gain in Ep (lifting a mass) 231 Electrical Case In order to return the charge in Fig.13.2 to its original position (i.e. from B back to A) we have to exert a force QE on the charge to balance the force exerted on it by the electric ﬂeld. An amount of work QEs is done by us. In the process, the charge gains electrical potential energy, Energy is conserved! QEs = Gain in Ep (charge moved against an electric ﬂeld) In summary, when an object moves under the inuence of a ﬂeld, the ﬂeld does work and potential energy is transferred into kinetic energy. Potential energy is lost, while kinetic energy is gained. When an object is moved against a ﬂeld we have to do work and the object gains potential energy. Worked Example 76 Work done and energy transfers in a ﬂeld Question: A charge of +5nC is moved a distance of 4 cm against a uniform electric ﬂeld of magnitude 2 1012N:C¡1 from A to B 4cm +5nC A - - - - - - - (a) Calculate the work done in moving the charge from A to B. (b) The charge is now released and returns to A. Calculate the kinetic energy of the charge at A. Answer: (a) Step 1 : (NOTE TO SELF: step is deprecated, use westep instead.) We are given the values of the charge, the ﬂeld and the distance the charge must move. All are in the correct units. Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Since the charge is positive we have to do work to move it from A to B (since this is against the ﬂeld). This work is given by, W = QEs 10¡9)(2 = (5 £ = 400 J 1012)(0:04) £ (b) 232 Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) When released the charge moves under the inuence of the electric ﬂeld and returns to A. Work is now done by the ﬂeld and the work done is equal to the kinetic energy gained, Gain in Ek = QEs 10¡9)(2 = (5 £ = 400 J 1012)(0:04) £ Since the charge
started at rest, the gain in kinetic energy is the ﬂnal kinetic energy, Eat A k = 400 J 13.5.2 Electrical Potential Diﬁerence Consider a positive test charge +Q placed at A in the electric ﬂeld of another positive point charge. + +Q A B The test charge moves towards B under the inuence of the electric ﬂeld of the other charge. In the process the test charge loses electrical potential energy and gains kinetic energy. Thus, at A, the test charge has more potential energy than at B { A is said to have a higher electrical potential than B. The potential energy of a charge at a point in a ﬂeld is deﬂned as the work required to move that charge from inﬂnity to that point. The potential diﬁerence between two points in an electric ﬂeld is deﬂned as the work required to move a unit positive test charge from the point of lower potential to that of higher potential. If an amount of work W is required to move a charge Q from one point to another, then the potential diﬁerence between the two points is given by, V = W Q unit : J:C¡1 or V (the volt) 233 From this equation it follows that one volt is the potential diﬁerence between two points in an electric ﬂeld if one joule of work is done in moving one coulomb of charge from the one point to the other. Worked Example 77 Potential diﬁerence Question: A positively charged object Q is placed as shown in the sketch. The potential diﬁerence between two points A and B is 4 10¡4 V. £ +Q A B (a) Calculate the change in electrical potential energy of a +2nC charge when it moves from A to B. (b) Which point, A or B, is at the higher electrical potential? Explain. (c) If this charge were replaced with another of charge -2nC, in what way would its change in energy be aﬁected? Answer: (a) The electrical potential energy of the positive charge decreases as it moves from A to B since it is moving in the direction of the electric ﬂeld produced by the object Q. This loss in potential energy is equal to the work done by the ﬂeld, Loss in Electrical Potential Energy = W = V Q (Since V = W Q ) = (4 = 8 £ 10¡4)(2 10¡13 J £ 10¡9) £ (b) Point A is at the higher electrical potential since work is required by us to move a positive test charge from B to A. (c) If the charge is replaced by one of negative charge, the electrical potential energy of the charge will increase in moving from A to B (in this case we would have to do work on the charge). As an example consider the electric ﬂeld between two oppositely charged parallel plates a distance d apart maintained at a potential diﬁerence V . 234 + + + + + + + P d O +Q - - - - - - - This electric ﬂeld is uniform so that a charge placed anywhere between the plates will experience the same force. Consider a positive test charge Q placed at point O just oﬁ the surface of the negative plate. In order to move it towards the positive plate we have to apply a force QE. The work done in moving the charge from the negative to the positive plate is, W = F s = QEd; but from the deﬂnition of electrical potential, Equating these two expressions for the work done, W = V Q: QEd = V Q; E = V d : and so, rearranging, Worked Example 78 Parallel plates Question: Two charged parallel plates are at a distance of 180 mm from each other. The potential diﬁerence between them is 3600 V as shown in the diagram. + + + + + + + d = 180mm X Y V = 3 600V - - - - - - - 235 + + + + + + + d -Q Fup=QE Fdown=mg - - - - - - - Figure 13.3: An oil drop suspended between oppositely charged parallel plates. 10¡9 C, is (a) If a small oil drop of negligible mass, carrying a charge of +6:8 placed between the plates at point X, calculate the magnitude and direction of the electrostatic force exerted on the droplet. (b) If the droplet is now moved to point Y, would the force exerted on it be bigger, smaller or the same as in (a)? Answer: (a) Step 1 : (NOTE TO SELF: step is deprecated, use westep instead.) First ﬂnd the electric ﬂeld strength between the plates, £ E = = V d 3600 0:180 = 20000 N:C¡1 from the positive to the negative plate Step 2 : (NOTE TO SELF: step is deprecated, use westep instead.) Now the force exerted on the charge at X is, F = QE = (6:8 = 1:36 £ £ 10¡9)(20000) 10¡4 down (b) Step 3 : (NOTE TO SELF: step is deprecated, use westep instead.) The same. Since the electric ﬂeld strength is uniform, the force exerted on a charge is the same at all points between the plates. 13.5.3 Millikan’s Oil-drop Experiment Robert Millikan measured the charge on an electron by studying the motion of charged oil drops between oppositely charged parallel plates. Consider one such negative drop between the plates in Fig.13.3. Since this drop is negative, the electric ﬂeld exerts an upward force on the drop. In addition to this upward force, gravity exerts a downward force on the drop. Millikan adjusted 236 the electric ﬂeld strength between the plates by varying the potential diﬁerence applied across the plates. In this way, Millikan was able to bring the drops to rest. At equlibrum, Since E = V d , and, therefore, Fup = Fdown QE = mg Q V d = mg; Q = mgd V Millikan found that all drops had charges which were multiples of 1:6 become charged by gaining or losing electrons, the charge on an electron must be The magnitude of the electron’s charge is denoted by e, £ 10¡19 C. Since objects 10¡19 C. 1:6 ¡ £ e = 1:6 10¡19 C £ Worked Example 79 Charge Question: A metal sphere carries a charge of +3:2 to lose to attain its charge? Answer: Since the sphere is positive it lost electrons in the process of charging (when an object loses negative charges it is left positive). In fact, it lost, 10¡8 C. How many electrons did it have £ 10¡8 10¡19 = 2 £ 3:2 1:6 £ £ 1011 electrons Worked Example 80 Millikan oil-drop experiment Question: In a Millikan-type experiment a positively charged oil drop is placed between two horizontal plates, 20 mm apart, as shown. 237 - - - - - - - + + + + + + + The potential diﬁerence across the plates is 4000V. The drop has a mass of 1:2 10¡14kg and a charge of 8 (a) Draw the electric ﬂeld pattern between the two plates. (b) Calculate: 10¡19C. £ £ 1. the electric ﬂeld intensity between the two plates. 2. the magnitude of the gravitational force acting on the drop. 3. the magnitude of the Coulomb force acting on the drop. (c) The drop is observed through a microscope. What will the drop be seen to do? Explain. (d) Without any further calculations, give two methods that could be used to make the drop remain in a ﬂxed position. Answer: (a 4000 0:02 = 2 £ 105 V:m¡1 up (b) 1. 2. Fgrav = mg = (1:2 = 1:2 10¡14)(10) 10¡13N £ £ 238 3. FCoulomb = QE = (8 = 1:6 £ £ 10¡19)(s 10¡13N 105) £ (c) Since Fup > Fdown, the drop accelerates upwards. (d) The Coulomb force can be decreased by decreasing the electric ﬂeld strength between the plates. Since E = V d , this can be done either by increasing d or decreasing V . 13.6 Important Equations and Quantities Units Quantity charge force mass acceleration radial distance electric ﬂeld work potential diﬁerence Unit Symbol q or Q C (Coulomb) N (Newton) \| \| \| A:s kg:m s2 kg m s2 m N=C or V =m kg:m A:s3 2 kg:m s2 2 kg:m A:s3 ¡!F m ¡!a r ¡!E W V J V (Volt) S.I. Units or \| or kg:m:s¡2 or \| or m:s¡2 or \| or kg:m:A¡1:s¡3 or kg:m2:s¡2 or kg:m2:A¡1:s¡3 Direction \| X \| X \| X \| \| Table 13.1: Units used in Electrostatics 239 Chapter 14 Electricity Warning: We believe in experimenting and learning about physics at every opportunity, BUT playing with electricity can be EXTREMELY DANGEROUS! Do not try to build home made circuits without someone who knows if what you are doing is safe. Normal electrical outlets are dangerous. Treat electricity with respect in your everyday life. You will encounter electricity everyday for the rest of your life and to make sure you are able to make wise decisions we have included an entire chapter on electrical safety. Please read it - not only will it make you safer but it will show the applications of many of the ideas you will learn in this chapter. 14.1 Flow of Charge The normal motion of "free" electrons in a conductor has no particular direction or speed. However, electrons can be inuenced to move in a coordinated fashion through a conductive material. This motion of electrons is what we call electricity, or electric current. This is in contrast to static electricity, which is an unmoving accumulation of electric charge. Just like water owing through the emptiness of a pipe, electrons are able to move between the atoms of a conductor. The conductor may appear to be solid to our eyes, but any material composed of atoms is mostly empty space! The liquid-ow analogy is so ﬂtting that the motion of electrons through a conductor is often referred to as a "ow." As each electron moves through a conductor, it pushes on the one ahead of it. This push of one electron on another makes all of the electrons move together as a group. The motion of the each electron in a conductor may be very slow. However, the starting and stopping of electron ow through a conductor is virtually instantaneous from one end of it to the other. As an analogy consider a tube ﬂlled end-to-end with marbles: Tube Marble Marble The tube is full of marbles, just as a conductor is full of free electrons. If a single marble is suddenly inserted into this full tube on the left-hand side, another marble will immediately try to exit the tube on the right. Even though each marble only traveled a short distance, the 240 transfer of motion through the tube is virtually instantaneous from the left end to the right end. The nearly instantaneous transfer of motion through the tube occurs no matter how long the tube is. With electricity, the overall eﬁect from one end of a conductor to the other is eﬁectively instantaneous. Each individual electron, though, travels through the conductor at a much slower pace. If we want electrons to ow in a certain direction to a certain place, we must provide the proper path for them to move. A path for electrons must be provided just as a plumber must i
nstall piping to get water to ow where he or she wants it to ow. Wires made of highly conductive metals such as copper or aluminum are used to form this path. This means that there can be electric current only where there exists a continuous path of conductive material (wire) providing a path for electrons. In the marble analogy, marbles can ow into the left-hand side of the tube only if the tube is open on the right-hand side for marbles to ow out. If the tube is blocked on the right-hand side, the marbles will just "pile up" inside the tube. Marble "ow" will not occur if the tube is blocked. The same holds true for electric current: the continuous ow of electrons requires there be an unbroken path. Let’s look at a diagram to illustrate how this works: A thin, solid line (as shown above) is the conventional symbol for a continuous piece of wire. The wire is made of a conductive material, such as copper or aluminum. The wire’s constituent atoms have many free electrons which can easily move through the wire. However, there will never be a continuous ow of electrons within this wire unless they have a place to come from and a place to go. Let’s add an hypothetical electron "Source" and "Destination:" Electron Source Electron Destination Now, with the Electron Source pushing new electrons into the wire on the left-hand side, electron ow through the wire can occur (as indicated by the arrows pointing from left to right). However, the ow will be interrupted if the conductive path formed by the wire is broken: Electron Source no flow! no flow! (break) Electron Destination Air is an insulator that impedes the ow of electrons. An air gap separates the two pieces of wire, the path has now been broken, and electrons cannot ow from Source to Destination. This is like cutting a water pipe in two and capping oﬁ the broken ends of the pipe: water can’t ow if there’s no exit out of the pipe. If we were to take another piece of wire leading to the Destination and connect it with the wire leading to the Source, we would once again have a continuous path for electrons to ow. The two dots in the diagram indicate physical (metal-to-metal) contact between the wire pieces: Electron Source no flow! (break) Electron Destination 241 Now, we have continuity from the Source, to the newly-made connection, down, to the right, and up to the Destination. Please take note that the broken segment of wire on the right hand side has no electrons owing through it. This is because it is no longer part of a complete path from Source to Destination. It is interesting to note that no "wear" occurs within wires due to this electric current. This is in contrast to water-carrying pipes which are eventually corroded and worn by prolonged ows. Electrons do encounter some degree of friction as they move and this friction can generate heat in a conductor. This is a topic we’ll discuss later. 14.2 Circuits In order for the Source-and-Destination scheme to work, both would have to have a huge reservoir of electrons in order to sustain a continuous ow! Using the marble-and-tube analogy, the source and destination buckets would have to be large reservoirs to contain enough marble capacity for a "ow" of marbles to be sustained. The answer to this paradox is found in the concept of a circuit: a never-ending looped pathway for electrons. If we take a wire, and loop it around so that it forms a continuous pathway, we have the means to support a uniform ow of electrons without having to resort to huge reservoirs. electrons can flow in a path without beginning or end, continuing forever! A marble-andhula-hoop "circuit" Each electron advancing clockwise in this circuit pushes on the one in front of it, which pushes on the one in front of it, and so on. A circuit is just like a hula-hoop ﬂlled with marbles. Now, we have the capability of supporting a continuous ow of electrons indeﬂnitely without the need for reservoirs. All we need to maintain this ow is a continuous means of motivation for those electrons. This topic will be addressed in the next section of this chapter. It must be realized that continuity is just as important in a circuit as it is in a straight piece of wire. Just as in the example with the straight piece of wire between the electron Source and Destination, any break in this circuit will prevent electrons from owing through it: 242 no flow! continuous electron flow cannot occur anywhere in a "broken" circuit! (break) no flow! no flow! An important principle to realize here is that it doesn’t matter where the break occurs. Any discontinuity in the circuit will prevent electron ow throughout the entire circuit. no flow! continuous electron flow cannot occur anywhere in a "broken" circuit! no flow! (break) no flow! A combination of batteries and conductors with other components is called an electric circuit or circuit. The word circuit implies that you return to your starting point and this is an important property of electric circuits. They must contain a closed loop before charge can ow. The simplest possible circuit is a battery with a single conductor. Now how do we form a closed loop with these two components? The battery has two terminals (connection points). One is called the positive terminal and one the negative. When we describe charge owing we consider charges moving from the positive terminal of the battery around the conductor and back into the battery at the negative terminal. As much charge ows out of the positive terminal as ows into the negative terminal. Because of this, there is no build up of charge in the battery. The battery does work on the charges causing them to move round the circuit. We have covered the topics of batteries and circuits but we need to draw these things to help keep the ideas clear in our minds. To do this we need to agree on how to draw things so that other people can understand what we are doing. We need a convention. 243 14.3 Voltage and current As was previously mentioned, we need more than just a continuous path (circuit) before a continuous ow of electrons will occur. We also need some means to push these electrons around the circuit. Just like marbles in a tube or water in a pipe, it takes some kind of inuencing force to initiate ow. With electrons, this force is the same force at work in static electricity: the force produced by an imbalance of electric charge. The electric charge diﬁerence serves to store a certain amount of energy. This energy is not unlike the energy stored in a high reservoir of water that has been pumped from a lower-level pond: Reservoir Energy stored Water flow Pump Pond The inuence of gravity on the water in the reservoir creates a force that attempts to move the water down to the lower level again. If a suitable pipe is run from the reservoir back to the pond, water will ow under the inuence of gravity down from the reservoir, through the pipe: 244 Reservoir Energy released Pond It takes energy to pump that water from the low-level pond to the high-level reservoir. The movement of water through the piping back down to its original level constitutes a releasing of energy stored from previous pumping. If the water is pumped to an even higher level, it will take even more energy to do so, and so more energy will be stored. The more energy is stored the more energy is released if the water is allowed to ow through a pipe back down again: 245 Energy stored Reservoir Pump Pond Reservoir Energy released More energy stored More energy released Pump Pond Electrons are not much diﬁerent. If we "pump" electrons away from their normal "levels," we create a condition where a force exists as the electrons seek to re-establish their former positions. The force attracting electrons back to their original positions is analogous to the force gravity exerts on water in the reservoir. Just as gravity tries to draw water down to its former level, the force exerted on the electrons attracts them back to their former posisions. Just as the pumping of water to a higher level results in energy being stored, "pumping" electrons to create an electric charge imbalance results in a certain amount of energy being stored in that imbalance. Providing a way for water to ow back down from the heights of the 246 reservoir results in a release of that stored energy. Similarly, providing a way for electrons to ow back to their original "levels" results in a release of stored energy. When the electrons are poised in that static condition (just like water sitting still, high in a reservoir), the energy stored there is called potential energy. It is given that name because it has the possibility (potential) of release that has not been fully realized yet. This potential energy, stored in the form of an electric charge imbalance, can be expressed as a term called voltage. Technically, voltage is a measure of potential energy per unit charge of electrons, or something a physicist would call speciﬂc potential energy. Deﬂned in the context of static electricity, voltage is the measure of work required to move a unit charge from one location to another. This work is against the force which tries to keep electric charges balanced. In the context of electrical power sources, voltage is the amount of potential energy available (work to be done) per unit charge, to move electrons through a conductor. Voltage is an expression of potential energy. As such, it represents the possibility or potential for energy release as the electrons move from one "level" to another. Because of this, voltage is always referenced between two points. Consider the water reservoir analogy: Reservoir Drop Drop Location #1 Location #2 Because of the diﬁerence in the height of the drop, there’s potential for much more energy to be released from the reservoir through the piping to location 2 than to location 1. The principle can be intuitively understood in dropping a rock: which results in a more violent impact, a rock dropped from a height of one
foot, or the same rock dropped from a height of one kilometre? Obviously, the drop of greater height results in greater energy released (a more violent impact). We cannot assess the amount of stored energy in a water reservoir simply by measuring the volume of water. Similarly, we cannot predict the severity of a falling rock’s impact simply from knowing the weight of the rock: in both cases we must also consider how far these masses will drop from their initial height. The amount of energy released by allowing a mass to drop is relative to the distance between its starting and ending points. Likewise, the potential energy available for moving electrons from one point to another is relative to those two points. Therefore, voltage is always expressed as a quantity between two points. Interestingly enough, the analogy of a mass potentially "dropping" from one height to another is such an apt model that voltage between two points is sometimes called a voltage drop. 247 Voltage can be generated in many ways. Chemical reactions, radiant energy, and the inuence of magnetism on conductors are a few ways in which voltage may be produced. Respective examples of these three sources of voltage are batteries, solar cells, and generators. For now, we won’t go into detail as to how each of these voltage sources works. The important thing is that we understand how voltage sources can be applied to create electron ow in a circuit. Let’s take the symbol for a chemical battery and build a circuit step by step: 1 - + 2 Battery Any source of voltage, including batteries, have two points for electrical contact. In this case, we have point 1 and point 2 in the above diagram. The horizontal lines of varying length indicate that this is a battery. The horizontal lines further indicate the direction which this battery’s voltage will try to push electrons through a circuit. The horizontal lines in the battery symbol appear separated, and so make it appear as if the battery is unable to serve as a path for electrons to move. This is no cause for concern: in real life, those horizontal lines represent metallic plates immersed in a liquid or semi-solid material that not only conducts electrons, but also generates the voltage to push them along by interacting with the plates. Notice the little "+" and "-" signs to the immediate left of the battery symbol. The negative (-) end of the battery is always the end with the shortest dash, and the positive (+) end of the battery is always the end with the longest dash. By convention, electrons are said to be "negatively" charged, so the negative end of a battery is that end which tries to push electrons out of it. Likewise, the positive end is that end which tries to attract electrons. With the "+" and "-" ends of the battery not connected to anything, there will be voltage between those two points. However, there will be no ow of electrons through the battery, because there is no continuous path for the electrons to move. 248 Water analogy Reservoir Electric Battery No flow 1 - + 2 Battery No flow (once the reservoir has been completely filled) Pump Pond The same principle holds true for the water reservoir and pump analogy: without a return pipe back to the pond, stored energy in the reservoir cannot be released in the form of water ow. Once the reservoir is completely ﬂlled up, no ow can occur, no matter how much pressure the pump may generate. There needs to be a complete path (circuit) for water to ow from the pond, to the reservoir, and back to the pond in order for continuous ow to occur. We can provide such a path for the battery by connecting a piece of wire from one end of the battery to the other. Forming a circuit with a loop of wire, we will initiate a continuous ow of electrons in a clockwise direction: 249 Electric Circuit Battery 1 - + 2 electron flow! Water analogy Reservoir water flow! water flow! Pump Pond So long as the battery continues to produce voltage and the continuity of the electrical path isn’t broken, electrons will continue to ow in the circuit. Following the metaphor of water moving through a pipe, this continuous, uniform ow of electrons through the circuit is called a current. So long as the voltage source keeps "pushing" in the same direction, the electron ow will continue to move in the same direction in the circuit. This single-direction ow of electrons is called a Direct Current, or DC. Because electric current is composed of individual electrons owing in unison through a conductor, just like marbles through a tube or water through a pipe, the amount of ow throughout a single circuit will be the same at any point. If we were to monitor a cross-section of the wire in a single circuit, counting the electrons owing by, we would notice the exact same quantity per unit of time as in any other part of the circuit. The same quantity would be observed regardless 250 of conductor length or conductor diameter. If we break the circuit’s continuity at any point, the electric current will cease in the entire loop. Futhermore, the full voltage produced by the battery will be manifested across the break, between the wire ends that used to be connected: 1 - + 2 no flow! Battery no flow! - (break) + voltage drop Notice the "+" and "-" signs drawn at the ends of the break in the circuit, and how they correspond to the "+" and "-" signs next to the battery’s terminals. These markers indicate the direction that the voltage attempts to push electron ow. Remember that voltage is always relative between two points. Whether a point in a circuit gets labeled with a "+" or a "-" depends on the other point to which it is referenced. Take a look at the following circuit, where each corner of the loop is marked with a number for reference: 1 - + 4 no flow! 2 - Battery (break) + 3 no flow! With the circuit’s continuity broken between points 2 and 3, voltage dropped between points 2 and 3 is "-" for point 2 and "+" for point 3. Now let’s see what happens if we connect points 2 and 3 back together again, but place a break in the circuit between points 3 and 4: 251 1 - + 4 no flow! 2 Battery no flow! + - (break) 3 With the break between 3 and 4, the voltage drop between those two points is "+" for 4 and "-" for 3. Take special note of the fact that point 3’s "sign" is opposite of that in the ﬂrst example, where the break was between points 2 and 3 (where point 3 was labeled "+"). It is impossible for us to say that point 3 in this circuit will always be either "+" or "-", because the sign is not speciﬂc to a single point, but is always relative between two points! 14.4 Resistance The circuit in the previous section is not a very practical one. In fact, it can be quite dangerous to directly connect the poles of a voltage source together with a single piece of wire. This is because the magnitude of electric current may be very large in such a short circuit, and the release of energy very dramatic (usually in the form of heat). Usually, electric circuits are constructed in such a way as to make practical use of that released energy, in as safe a manner as possible. One practical and popular use of electric current is for the operation of electric lighting. The simplest form of electric lamp is a tiny metal "ﬂlament" inside of a clear glass bulb, which glows white-hot ("incandesces") with heat energy when su–cient electric current passes through it. Like the battery, it has two conductive connection points, one for electrons to enter and the other for electrons to exit. Connected to a source of voltage, an electric lamp circuit looks something like this: Battery - + electron flow electron flow Electric lamp (glowing) As the electrons work their way through the thin metal ﬂlament of the lamp, they encounter more opposition to motion than they typically would in a thick piece of wire. This opposition to 252 electric current depends on the type of material, its cross-sectional area, and its temperature. It is technically known as resistance. It serves to limit the amount of current through the circuit with a given amount of voltage supplied by the battery. The "short circuit" where we had nothing but a wire joining one end of the voltage source (battery) to the other had not such a limiting resistance. Interesting Fact: Materials known as conductors have a low resistance, while insulators have a very high one. When electrons move against the opposition of resistance, "friction" is generated. Just like mechanical friction, the friction produced by electrons owing against a resistance manifests itself in the form of heat. The concentrated resistance of a lamp’s ﬂlament results in a relatively large amount of heat energy dissipated at that ﬂlament. This heat energy is enough to cause the ﬂlament to glow white-hot, producing light. The wires connecting the lamp to the battery hardly even get warm while conducting the same amount of current. This is because of their much lower resistance due to their larger cross-section. As in the case of the short circuit, if the continuity of the circuit is broken at any point, electron ow stops throughout the entire circuit. With a lamp in place, this means that it will stop glowing: no flow! - Battery - + no flow! + (break) voltage drop Electric lamp (not glowing) no flow! As before, with no ow of electrons the entire potential (voltage) of the battery is available across the break, waiting for the opportunity of a connection to bridge across that break and permit electron ow again. This condition is known as an open circuit, where a break in the continuity of the circuit prevents current throughout. All it takes is a single break in continuity to "open" a circuit. Once any breaks have been connected once again and the continuity of the circuit re-established, it is known as a closed circuit. What we see here is the basis for switching lamps on and oﬁ by switches. Because any break in a circuit’s continuity results in current stopping throughout the entire circuit,

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