95 21 Physics Term 3 STPM Chapter 21 Sound Waves 8. The table below gives the intensity level of various types of sound. Types of sound Intensity × 10–12 W m–2 Intensity level/dB Launching a rocket 1014 140 1013 130 Jet plane 1012 120 Disco 1011 110 Thunder 1010 100 109 90 Train 108 80 107 70 Shout 106 60 105 50 Talking 104 40 Ticking of a clock 103 30 Whispering 102 20 Breathing 10 10 1 0 Example 10 The intensity level of sound on the surface of a loudspeaker of area 75 cm2 is 95 dB. (a) What is the sound intensity on the surface of the loudspeaker? (b) What is the power of sound emitted by the loudspeaker? Solution: (a) Using intensity level = log10 I I0 (b) Power = Intensity × Area 95 dB = 9.5 B = log10 I 10–12 = (3.16 × 10–3)(75 × 10–4) I = 10–12 × 109.5 = 2.37 × 10–5 W = 3.16 × 10–3 W m–2 Quick Check 5 1. At a distance of 0.50 m from a point source of sound, the intensity level is 60 dB. What is the emissive power of the source? A 1.1 × 10–6 W C 2.0 × 10–6 W B 2.4 × 10–6 W D 3.1 × 10–6 W 2. A point source emits sound uniformly in all directions with a power of 22.0 W. Find the intensity level of sound at a distance of (a) 20.0 m, (b) 50.0 m from the source. [Hearing threshold intensity, I0 = 1.0 × 10–12 W m–2.] 3. The engine of a plane emits sound at a rate 250.0 kW. The accepted safety standard sets the maximum safe sound intensity level at 100 dB. Determine (a) the intensity level of sound from the plane engine at distance of 200.0 m, (b) the minimum distance from the plane for safe sound intensity level. Hence explain the use of ear muffs by airport workers. [Hearing threshold I0 is 1 × 10–12 W m–2.]
96 21 Physics Term 3 STPM Chapter 21 Sound Waves 4. The intensity level in a factory is 100 dB. How much sound energy is incidenting on per square metre of a wall in 1.0 hour? [Hearing threshold I0 is 1 × 10–12 W m–2.] 5. When a note is emitted from a musical instrument, the sound intensity level at a distance of 5.0 m from the instrument is 50 dB. Three of the same musical instrument emit note of the same frequency in phase. What is the sound intensity level at a distance of 5.0 m from the instruments? [Hearing threshold I0 is 1 × 10–12 W m–2.] 21.4 Beats Students should be able to: • use the principle of superposition to explain the formation of beats • use the formula for beat frequency f = f 1 − f 2 ; Learning Outcomes 1. Superposition of two waves of the same frequency, same amplitude but travelling in opposite directions produces a standing wave. 2. What happens when two sound waves of the same amplitude but of slightly different frequencies are superposed? The combined wave produces fl uctuation in the loudness of the sound detected by a stationary observer. 3. The intensity of sound increases, then decreases and increases again, and so on. This phenomenon is called beats. 4. Beats may be explained as the periodic variation in the intensity of a sound produced by the superposition of two notes of slightly different frequencies but of the same amplitude. (a) Audio frequency generator, frequency ƒ1 Displacement, y1 Loudspeaker +A −A 2T t 2T 2T 0 (b) Audio frequency generator, frequency ƒ2 (b) Audio frequency generators, frequency ƒ1 and ƒ2 Displacement, y2 +A −A t T T T 2 0 Resultant displacement, y = y1 + y2 +2A −2A t 0 Loudspeaker Loudspeaker T 2 T Figure 21.13 2009/P1/Q15, 2011/P2/Q10, 2011/P1/Q13, 2013/P3/Q6, 2014/P3/Q6, 2015/P3/Q16 VIDEO Beats
97 21 Physics Term 3 STPM Chapter 21 Sound Waves 5. Figure 21.13(a) shows an audio frequency generator with its frequency f 1 connected to a loudspeaker. The sound emitted has a frequency of f 1 , and its displacement-time graph is as shown in figure 21.13. 6. When the frequency of the audio frequency generator is changed to f 2 which is slightly lower than f 1 , the displacement-time graph is as in figure 21.12(b). 7. Figure 21.13(c) shows two audio frequency generators with frequencies fixed at f 1 and f 2 connected in parallel with the loudspeaker. Beats with displacement-time graph as shown is produced. 8. At time t = 0, y1 = +A, and y2 = – A, the resultant displacement is y = y1 + y2 = 0. Hence there is no sound heard. At time t = T/2, y1 = +A, and y2 = + A, the resultant displacement is y = y1 + y2 = +2A. Hence there is a loud sound heard. At time t = T, y1 = +A, and y2 = – A, the resultant displacement is y = y1 + y2 = 0. Hence there is no sound heard again. 9. The process is repeated. At time t = 3T/2, a loud sound is heard again, and at t = 2T , no sound is heard again. 10. The time interval between two successive loud sounds or two successive silences is the period T of the beats. 11. In the time interval between t = 0 and t = T, there are 8 complete cycles in Figure (a), and there are 7 complete cycles in Figure (b), which is one cycle less. 12. In general, in the time interval T which equals the period of the beats, number of complete cycles of y1 is f 1 T and number of complete cycles of y2 is f 2 T, hence (f 1 T – f2 T) = 1 cycle f 1 – f 2 = 1 T = f Hence beats frequency, f = f 1 – f 2 = difference in the frequencies of y1 and y2 . 13. The beats frequency f which is equal to the difference between the frequencies of the two waves can also be deduced mathematically. 14. Suppose the two waves are represented by the equations y1 = A sin (2π f 1 t) and y2 = A sin (2π f 2 t). Using the principle of superposition of waves, the resultant wave is represented by y = y1 + y2 = A [sin (2πf 1 t) + sin (2πf 2t)] = 2A sin π(f 1 + f 2 )t cos π(f 1 – f 2 )t = A’ sin π(f 1 + f 2 )t where the resultant amplitude A’ = 2A cos π(f 1 – f 2 )t The resultant amplitude A’ = 0 when cos π(f 1 – f 2 )t = 0 π(f 1 – f 2 )t = π 2 rad, 3π 2 rad, 5π 2 rad, … t = 1 2(f 1 – f 2 ) , 3 2(f 1 – f 2 ) , 5 2(f 1 – f 2 ), …
98 21 Physics Term 3 STPM Chapter 21 Sound Waves Time interval between zero amplitude (no sound is heard) T = 3 2(f 1 – f 2 ) – 1 2(f 1 – f 2 ) = 1 (f 1 – f 2 ) Hence beats frequency, f = 1 T = f1 – f 2 15. From the equation for the resultant wave, y = A’ sin π(f 1 + f 2 )t = A’ sin 2π (f 1 + f 2 ) 2 t and comparing with the wave equation y = A sin 2πft, frequency of the resultant wave = (f 1 + f 2 ) 2 16. Beats can be used to tune a musical instrument such as a piano or a guitar. (a) To tune a string of the piano which is out of tune, a tuning fork which has the same frequency as the string is used. (b) Beats will be heard if the frequency of vibration of the string differs from that of the tuning fork. (c) The tension of the piano string is then adjusted until no beats are heard. 17. Beats is used to determine the unknown frequency of a musical note. (a) If f 1 = unknown frequency of the note and f 2 = known frequency of a tuning fork. When both the note and tuning fork are sounded together, the beat frequency f is noted. Hence, f = f 1 – f 2 if f 1 > f 2 .............................. a or f = f 2 – f 1 if f 2 > f 1 .............................. b (b) To decide which of the equations a or b is correct, a little plasticine is stuck on the tuning fork. This causes the frequency f 2 of the tuning fork to be slightly lower. (c) The note and tuning fork are again sounded together. (i) If the beat frequency is increased, then equation a is true. (ii) If the beat frequency is lower, then equation b is true. Example 11 A microphone connected to a cathode ray oscilloscope (CRO) is used to display the waveform produced by two signal generators. The time base of the CRO is fi xed such that a complete sweep across the screen takes 50 ms. The trace obtained on the screen is as shown on the right. Calculate (a) the period of the resultant wave. (b) the beats frequency. (c) the frequencies of the two signal generators. Solution: (a) From the trace shown, the number of complete oscillations is 50 ms = 10 Period, T1 = Time for one complete oscillation = 5 ms
99 21 Physics Term 3 STPM Chapter 21 Sound Waves (b) Period of beats = Time between two successive minimum T = 1 2 × (50 ms) = 25 × 10–3 s Beats frequency, f = 1 T = 1 25 × 10–3 = 40 Hz (c) Frequency of resultant wave = 1 T1 = 1 5 × 10–3 = 200 Hence, f 1 + f 2 2 = 200 f1 + f 2 = 400 .............................. a Also, beat frequency = f 1 – f 2 = 40 .............................. b a + b 2f 1 = 440 f1 = 220 Hz a – b 2f 2 = 360 f2 = 180 Hz Quick Check 6 1. Which wave phenomenon produces beats? A Refraction C Dispersion B Diffraction D Superposition 2. Two sources of sound have frequencies f 1 and f 2 respectively. What is the period of the beats being heard? A f 1 – f 2 C 1 f 1 – f 2 B 1 f 1 + f 2 D f 1 + f 2 2 3. Two loudspeakers are arranged as shown in the fi gure below. Each loudspeaker is connected to a separate frequency signal generator. P In order for beats to be heard at P, the frequency signal generators must be adjusted so that their frequencies A are the same and in phase. B differ slightly and of the same amplitude. C are the same and out of phase by π radians. D differ so that one is twice the other. 4. When a note is sounded together with a tuning fork of frequency 512 Hz, beats of frequency 5 Hz is heard. When the same note is sounded together with another tuning fork of frequency 516 Hz, the beat frequency is less than 5 Hz. What is the frequency of the note? A 511 Hz B 514 Hz C 517 Hz D 521 Hz 5. The frequency of a tuning fork P is 256 Hz. The frequency of another tuning fork Q is unknown. When both P and Q are sounded together, four beats are heard per second. When a small load is attached to Q, the beat frequency is 2 Hz. What is the original frequency of Q?
100 21 Physics Term 3 STPM Chapter 21 Sound Waves 6. Two notes of wavelength 0.500 m and 0.505 m produce 10 beats in 3.0 seconds in a gaseous medium. What is the speed of sound in the gas? 7. The fi gure below shows the resultant wave produced when two waves of the same amplitude but different frequencies, f 1 and f 2 (f 1 > f 2 ) are superposed. (a) Write an expression for T in terms of f 1 and f 2 . (b) What is the ratio of f 1 f 2 ? 21.5 Doppler Effect Students should be able to: • describe the Doppler effect for sound, and use the derived formulae (for source and/or observer moving along the same line). Learning Outcome 1. A stationary observer will notice that the pitch of an approaching siren of an ambulance is higher than when it is moving away. 2. The apparent change in the pitch or frequency of a sound when there is relative motion between the source and the observer is known as Doppler effect. 3. Source approaching stationary observer P S S us us P (a) (b) Figure 21.14 (a) Figure 21.14(a) shows a stationary source of sound S producing waves of frequency f. If the speed of the sound wave is v, then the wavelength λ = v f (b) Figure 21.14(b) shows the same source S moving with a velocity uS towards the stationary observer P. In 1 second, f wavefronts are produced. The distance travelled by the fi rst wavefront in 1.0 s is v. The fth wavefront is produced after the source has moved a distance uS towards the observer P. Hence, the distance between the fi rst and fth wavefront is (v – uS ) (c) The apparent wavelength λ′ = Distance between f wavefronts f = (v – uS ) f The apparent frequency, f ′ = v λ′ f′ = ( v v – uS ) f > f 2008/P1/Q14, 2011/P2/Q3, 2012/P1/Q14, 2013/P3/Q6, 2014/P3/Q5 2015/P3/Q5, 2017/P3/Q18 INFO Doppler Effect
101 21 Physics Term 3 STPM Chapter 21 Sound Waves (d) When the source approaches the stationary observer, the apparent frequency detected by the observer is higher than the actual frequency. This explains for the higher pitch of the approaching siren of ambulance. 4. Source moving away from a stationary observer P S us us Figure 21.15 (a) Figure 21.15 shows a sound source which produces sound of frequency f moving away from a stationary observer with a velocity of uS . (b) In 1.0 s, f wavefronts are produced. Distance moved by the fi rst wavefront is v. The fth wavefront is produced when the source has moved a distance uS from its initial position. Hence, distance between f wavefronts is (v + uS ). (c) The apparent wavelength λ″ = v + uS f The apparent frequency detected by the observer f ″ = v λ″ f ″ = ( v v + uS ) f < f Example 12 The frequency of an ambulance’s siren is 600 Hz. Calculate the apparent frequency heard by a stationary observer (a) when the ambulance approaches him with a speed of 30 m s–1. (b) when the ambulance moves away with a speed of 30 m s–1. (Speed of sound = 330 m s–1) Solution: (a) When the ambulance approaches the observer, the apparent frequency, f ′ = ( v v – uS)f = ( 330 330 – 30) 600 = 660 Hz (b) When the ambulance moves away from the observer, apparent frequency, f ″ = ( v v + uS)f = ( 330 330 + 30) 600 = 550 Hz Exam Tips Apparent frequency increases when source approaches stationary observer, but decreases when source moves away from the observer.
102 21 Physics Term 3 STPM Chapter 21 Sound Waves 5. Observer approaches a stationary source (a) Figure 21.16 shows an observer P approaching a stationary source S with a velocity uO. The wavelength remains unchanged, λ = v f but the velocity of the wave relative to the observer is (v – (–uO)) = (v + uO). (b) Hence, the apparent frequency f ′ = Relative velocity λ = v + uO ( v f ) f′ = ( v + uO v ) f > f 6. Observer moving away from a stationary source P S O Figure 21.17 (a) Figure 21.17 shows the observer P moving away from a stationary source with a velocity of uO. The relative velocity of the wave relative to the observer is (v – uO). (b) Hence, the apparent frequency f ″ = Relative velocity λ = v – uO ( v f ) f ″ = ( v – uO v ) f < f When the observer moves away from the stationary source, the apparent frequency decreases. Example 13 The whistle from a stationary policeman at a junction emits sound of frequency 1 000 Hz. If the speed of sound is 330 m s–1, what is the frequency of the sound heard by a passenger inside a car moving with a speed of 20 m s–1 (a) towards the junction? (b) away from the junction? S O Figure 21.16
103 21 Physics Term 3 STPM Chapter 21 Sound Waves Solution: (a) When the car approaches the source, apparent frequency, f′ = ( v + uO v ) f = ( 330 + 20 330 ) 1 000 = 1 060 Hz (b) When the car moves away from the source, apparent frequency, f ″ = ( v – uO v ) f = ( 330 – 20 330 ) 1 000 = 939 Hz 7. When both the source and observer move, the formula for the apparent frequency is given by f′ = ( v uO v uS ) f where +uO, if the observer approaches the source. – uO, if the observer moves away from the source. – uS , if the source approaches the observer. + uS , if the source moves away from the observer. Example 14 A car is travelling at 25 m s–1 along a straight road when it is overtaken by a police patrol car travelling at 50 m s–1. If the frequency of the patrol car’s siren is 1 000 Hz, what is the frequency of the siren heard by the driver of the car (a) when the patrol car is behind him? (b) when the patrol car is in front of him? (Speed of sound = 330 m s–1) Solution: S O O S (a) When the source is behind the observer, • the observer O is moving away from the source S, hence, use –uO. • the source S is approaching the observer, hence, use –uS in the equation f ′ = ( v uO v uS ) f Apparent frequency, f ′ = ( v – uO v – uS ) f = ( 330 – 25 330 – 50 ) 1 000 = 1 089 Hz Exam Tips • Apparent frequency f increases when observer and source get closer. • Apparent frequency f decreases when separation between observer and source increases.
104 21 Physics Term 3 STPM Chapter 21 Sound Waves (b) S O When the source is in front of the observer, • the observer O is approaching the source S, hence, use +uO. • the source S is moving away from the observer O, hence use +uS . in the equation f′ = ( v uO v uS ) f Apparent frequency, f′ = ( v + uO v + uS ) f = ( 330 + 25 330 + 50 ) 1 000 = 934 Hz Quick Check 7 1. A police car with its siren on travels at 20 m s–1 in the opposite direction of a car travelling at 10 m s–1 along a straight road. The frequency of the siren is 600 Hz. The speed of sound is 340 m s–1. (a) What is the velocity of sound from the police car relative to the car? (b) What is apparent frequency of the siren heard by the driver of the car? 2. A source of sound moves towards a stationary observer with a speed onequarter that of sound. If the frequency from the source is 600 Hz, what is the apparent frequency heard by the observer? 3. A train moves away from an observer standing on the platform with a velocity of 20 m s–1. The siren emitted by the train has a frequency of 1 200 Hz. (a) What is the frequency of the sound heard by the observer? (b) What is the frequency of the sound heard by a passenger in the train? (Speed of sound = 330 m s–1) 4. A police patrol car moves with a speed of 30 m s–1 has its siren on. The sound emitted is at frequency 1 000 Hz. What is the apparent frequency of the sound heard by the driver of a car which is approaching the patrol car with a speed of 30 m s–1? (Speed of sound = 330 m s–1) 5. A car travelling in a straight line at 30 m s–1 approaches a stationary source which emits sound of frequency 5 000 Hz. If the speed of sound is 330 m s–1, calculate (a) apparent frequency received by the car. (b) apparent frequency refl ected from the car. 6. A whistle which emits sound of frequency 2 000 Hz moves away from an observer towards a cliff with a speed of 20 m s–1. (a) What is the frequency of the sound that the observer hears directly from the whistle? (b) What is the frequency of the sound refl ected from the cliff and heard by the observer? (c) What is the frequency of the beats detected by the observer? (d) Do we hear the beat? (Speed of sound = 330 m s–1) 7. A whistle which emits sound of frequency 540 Hz moves in a circle of radius 1.0 m with an angular velocity of 5.0 rad s–1. Calculate (a) the minimum frequency, (b) the maximum frequency, being heard by a distant observer. (Speed of sound = 330 m s–1)
105 21 Physics Term 3 STPM Chapter 21 Sound Waves Important Formulae 1. Propagation of sound wave Displacement y with distance x: y = y0 sin (ωt − kx) Variation of pressure Dp with distance x: Dp = Dp0 sin (ωt − kx – π 2 ) 2. Sources of sound Standing wave in a stretched string, fundamental frequency f 0 = 1 2l T µ Frequencies obtainable: f 0, 2f 0, 3f 0, 4f 0 , … Standing wave in pipe opens at one end fundamental frequency f 0 = v 4l Frequencies obtainable: f 0 , 3f 0 , 5f 0 , 7f 0 , … (odd harmonics only) Standing wave in pipe opens at both ends fundamental frequency f 0 = v 2l Frequencies obtainable: f 0, 2f 0, 3f 0, 4f 0 ,… 3. Intensity level of Sound = 10 log I I0 dB 4. Beat frequency f = f1 – f 2 Frequency of resultant wave = 1 2 (f 1 + f2 ) 5. Doppler Effect: apparent frequency f’ = ( v uO v uS )f STPM PRACTICE 21 1. The fi gure shows standing wave along a string fi xed at both ends. Which statement is not correct? A Vibration of the string produces sound. B There are nodes at the fi xed ends of the string C The standing wave is the result of superposition of waves refl ected from the ends D The sound produced is of the same wavelength as that of the standing wave 2. A siren from a stationary ambulance has a frequency of 600 Hz. A car approaches the ambulance with a speed of 20 m s–1. What is the frequency heard by the driver of the car? [Speed of sound in air = 340 m s–1] A 564 Hz B 567 Hz C 635 Hz D 638 Hz 3. A sound wave of frequency 500 Hz travels in air. What is the separation between successive compressions in air? [Speed of sound in air = 340 m s–1] A 0.34 m C 0.77 m B 0.68 m D 1.47 m 4. Superposition of two notes of different frequencies produces the resultant displacement-time graph shown below. Resultant displacement, y / cm 30 60 t / ms +20 -20 0 What is the beats frequency? A 7.5 Hz C 30.0 Hz B 15.0 Hz D 33.3 Hz 5. A sound source emits sound uniformly in all directions. At a distance r from the source, the intensity is I, and the amplitude of vibration is A. What is the intensity, and amplitude of vibration at a distance of 3r from the source?
106 21 Physics Term 3 STPM Chapter 21 Sound Waves Intensity Amplitude A 1 9 I 1 9 A B 1 9 I 1 3 A C 1 3 I 1 9 A D 1 3 I 1 3 A 6. When sound wave of a fixed frequency travels through air, the variation of displacement, y with distance, x along the direction of propagation is as shown in the graph below. 0 y x Which graph is the correct graph of pressure change, Dp against distance, x? A 0 Δp y x B 0 χ Δp y C 0 Δp y x D 0 Δp y χ 7. A stationary boat emits a sonar wave of frequency f to a school of fish which is moving with a velocity u towards the boat. The sonar wave is then reflected by the school of fish. If v is the velocity of the sonar wave in water, the frequency of the wave reflected by the school of fish is A ( v – u v + u )f C ( v + u v – u )f B ( v v – u )f D ( v + u v )f 8. The variation of displacement y of an air molecule with distance x along the direction of propagation of a sound at an instant is shown by the graph below. 0 y 2 x λ λ − Which graph shows the variation of changing pressure Dp with distance x of the air molecule at the same instant? A 2 0 x λ λ − ∆p B 2 0 x λ λ − ∆p C 2 0 x λ λ − ∆p D 2 0 x λ λ − ∆p 9. Two identical sources of sound S1 and S2 sending out sound waves of frequency 500 Hz to observer P as shown in the diagram below. S1 P S2 If the observer is moving towards S2 at a speed of 1 10 of the speed of sound, what is the beat frequency detected by the observer?
107 21 Physics Term 3 STPM Chapter 21 Sound Waves A 0 Hz C 450 Hz B 100 Hz D 550 Hz 10. A police car travelling with a speed v has its siren on and is approaching a car which is travelling in the opposite direction with a speed of 60 km h-1. The frequency of the siren is 1.0 kHz, but the frequency heard by the driver of the car is 1.1 kHz. What is the speed v of the police car? [Speed of sound in air is 330 m s-1.] A 1554 km h-1 C 83 km h-1 B 54 km h-1 D 93 km h-1 11. A car approaches a stationary police car which has its siren on. The frequency of the siren is f. Which statement about the frequency heard, and the intensity of sound is correct? A Frequency heard is f, intensity increases. B Frequency heard is lower than f, intensity increases. C Frequency heard is higher than f, intensity increases. D Frequency heard is higher than f, intensity remains constant. 12. Which statement is the correct comparison between sound waves and electromagnetic waves? Sound waves Electromagnetic waves A Cannot be plane- Can be planepolarised polarised B Produce Doppler Do not produce effect Doppler effect C Require a Require a material medium material medium D Transverse waves Longitudinal waves 13. Which of the followings about a standing wave is true? A Between successive nodes, oscillations of particles are not in phase. B Between successive nodes, particles have different amplitudes. C Neighbouring particles have the same maximum velocity. D Neighbouring particles have different frequencies of oscillation. 14. The successive resonant frequencies of a pipe are 750 Hz, 1 250 Hz and 1 750 Hz. Which of the following deduction about the pipe and the fundamental frequency is correct? A Pipe closes at one end, fundamental frequency is 250 Hz. B Pipe opens at both ends, fundamental frequency is 750 Hz. C Pipe closes at one end, fundamental frequency is 750 Hz. D Pipe opens at both ends, fundamental frequency is 250 Hz. 15. A resonant tube of length 100 cm is filled with water. A tuning fork at frequency 512 Hz is held at the opened end of the tube, and water slowly drained off from the other end. If the speed of sound in air is 340 m s–1, how many times does resonance occur? A 1 C 3 B 2 D 4 16. What is the intensity of sound which has an intensity level of 100 dB? A 10–11 W m–2 C 10–2 W m–2 B 10–10 W m–2 D 10–1 W m–2 17. The length of a pipe opened at both ends is 35.5 cm. If the speed of sound is 330 m s–1 and each end of the pipe has an end correction of 1.0 cm, what is the frequency of the first overtone? A 220 Hz C 660 Hz B 440 Hz D 880 Hz 18. An observer moves away at a constant speed from a source which emits sound of a constant frequency f and wavelength λ. Which of the following is true about the apparent frequency and wavelength of the sound being heard? Apparent frequency Wavelength A Lower than f Same as λ B Lower than f Smaller than λ C Higher than f Smaller than λ D Higher than f Bigger than λ 19. A standing wave is represented by the equation y = 5 sin π 8 x cos 50 πt, where x and y are in cm, and t in second. What is the distance between successive antinodes? A 4.0 cm C 12.0 cm B 8.0 cm D 25.0 cm
108 21 Physics Term 3 STPM Chapter 21 Sound Waves 20. Three successive overtones emitted by a pipe are at frequencies 210 Hz, 350 Hz and 490 Hz. What is the fundamental frequency? A 35 Hz C 140 Hz B 70 Hz D 210 Hz 21. A pipe open at one end is 23.0 cm long. Draw diagrams to show the two lowest mode of vibration of the standing wave set up in the pipe. Calculate the frequencies of the standing waves. [Speed of sound in air = 340 m s–1] 22. (a) When someone blows across the open end of a test tube, a standing wave is formed in the test tube. Explain why a sound is heard even though the standing wave is not propagated outside the test tube. (b) Compare the fundamental frequencies of the notes produced by a pipe closed at one end, and a pipe of the same length but opened at both ends. 23. An ambulance with its siren emitting a sound of frequency 600.0 Hz moves with a speed of 30.0 ms–1. [Assume that the speed of sound is 345 m s–1.] (a) Calculate the frequency of the siren heard by a stationary pedestrian who is behind the ambulance. (b) Calculate the frequency of the siren heard by a driver of a car moving with a speed of 20.0 m s–1 away from the ambulance in the opposite direction. 24. (a) Explain the term Doppler effect. (b) The siren of a police car emits sound wave which is represented by the equation y = (5.0 × 10–3) sin (5200t – kx) where x and y are in m and t in seconds. The speed of sound in air is 340 m s-1. (i) What is the frequency of the sound? (ii) What is its wavelength? (iii) Find the value of k. (c) The police car with its siren on approaches a tall building with a speed of 20 m s–1. (i) What is the frequency reflected from the building? (ii) What is the frequency of the reflected sound incident on the approaching police car? (iii) Find the frequency of the beat detected by the driver of the police car? 25. A radio station transmits a radio wave of frequency 100 MHz. The radio wave is detected by a receiver via two different paths, a direct path and a reflected path from a nearby wall. The distance travelled by the radio wave for the direct path and the reflected path are 1500 m and 1512 m respectively. Explain whether the interference which occurs at the receiver is constructive or destructive interference. 26. (a) Two vibrating strings P and Q produce notes of frequencies 256 Hz and 258 Hz respectively. What is the frequency of the beats produced? (b) A string which is fixed at the ends has a length of 65.0 cm. The fundamental note produced has a frequency of 258 Hz. (i) Calculate the speed of transverse waves in the string. (ii) The tension in the string is then increased, discuss any change in the frequency of the note produced. 27. (a) Explain how standing wave is produced from two progressive waves. (b) A standing wave along a stretched string is represented by the equation y = 4.0 cos 0.20πx sin 60πt, where x and y are in cm and t in seconds. (i) Find the distance between successive nodes. (ii) Determine the speed of progressive waves in the string. (c) A vibrating tuning fork of frequency 512 Hz is held above a tube filled with water. Water in the tube is slowly run off from the bottom of tube. Successive resonance is produced when the lengths of the air column in the tube are 15.5 cm and 47.8 cm. (i) Draw diagrams to show the standing waves in the air column at resonance. (ii) Calculate the speed of sound in the air column and the end correction.
109 21 Physics Term 3 STPM Chapter 21 Sound Waves 28. (a) What is meant by (i) intensity of sound (ii) intensity level of sound? (b) Sound is emitted from a jet engine at a rate of 300 kW uniformly in all directions. The accepted safe sound intensity level is 100 dB. (i) What is the sound intensity level at a distance of 120 m from the jet engine? (ii) What is the minimum distance from the jet engine for the sound intensity level to be safe? [Threshold of hearing is 1.0 × 10–12 W m-2] (c) A guitar string of length 0.55 m has a fundamental frequency of 146 Hz. The tension in the string is 2.50 N. (i) Find the mass per metre length of the string. (ii) What is the frequency of the first overtone? (iii) When this string is sounded together with another string of the same length and material but thicker, six beats are heard every 2.0 s. What is the fundamental frequency of the second string if the strings emit their respective fundamental note? 29. Two sound waves have frequencies 880 Hz and 884 Hz. The intensity of each wave at a certain point is 6.0 × 10–9 W m–2. (a) What is the frequency of the beat produced? (b) Calculate the maximum intensity level of the beats. (Minimum sound intensity which can be detected is 1.0 × 10–12 W m–2) 30. A pipe opened at both ends has fundamental frequency 550 Hz. If the speed of sound in air is 330 m s–1, calculate (a) length of the pipe (b) the positions of the antinodes along the pipe for the second overtone. 31. (a) Describe the mechanism of sound propagation in a solid. (b) Differentiate between intensity of sound and intensity level of sound. The intensity level of sound at a point P which is 50 m from a point source is 60 dB. Calculate (i) the intensity of sound at the point P. (ii) the intensity level at another point Q which is 100 m from the source of sound. [Threshold of hearing is 1.0 × 10–12 W m–2] 32. A vibrator produces standing waves along a string as shown in the figure. If the frequency of the vibrator is 50 Hz, what is the speed of transverse wave along the string? 33. A suspension bridge will be built across a river where the wind is found to gust at 4.0 s interval. The speed of transverse wave along the span of the bridge is estimated to be 320 m s–1. What is the length of the bridge if it resonates at its fundamental frequency? 34. Sketch the four lowest modes of vibration of a wire stretched between two fixed points. If the fundamental frequency of the string is 135 Hz, what are the other three resonant frequencies? 35. A guitar player changes the frequency of the note produced by a guitar string by pressing his fingers at different points along the string. The fundamental frequency of a string is 264 Hz. What are the frequencies of the fundamental note, first overtone and second overtone if the guitar player presses the wire 1 4 of the way from one end? 36. Calculate the intensity of sound if the intensity level is (a) 50 dB, (b) 112 dB. 37. The intensity level at a distance of 20 m from a loudspeaker is 102 dB. If the sound from the loudspeaker radiates uniformly over a hemisphere in front of it, what is the power of the sound transmitted from the speaker? 38. (a) State the necessary conditions for the establishment of well-defined standing wave. (b) A vibrating loudspeaker is held at the opened end of a resonant tube and the
110 21 Physics Term 3 STPM Chapter 21 Sound Waves length of the air column in the tube is increased until resonant occurs. Explain resonance in the air column with reference to the conditions stated in (a). (c) Such a tube resonates with the same frequency of sound from the loudspeaker for a number of different lengths of the air column. (i) Explain this with the aid of suitable diagrams. (ii) Deduce a general expression for these lengths l n in terms of f, the frequency of the sound and v, the speed of sound. (d) A loudspeaker which emits sound of constant frequency resonates with a tube which is closed at one end. The two shortest resonant lengths l 1 and l 2 are measured at two different temperatures. The results are as shown below. Temperature/°C l 1 /cm l 2 /cm 0 31.0 97.0 20.0 32.2 100.6 (i) Calculate the end correction of the tube. (ii) Confirm, or otherwise, that the speed of sound is proportional to the square root of the absolute temperature. 39. (a) Explain the meaning of Doppler effect. (b) In the fi gure on the right, a source of sound of frequency f moves with constant speed u towards the right. P, Q and R are the positions of the source at time t = 0, t = 1 s, and t = 2 s. W1 , W2 are the wavefronts from the points P and Q at time t = 2 s. If v is the speed of sound in air, (i) what is the radius of the wavefront W1 and W2 ? P Q R O W W (ii) what is the distance between the points P and Q? (iii) hence, fi nd the expression for d, the distance between W1 and W2 along the line PQRO in terms of v and u. (iv) how many wavefronts are there in the distance d? (v) what is the apparent frequency f ′ heard by an observer at O? What is the phenomenon represented by the expression for f′? 40. (a) (i) A vibrator at a point O produces circular wavefronts. Explain why the amplitude of the wave decreases as the distance of the wavefront from O increases. (ii) At a distance r from O, the amplitude of the wavefront is a. Calculate the amplitude of the wavefront at a distance 2r from O. Explain your answer. (b) The intensity level of a sound is 50 dB. If the intensity of the sound is doubled, what is the new intensity level? 41. The fi gure shows two sound sources close to each other and emit sound of constant amplitude A and frequency 500 Hz. P is a stationary observer close to the two sources. S S P (a) Sketch a graph to show the variation of displacement with time for the resultant wave at P. (b) If P moves away from S1 and S2 at 20 m s–1, determine the frequency and wavelength of the sound heard by P. (c) The frequency of S2 is changed to 510 Hz, but the frequency of S1 remains at 500 Hz. Explain quantitatively the sound heard (i) when P is stationary. (ii) when P moves at 20 m s–1 away from S 1 and S 2 . S k e t c h t h e displacement-time graph for the resultant wave for (c)(i). (Speed of sound in air = 330 m s–1)
111 21 Physics Term 3 STPM Chapter 21 Sound Waves 1 1. A 2. C 3. D 4. A 5. D 6. D 7. f 0 = c 2L f0 = c L 8. (a) f = v λ λ = 0.80 m = 375 Hz (b) 125 Hz; 250 Hz 9. (a) B : Stationary C : Maximum amplitude D : In phase with C, but smaller amplitude E : Same amplitude as C, but antiphase. (b) λ = L 2 2 1. 125 Hz, 375 Hz 2. Depth = λ 4 = 12.5 m 3. 3f, 5f, 7f, ... 4. 475 m s–1 (λ = 0.19 m) 5. Standing waves set up between X and Y. Intensity high at antinodes, low at nodes. 20 Hz (λ = 0.3 m) 3 1. λ = 32 cm 39 7 ≈ 5 c = 1 cm 7 + c = λ 4 l 1 + c = 3 4 λ 39 + c = 5 4 λ l1 = 23 cm l2 + c = 7 4 λ , l 2 = 55 cm 2. (a) 330 m s–1 0.30 + c = λ 4 (b) 3.0 cm 0.96 + c = 3 4 λ 3. 321 m s–1 4. 0.125 m, 0.375 m, 0.625 m 5. (a) Refer to page 87 (b) Displacement-distance graph. 6. 340 m s–1, f = v 4l , plot graph of f against 1 l . 4 1. D 2. B 3. D 4. B 5. B 6. Opened pipe : 0.55 m Closed pipe : 0.413 m 7. (a) 100 Hz (b) (i) no change (ii) f increases because v increases 8. 367 Hz, 734 Hz, 1 101 Hz, 1 468 Hz 5 1. D: 6.0 dB = 10 and lg( I I0 ) I I0 = 106.0 P = (4πr2 )I 2. (a) Intensity I = Power Area = P 4πr2 When r = 20.0 m, I = 22.0 4π(20.0)2 W m–2 = 4.377 × 10-3 W m–2 Intensity level = 10 log10 I I0 dB = 10 log10 4.377 × 10–3 1.0 × 10–12 dB = 96.4 dB (b) When r = 50.0 m, I = 22.0 4π(50.0)2 W m–2 = 7.003 × 10–4 W m-2 Intensity level = 10 log10 I I0 dB = 10 log10 7.003 × 10–4 1.0 × 10–12 dB = 88.5 dB 3. (a) Intensity of sound from the jet engine at a distance of 200 m, I = 250.0 × 103 4π(200)2 W m–2 Intensity level of sound = 10 log10 (I I0 ) dB = 10 log10 ( 250.0 × 103 (4π × 2002 )(1.0 × 10–2)) dB = 117 dB (b) 100.0 dB = 10 log10 (I I0 ) dB I I0 = 1010 I = (1010)(10-12) W m–2 = 250.0 × 103 (4π × r2 ) Minimum distance, r = 250.0 × 103 4π(10–2) m = 1.41 km ANSWERS
112 21 Physics Term 3 STPM Chapter 21 Sound Waves Airport workers wear ear muff to prevent damage to the ear due to the high sound intensity level from planes engines. 4. 100 dB = 10 log10 I I0 dB 10 = log10 I 10.0 × 10–12 I = 1010 (1.0 × 10–12) W m–2 = 1.0 × 10–2 W m-2 Energy incident on a wall of 1 m2 in 1 hour = (3600)( 1.0 × 10–2) J = 36.0 J 5. 50 dB = 10 log10 I I0 dB I = 105 (1.0 × 10-12) W m-2 = 1.0 × 10–7 W m–2 ∝ A2 (A = amplitude) Three sources of sound are in phase. Constructive superposition occurs. Amplitude of resultant sound wave = 3A Resultant intensity, I’ ∝ (3A)2 = 9A2 I’ = 9(1.0 × 10–7)W m–2 Intensity level = 10 log10 9.0 × 10–7 1.0 × 10–12 dB = 59.5 dB 6 1. D 2. C 3. B 4. C 5. 260 Hz 6. 168 m s–1 f = v λ 7. (a) T = 1 f 1 – f 2 (b) f 1 f 2 = 6 5 f 1 + f 2 2 = 5.5 T 7 1. (a) 340 – (–10) = 350 m s–1 (b) f ' = ( c + uO c – uS ) f = 636 Hz 2. 800 Hz f ′= ( v v – uS )f 3. (a) 1 131 Hz (b) 1 200 Hz f ′ = ( v v + uS )f 4. 1 200 Hz f ′ = ( v + uO v – uS ) f 5. (a) 5 455 Hz f ′ = ( v + uO v ) f (b) 6 001 Hz f ″= ( v v – uS ) f′ 6. (a) 1 886 Hz f ′ = ( v v + uS ) f (b) 2 129 Hz f ″ = ( v v – uS )f (c) 243 Hz f = f ″ – f ′ (d) Can, but unable to hear the maximum distinctly. 7. (a) 532 Hz us = w f ′ = ( v v + uS ) f (b) 548 Hz f ″ = ( v v – uS ) f STPM Practice 21 1. D: Same frequency but different wavelength 2. C: f = ( v + u0 v )f = ( 340 + 20 340 )(600) Hz = 635 Hz 3. B: Separation between compressions = λ = v f = 340 500 m = 0.68 m 4. D: Beats frequency = 1 30 × 10–3 Hz = 33.3 Hz 5. B: Intensity, I ∝ A2 ∝ 1 r2 , I ∝ A2 1 ∝ 1 (3r) 2 I= 1 9 I and A1 = 1 3 A 6. B: Dp lags behind y by ¼-cycle 7. C: Frequency of wave detected by fish, f 1 = ( v + u v )f For the reflected wave, the shoal acts as source, frequency reflected, f 2 = ( v v – u)f 1 = ( v v – u) ( v + u v )f = ( v + u v – u)f 8. D: Maximum pressure change at x = λ 2 9. B: Apparent frequency from S1 , f 1 = ( c – v0 c )(500 Hz) Apparent frequency from S2 , f 2 = ( c + v0 c )(500 Hz) Beat frequency = f 2 – f 1 = 2( 0.10c c )(500 Hz) = 100 Hz 10. B: f’ = ( c + vo c – vs )f = 1100 Hz v = vs = 15 m s-1 = 54 km h-1 11. C 12. A 13. B 14. A 15. C 16. C 17. D 18. A 19. B 20. B
113 21 Physics Term 3 STPM Chapter 21 Sound Waves 21. 23.0 cm Fundamental frequency, f 0 = v λ = 340 (4 × 0.230) = 370 Hz First overtone, f 1 = 3f 0 = 1 110 Hz 22. (a) The standing wave in the test tube causes the air around it to vibrate with the same frequency. Sound wave then propagates from the test tube. (b) For a pipe closed at one end, length l = λ 4 Fundamental frequency, f 0 = v λ = v 4l For a pipe open at both ends, length l = λ 2 Fundamental frequency, f 0 = v λ = v 2l = twice fundamental frequency of pipe closed at one end. 23. (a) Frequency heard, f’ = ( v v + us )f = ( 345 345 + 30.0)(600) Hz = 552 Hz (b) Frequency heard, f’ = ( v – u0 v + us )f = ( 345 – 20.0 345 + 30.0)(600) Hz = 520 Hz 24. (a) Doppler effect is the apparent change in the frequency of sound heard by an observer when there is relative motion between the observer and the source. (b) (i) Compare y = (5.0 × 10–3) sin (5200t – kx) with y = A sin (2 πft – 2π λ x) 2πf = 5200, frequency f = 828 Hz (ii) Wavelength, λ = v f = 340 828 m = 0.411 m (iii) k = 2π λ = 2π 0.411 rad m–1 = 15.3 rad s–1 (c) (i) Frequency reflected, f 1 = frequency received by the building = ( v v – us )f = ( 340 340 – 20)(828) Hz = 880 Hz (ii) uo = 20 m s–1, f 2 = ( v + u0 v )f 1 = ( 340 + 20 340 ) (880) Hz = 932 Hz (iii) Beat frequency = f 2 – f = (932 – 828) Hz = 104 Hz 25. Wavelength, λ = c f = 300 × 108 100 × 106 m = 3.0 m There is a phase shift that is equivalent to a path difference of λ 2 when radio wave is reflected. Total path difference between the reflected and direct paths = (1512 – 1500)m + λ 2 = 4λ + λ 2 = 4 1 2 λ Hence there is destructive interference. 26. (a) 2 Hz (b) (i) λ 2 = 0.65 m v = f0 λ = 335 m s–1 (ii) f 0 = v 2l T µ T increases, f 0 increases. 27. (a) Refer to page 60 (b) (i) At nodes, 4.0 cos 0.20πx = 0 x = 2.5 cm, 7.5 cm Distance between nodes = (7.5 – 2.5) cm = 5.0 cm (ii) 2πf = 60π v = f λ= (30)(2 × 5.0) = 300 cm s–1 (c) (i) 15.5 cm 47.8 cm c c λ = 2(47.8 – 15.5) = 64.6 cm v = f λ= 331 m s–1 (ii) 15.5 + c = 64.6 4 cm c = 0.65 cm 28. (a) (i) Intensity of sound = sound energy per second incident normally on a unit area. (ii) Intensity level of sound = 10 log I I0 dB, where I = intensity, I0 = threshold intensity (1 × 10–12 W m–2) (b) (i) Intensity I = P 4πr2 Intensity level = 10 log I I0 dB
114 21 Physics Term 3 STPM Chapter 21 Sound Waves = 10 log P I 0 (4πr2 ) dB = 10 log 300 × 103 (10–12)(4p(120)2 ) dB = 122 dB (ii) 100 dB = 10 log I 10–12 dB I = 1 × 10–2 W m–2 = P (4πr2 ) Distance, r = 300 × 103 4π(1 × 10–2) m = 1.55 km (c) (i) Fundamental frequency, f 0 = 1 2l T µ Mass m–1, µ= T 4l 2 f 0 = 2.50 4(0.55)2 (146) kg m–1 = 0.0142 kg m-1 (ii) Frequency of first overtone = 2f 0 = 2(146) Hz = 292 Hz (iii) Second string is thicker, hence value of µ is bigger. Fundamental frequency f 0 < f0 = 146 Hz Beat frequency = f 0 – f 0 = 6.0 2 f 0 = (146 – 3) Hz = 143 Hz 29. (a) 4 Hz f = f 1 – f 2 (b) 43.5 dB amplitude A′ = A + A I ∝ A2 l′ = 4I 30. (a) 0.30 m l = λ 0 2 (b) A A A 0.10 m 0.10 m 0.10 m A 31. (i) 60 dB = 10 log( I 1.0 × 10–12 ) I = 1.0 × 10–6 W m–2 (ii) I' = ( 50 100 )2 (1.0 × 10–6 ) W m–2 = 2.5 × 10–7 W m–2 Intensity level = 10 log( 2.5 × 10–7 1.0 × 10–12 ) dB = 54 dB 32. 25 m s–1 v = f λ λ = 0.50 m 33. l = λ 0 2 = 640 m λ = v f , f = 1 4 Hz 34. f 0 = 135 Hz f 1 = 270 Hz f 2 = 405 Hz f 3 = 540 Hz 35. 352 Hz f 0 = v 2l , l′ = 3 4 l 704 Hz 1 056 Hz 36. (a) 10–7 W m–2 (b) 0.158 W m–2 37. 39.7 W Power = Intensity × Area 38. (a) Superposition of two progressive waves of the same frequency, same amplitude but travelling in opposite directions. (b) (ii) l n = v f (n + 1) 4 , n = 0, 1, 2, 3, … (c) Refer to page 87 (d) (i) 2.0 cm (ii) v20 v0 = 1.036 = 293 273 39. (a) (i) 2v, v (iv) f (ii) u (v) f′ = ( v v – u ) f (iii) d = 2v – (v + u) = v – u 40. (a) (i) Circumference of wavefront increases. The same energy is distributed over a longer length. (ii) A1 = A 2 I ∝ E 2πr ∝ A2 I1 ∝ E 2π (2r) ∝ A1 2 (b) 5.30 dB I1 = 10–7 W m–2 I2 = 2 × 10–7 W m–2 41. (a) Displacement 0 Time 2A – 2A (b) 470 Hz f ′= ( v – uO v ) f λ = 0.66 m λ = v f (c) (i) Beats frequency = 10 Hz (ii) Beats frequency f 1 ′ = 470 Hz, = 9 Hz f 2 ′ = 479 Hz
CHAPTER GEOMETRICAL OPTICS 22 Concept Map Spherical Mirrors F C f r C F f r r = 2f 1 f = 1 u + 1 v m = v u Refraction At Curved Surfaces n1 u r v O C n2 I n1 u + n2 v = |n1 – n2 | r Thin lenses Lensmaker’s Equation f F f F 1 f = 1 u + 1 v r1 r2 n2 n1 n1 1 f = ( n2 n1 – 1)( 1 r1 + 1 r2 ) 1 f = (n – 1)( 1 r1 + 1 r2 ) Geometrical Optics 115 Bilingual Keywords Centre of curvature: Pusat kelengkungan Concave mirror: Cermin cekung Convex mirror: Cermin cembung Focus: Fokus Linear magnifi cation: Pembesaran Linear Pole: Kutub Principal axis: Paksi utama Spherical mirror: Cermin sfera
Physics Term 3 STPM Chapter 22 Geometrical Optics 116 22 22.1 Spherical Mirrors Learning Outcomes Students should be able to: • use the relationship f = r 2 for spherical mirrors • draw ray diagrams to show the formation of images by concave mirrors and convex mirrors • use the formula 1 u + 1 v = 1 f for spherical mirrors 1. Laws of refl ection (a) The incident ray, the refl ected ray, and the normal, all lie on the same plane. (b) Angle of refl ection = angle of incidence ∠r = ∠i Normal Plane mirror Incident ray Reflected ray i r Figure 22.1 2. Refl ection at curve mirrors obeys the laws of refl ection. 3. Figure 22.2 (a) shows a concave mirror and Figure 22.2 (b) shows a convex mirror. These are spherical mirrors, being sections of a spherical glass shell centre at C with the relevant surfaces silvered. C F P f r P F C f r (a) Concave mirror (b) Convex mirror Figure 22.2 4. Terms related to curve spherical mirrors. (a) Centre of curvature C, of a curve mirror is the centre of the sphere of which the mirror is part. (b) Radius of curvature, r is the radius of the sphere of which the mirror is part. (c) Pole, P of a curve mirror is the point at the centre of the mirror. (d) Principal axis is the straight line through the centre of curvature C and pole P of the mirror. INTRODUCTION 1. In this chapter you will learn about refl ection of light and its application in curve mirrors, and refraction of light and its application in lenses. 2012/P3/Q7, 2014/P3/Q19(a), 2015/P3/Q7, 2016/P3/Q19, 2017/P3/Q7 INFO Uses of Concave Mirror and the Convex Mirror
Physics Term 3 STPM Chapter 22 Geometrical Optics 117 22 (e) Focus, F of a concave mirror is the point on the principal axis where rays parallel and close to the principal axis pass after refl ection. The focus F of a convex mirror is the point on the principal axis where rays parallel to the principal axis appear to diverge from after refl ection. Focal length, f of a curve mirror is the distance between the pole P and the focus F. The focal length of a concave mirror is assumed positive, and that of a convex mirror is negative. (f) Focal plane is the plane at the focus and perpendicular to the principal axis. Rays which are not parallel to the principal axis of a concave mirror are brought to a focus on the focal plane. 5. Relation between focal length f and radius of curvature r. A B P C F f θ θ θ r A B P F C f θ θ θ θ r (a) Concave mirror (b) Convex mirror Figure 22.3 Figure 22.3 (a) shows a ray AB close and parallel to the principal axis of a concave mirror. The normal to the point of incident B passes through the centre of curvature C of the mirror. The ray is refl ected and passes through the focus F of the mirror such that angle of refl ection = angle of incidence ∠ ABC = ∠ CBF = θ Also ∠ ABC = ∠ BCF (alternate angles) Hence ∠ CBF = ∠ BCF Triangle BCF is an isosceles triangle Since the point B is close to P, FP = FB = FC Hence FP = 1 2 PC (PC = r) f = 1 2 r The same relationship between f and r can be deduced for a convex mirror, using Figure 22.3 (b). Example 1 The radius of curvature of a spherical concave mirror is 20.0 cm, what is the focal length of the mirror? Solution: Using f = 1 2 r Focal length, f = 1 2 × 20.0 = 10.0 cm
Physics Term 3 STPM Chapter 22 Geometrical Optics 118 22 Formation of Images by Spherical Mirrors 1. By drawing a ray diagram accurately to scale, we can determine the position and characteristics of images formed by a curve mirror. 2. By drawing two out of the three rays, shown in Figure 22.4 from the same point of the object we can locate the position of the image. In a ray diagram, the curve mirror is represent by a straight line perpendicular to the principal axis. Use graph paper for your ray diagrams. O I F C fi fi ff ffl ffl Convex mirror O C F I fi fi ff ff ffl ffl Concave mirror (a) (b) Figure 22.4 Concave mirror Convex mirror Ray 1: Parallel to principal axis after reflection passes focus F. Ray 1: Parallel to principal axis, after reflection, diverges from focus F. Ray 2: Passes centre of curvature C, reflected back along the same path. Ray 2: Directed towards centre of curvature C, reflected back along the same path. Ray 3: Passes focus, F reflected parallel to principal axis. Ray 3: Directed towards focus F, reflected parallel to principal axis. 3. Images formed by a concave mirror The following ray diagrams shows the formation of images by a concave mirror of focal length 10.0 cm for various object distance u. (a) O F I Concave mirror Object distance u = 30.0 cm > 2f. Image real, diminished and inverted. Image distance, v = 15.0 cm. A real image is formed when the reflected rays intersect. A real image can be formed on a screen. Figure 22.5
Physics Term 3 STPM Chapter 22 Geometrical Optics 119 22 (b) O F I Concave mirror Object distance u = 20.0 cm = 2f = r radius of curvature. Image is real, same size as object and inverted. Image distance v = 2f which is the same as the object distance. Hence the object and image are at the same place. Figure 22.6 (c) O F I Concave mirror Object distance u = 15.0 cm, which is between f and 2f. Image is real, magnified and inverted. Image distance v = 30.0 cm. Figure 22.7 (d) C I F Concave mirror Image at infinity Object distance u = f. The rays from the object after reflection are parallel. Hence the image is inverted and at infinity. Figure 22.8 (e) C O F Concave mirror Infinity Conversely, when the object is at infinity, the image is formed at the focus. Figure 22.9 Exam Tips The image characteristics, real and inverted form a pair. The other pair of image characteristics is: virtual and upright.
Physics Term 3 STPM Chapter 22 Geometrical Optics 120 22 (f) C F Concave mirror O I Object distance u = 7.5 cm < f. Image is virtual, magnified and upright. Image distance v = 30.0 cm behind the mirror. A virtual image cannot be formed on a screen. Figure 22.10 Note that rays cannot pass through a mirror. Lines behind the mirror represent virtual rays and are drawn as broken lines. A virtual image is also drawn with broken line. Concave minors in this arrangement are used as shaving-mirrors, make-up-mirrors and mirrors used by dentists. 4. Image formed by a convex mirror The image formed by a convex mirror is always virtual, upright and diminished for a real object in front of the mirror, irrespective of the object distance (Figure 22.11). O Convex mirror I F C Figure 22.11 5. Convex mirrors are used as rear-mirrors or wing-mirrors of a motor vehicle. As the object approaches from the back, the driver sees an upright, diminished image that becomes bigger. 6. The main advantage of the convex mirror is its wide field of view as shown in Figure 22.12. Normal Normal Centre of curvature, C Driver's eye Wide field of view Convex mirror Figure 22.12
Physics Term 3 STPM Chapter 22 Geometrical Optics 121 22 Spherical Mirror Formula I F O v u F O I v u (a) Concave mirror (b) Convex mirror Figure 22.13 1. Figure 22.13 shows an object O at a distance u in front of (a) a concave spherical mirror, and (b) a convex spherical mirror. The image is formed at I, which is at a distance of v from the mirrors. 2. The object distance and the image distance v are related to the focal length f of the spherical mirrors by the spherical mirror formula: 1 u + 1 v = 1 f 3. Real is positive sign convention for spherical curve mirror formula 1 u + 1 v = 1 f • u is positive if the object is real u is negative if the object is virtual • v is positive if the image is real v is negative if the image is virtual • f is positive for concave mirror f is negative for convex mirror 4. An object O is consider real, if rays diverges from it as shown in Figure 22.13. According to the real is positive convention, the object distance is positive. u Real image, I Concave mirror Virtual object, O Figure 22.14 5. Figure 22.14 shows rays from an object incident on a concave mirror. These rays converges at a point O behind the mirror. The point O is considered as the virtual object, and its distance u from the mirror is given as negative value according to the real is positive convention. 6. No rays originate from a virtual object. 7. In figure 22.14, the rays converges at I after reflection. The image at I is real.
Physics Term 3 STPM Chapter 22 Geometrical Optics 122 22 Example 2 A concave mirror has radius of curvature 20.0 cm. An object is placed 5.0 cm from the mirror. Find the position of the image. Solution: Focal length, f = r 2 = + 20.0 2 cm = + 10.0 cm Object distance, u = + 5.0 cm Using 1 u + 1 v = 1 f 1 v = 1 10 – 1 5 = – 1 10 Image distance, v = –10.0 cm The image is virtual and 10.0 cm behind the mirror. Example 3 10.0 cm P The fi gure shows rays incident on a concave mirror of focal length 20.0 cm. Sketch a diagram to show the path of the rays after refl ection. Mark clearly suitable distances on your diagram. Solution: f = +20.0 cm Since the rays converges at P, P is a virtual object, and u = –10.0 cm. Using 1 u + 1 v = 1 f 1 v = 1 20.0 – ( 1 –10.0) v = 6.67 cm Exam Tips In solving problems involving 1 u + 1 v = 1 f , when writing down the values of u, v, or f, pause to think of the sign + or – for each value of u, v, or f. P 6.67 cm 10.0 cm I
Physics Term 3 STPM Chapter 22 Geometrical Optics 123 22 Example 4 A convex mirror of focal length 10.0 cm produces a real image 5.0 cm from the mirror. Calculate the object distance and draw a ray diagram to show the formation of the image. Solution: Using the real–is–positive sign convention, focal length of convex mirror, f = –10.0 cm. Since the image is real, image distance, v = +5.0 cm Using 1 u + 1 v = 1 f 1 u = 1 –10.0 – 1 5.0 Object distance, u = –3.33 cm. Since the object distance is negative, the object is virtual. Linear Magnifi cation 2008/P1/Q39, 2009/P1/Q40 θ θ u v O P B A I ho hi θ θ θ u v O P B A I ho hi (a) Concave mirror (b) Convex mirror Figure 22.15 1. Figure 22.15 (a) and Figure 22.15 (b) shows a ray AP from the point A of an object incident at the pole P of a concave mirror and convex mirror respectively. 2. The ray is refl ected such that the angle of refl ection is equal to the angle of incidence θ. 3. For the concave mirror the image IB is real, whereas for the convex mirror the image IB is virtual. tan ∠ IPB = tan ∠ APO IB IP = OA OP IB OA = IP OP hi ho = v, image distance u, object distance ............................. ① 4. The linear magnifi cation of an image m = height of image, hi height of object, ho From equation ①, m = hi ho = v u O 5.0 cm 3.33 cm I Real image Virtual object
Physics Term 3 STPM Chapter 22 Geometrical Optics 124 22 5. Using the curve mirror equation. 1 u + 1 v = 1 f ..............................➁ ➁ × v v u + 1 = v f Hence, m = v u = v f – 1 ➁ × u : 1 + u v = u f 1 m = u v = u f – 1 Example 5 The diameter of the image of the moon formed by a concave mirror used in telescope is 28 mm. The radius of curvature of the concave mirror is 6.0 m. (a) What is the angle subtended by the image at the mirror? (b) Calculate the diameter of the moon if its distance from the Earth is 3.8 × 108 m. Solution: v I θ θ 28 mm (a) Radius of curvature, r = 6.0 m Focal length, f = r 2 = 3.0 m Since the moon is far away from the Earth, object distance u = fi. Hence the image is formed at the focal plane of the mirror, and v = f = 3.0 m. Angle subtended by image at the mirror is θ. Since θ is small θ < tan θ = (28 × 10–3) 3.0 = 9.3 × 10–3 rad (b) If D = diameter of the moon. Angle subtended by the moon at the mirror = angle subtended by the moon’s image D 3.8 × 108 = 9.3 × 10–3 rad D = 3.5 × 106 m
Physics Term 3 STPM Chapter 22 Geometrical Optics 125 22 Example 6 A concave mirror produces a virtual image of height twice that of the object which is at a distance 5.0 cm from the mirror. Calculate the radius of curvature of the mirror. Solution: Linear magnifi cation, m = v u –2 = v 5.0 m = –2, negative because the image is virtual. Hence, image distance, v = –2(5.0) = –10.0 cm Using 1 u + 1 v = 1 f 1 f = 1 +5.0 + 1 –10.0 = 1 10.0 f = 10.0 cm Radius of curvature of mirror, r = 2f = 20.0 cm Example 7 A concave mirror of focal length 20.0 cm produces an image whose area is four times that of the object. Discuss the possible image distances and the characteristics of each image. Solution: Area of image = 4 × (area of object) ∴ Length of one side of image = 2 × (length of the corresponding side). Hence, linear magnifi cation, m = ff2 Case 1 When m = v u = + 2, image is real u = v 2 Using 1 u + 1 v = 1 f 2 v + 1 v = 1 20.0 Image distance, v = 60.0 cm Image is real, inverted and magnifi ed. Case 2 When m = v u = –2, image is virtual u = – v 2 Using 1 u + 1 v = – 1 f – 2 v + 1 v = 1 20.0 v = –20.0 cm Image is virtual, upright and magnifi ed.
Physics Term 3 STPM Chapter 22 Geometrical Optics 126 22 Example 8 A concave mirror of focal length 10.0 cm is placed in a beaker containing water to a depth of 4.0 cm. An optical pin is moved above the water on top of the mirror until there is no parallax between the optical pin and its image. Calculate the distance of the optical pin from the surface of the water. (Refractive index of water = 4 3 ) Solution: f = 10.0 cm Hence, r = 2f = 20.0 cm When there is no parallax between the image and the optical pin, then the object and image are at the same place. Rays from the object must be incident normally at the concave mirror and refl ected back along the same path. Hence the ray diagram is as shown in the fi gure. Distance of centre of curvature C from the water surface = 20.0 – 4.0 = 16.0 cm If x = distance of optical pin O from the water surface. Then refractive index of water = 16.0 x 4 3 = 16.0 x x = 12.0 cm Quick Check 1 O C x I 16.0 cm 20.0 cm 4.0 cm 1. An object is 2.5 cm from a concave mirror of radius of curvature 10.0 cm. Where is the image and what are its characteristics? 2. A curve mirror of focal length +20.0 cm is at a distance of 8.0 cm from an object. Find the position and characteristics of the image produced. 3. When an object is 20.0 cm from a concave mirror, an real image three times the height of the object is formed. Calculate the radius of curvature of the concave mirror. 4. A concave of mirror has a radius of curvature of 30.0 cm. Find the position and height of the image when the distance of an object of height 5.0 cm from the mirror is (a) 50.0 cm, (b) 30.0 cm, (c) 10.0 cm. Draw ray diagrams to illustrate the formation of the images. 5. A convex mirror has radius of curvature 20.0 cm. An object is 14.0 cm from the mirror. (a) Where is the image? (b) What is the image linear magnifi cation? 6. A curve mirror forms an upright image twice the height of the object. The object distance is 10.0 cm. (a) Discuss whether the curve mirror is concave or convex, and where the image is real or virtual. (b) Calculate the radius of curvature of the mirror. 7. A spherical mirror forms an image half the height of the object on a screen 10.0 cm from the mirror.
Physics Term 3 STPM Chapter 22 Geometrical Optics 127 22 (a) Discuss whether the spherical mirror is convex or concave? (b) What is the radius of curvature of the mirror? 8. In an experiment to determine the focal length of a concave mirror, an optical pin is used as an object. The position of the optical pin in front of the concave mirror is adjusted until there is non-parallax between the optical pin and its image formed by the mirror as seen by the eye as shown in the fi gure below. When this occurs the distance of the optical pin from the concave mirror is 40.0 cm. Concave mirror Optical pin Eye Image 40.0 cm (a) Explain the term non-parallax. (b) Calculate (i) the focal length, (ii) the radius of curvature of the concave mirror. (c) Draw two rays from the tip of the optical pin to the concave mirror and the refl ected rays. 9. Figure (a) shows an illuminated point object 15.0 cm from a convex lens. An image is formed by the lens on a screen P which is 30.0 cm from the lens. A convex mirror of focal length f is moved between the convex lens and screen P until the image formed by the lens and mirror is at the same position as the object. Convex lens Object 15.0 cm 30.0 cm P Screen Convex lens Convex mirror Object 15.0 cm 10.0 cm P Screen (a) Draw in figure (b) the path of the two rays from the object to show the formation of the image at the position of the object after passing through the convex lens and refl ected from the convex mirror. (b) Deduce the focal length of the convex mirror. 22.2 Refraction at Spherical Surfaces Learning Outcome Students should be able to: • use the formula n1 u + n2 v = n2 – n1 r for refraction at spherical surfaces I h u r v O P N C i A 1 n1 n2 i 2 α β γ Figure 22.16 2011/P2/Q10
Physics Term 3 STPM Chapter 22 Geometrical Optics 128 22 1. Figure 22.16 shows a point object O in a medium of refractive index n1 at a distance u from a spherical surface of radius of curvature r. The refractive index of the medium on the other side of the spherical surface is n2 where n2 > n1 . For example, the medium of refractive index n1 is air, and that of n2 is glass. 2. The point C is the centre of curvature of the spherical surface, and P is the pole of the spherical surface. Hence CP = r. 3. The ray from the point object O is incident normally at P and enter the second medium undeflected. 4. Another ray OA incident at A which is close to P is refracted towards the normal AC. The angle of refraction i 2 is smaller than the angle of incidence i1 . 5. The two rays in the second medium intersect and produce the image at I, distance v from P. 6. The object distance u, image distance v and the radius of curvature of the spherical surface r are related by the formula n1 u + n2 v = n2 – n1 r 7. Sign Convention : for the formula n1 u + n2 v = |n2 – n1| r Table 22.1 Real-is-positive sign convention Quantity Sign Positive (+) Negative (–) u Real object Virtual object v Real image Virtual image r r C n1 n2 n1 < n2 Spherical surface convex to less dense medium or C in denser medium C r n1 n2 n1 < n2 Spherical surface concave to less dense medium or C in less dense medium The difference of the refractive indices is obtained by substracting the small value from the larger value, hence it is written as |n2 – n1|. The formula becomes n1 u + n2 v = |n2 – n1| r
Physics Term 3 STPM Chapter 22 Geometrical Optics 129 22 Example 9 A sphere of radius 10.0 cm is made of glass of refractive index 1.50. An object is placed at 40.0 cm from the centre of the sphere. Find the position of the image produced due to refraction (a) by the fi rst refracting surface, (b) by both the refracting surfaces of the sphere. Draw ray diagrams to show the formation of the image in (a), and (b). Solution: (a) u = 30.0 cm v = 90.0 cm 10.0 cm C P O n1 = 1.00 I1 n2 = 1.50 The object distance, u = 40.0 – 10.0 = 30.0 cm r = +10.0 cm r positive because the centre of curvature C is in glass of n2 = 1.50, the optically denser medium. Using n1 u + n2 v = n2 ~ n1 r 1 30.0 + 1.5 v = 1.50 – 1.00 +10.0 1.5 v = 0.50 10 – 1 30 v = +90.0 cm The image I1 is real and 90.0 cm from the point P as shown in the ray diagram. 70.0 cm 90.0 cm C P Q O n1 = 1.00 n1 = 1.00 I I 1 n2 = 1.50 Second refracting surface First refracting surface (b) The image I1 , formed by the fi rst refracting surface is in glass (n2 = 1.50) and (90.0 – 20.0) cm = 70.0 cm from the point Q of the second refracting surface. I1 acts as a virtual object for the second refraction surface, and u = –70.0 cm, n1 ′ = 1.50 After refraction light emerges into air (n2 ′ = 1.00)
Physics Term 3 STPM Chapter 22 Geometrical Optics 130 22 Using n1 ′ u + n2 ′ v = n1 ′ – n2 ′ r 1.50 –70.0 + 1 v ′ = 1.50 – 1.00 10.0 1 v ′ = 0.50 10.0 + 1.50 70.0 v ′ = 14.0 cm The fi nal image I is real and 14.0 cm from the point Q as shown in the ray diagram. Example 10 A small strip of paper is pasted on one side of a glass sphere of diameter 10.0 cm. The paper is then view from the opposite surface of the sphere. Where is the image? (refractive index of glass = 1.50) Solution: The strip of paper on the glass surface acts as an object O in glass (n1 = 1.50), and object distance, u = 10.0 cm. The ray from O are then refracted into air (n2 = 1.00). r = + 1 2 (10.0 cm) = +5.0 cm Using n1 u + n2 v = n1 – n2 r 1.50 10.0 + 1 v = 1.50 – 1.00 5.0 1 v = 0.50 5.0 – 1.50 10.0 v = –20.0 cm Image is virtual and at a distance of 20.0 cm from P. Quick Check 2 u = 10.0 cm v = 20.0 cm C O P n1 = 1.50 I n2 = 1.00 1. Find the position of the image formed by a spherical glass surface (refractive Index = 1.50) when a point object is (a) 20.0 cm from the convex spherical glass surface of radius of curvature 10.0 cm, (b) 50.0 cm from the convex spherical glass surface of radius of curvature 20.0 cm, (c) 40.0 cm from the concave spherical glass surface of radius of curvature 50.0 cm. 2. A point object is 25.0 cm from the centre of a glass sphere of radius 5.0 cm. The refractive index of glass is 1.50. Find the position of the image formed due to refraction by (a) the fi rst spherical glass surface, (b) the fi rst and second refractive surfaces of the sphere. 3. (a) Find the position of the image of a distant object formed by a glass sphere of radius 5.0 cm. The refractive index of glass is 1.50. (b) Draw a ray diagram to show how the image is formed. 4. A sphere of radius 12.0 cm is made of material of refractive index 4 3 . A speck is 4.0 cm from the centre of the sphere is viewed along the diameter that passes through the speck
Physics Term 3 STPM Chapter 22 Geometrical Optics 131 22 (a) from the near side (b) from the far side of the sphere. Find the position of the image for (a) and (b) above. 5. The thickness of a plano-convex lens is 3.0 cm. When placed on a horizontal table with its plane surface on the table, the apparent thickness is 2.25 cm. Calculate the radius of curvature of the convex surface of the lens. (Refractive index of glass = 1.50) 22.3 Thin Lenses Learning Outcomes Students should be able to: • use the formula n1 u + n2 v = n2 – n1 r to derive the thin lens formula 1 u + 1 v = 1 f and lensmaker’s equation 1 f m + ( n1 nm – 1)( 1 r1 + 1 r2 ) • use the thin lens formula and lensmaker’s equation. 1. A lens has two refractive surfaces. Rays from an object are refracted by the surfaces of the lens and an image is formed. 2. In this section, we will discuss action of thin lenses whose thickness is negligible compared to the object distance or image distance. 3. A converging or convex lens is thicker at the centre than at the edge. Figure 22.17 shows three types of convex lens. 4. Diverging or concave lenses are thicker at the edge than at the centre. Figure 22.18 shows three types of concave lens. Biconvex Plano-concave Diverging meniscus Figure 22.18 5. A convex lens or a concave lens has two surfaces of radius of curvature r1 and r2 and centre at centres of curvature C1 and C2 respectively (Figure 22.19) C1 C2 Principal axis r1 r2 C1 C2 Principal axis r1 r2 Figure 22.19 2012/P1/Q38,Q39, 2013/P3/Q8, 2014/P3/Q7,8,19(b) Figure 22.17 Biconvex Plano-convex Converging meniscus 2015/P3/Q19, 2016/P3/Q7,Q8, 2017/P3/Q19 INFO Nearsightedness and its Correction
Physics Term 3 STPM Chapter 22 Geometrical Optics 132 22 6. The straight line that passes through the centres of curvature of the surfaces of a lens is known as the principal axis. (a) (b) f F Principal axis f F F f Principal axis Figure 22.20 7. The focus F of a convex lens is the point on the principal axis where rays which are parallel and close to the principal axis converge after passing through the lens. The distance of the focus F from the centre of the lens is the focal length f of the lens. The focal length of a convex lens is assumed positive. 8. A lens has two foci, one on each side of the lens (Figure 22.21 (b)). The distances of the two foci from the lens is the same, that is the focal length. (a) (b) f F F f F f Figure 22.21 9. The focus F of a concave lens is the point on the principal axis where rays which are parallel to the principal axis seem to diverge from after passing through the lens. (Figure 22.21) The focal length of a concave lens is assumed negative. 10. The power of lens = 1 focal length in metres Unit for power of a lens is diopter (D). If the focal length of a convex lens, f = +10 cm = +0.10 m Power of lens = 1 +0.10 = +10 D If the focal length of a concave lens, f = –20 cm = –0.20 m power of concave lens = 1 –0.20 = –5.0 D
Physics Term 3 STPM Chapter 22 Geometrical Optics 133 22 11. When two lenses of power 1 f 1 and 1 f 2 are in contact, the power of the combination is 1 f = 1 f 1 + 1 f 2 . (Figure 22.22) (b) f 1 = 0.30 m 1 f = 1 0.30 + 1 0.15 = 10 D f = 0.10 m (b) f 1 = 0.30 m 1 f = 1 0.30 + 1 –0.15 = –3.33 D f = –0.30 m ƒ2 = 0.15 m ƒ1 = 0.30 m ƒ2 = 0.15 m ƒ1 = 0.30 m Figure 22.22 12. The focal plane of a convex lens is the plane which passes through the focus F and is perpendicular to the principal axis. Rays which are parallel to each other but not necessary parallel to the principal axis are brought to a focus on the focal plane. (Figure 22.23) f F Focal plane Figure 22.23 13. The optical centre P of a lens is the point where a ray through the point is undeflected. (Figure 22.24). (a) (b) P : Optical centre P P Figure 22.24 14. The following ray diagrams show the formation of images by a convex lens. (a) Object distance u > 2f Image is real, inverted and diminished. Image distance u is between f and 2f O F F I u v Object Image Figure 22.25
Physics Term 3 STPM Chapter 22 Geometrical Optics 134 22 (b) Object distance u = 2f Image is real inverted and same size as the object. Image distance v = u, object distance = 2f When u = 2f, the distance between the object and the real image formed by a convex lens is minimum. Minimum distance between object and image OI = u + v = 4f (c) Object distance between f and 2f. O F F I u v Object Image Figure 22.27 The image is real, inverted and magnified. Image distance is more that 2f. (d) Object distance u = f F O F u Object Figure 22.28 Image is at infinity, inverted and magnified. Conversely when the object is at infinity, the image formed by the convex lens is at the focus of the lens. (e) Object distance u < f. I F O Image Object F Figure 22.29 The image is virtual, upright and magnified. The convex lens acts as a magnifying glass. 15. When an object is placed in front of a concave lens, the image formed is always virtual, upright diminished and on the same side of the lens as the object (Figure 22.30). Figure 22.26 O F F I u v Object Image
Physics Term 3 STPM Chapter 22 Geometrical Optics 135 22 F F O I Object Figure 22.30 Thin Lens Formula and Lensmaker’s Equation 1. The thin lens formula relates the object distance, u, image distance v and focal length f. It can be derived using the formula. n1 u + n2 v = n2 – n1 r Object (Virtual object) P u v O A B I v I r1 n1 n1 n2 r2 Figure 22.31 2. Figure 22.31 shows a thin convex lens of refractive index n2 in a medium of refractive index n1 . The radii of curvature of the lens surfaces are r1 and r2 . The object O is at a distance u from the lens. 3. Since the lens is thin, the object distance u from the optical centre P of the lens can be assumed to be equal to the object distance from the fi rst refracting surface of the lens of radius of curvature r1 . 4. Refraction of light from the object O by the surface of radius of curvature r1 , produces the image I′ at a distance v ′ from the lens. Using n1 u + n2 v = n2 – n1 r n1 u + n2 v’ = n2 – n1 r1 ............................. ① 5. The image I′ acts as a virtual object for the second refracting surface of radius of curvature r2 . Hence n2 –v’ + n1 v = n2 – n1 r2 ..............................➁ ① + ➁ : n1 v + n1 u = (n2 – n1 ) ( 1 r1 + 1 r2 ) 1 u + 1 v = ( n2 n1 – 1) ( 1 r1 + 1 r2 ) .............................➂ Info Physics Divers wear goggles so that the focal lengths of the eyes lenses do not change. Without goggles, when in water, the medium in front of the eyes lenses is water (not air). Focal lengths of lenses changed.
Physics Term 3 STPM Chapter 22 Geometrical Optics 136 22 6. From the defi nition of focal length f on a convex lens, when u = ∞, v = f. From equation ➂: 1 ∞ + 1 f = ( n2 n1 – 1) ( 1 r1 + 1 r2 ) ............................. ➃ From equation ➂ and ➃, 1 u + 1 v = 1 f .............................➄ This equation is known as the thin lens formula. 7. From equation ➃, 1 f = ( n2 n1 – 1) ( 1 r1 + 1 r2 ) The focal length f of a lens depends on the medium the lens is in. For this reason a diver is not able to see clearly distant objects unless he uses a pair of goggles. 8. If the lens is in air, refractive index of air, n1 = 1.00, hence 1 f = (n – 1) ( 1 r1 + 1 r2 ) where n = refractive index of the material of the lens. This formula is known as the lens-maker formula. Example 11 The surfaces of a biconvex lens have the same radius curvature 10.0 cm. The lens is made of glass of refractive index 1.50. Calculate the focal length of the lens (a) in air, (b) in water of refractive index 1.33. Solution: (a) Using 1 f = (n – 1) ( 1 r1 + 1 r2 ) = (1.50 – 1) ( 1 10 + 1 10) = 0.10 f = 10.0 cm (b) n1 = 1.33, n2 = 1.50 Using 1 f = ( n2 n1 – 1) ( 1 r1 + 1 r2 ) = ( 1.50 1.33 – 1) ( 1 10 + 1 10) f = 39.1 cm Example 12 Calculate the focal length of the following lenses which are made of glass of refractive index 1.50. (a) A plano–convex lens, with a surface having a radius of curvature 10.0 cm. (b) A converging meniscus of radii of curvature 20.0 cm and 10.0 cm. (c) A biconcave lens having equal radii of curvature 10.0 cm.
Physics Term 3 STPM Chapter 22 Geometrical Optics 137 22 Solution: (a) Using the lens maker formula 1 f = (n – 1) ( 1 r1 + 1 r2 ) = (1.50 – 1) ( 1 10.0 + 1 ∞ ) f = +20.0 cm (b) r1 = –20.0 cm, negative because the surface is concave to air, the less dense medium r2 = +10.0 cm, positive because the surface is convex to air. Using 1 f = (n – 1) ( 1 r1 + 1 r2 ) = (1.50 – 1) ( 1 –20.0 + 1 10.0) f = 40.0 cm (c) Both the surfaces are concave to air, hence r1 = r2 = –10.0 cm. Using 1 f = (n – 1) ( 1 r1 + 1 r2 ) = (1.50 – 1) ( 1 –10.0 + 1 –10.0) f = –10.0 cm Example 13 A beam of light is directed to a point on the principal axis 6.0 cm behind a concave lens of focal length 18.0 cm. Find the position of the image. Solution: For the concave lens, f = –18.0 cm. The point O is a virtual object, u = –6.0 cm. Using 1 u + 1 v = 1 f 1 v = 1 –18.0 – 1 –6.0 v = 9.0 cm The image is real and is at the point I as shown in the fi gure. r1 = 10 cm r2 = fi r1 = –20.0 cm r2 = 10.0 cm r1 = –10.0 cm r2 = –10.0 cm n =1.50 6.0 cm Virtual object O Image I
Physics Term 3 STPM Chapter 22 Geometrical Optics 138 22 Example 14 A thin converging lens L is held 15.0 cm above an object O in a beaker. The image is at a distance 12.0 cm above the lens. When the beaker is fi lled with a liquid to a depth of 7.5 cm, the image of O is 14.0 cm above the lens. Calculate the refractive index of the liquid. Solution: When the beaker is empty, u = 15.0 cm, v = 12.0 cm Using 1 f = 1 u + 1 v = 1 15.0 + 1 12.0 ............................. ① When the depth of liquid in the beaker is 7.5 cm, the path of rays from the object O through the lens is as shown in the fi gure. v = 14.0 cm Using 1 f = 1 u + 1 v , and from ①, 1 15.0 + 1 12.0 = 1 u1 + 1 14.0 u1 = 12.73 cm Hence apparent depth of liquid, d = 12.73 – 7.5 = 5.23 cm Refractive index of liquid, n = real depth apparent depth = 7.5 5.23 = 1.43 Quick Check 3 O L I 7.5 cm 7.5 cm 14.0 cm d Ofi O L u1 1. A lens forms a real image of an object. The distance of the object from the lens is x, and the distance of the image from the lens is y. The variation of y with x is as shown by the graph below. y / cm 0 X / cm 20 20 40 60 80 40 60 80 From the graph, it can be deduced that the lens is A converging of focal length 10 cm B converging of focal length 20 cm C diverging of focal length 10 cm D diverging of focal length 20 cm 2. The focal length of a convex lens is 1 m. The minimum distance between a real object and the real image formed by the lens is A 1 m B 2 m C 4 m D 8 m
Physics Term 3 STPM Chapter 22 Geometrical Optics 139 22 3. An luminous object and a screen are placed 1.2 m apart. The lens which is able to produce an inverted image of the same size as the object on the screen is A a converging lens of focal length 0.3 m B a diverging lens of focal length 0.3 m C a converging lens of focal length 0.6 m D a diverging lens of focal length 0.6 m 4. A converging lens of focal length 0.3 m is placed between an object and a screen separated by a distance of 1 m. By varying the position of the lens it is possible to form on the screen A two real and inverted images B one real and inverted image C one virtual and upright image D no image at all 5. A converging lens of focal length f is placed 0.3 m from an object and an image is produced on a screen 0.9 m from the lens. With the object and screen fixed, an image of the object can also be formed on the screen by placing a converging lens of focal length A 3f at a distance 0.3 m from the screen B f at a distance 0.3 m from the screen C 3f at a distance 0.1 m from the screen D f at a distance 0.1 m from the screen 6. Two thin converging lenses of focal length f 1 and f 2 are placed in contact with each other. The focal length of the combination is A (f1 f2 ) 1 2 C f 2 (f 1 + f 2 ) f 1 B 1 2 (f 1 + f 2 ) D f 1 f 2 (f 1 + f 2 ) 7. A luminous object is placed 100 cm from a screen. A convex lens of focal length 25 cm is placed between the object and the screen to produce an image on the screen. The distance of the lens from the screen is A 25.0 cm C 50.0 cm B 33.3 cm D 75.0 cm 8. Lenses P and Q have the same focal length 10 cm. Parallel rays are incident on the lens P. In which of the following diagrams the rays that emerge from lens Q diverge? A PQ B P Q 15 cm C P Q 5 cm D P Q 20 cm 9. Two converging lenses P and Q have focal lengths of 10 cm and 5 cm respectively are arranged coaxially 40 cm apart. An object is placed 15 cm from lens P as shown in the figure. P Q 15 cm Object 40 cm Which of the following gives the final image position and its characteristic? A Image real, 10 cm from lens Q B Image virtual, 10 cm from lens Q C Image real, 6 cm from lens Q D Image virtual, 6 cm from lens Q 10. Rays parallel to the principal axis are incident or a convex lens of focal length 10 cm. A concave lens of focal length 5 cm is placed 5 cm behind the convex lens.
Physics Term 3 STPM Chapter 22 Geometrical Optics 140 22 When the concave lens is moved closer to the convex lens, the rays that emerge from the concave lens A remain parallel B change from parallel to diverging C change from converging to diverging D change from diverging to converging 11. A diverging lens and a converging lens, both of focal length 5.0 cm are separated by a distance of 5.0 cm. Parallel rays from a distant object are incident on the diverging lens as shown in the diagram. 5.0 cm The final image formed by the lenses is A at infinity B virtual and at the midpoint between the lenses C real and 5.0 cm on the right of the converging lens D real and 10.0 cm on the right of the converging lens 12. The diagram shows the actual cross section of a converging lens of focal length f 1 and a diverging lens of focal length f 2 in contact with each other. The effective focal length of the combined lens is F. f 1 f 2 Which statement about the combined lens is correct? A Combined lens is converging, and F is less than f 1 . B Combined lens is converging, and F is greater than f 1 . C Combined lens is diverging, and F is less than f 2 . D Combined lens is diverging, and F is greater than f 2 . 13. An object O is at a distance x1 from the principal focus F1 of a converging lens. The image is formed at I, a distance of x2 from the principal focus F2 . x1 x2 F1 F2 O I The focal length f of the lens is A f = x1 x2 C f = x1 x2 x1 + x2 B f = x2 1 + x2 2 D f = x1 x2 x2 – x1 14. The near point and far point of a person with normal eyesight are at 25.0 cm and at infinity from the eyes. The near point of a particular long-sighted person is 50.0 cm and his far point is at infinity. What is the focal length of corrective lens for this person to be able to read a book held 25.0 cm from the eyes? A +16.7 cm C – 16.7 cm B +50.0 cm D – 50.0 cm 15. A plano-convex lens is made of glass of refractive index 1.50. The radius of curvature of the curved surface is 20.0 cm. What is the distance x of an object from the lens when the image formed is at a distance x behind the lens? A 20.0 cm C 60.0 cm B 30.0 cm D 80.0 cm 16. (a) An object is 15.0 cm from a converging lens of focal length 5.0 cm. (i) Use a suitable scale, draw a ray diagram to show the formation of the image. (ii) Deduce from the ray diagram the image distance, the linear magnification and the characteristics of the image. (b) An object is 15.0 cm from a convex spherical mirror of radius of curvature 20.0 cm. (i) Where is the image? (ii) If the height of the object is 5.0 cm, what is the height of the image? (c) A lens formed an image which is half the height of the object on a screen. The distance between the object and screen is 60.0 cm.
Physics Term 3 STPM Chapter 22 Geometrical Optics 141 22 Determination of Focal Length of a Convex Lens O u v Object Convex lens Screen Image I Figure 22.32 1. Figure 22.32 shows the arrangement used to determine the focal length of a convex lens. An illuminated object such as a lighted bulb is placed at a distance u from the lens. The position of the image is obtained by moving a screen on the other side of the lens, a sharp image is obtained on the screen. The object distance u, and image distance v are measured. 2. The procedure is repeated for other values of u. 3. The following graphs may be plotted to determine the focal length f of the convex lens. (a) Graph of v against u (b) Graph of 1 v against 1 u . v v = u u 2f 2f 0 P (2f, 2f) 1 – v 1 – f 0 1 – u 1 – f Figure 22.33 Figure 22.34 Image distance, v = u, object distance when v = u = 2f From the coordinates of the point P, the value of f can be determined. (i) State the type of the lens used. (ii) Determine the distance of the screen from the lens. 17. A man wears a pair of spectacle which has lenses of the same focal length. Each of the lens is a diverging meniscus of radii of curvature 20.0 cm and 30.0 cm respectively. (a) Calculate the focal length of the lens (i) in air, (ii) in water of refractive index 1.33. (b) Explain whether the man can see clearly using his spectacles when in water? (Refractive index of material of lens = 1.50) 18. An object is placed 80.0 cm in front of a diverging lens of focal length 20.0 cm. Find the position and characteristics of the image. 19. An object is placed at the bottom of a beaker. A convex lens of focal length 10.0 cm is held 15.0 cm above the object. Water of refractive index 4 3 is poured into the beaker to a depth of 8.0 cm. Find the position of the image. 20. A converging lens of focal lens 25.0 cm and a diverging lens of focal lens 5.0 cm are arranged coaxially and separated by a distance of 20.0 cm. A point object is placed 50 cm from the converging lens on the axis. Find the position of the final image and describe its characteristics. Illustrate your answer with a ray diagram.
Physics Term 3 STPM Chapter 22 Geometrical Optics 142 22 From the thin lens equation 1 u + 1 v = 1 f When 1 u = 0, 1 v = 1 f When 1 v = 0, 1 u = 1 f From the intercepts on the horizontal and vertical axis, the value of f may be obtained. (c) Graph of (u + v) the distance between object O and image I against the object distance u. OI = (u + v) u 4f 2f 0 P (2f, 4f ) Figure 22.35 The minimum distance between the object O and its real image is 4f, and this occurs when u = v = 2f. Hence from the coordinate of P, the minimum point on the graph, the value of f can be determined. (d) Graph of linear magnification m against image distance v. From 1 u + 1 v = 1 f v u + 1 = v f m = v u = v f – 1 Gradient of graph = 1 f and when m = 0, v f – 1 = 0 v = f Hence the value of f can be determined from the gradient of the graph or from the intercept on the horizontal axis. 4. Figure 22.37 shows an object at O in front of a convex lens and its real image at I. By the principle of reversibility of light, if an object is placed at the point I, an image will be formed at the point O. The points O and I are known as conjugate points. 5. When the object is at O, and the image at I. Object distance = a, image distance = b. When the object is at I, the image is at O, then object distance = b, image distance = a. Hence for a pair of conjugate points, the values of the object distance and image distance may be interchanged. a b O Image Object I Figure 22.37 Figure 22.36 v f 0 m = v – u Gradient = 1 – f
Physics Term 3 STPM Chapter 22 Geometrical Optics 143 22 Example 15 A luminous object is placed 36.0 cm from a screen. There are two possible positions P and Q between the object and the screen for a convex lens of focal length 8.0 cm to produce a sharp image on the screen. Calculate the distance of P, and Q from the object. Solution: Let x = distance of lens from object O. Then object distance, u = x and image distance, v = 36 – x f = 8.0 cm Using 1 u + 1 v = 1 f 1 x + 1 (36 – x) = 1 8 (36 – x) + x x(36 – x) = 1 8 x2 – 36x + 36 × 8 = 0 x = 36 ± 362 – 4(36 × 8) 2 fiffffffffffffffffffffff = 24.0 cm or 12.0 cm Hence distance of P from object = 12.0 cm distance of Q from object = 24.0 cm Example 16 A student is given two thin converging lenses L1 and L2 . In experiment carried out using the lenses, the student measures the distance r of the object O from the lens (or combination of lenses) and the corresponding image distance s using (i) lens L1 only, (ii) lenses L1 and L2 in contact with each other as shown in the fi gures below. L1 O Object Image r I s L1 L2 O Object Image r I s Experiment (i) Experiment (ii) From the results of experiment (i) a graph of 1 s against 1 r was plotted, and from the results of experiment (ii) a graph of rs against (r + s) was plotted as shown below. x cm (36 – x) cm 36 cm P Q O I
Physics Term 3 STPM Chapter 22 Geometrical Optics 144 22 0 0.025 0.025 0.050 0.075 0.100 0.050 0.075 0.100 1 – cm–1 r 1 – cm–1 s / / Graph (i) Graph (ii) 20 25 160 140 180 200 220 30 35 40 rs / cm2 (r + s) cm (a) From graph (i), determine the value of the focal length of lens L1 . Explain your calculations clearly. (b) Determine the gradient of graph (ii). Hence calculate the value of the focal length of the combination of lenses L1 and L2 . Explain your calculations. (c) Calculate the focal length of lens L2 . Solution: (a) Object distance, u = r Image distance, v = s using 1 u + 1 v = 1 f 1 r + 1 s = 1 f when 1 r = 0, 1 s = 1 f = 0.100 cm–1 (from the graph (i)) f = 10.0 cm (b) From the equation 1 r + 1 s = 1 f r + s rs = 1 f f = rs r + s = gradient of graph (ii) Hence focal length of combination f = 220 – 140 37.5 – 23.5 = 5.71 cm (c) For two lenses of focal length f 1 and f 2 in contact, the resultant focal length f is given by 1 f = 1 f 1 + 1 f 2 Hence 1 5.71 = 1 10.0 + 1 f 2 Focal length of lens L2 f 2 = 13.3 cm Exam Tips The equation 1 f = 1 f 1 + 1 f 2 is only applicable for two lenses in contact