11 Superposition

October 12, 2017 | Author: Ronnie Quek | Category: Diffraction, Interference (Wave Propagation), Wavelength, Coherence (Physics), Waves
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11 Superposition...


Raffles Institution Physics Department Revision Package H2 Prelims 2010


Compiled by: BYH & HCB


1 1

MCQs 1.

[AJC/H2/Prelims 2010/P1/22] Fig. 1 shows a transverse wave moving to its right. Fig. 2 shows a stationary wave on a string stretched between two points X and Y. Positions of particles P, Q, R, S, T, A, B, C, D and E are indicated in the two figures.

Which of the following statement regarding the particles is not correct? Fig. 1 Fig. 2 A P and S are in antiphase. A and C are in antiphase. B P and Q are in phase. A and E are in phase. C Q and S are antiphase. B and E are antiphase. D R and T are out of phase. B and D are out of phase. 2.

[AJC/H2/Prelims 2010/P1/23] Microwaves of wavelength 4.0 cm are produced by two microwave transmitters P and Q operating in phase. Point X is 1.0 m from transmitter P and 0.5 m from transmitter Q as shown in the figure. Microwaves from transmitter P arrives at point X with intensity I and amplitude of oscillation A while the microwaves from transmitter Q arrives at point X with intensity 4I. Determine the resultant intensity at point X in terms of I.

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[CJC/H2/Prelims 2010/P1/22] A boy blows gently across the top of a piece of glass tubing the low end of which is closed by his finger so that the tube gives its fundamental note of frequency, f. While blowing, he removes his finger from the lower end. The note he then hears will have a frequency of approximately A ¼f B ½f C 2f D 4f


[CJC/H2/Prelims 2010/P1/23] Two loudspeakers L1 and L2, driven by a common oscillator and amplifier, are set up as shown. As the frequency of the oscillator increases from zero, the detector at D recorded a series of maximum and minimum signals. At what frequency is the first maximum observed? (Speed of sound = 330 m s-1) 40 m


D 9m

L2 A 5.

165 Hz


330 Hz


495 Hz


660 Hz

[Dunman High/H2/Prelims 2010/P1/21] The diagram below illustrates an experimental arrangement that produces interference fringes with a double slit. Thin glass plate


S1 Screen Monochromatic light source S2 Y When slit S1 is covered with a very thin plate of glass as shown, A the separation of the fringe increases. B the separation of the fringe decreases. Page 2 of 75


the fringe pattern moves towards X. the fringe pattern moves towards Y.


[HCI/H2/Prelims 2010/P1/21] A tuning fork of pitch 2.4 kHz is placed near the mouth of a long cylindrical container which is fully filled with water. Water is slowly drained from the bottom of the container. Resonance is first heard when the water level had dropped 3.4 cm below the mouth of the container. At which position of water below the mouth of the container will the next resonance be heard? Assume the speed of sound in air = 3.4 × 102 m s-1. A 7.1 cm B 8.0 cm C 10.2 cm D 10.5 cm


[HCI/H2/Prelims 2010/P1/19] A guitar string of length L is stretched between two fixed points P and Q and made to vibrate transversely as shown in the figure.



s B A

L Two particles A and B on the string are separated by a distance s. The maximum kinetic energies of A and B are KA and KB respectively. Which of the following gives the correct phase difference and maximum kinetic energies of the particles? Phase difference Maximum kinetic energy



 3s    360  2L



 3s    360  2L








[HCI/H2/Prelims 2010/P1/20] A beam of red light of wavelength 710 nm is incident normally on a diffraction grating. The angular separation between the two second order maxima is 60°. What is the resolution of the grating? A 3.5 × 105 lines per millimetre B 3.5 × 105 lines per metre C 6.1 × 105 lines per millimetre D 6.1 × 105 lines per metre Page 3 of 75


[IJC/H2/Prelims 2010/P1/20] A stationary wave has a series of nodes. The distance between the first and the sixth node is 30.0 cm. What is the wavelength of the sound wave? A 5.0 cm B 6.0 cm C 10.0 cm D 12.0 cm

10. [IJC/H2/Prelims 2010/P1/22] A narrow beam of monochromatic light is incident normally on a diffraction grating. Thirdorder diffracted beams are formed at angles of 40° to the original direction. What is the highest order of diffracted beam produced by this grating? A 3rd B 4th C 5th D 6th

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11. [IJC/H2/Prelims 2010/P1/21] In an interference experiment, two slits are illuminated with white light through a single-slit lying on the perpendicular bisector through the double-slit.

What is seen on the screen? A The central fringe is black with black and white fringes on each side. B The central fringe is black with coloured continuous spectrum on each side. C The central fringe is white with black and white fringes on each side. D The central fringe is white with coloured continuous spectrum on each side. 12. [MI/H2/Prelims 2010/P1/20] A stationary sound wave has a series of nodes. The distance between the first and the sixth antinode is 30.0 cm. What is the wavelength of the sound wave? A 5.0 cm B 6.0 cm C 10.0 cm D 12.0 cm 13. [MI/H2/Prelims 2010/P1/21] Monochromatic light with a wavelength of λ is incident normally on a diffraction grating. The angle between the directions of the two second-order diffracted beams is α. What is the spacing of the lines on the grating? 2  2       A B C D sin  sin  sin sin  2  2 14. [MJC/H2/Prelims 2010/P1/17] A microwave source S is placed in front of a detector D, and a metal reflecting screen R is placed beyond D such that its plane is perpendicular to the line joining S to D. As the detector is moved slowly away from the source, it registers a series of maxima and minima.




It is observed that the detector moved through a distance of 5.6 cm between the first and fifth minimum. What is the frequency of the microwaves in GHz? A 5.4 B 10.7 C 13.4 D 27.5

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15. [MJC/H2/Prelims 2010/P1/18] In a two-slit interference experiment, one slit transmits twice the amplitude of the other slit. If the maximum intensity of the interference pattern is I0, the minimum intensity in the pattern would be A zero B I0/9 C I0/4 D I0/2 16. [NJC/H2/Prelims 2010/P1/23] A vibrating string fixed at two ends has a fundamental frequency of 250 Hz. When the same string is vibrating in 750 Hz, A it has 4 nodes. B its wavelength is 3 times the original wavelength. C it is vibrating in the second mode. D its wave speed has increased by 3 times. 17. [NJC/H2/Prelims 2010/P1/24] Which of the following methods would result in no change in fringe separation in a double slit experiment, as shown below?

Single slit A B C D

Double slit

Conduct the double slit experiment in water. Use light of the same intensity but of different frequency. Keep the distance between the single slit and the screen constant, but move the double slit closer to the single slit. Increasing the slit width of the double slits.

18. [NYJC/H2/Prelims 2010/P1/19] A sound wave of period T is traveling in a gas at a speed of 330 m s -1. 5 4 When the phase difference between two points is radians, the distance between the two points is 1.05 mm. Calculate T. A 1.96 × 105 s B 3.14 × 105 s C 5.03 × 105 s D 5.09 × 10-6 s

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19. [NYJC/H2/Prelims 2010/P1/20] A student blows at one end of a drinking straw of length 10.0 cm while the other end is dipped inside his drink for 6.0 cm. What is the lowest frequency of sound produced? [Speed of sound in air = 320 m s-1] A 800 Hz B 1600 Hz C 2000 Hz D 4000 Hz 20. [NYJC/H2/Prelims 2010/P1/21] When coherent monochromatic light falls on a double slits, interference pattern is observed on a screen some distance from the slits. The fringe separation can be increased by A decreasing the distance between the screen and the slits. B increasing the distance between the slits. C using monochromatic light of lower frequency. D immensing the whole set up in water. 21. [NYJC/H2/Prelims 2010/P1/22] A diffraction grating has 500 lines per mm and is illuminated normally by monochromatic light of wavelength 600 nm. The total number of bright lines seen on the screen is A 5 B 7 C 8 D 9 22. [PJC/H2/Prelims 2010/P1/21] The length l of an air column is slowly increased from zero while a note of constant frequency is produced by a tuning fork placed in front of it.

air column

tuning fork l

When l reaches 20 cm the sound increases greatly in volume. What is the wavelength of the sound wave produced by the tuning fork? A 20 cm B 40 cm C 80 cm D

100 cm

23. [RVHS/H2/Prelims 2010/P1/21] A narrow beam of monochromatic light falls at normal incidence on a diffraction grating. Second-order diffracted beams are formed at angles of 20° to the original direction. What is the number of diffracted beams produced by this grating? A 5 B 6 C 10 D 11

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24. [RI/H2/Prelims 2010/P1/20] Some fine sand particles are present in a long transparent tube. A speaker is placed at the end of the tube, and the frequency of the sound emitted is varied until the fine sand settles into a series of small heaps. The diagram below shows a section of the tube and some of the heaps that were formed. L Which of the following statements is true? A The air molecules are vibrating vertically. B The wavelength of the sound is given by L. C The air pressure where the heaps are is the lowest. D The positions of the heaps show the positions of the displacement nodes. 25. [RI/H2/Prelims 2010/P1/21] A horizontal steel wire is fixed at one end and is kept under tension by means of weights suspended over a pulley. The length of wire between the fixed end and the pulley is 1.0 m. Magnets are placed near the centre of the wire, and an alternating voltage supply is connected to the wire between the fixed end and the pulley. Standing waves are formed when the voltage supply is turned on. Five antinodes are observed on the wire. 1.0 m S Fixed end


N weights

Given that the speed of the wave on the wire is 24 m s -1, what is the frequency of the voltage supply? A 48 Hz B 60 Hz C 96 Hz D 120 Hz 26. [RI/H2/Prelims 2010/P1/22] In a diffraction grating experiment, the first order image of a 438 nm blue light occurred at an angle of 16.2°. A second order coloured light was observed at 47.4°. What is the wavelength of this coloured light? A 578 nm B 631 nm C 637 nm D 696 nm 27. [SAJC/H2/Prelims 2010/P1/20] Light of wavelength 550 nm is incident normally on a diffraction grating having 400 lines per millimetre. What is maximum number of bright fringes that can be observed? A 4 B 5 C 9 D 11

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28. [SAJC/H2/Prelims 2010/P1/21] Two wave generators S1 and S2 produce water waves of wavelength 0.5 m. A detector is placed at position X, 3 m from S1 and 2 m from S2 as shown in the diagram below. Each generator produces a wave of amplitude A at X when operated alone. The generators are operating together and producing waves which have a constant phase difference of π radians. What is the resultant amplitude at X? X 3m


S2 S1 A








29. [TJC/H2/Prelims 2010/P1/22] Monochromatic light of wavelength 4.0 × 10 -7 m passes through two narrow slits and produces light and dark fringes on a screen. What is the separation of the slits such that the angular separation between the two first order bright fringes is 4.00 × 10-4 rad? A 0.5 × 10-3 m B 1.0 × 10-3 m C 1.50 x 10-3 m D 2.0 × 10-3 m 30. [TJC/H2/Prelims 2010/P1/23] A diffraction grating ruled with 5000 lines per cm is illuminated with white light. If the wavelength for yellow light and violet light are 600 nm and 400 nm respectively, which one of the following statements is NOT correct? A The central image is white. B The red end of the first-order spectrum is closer to the central image than the violet end of the first-order spectrum. C The second-order image of the yellow light coincides with the third-order image of violet light. D There is no fourth-order image for yellow light. 31. [YJC/H2/Prelims 2010/P1/19] Coherent monochromatic light illuminates two narrow parallel slits and the interference pattern which results is observed on a screen some distance beyond the slits. Which of the following will not affect the fringe separation of the interference pattern? A Using monochromatic light of higher frequency B Increasing the distance between the screen and the slits C Increasing the width of each slit D Increasing the distance between the slits

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32. [VJC/H2/Prelims 2010/P1/20] Two wave generators S1 and S2 produce water waves of wavelength 2.0 m. They are placed 6.0 m apart as shown and are operated in phase. A sensor D which measures the amplitude of water waves is 7.0 m away from S1 as shown in the diagram below. S1

6.0 m


7.0 m

D The shortest distance D could be moved along the straight line S 1D in order to detect large amplitude of the resultant wave motion is A 1.0 m towards S1 B 3.0 m towards S1 C 1.0 m away from S1 D 3.0 m away from S1 35. [VJC/H2/Prelims 2010/P1/21] Light from a laser is directed normally at a diffraction grating as shown in the figure below. The diffraction grating is situated at the centre of the circular scale, marked in degrees. The readings on the scale for the second order diffracted beams are 140° and 166°. The wavelength of the laser light is 500 nm.

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166° laser 140°

What is the spacing of the slits of the diffraction grating? A

1.14 × 10-6 m


2.22 × 10-6 m


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2.28 × 10-6 m


4.45 × 10-6 m

36. [YJC/H2/Prelims 2010/P1/20] A thin copper rod is clamped at one end and made to vibrate by a driving force of variable frequency applied to the free end. At specific frequencies, it is found that the rod resonates. Which of the following diagram is incorrect, where N and A represent a nodal and antinodal position respectively? clamp A


A clamp





clamp C





clamp D







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Structured Questions 1. [AJC/H2/Prelims 2010/P2/4] A diffraction grating with 300 lines per millimetre is being used in a typical light experiment. Different types of light are allowed to fall normally on the diffraction grating and the resultant pattern to be studied is formed on a screen placed at distance D from the grating as shown in Figure 1. The distance D is 3.00 m.

Figure 1 a. The first light source to be studied is a monochromatic light source of wavelength 650 nm. i. Explain the meaning of the term “monochromatic”.


ii. State the reason why the light rays can be assumed to be parallel.


iii. Calculate the maximum number of maxima.


b. The next experiment is of light from a low pressure sodium lamp. Light from the lamp consists mostly of two wavelengths, 588.99 nm and 589.59 nm. i. Instead of the expected two spectral lines, only one spectral line is observed. Explain quantitatively why this has happened. [2] ii. Suggest an improvement to the experimental set-up to help overcome this problem. [1] [Total: 8]

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2. [AJC/H2/Prelims 2010/P3/6] a. i. Explain the meaning of the term diffraction.


ii. A ripple tank experiment is used to observe the appearance of plane water waves passing through gaps. Sketch on Figure 2 and Figure 3 the diffraction of the water waves. [2]

Figure 2 Figure 3 iii. A band is practising in the band room down the corridor with the door slightly ajar. Explain why a student walking along the corridor can hear the band but cannot see the band. [1] b. A student was standing 5.50 m away a loudspeaker S 1 placed at point X. The loudspeaker S1 is transmitting with a power output of 20.0 W. i. Assuming the loudspeaker S 1 is radiating uniformly in all directions and sound waves from the loudspeaker strikes the surface of the student’s ear perpendicularly, calculate the power intercepted by the student’s ear with an effective area of 2.0 × 10–3 m2. [2] ii. The actual power received by the student’s ear is 4.36 × 10 –2 W. Suggest why the actual power received is greater than that calculated in b.i.. [1] c. A second loudspeaker S 2 placed at point Y is 1.2 m from loudspeaker S 1 as shown in Figure 4. The sound waves from the two loudspeakers have frequency 2.75 kHz and speed 330 m s–1.

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Figure 4 i. Explain what is meant by the principle of superposition with reference to the sound waves emitted from the two loudspeakers. [2] ii. Show that the wavelength of the sound waves is 0.12 m.


iii. The loudspeaker S1 emits a signal that arrives at point A with intensity I and the loudspeaker S2 emits a signal that arrives at point A with intensity 2I. The student now stands on the centre line at point A and hears a sound of maximum intensity. As the student moves from point A to point B, the intensity varies between maximum and minimum values. At point B, the distance S1B is 3.82 m and S2B is 4.12 m. Determine the number of high intensity regions that are found between points A and B. Do not include the maximum at point A. [2] iv. State one condition that allowed the student to hear the maxima and minima intensity pattern. [1] v. Determine the intensity of the sound at point B in terms of I.


vi. Now, the two loudspeakers are placed closer together along the line joining X and Y. Without any calculations, state the difference that would be detected by the student as she walks from point B back to point A. [1] d. Keeping the same frequency, the two loudspeakers S 1 and S2 are now placed facing each other, at a distance d apart as seen in Figure 5. A microphone is placed at Z, midway between loudspeakers S1 and S2.

Figure 5 i. Explain why the microphone detects maximum signal at point Z.


ii. As the microphone moves from S1, a position of maximum intensity, to S 2, also a position of maximum intensity, along the line joining S 1S2, it detects 6 positions of minimum intensity. Determine the value of d. [2] [Total: 20] Page 15 of 75

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3. [CJC/H2/Prelims 2010/P3/7] a. i. State a property that can be used to differentiate a transverse wave from a longitudinal wave. [1] ii. State one feature of electromagnetic waves which is common across the whole spectrum. [1] b. Two-source interference fringes using light can only be obtained if light from the two sources is coherent. Explain i. the meaning of the term coherent;


ii. why, in practice, interference fringes can be seen only if light from a single source is split into two. [2] c. In an experiment on superposition, light from a laser is incident normally on a double slit, and the interference pattern is observed on a screen situated a distance D from the slits. The fringe spacing x is measured for a number of different values of D and the graph is plotted, as shown in Figure 6.

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x / mm 20.0 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0






3.0 D/m

Figure 6

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i. Determine the gradient of the graph shown in Figure 6.


ii. Use your result in c.i. to determine the slit separation a, given that the wavelength of the light is 600 nm. [2] iii. The experiment is repeated with the frequency of the light slightly decreased. State what will happen to 1. the interference pattern;


2. the graph shown in Figure 6.


d. Give reasons for the following. i. Sound waves and water waves can go round corners but light waves seem to travel only in straight lines. [2] ii. Figure 7 represents a stationary sound wave in a pipe. This figure looks like a transverse wave although sound waves are longitudinal waves. [2]





Figure 7 e. In a simple experiment to find out the wavelength of monochromatic red light emitted by a laser, a fine beam of red laser light is shone through a diffraction grating as shown below in Figure 8.


1.00 m


0.46 m

2.00 m

Figure 8 The diffraction grating has 300 000 rulings per meter and is set so that its plane is normal to the incident light. Bright spots are observed at 0.46 m and 1.00 m from the central spot on a screen, which is 2.00 m from the grating. i. From the first-order diffracted light, estimate the wavelength of the laser light. Page 19 of 75


ii. State and explain an advantage of obtaining the wavelength of the laser light by using the second-order diffracted light rather than the first-order diffracted light. [2] [Total: 20] 4. [Dunman High/H2/Prelims 2010/P2/3] The diffraction pattern due to a single light source on a single slit is shown in Figure 9 below.



screen Figure 9 a. Explain what is meant by the term diffraction.


Two additional slits were placed between the single slit and the screen as shown in Figure 10 and an interference pattern is observed on the screen.



screen Figure 10 b. i. Explain what is meant by term coherent.

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ii. The slits are 3.5 × 10 6 m apart and the wavelength of the light is 0.65 × 10 6 m. Calculate the fringe spacing x if the screen is placed 2.5 m away from the double slits. [2] iii. State and explain what will happen to the interference pattern if the single slit is now removed. [2] [Total: 6] 5. [HCI/H2/Prelims 2010/P3/3] A researcher is investigating a cubic crystal with x-rays. The x-rays are incident at an angle  with the crystal surface and the crystal has a lattice spacing a. He is looking at reflection from parallel planes of atoms in a thin film deposit of the material, as shown in Figure 11. Constructive interference occurs when the path difference between radiations reflected off adjacent layers is an integer product of the wavelength of the radiation.

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Reflected radiation

incident parallel beams of radiation

Top-most layer of atoms Second layer of atoms


Figure 11 a. Derive an expression of the path difference, x, between the radiation reflected from two  adjacent layers of atoms, in terms of and a. [1] b. i. When the x-rays of wavelength 0.165 nm are used, a strong first order maximum   23.5 occurs as the beam makes an angle of with the top-most plane. Calculate the lattice spacing a. [3] ii. The beams indicated in Figure 11 depict the conditions for first order maxima. Sketch, in the same figure, the beams of incident and reflected radiations corresponding to the second order maxima. [1] Page 22 of 75

c. It is suggested that electron beam can be used in place of X-rays to carry out this diffraction experiment. i. Calculate the maximum wavelength of X-rays which can be used to probe the crystal lattice. [2] ii. Hence, find the minimum speed which the electrons must possess to be used for electron diffraction. [3] [Total: 10]

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6. [IJC/H2/Prelims 2010/P2/5] A student sets up the apparatus shown in Figure 12 to demonstrate a two slit interference pattern on the screen. The set-up is modelled after Young’s double slit experiment. The slits S1 and S2 are of the same width. Single slit Double slit

laser beam


S1 S2



D Figure 12 a. Explain why the single slit S o is not necessary in this particular set-up.


b. The laser beam has a wavelength of 630 nm. The separation d is 1.0 mm and the distance D is 2.5 m. Determine the separation of the fringes on the screen. [2] c. Describe and explain what would be observed on the screen if the slit S 1 is made narrower by half compared to slit S 2 while maintaining the same separation d. [2] d. Explain how the result of this experiment provided evidence that light must exhibit wave properties. [2] [Total: 7]

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7. [MI/H2/Prelims 2010/P3/4] Figure 13 below shows an arrangement used to determine the wavelength  of monochromatic light emitted by a laser.

Figure 13 S1 and S2 are slits that are at right angles to the plane of this page. When illuminated by light from the laser, they form coherent sources of light. An interference pattern is formed on the  screen, from which measurements can be taken to determine . a. Explain what is meant by interference of two coherent sources.


b. i. Describe briefly the interference pattern produced using the arrangement shown in Figure 13 above. [2] ii. Describe the roles played by diffraction and interference in the production of the interference pattern. [2] iii. Calculate the wavelength

of light emitted by the laser, given the following values:

D = 2.0 m a = 0.5 mm fringe spacing = 2.5 mm.


c. The interference pattern is easier to observe when the amplitudes of the waves from the two coherent sources are similar. Explain why this is so.

[2] [Total: 10]

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8. [MJC/H2/Prelims 2010/P2/1] a. State the Principle of Superposition.


b. Two sinusoidal transverse waves W 1 and W2, of the same type, are incident simultaneously on a point P. The amplitude of W 2 is the same as the amplitude of W 1. The frequency of W2 is half the frequency of W 1. At a certain instant (time t = 0) at P, both waves have zero displacement and then both displacements increase in the same direction. Figure 14a is a graph of displacement at P against time for wave W 1. displacement (wave W1) time a

displacement (wave W2) time b

displacement (resultant wave) time c

Figure 14 i. On Figure 14b, sketch a graph of displacement against time for wave W 2.


ii. On Figure 14c, sketch a graph to show the resultant wave produced by the superposition at P of waves W 1 and W2. [3] iii. The frequency of wave W 1 is 4.2 × 1015 Hz. Determine the frequency of the resultant wave produced by the superposition of waves W 1 and W2. [2] iv. Explain why it is incorrect to say that waves W 1 and W2 are coherent.

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c. Blue light of wavelength 485.6 nm from a star is incident normally on a diffraction grating. The light is diffracted into a number of beams as shown in Figure 15. second order first order zero order first order second order grating Figure 15 The angular separation of the two second order beams is 45.7°. Calculate the number of lines per millimetre on the grating. [4] [Total: 13]

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9. [NJC/H2/Prelims 2010/P3/6] a. State what is meant by coherent waves.


b. i. Two coherent sources of sound waves are located at position X and Y as shown in Figure 16 below. The sources have zero phase difference. An observer stands at position O. If the frequency of the sound wave is 660 Hz, with suitable calculations, determine whether or not the observer experiences constructive or destructive interference. (Take the speed of sound to be 330 m s -1) [3] Sound source at X

Sound source at Y


12 m

Observer at O Figure 16 ii. The source at position Y is slowly moved to the right until it eventually reaches position Y’, as shown in Figure 17. Describe what is experienced by the observer at O while the source is being moved . [2] Sound source at X 9m

Sound source at Y’ 9m

12 m

Observer at O Figure 17 iii. Would you expect the observer to hear complete silence when there is destructive interference? Explain. [1]

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c. The setup in b.i. has been changed by replacing one of the sound sources with a reflecting plane (as shown in Figure 18). With suitable calculations, describe what is experienced by observer at O. (Note that reflected wave from a hard surface undergoes a phase change of  rad with respect to the incident wave.) [3] Sound source at Y 8m

12 m Reflecting plane

Observer at O Figure 18 d. Explain what is meant by diffraction of a wave.


e. A simplified model of the way the human eye responds to light of different wavelengths incident normally on a diffraction grating of spacing d = 2.5 µm is as follows: Light:

Perceived as:

Single wavelength between 0.40 µm and 0.50 µm


Single wavelength between 0.50 µm and 0.60 µm


Single wavelength between 0.60 µm and 0.70 µm


Determine whether there is any overlapping between the first order and second order spectra. [3] f. The spectrometer setup below shows how light from a collimator is made to fall normally on a diffraction grating.

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The telescope can be used to locate the second order bright fringes of any particular wavelength,

1 at angular positions

2 and

. Sodium vapour lamp of wavelength 589.3

2 1  nm is first used and the angle between and , is shown in the table below. The sodium vapour light is then replaced by a discharge tube containing a mixture of gases  and the values are recorded in table below for two pairs of second order bright fringes. Gas

   2  1

Sodium Unknown 1 Unknown 2

90.033° 71.367° 93.667°

i. State the purpose of placing a single slit before the light source.


ii. Identify the gases (unknown 1 and unknown 2) in the tube by using the data in the table below which shows the wavelength of the spectral lines emitted by various gases. [4] Gas Helium Carbon dioxide Hydrogen Oxygen

Wavelength / nm 668 608 486 441 [Total: 20]


[SAJC/H2/Prelims 2010/P3/2 (part)]

During the 2010 FIFA World Cup, the Vuvuzela (refer to Figure 19), the South African musical horn, was badly received by some television viewers as the playing of the horns by the stadium spectators created a constant annoying humming sound for the viewers.

Figure 19 Audio engineers have suggested the use of noise-cancelling technology to generate tones of approximately 235 Hz, 470 Hz and 705 Hz to cancel out the humming sounds. By considering the Vuvuzela to be a 70 cm long pipe with two open ends and taking the speed of sound to be -1 330 m s , explain the choice of the frequencies used and suggest how this method could cancel out the humming sounds. [4] Page 31 of 75

[Total: 4]

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[NYJC/H2/Prelims 2010/P2/4]

a. State two conditions that must be satisfied in order to obtain observable interference patterns. [2] The apparatus shown in Figure 20 below (not to scale) is used to demonstrate two-source interference. Screen

Double slit Light, wavelength



Figure 20 b. The separation of the two slits in the double slit arrangement is a and the interference fringes are viewed on a screen at a distance D from the double slits. When light of wavelength  is incident on the double slit, the separation of the bright fringes on the screen is x. i. Write the equation that links the quantities described in the above paragraph, and state the assumption made in the use of that equation. [2] ii. The slits are separated by a distance of a, with the screen at a distance of 1.00 m from the plane of the slits. The slits are illuminated by monochromatic light of wavelength 589.3 nm traveling perpendicular to the plane of the slits. It was observed that the distance between the two 4 th-order bright fringes is 20 mm. Calculate the separation of the slits, a. [2] c. i. Explain why the central fringe is always a bright one.


ii. Explain why an experiment using two separate sources of light will not show interference. [1] [Total: 8]

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[PJC/H2/Prelims 2010/P2/3]

a. A wine glass can be shattered through resonance by maintaining a certain frequency of high intensity sound wave. Figure 21 shows the side view of a wine glass vibrating in response to such a sound wave. On Figure 22, sketch a possible standing wave pattern on the rim of the glass as seen from the top. [2]

Figure 21 b. The speed v of a progressive wave is given by the expression

Figure 22

v  f . A stationary wave does not have a speed. By reference to the formation of a stationary

f wave, explain the significance of the product

for a stationary wave.

c. Explain what is meant by diffraction of a wave.

[3] [2]

d. A narrow beam of coherent light of wavelength 589 nm is incident normally on a diffraction grating having

4.00 102

lines for every 1 mm.

i. Determine the number of orders of diffracted light that are visible on each side of the zero order. [2] ii. A student suspects that there are in fact two wavelengths of light in the incident beam, one at 589.0 nm and the other at 589.6 nm. 1. State the order of diffracted light at which the two wavelengths are most likely to be distinguished. [1] 2. The minimum angular separation of the diffracted light for which two wavelengths may be distinguished is 0.10°. By means of suitable calculations, explain whether the student can observe the two wavelengths as separate images. [2] [Total: 12] Page 34 of 75

Page 35 of 75


[RI/H2/Prelims 2010/P3/6]

a. State two conditions for observable interference of two waves.


b. In an aircraft landing system, it is important to guide the aircraft along the centre-line of the runway prior to landing. In a simple landing system, rows of light guides are lined along the runway to help guide the pilot. The minimum power of light that can be detected by the human eye of area 0.50 cm2 is about 2.5 × 10-11 W. If an aircraft is 12 km away from the runway, find the required power of one light guide such that it is observable by the pilot. Assume that the light guide is a point source and that there are no energy losses. [3] c. In another type of landing system, aircrafts are guided using interference of radio waves. Figure 23 shows two radio wave emitters P and Q positioned 50 m apart at the end of the runway. The two emitters emit radio waves of frequency f1 in phase. The aircraft can be guided by searching for the strong signal radiated along the lines of constructive interference, also known as anti-nodal lines. To ensure that the aircraft is along the centre-line of the runway, the aircraft needs to “lock on” to the central antinodal line. Top view (figure not to scale) anti-nodal lines P

Q runway

Figure 23 i. Suggest why radio waves are used instead of waves of shorter wavelengths (e.g. microwaves, etc.). [2] ii. Explain why the entire centre-line will always be an anti-nodal line.

Page 36 of 75


d. One particular aircraft at a vertical height of 480 m strays off the centre-line as shown in Figure 24. Figure 25 shows the radio wave signals from P and Q detected by the aircraft in this position.

Page 37 of 75

480 m P Q

Page 38 of 75

180 m Figure 24 Signal detected by aircraft

4800 m

Signal A time

50 m

Signal B

Figure 25 i. The source of signal B is emitter P. Using Figure 25, explain why this is so.


ii. State the phase difference between signals A and B.


iii. Hence determine the frequency f1 of the radio wave used.


e. As an additional precaution to prevent the aircraft from “locking on” to the wrong antinodal line, the emitters can simultaneously emit another radio wave of a different

f1 f2 frequency f2. However, for this precaution to work, the ratio of the two frequencies

should not be an integer ratio (e.g.

1 2


2 3


4 3

, etc.).

i. Explain how this precaution can prevent the aircraft from “locking on” to the wrong anti-nodal line. [1] ii. Explain why the ratio of the two frequencies should not be an integer ratio.


f. Suggest one advantage and one disadvantage of the wave-interference system over the light guide system in guiding aircrafts to land safely. [2] [Total: 20] 14.

[RVHS/H2/Prelims 2010/P3/4]

Figure 26 shows a pair of identical loudspeakers A and B placed 2.00 m apart and emitting coherent sound waves of frequency 470 Hz. An observer walks from X to Y. The perpendicular distance between the sources and XY is 12.0 m. As he walks, he hears sound of maximum intensity at P, followed by minimum intensity at Q and the next maximum intensity at R. R is 4.50 m away from P.

Page 39 of 75

(diagram n

Figure 26 a. Explain why the observer hears sound of maximum and minimum intensity as he moves from X to Y. [2] b. i. AR is 12.5 m, show that BR is 13.2 m to 3 significant figures.


ii. Determine the wavelength of the sound.


iii. Determine the speed of the sound.


c. The power of the loudspeakers A and B are identical. Suggest why the intensity at Q is not zero. [3] [Total: 10]

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[TJC/H2/Prelims 2010/P3/6]

a. What do you understand by interference?


b. Figure 27 shows two loudspeakers S 1 and S2 connected to the same sound source such that they emit sound waves of the same intensity and wavelength. A sound detector is placed at point P such that S 1P = S2P initially. X


P S2 Figure 27 i. As the loudspeaker S 1 is moved slowly away from P along the line PS 1 towards X, the sound detected at P fluctuates in intensity. Explain this observation. [3] ii. As the loudspeaker S 1 is moved towards X through a distance of 0.082 m, the intensity of the sound detected at P decreases from a maximum to a minimum. Calculate the wavelength of the sound emitted by the sources. [2] iii. If S1 remains at point X and the frequency f of the sound emitted from both loudspeakers is now gradually changed to 4100 Hz, the sound intensity detected at P increases from the minimum in b.ii. to a maximum. Estimate a value for the speed of sound. [3] c. In another experiment to determine the speed of sound, a long tube, fitted with a tap, is filled with water. A tuning fork is sounded above the top of the tube as the water is allowed to run out of the tube, as shown in Figure 28.

Figure 28 Page 41 of 75

Figure 29

A loud sound is first heard when the water level is as shown in Figure 28, and then again when the water level is as shown in Figure 29. Figure 28 illustrates a stationary wave produced in the tube. i. Explain the formation of a stationary wave in the tube.


ii. Explain, by reference to resonance, why the loudness of the sound changes as the water level changes. [3] iii. On Figure 29, sketch the form of the stationary wave set up in the tube.


iv. The frequency of the fork is 512 Hz and the difference in the height of the water level for the two positions where a loud sound is heard is 32.4 cm. Calculate the speed of the sound in the tube.


v. The length of the column of air in the tube in Figure 28 is 15.7 cm. Suggest where the antinode of the stationary wave produced in the tube in Figure 28 is likely to be found. [2] [Total: 20] 16.

[YJC/H2/Prelims 2010/P3/6 (part)]

White light has a wavelength range from 400 nm to 750 nm. A diffraction grating with 6 × 105 lines per metre is placed at right angles to a ray of white light and produces the first and second order spectra as shown in Figure 30.

Page 42 of 75

Second order spectrum

White light


First order spectrum

Figure 30 a. Show, by calculation, that the angle  is greater than .


b. Show, by calculation, whether the second order spectrum overlaps with the third order spectrum. [3] c. State two advantages of analysing the light in the first order spectrum.


d. State what would be seen at A.

[1] [Total: 10]

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[VJC/H2/Prelims 2010/P3/2]

a. The figure below shows a thin taut wire held horizontally by two supports placed 0.40 m apart.

When the wire is plucked at its centre, a standing wave is formed and the wire vibrates in its fundamental mode of frequency 50 Hz. i. Explain why a standing wave is formed between the supports.


ii. Determine the speed of the wave in the wire.


iii. Sketch the next 2 higher modes which the string can vibrate in and hence determine their corresponding frequencies. [3] b. The wire is then connected to an a.c. source in a closed circuit and a magnet is brought near to the wire as shown in the next figure below. This causes the wire to vibrate in its fundamental mode with a large amplitude. When the movable support is shifted from its position, the amplitude of vibration decreases abruptly.

Page 44 of 75

Page 45 of 75

i. Explain the change in amplitude of the wire’s vibration when the movable support is shifted. Hence, deduce the frequency of the a.c. source. [3] ii. Suggest two ways that the same wire can be made to resonate with a fundamental frequency of 100 Hz. [2] [Total: 11]

Page 46 of 75


Solutions and Answers The suggested solutions and answers below are only for your reference, and have not been verified to be error-free. If you have any doubt please discuss with your tutor to clarify.

MCQs 1 D

2 B

3 C

4 B

5 C

6 D

7 C

8 B

9 D

10 B

11 D

12 D

13 C

14 B

15 B

16 A

17 D

18 D

19 C

20 C

21 B

22 C

23 D

24 D

25 B

26 A

27 C

28 A

29 D

30 B

31 C

32 C

33 D

34 B

Structured Questions 1 [AJC/H2/Prelims 2010/P2/4] a. i. Monochromatic means the light source is made of a single wavelength or frequency. [1] ii. Distance D is much longer than distance between 2 adjacent slits of the grating (i.e. grating spacing d). [1] Common mistakes: 

D large, hence small

, hence parallel.

 Invalid because for diffraction grating, 

Grating spacing

d 

0    90

, hence

is not small.

, hence no observable diffraction.

 Invalid as observable diffraction happens when you compare aperture size and not grating spacing d.

sin  iii. For diffraction grating, we have

n d


For maximum order, 1 103 m n d 300 sin  1  1 n    5.128 nmax  5 d  650 109 m

5 2 1 11 Hence, maximum number of maxima is Page 47 of 75



Common mistakes: 

Confusion between “maximum number of maxima” and “maximum order”

Page 48 of 75

b. 

i. For 588.99 nm & 1st order,

 9   1  588.99  10 m     sin1    10.1774 3 1 10 m     300 

For 589.59 nm & 1st order,

 9   1  589.59  10 m     sin1    10.1879 3 1 10 m   300  

The difference between the 2 angles is very small , hence we cannot be able to differentiate between / resolve the 2 spectral lines. [2] Common mistakes: 

Explanation by experiment)









 Irrelevant experiment 

Explanation via absorption spectrum principle where photons of only one corresponding energy level are suitable for absorption by the sodium atoms. The de-excitation of Na atoms causes emission of photons in all directions, hence, the corresponding spectral line is relatively darker, explaining why one spectral line can be seen.  Irrelevant as this is the case of an emission spectrum.  The “absorption of photons of 1 of the 2 mentioned wavelengths” is not possible as these photons are given out by the Na atoms themselves.

Difference between the energies corresponding to the 2 wavelengths is very small  de-excitation of Na atoms gives photons of similar energy  one spectral line  Irrelevant to talk about photons emitted from excited Na atoms.

Significant number of candidates found maximum numbers of maxima for both wavelengths and concluded that since both maximum no. of maxima were about the same, hence, only one spectral line was seen.  Having the same number of maximum number of maxima does not necessarily imply both spectral lines overlap.

Significant number of candidates mentioned that difference between wavelengths was small, resulting in a small difference in angles of diffraction and hence the 2 spectral lines overlapped.  Lacking quantitative analysis

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 Also, even though small difference between wavelengths leads to small sin1 sin2 difference between and , student lacked explanation of link/relevance as to why small difference between angles of diffraction was resulted.

sin  




n  sin   for the same n and d d



1 sin1    1 1  1 2 sin 2 2 .  Ambiguous working/rationale 

 

Did student assume that as ASSUMPTION Did student state that

0    90

is small, hence

sin  


sin1  sin2  1  2

, so


Some mentioned the waves interfered constructively and destructively hence only one line was seen.  Each maximum (coloured line seen) is a result of constructive interference. Each dark region between adjacent maxima is a result of destructive interference. However, the dark regions have varying degrees of “darkness”, hence varying degrees of destructive interference. In this case, 2 coloured lines of wavelengths 588.99 nm and 589.59 nm are supposed to be seen where both are results of constructive interference.

Handful commented that the nth bright fringe overlapped with the (n + 1)th fringe  1 spectral line was seen.  Wrong to talk about fringe (for double slit)  Wrong concept

ii. Any one way [1] Change grating to a finer one or use a grating with more lines per millimetre (Do not accept mere stating of “decrease d”) Move the screen further away Use a light sensor based detector attached to a data logger to detect the 2 spectral lines that cannot be distinguish by our naked eyes Common mistakes: 

Terminologies  Smaller slits/apertures instead of smaller grating spacing Page 50 of 75

 Increase number of lines  ambiguous, needs to include “per millimetre”. Many did not spell out “no” and “mm”. Avoid writing short forms.  Increase diffraction grating / use smaller diffraction grating  lacking understanding or expertise in terminologies 

Large number suggested changing the light source to  a source with one wavelength / another light source where the difference of 2 wavelengths are larger  not solving the identified problem which is to to better distinguish the 2 wavelengths of sodium light  white light  not sure what students hoped to achieve since this does not help distinguish the 2 wavelengths of sodium light at all and further complicates the problem by introducing more wavelengths of light.

Some suggested increasing the pressure of lamp (especially for those who answered b.i. in terms of energies of absorption spectrum). [Total: 8]


[AJC/H2/Prelims 2010/P3/6]

a. i. Diffraction is the bending of waves around an object or spreading of waves through an aperture [1] where the linear dimensions of the object or aperture should be comparable to the wavelength of the waves . [1] ii. contrast between spreads for the 2 figures (basic understanding) [1] contrast between curvatures for the 2 figures, constant wavelength before and after gap for each figure, include drawing of wavefront at the apertures for each figure [1]

iii. The ratio of wavelength to the size of door gap is large (ratio > 1) for sound and small (ratio ≪ 1) for light waves. Hence, more significant diffraction is observed for sound waves than light. [1] b. i. Intensity at 5.50 m from loudspeaker S 1:

Page 51 of 75


Pradiated 20.0 W   0.052613 W m2  0.0526 W m2 (3 s.f.) [1] 2 2 4 r 4  5.50 m

Hence, power intercepted by the student’s ear:

Preceived  I  Aear   0.052613 W m2    2.0 103 m2   1.0523 104 W  1.05 104 W (3 s.f.) [1] ii. In reality, the signal is not transmitted uniformly in all directions but in a more directional manner towards the student . Thus power received would be much larger than that calculated in b.i.. [1] c. i. When the two sound waves from S1 and S2 meet at a point, the resultant displacement at that point is the vector sum of the separate displacements that each individual sound wave would cause at that point. [2]

v f   

v 330 m s1   0.120 m (3 s.f.) (shown) f 2.75 103 s1

ii. We have:


iii. At A, path difference = 0; at B, path difference = 4.12 m – 3.82 m = 0.30 m = 2.5λ. [1] Hence, number of maxima = 2. [1] iv. The two sound waves are coherent / have constant phase difference. [1] v. At A, path difference = 0 corresponds to a constructive interference (maximum amplitude of sound), so at B, path difference = 2.5λ corresponds to a destructive interference (a soft sound). Now we need to find how the amplitudes of sound from the two sources are related. We have, at A:



 A   2  2 A2  A1 2 IA1  A1

Hence, at B, the resultant amplitude is:

AB  A2  A1 

2  1 A1 [1]

Hence, intensity at B: 2

IB  AB    IA1  A1



 IB  0.172IA1  0.172I [1]

vi. Either: When separation between loudspeakers S 1 and S2 is decreased, spacing between consecutive high intensity regions increases. [1] Page 52 of 75

Or: The intensity of the high intensity regions will be smaller. Hence the maximum would be softer. (But the difference may not be significant and hence may not be effectively detected by our naked ear. Hence the first answer may be detected easier). [1] d. i. A stationary wave is formed between the 2 speakers and a displacement node occurs at point Z where there is maximum variation of pressure; maximum signal detected. [1] ii. We have: L – regions of louder volume/higher intensity S – regions of smaller volume/lower intensity

d  3  3 0.120 m 0.360 m (3 s.f.) So,

[2] [Total: 20]


[CJC/H2/Prelims 2010/P3/7]

a. i. A transverse wave can be polarised while a longitudinal wave cannot. [1] ii. Constant speed of

3.00 108 m s1

in vacuum. [1]

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b. i. Coherent waves are waves that have similar frequency with a constant phase difference. [2] ii. Light from a single source when split into two ensures that the two ‘new’ waves are coherent. Light waves from two different sources are never coherent as the two sources emit light waves differently due to the random nature of their emitting mechanisms. [2] c. x / mm 20.0 (4.500, 17.9)

18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0

(1.000, 4.0)











 17.9 4.0  103 m  13.9 103 m  3.9714 103  3.97 103 3.500 m  4.500 1.000 m

(3 s.f.)


D/m i. From graph, gradient 

[1] Page 54 of 75

ii. For double slit diffraction pattern, fringe separation x and slit separation a are given by


D a


From the plot of x against D, the gradient gives

 a

, so

 600 109 m  3.9714 103  a   1.5108 104 m 1.51 104 m (3 s.f.) a 3.9714 103


iii. The experiment is repeated with the frequency of the light slightly decreased. State what will happen to 1. The fringe separation will increase slightly. [1] 2. The gradient of the graph increases. / The graph becomes steeper. [1] d. i. The ratio of wavelength to the size of common corners is large (ratio > 1) for sound and water waves and small (ratio ≪ 1) for light waves. Hence, more significant diffraction (i.e. bending around corners) is observed for sound and water waves than light. [2] ii. Sound waves and water waves can go round corners but light waves seem to travel only in straight lines. [2] iii. Figure 7 shows the horizontal displacement of air particles from equilibrium positions against the horizontal distance along the tube. Air particles only vibrate horizontally, consistent with sound waves being longitudinal waves. It is the graph of their horizontal displacements that appears like the shape of a transverse wave. [2] e.

1  tan1 i. Angle of diffraction of the first-order diffracted light:

0.46 m  12.953  13 (2 s.f.) 2.00 m

Let d be the slit separation of the diffraction grating. We have, for the first-order diffraction:

dsin1   

1m  sin 12.953   7.4717 107 m 750 nm (2 s.f.) 300000


ii. The second-order fringe is further away from the central maximum, so the percentage error in the distance measured is smaller. This will result in a smaller percentage error propagated to the value of wavelength calculated. [2] [Total: 20] Page 55 of 75


[Dunman High/H2/Prelims 2010/P2/3]

a. Diffraction is the spreading of waves through an aperture or around an obstacle. [B1] b. i. Two wave sources of similar frequency are said to be coherent if they have a constant phase difference. [B1]

Page 56 of 75

ii. Fringe spacing: x

6 D  0.65 10 m   2.5 m  [M1] d 3.5 106 m  0.4643 m 0.46 m (2 s.f.) [A1]

iii. If the single slit is removed, the light passing through the double slit will not be coherent. [M1] Hence the interference pattern will not be observed. [A1] [Total: 6] 21.

[HCI/H2/Prelims 2010/P3/3]

a. Consider the geometry of the beams:

Hence, path difference is:

x  2asin


b. i.

x  n

. For first-order constructive interference, we have:

x   1  [M1]  2asin1    a 

0.165 109 m [M1] 2sin 23.5 

 2.06897 1010 m 2.07 1010 m (3 s.f.) [A1]

Page 57 of 75

ii. Second-order beams at a greater angle [B1] 2nd order

1st order

c. i. We have: 2asin  n 2asin 2a    as sin  1 n n 2a is maximum when n  1 is minimum [M1] n Hence,

max  2a  2  2.06897 1010 m  4.1379 1010 m 4.14 1010 m (3 s.f.) [A1] ii. Using de Broglie’s equation to relate electrons’ wavelength and their momentum, mv e 

h h  v [M1]  me h  [M1] memax 

6.63 1034 J s  1.76 106 m s1 31 10 9.11  10 kg  4.1379  10 m    

1.76 106 m s1 [A1] Hence, the minimum speed is

. [Total: 10]


[IJC/H2/Prelims 2010/P2/5]

a. The light source used is a laser and it is already coherent by nature. [B1] b. Fringe separation: 9 D  630 10 m   2.5 m x  [M1] d 1.0 103 m

 1.575 103 m 1.6 103 m (2 s.f.) [A1]

Page 58 of 75

Page 59 of 75

c. Since the separation of the slits is maintained, the separation of the fringes is unchanged. [B1] Because one slit is made narrower, the fringes are not so distinct or clearly observable (i.e. the contrast is lowered ). The amplitude of the light emerging from it will be less than that from the other slit. The amplitudes are not the same and hence, when the waves combine at the locations of destructive interference, there are no total cancellations of waves . [B1] d. The observation that there are bright and dark fringes on the screen is an indication that interference has taken place . [B1] Bright fringes are where constructive interference takes place while dark fringes are where destructive interference takes place. The summation of the two combined light rays is a direct consequence of the principle of superposition which applies to waves. Thus, light has wave properties. [B1]

[B1] [Total: 7] 23.

[MI/H2/Prelims 2010/P3/4]

a. The superposition of two waves meeting at a point in space [B1] to give a resultant wave whose amplitude is given by the Principle of Superposition [B1]. b. i. Alternating bright and dark fringes [B1] Bright fringes decreasing in intensity as order increases [B1] ii. Diffraction occurs at the slits where the beams spread out and overlap . [B1] Interference occurs when the 2 waves meet at different positions on the screen . Constructive interference occurs if both waves are in phase, and destructive if out of phase. [B1] iii. Let x be the fringe spacing. We have: 3 3 D xa  2.5 10 m   0.5 10 m x    [M1] a D 2.0 m

 6.25 107 m 6.3 107 m (2 s.f.) [A1] c. Constructive interference results in twice of original amplitude and destructive interference results in zero amplitude. [B1] Hence, better contrast for observation. [B1] [Total: 10] 24.

[MJC/H2/Prelims 2010/P2/1] Page 60 of 75

a. The Principle of Superposition states that when two waves of the same kind meet at a point in space, the resultant displacement at that point is the vector sum of the displacements that the two waves would separately produce at that point. [B2]

Page 61 of 75

b. i. A positive sine wave, i.e. correct phase at start [B1] Amplitude = 1 div [B1] Period = 8 div [B1] ii. Period = 8 div [B1] Two big peaks, two small peaks [B1] Approximate correct heights/times [B1]

iii. The frequency of wave W 1 is 4.2 × 1015 Hz. The frequency of wave W 2 is ½ × (4.2 × 1015 Hz) = 2.1 × 1015 Hz. [M1] Frequency of the resultant wave is the same as the wave with a lower frequency i.e. the frequency of resultant wave is 2.1 × 10 15 Hz. [A1] iv. They do not have a constant phase difference . [B1] OR They do not have the same wavelength /frequency. [B1] c. We have:

2 

45.7  22.85  22.9 (3 s.f.) [C1] 2

Let d be the grating space. We have:

dsin2  2

2  485.6 109 m 2  d  [M1] sin2 sin 22.85   2.501 106 m 2.50 106 m (3 s.f.) [M1] Hence, number of lines per millimetre:

1 mm  400 (3 s.f.) [A1] 2.501 103 mm [Total: 13]

Page 62 of 75


[NJC/H2/Prelims 2010/P3/6]

a. Two waves are said to be coherent when there is a constant phase difference between them. [1] b.


v 330 m s1   0.500 m (3 s.f.) f 660 Hz

i. Wavelength of the sound waves:

 9 m


  12 m  15 m 2

Path OX: Path difference:

15 m 12 m 3 m 6


Since the sources have zero phase difference , the path difference, being an integer multiple of the wavelength, shows that the waves are in phase at O , hence the observer experiences constructive interference . [3] ii. When the source is at Y’, the new path difference at O = 0 m, thus there is constructive interference. 6 As the source moves from initial position Y to Y’, the path difference goes from to 0. Thus the observer at O experiences 6 instances of soft sound (or 6 instances of 5.5 4.5 3.5 2.5 1.5 destructive interference) when the path difference is , , , , and 0.5 and 7 instances of loud sound (or 7 instances of constructive interference) 6 5 4 3 2 1 when the path difference is , , , , , and 0. [2]

iii. No. Amplitudes of the waves depend on their intensities which depend on the distances from the sources to the observer . Since the 2 waves travel different distances from the source to the observer, the amplitudes of the two waves will not be exactly the same , hence there will not be complete cancellation. [1]


 6 m


  8 m  20 m 2

c. Path of the reflected wave: Path difference between the wave coming directly from Y and the reflected wave: 20 m 12 m 8 m 16 Due to the π radians phase shift at reflection, the two waves are out of phase at O, and hence interfere destructively . The observer will thus hear a soft sound . [3]

Page 63 of 75

Sound source at Y 8m

Reflecting planeReflected wave 12 m

Observer at O d. Diffraction is the spreading of waves when the waves pass through a gap or around an obstacle. [2] e. In order to determine if there is overlap of the first and second spectra, we have to check the maximum angular deviation of the first spectra and the minimum angular deviation of the second spectra. In the first order spectra, red light with the longest wavelength will constitute the largest angular deviation. Hence, maximum angular deviation for the first-order spectra is: dsin1, max  1red, max  red, max  1 0.70 m    sin    16.3  16 (2 s.f.) d    2.5 m

 1, max  sin1 

Minimum angular deviation for the second-order spectra: dsin2, min  2 blue, min  2 blue, min  1 2 0.40  m   sin    18.7  19 (2 s.f.) d  2.5 m  

 2, min  sin1 

From the above calculation, angular deviation for 2nd order blue light > angular deviation for 1st order red light. There is no overlapping of 2nd order blue light and 1st order red light. Therefore, there is no overlapping between the first-order and second-order spectra. [3] f. i. It is to produce a coherent light when it pass through the collimator. [1] ii. Using

dsin  n

 ' sin '   sin

, for second-order spectra of various wavelengths,

 where

 2  1 2


Page 64 of 75

Na  For Na,

90.033  45.0165  45.017 (5 s.f.) 2

For unknown 1:  71.367  71.367   sin sin   2  2 unknown 1  Na    589.3 nm    485.99 nm 486.0 nm (4 s.f.) sin 45.0165  sin 45.0165  Hence, unknown 1 is hydrogen. For unknown 2:  93.667  93.667   sin sin   2  2   unknown 2  Na   589.3 nm   607.68 nm 607.7 nm (4 s.f.) sin 45.0165  sin 45.0165  Hence, unknown 2 is carbon dioxide. [4] [Total: 20]

Page 65 of 75


[SAJC/H2/Prelims 2010/P3/2 (part)]

Consider the Vuvuzela as a tube of length L with two open ends. For waves of speed v,

f resonance frequencies are given by

nv 2L

. f1 

At the fundamental frequency, n = 1, and so

1 330 m s1  236 Hz 2 0.70 m


The next few harmonics will be multiples of the fundamental frequency, hence f2 = 236 × 2 = 472 Hz and f3 = 236 × 3 = 708 Hz. [1] By generating sound at the same resonance frequencies as well as almost similar amplitudes [1] as the vuvuzela but 180° out of phase [1], destructive interference occurs and both sound waves cancel out. [Total: 4] 27.

[NYJC/H2/Prelims 2010/P2/4]

a. Any two [2] The sources must be coherent (for interference to be possible). The amplitude of the waves at the point of interference must be about the same (for interference pattern to be observable). The distance between the two sources is much larger than the wavelength of the waves emitted. b.

x i.

D a

D? a ; assumption:

for small-angle approximation to be valid [2]

 4  tan1 ii. Angular deviation of a 4 th-order bright fringe:

20 mm 2  0.57294  0.57 1.00 103 mm

We have: asin 4  4  a

4 589.3 109 m  2.357 104 m 0.24 mm (2 s.f.) [2] sin 0.57294 

c. i. Since the central fringe is equidistant from the double slits, path difference = 0 . Hence the waves from each slit must arrive in phase resulting in constructive interference. [1] Page 66 of 75

ii. Two separate sources of light will not be coherent and hence interference cannot take place. [1] [Total: 8]

Page 67 of 75


[PJC/H2/Prelims 2010/P2/3]

a. Stationary wave drawn around rim [2]

b. A stationary wave is formed by the superposition of two progressive waves of the same type with equal speed, frequency, wavelength and amplitude travelling in opposite directions. The waveform does not advance and there is no net energy transfer. explaining the formation of stationary waves [2]

f The product

in a stationary wave refers to the speed of the two progressive

fλ waves that superpose to form the stationary wave. explaining significance of


c. Diffraction is the spreading of waves through an aperture or around an obstacle. It is observable when the wavelength of the wave is of the same order of magnitude as the dimensions of the aperture or obstacle . [2] d. A narrow beam of coherent light of wavelength 589 nm is incident normally on a diffraction grating having d i. Slit separation:

4.00 102

lines for every 1 mm.

1 103 m  2.50 106 m (3 s.f.) 4.00 102


We have: dsin  n dsin d  since sin  1 for   90 for observable fringes   2.50 106 m  n  4.24 589 109 m  n

Hence, the maximum order observable is 4th . [1] ii. A student suspects that there are in fact two wavelengths of light in the incident beam, one at 589.0 nm and the other at 589.6 nm. 1. 4 th since the higher the order, the greater the separation between the fringes. [1] 2. For wavelength 589.0 nm, we have: Page 68 of 75

 2.50 10


m  sin  4  589.0 109 m

   70.459 For wavelength 589.6 nm, we have:

 2.510


m  sin '  4 589.6109 m

  '  70.624 calculation of angle for both wavelengths [1]

angularseparation  70.624  70.459  0.165  0.10 We thus have: This is greater than minimum required , therefore students will be able to observe the wavelengths as separate images . calculating the difference between the 2 angles and correct conclusion [1] [Total: 12] 29.

[RI/H2/Prelims 2010/P3/6]

a. Any two [B1] × 2 

The waves must be coherent.

The waves must have approximately the same amplitude.

The waves must be unpolarised or polarised in the same plane (for transverse waves).

The waves must interfere to give regions of maxima (constructive interference) and minima (destructive interference).

b. Intensity required at 12 km away: I

Peye Aeye

 2.5 10


 0.50 10



m 2

 5.0 107 W m2 (2 s.f.) [M1]

Consider the light guide: I

Plight A

 Plight  I  A   5.0 107 W m2    4  12000 m   



 904.78 W  900 W (2 s.f.) [A1]

c. i. Shorter wavelengths means the anti-nodal lines will be closer to one another. [B1] Hence, aircrafts may “lock on” to the wrong line of maxima / difficult to identify the central line of maxima / difficult to differentiate the lines of maxima. [B1] Page 69 of 75

ii. Since the two radio waves are in phase [B1], along centre-line, path difference is always zero / phase difference is always zero / P & Q are equidistant from any point on the centre-line. Hence constructive interference occurs. [B1] d. i. P is nearer to the aircraft , hence intensity (or amplitude) of signal should be higher. [B1]

ii. From Figure 24, phase difference =

iii. Since phase difference =

 rad 2

 rad 2

. [B1]

, path difference =

 4


2 2   180 m   480 m  4827.30 m  [M1] 2 2 2 Distance from Q to plane   4800 m   230 m   480 m  4829.42 m

Distance from P to plane 

 4800 m

4829.42 m 4827.30 m Hence,

f1  Thus,


    8.48 m [M1] 4

c  35.3 106 Hz [A1] 

e. i. If aircraft is on the central anti-nodal line , it should detect maximum signals from both frequencies / the maximum signal will be stronger . [B1] OR If the aircraft is on the wrong anti-nodal line , only one of the frequencies will show a strong signal . [B1] ii. If the ratio is an integer ratio, higher orders of maxima will still coincide/overlap [B1]. Hence the aircraft could still detect maximum signals from both frequencies even though it is not on the central anti-nodal line . [B1] f. Advantage: Can still work under low visibility conditions / Use of detector to align aircraft is more accurate than using visual inspection . [B1] Disadvantage: Possible interference of signals from other sources (e.g. radio stations, telecommunication base stations, etc) / It is more costly to install the emitters and receivers on every airplane . [B1] [Total: 20] 30.

[RVHS/H2/Prelims 2010/P3/4] Page 70 of 75

a. The distances of a point on XY from the two sources are different. When the difference in distance is integer multiples of wavelength apart , the wave meet in phase and constructive interference occurs. This results in maximum intensity. When the difference in distance is half-integer multiples of wavelength apart , the wave meet out-of-phase and destructive interference occurs. This results in minimum intensity. [2] b.


 12 m


  5.5 m  13.200 m 13.2 m (3 s.f.) (shown) 2

i. We have:


ii. Since a maximum is detected at P and next at R, the path difference from A and B at 1 R is . [1]

  BR AR 13.2 m 12.5 m 0.7 m (1 d.p.) [1] Thus, v  f    470 Hz   0.7 m  329 m s1  330 m s1 (2 s.f.) iii. We have:


Page 71 of 75

c. Distances to Q from A and B are different. Since intensity at a position from a point source is inversely proportional to the square of the distance between them, the intensities of the waves arriving at Q from A and B are different. Since intensity is proportional to the square of the amplitude , the amplitude of the waves arriving from A and B will be different. Q is a position with destructive interference without complete cancellation of the waves occurs. Hence the intensity at Q is not zero. [3] [Total: 10] 31.

[TJC/H2/Prelims 2010/P3/6]

a. It is a phenomenon in which waves from two or more coherent waves superpose with one another producing a resultant wave whose amplitude is given by the Principle of Superposition . [1] b. i. Sound from the 2 sources undergo interference. As the source S 1 moves, the path difference changes . [1] When the path difference between S1P and S2P is an integral number of the wavelength, constructive interference occurs at P and a maxima is detected. [1] When the path difference between S1P and S2P is an odd integral number of half wavelength, destructive interference occurs at P and a minima is detected. [1] ii. At destructive interference,

  0.082 m [1] 2    2 0.082 m 0.164 m [1]

path difference 

S1X    0.082 m

iii. At this new frequency, since constructive interference is detected at P, . [1] Hence,

v  f    4100 Hz   0.082 m [1]  336.2 m s1  340 m s1 (2 s.f.) [1] c. i. Waves travel down the tube and get reflected. [1] The incident and the reflected waves, both having the same amplitude, frequency and speed travelling in opposite directions superpose to form standing wave . [1] ii. Air column in tube has a natural frequency of vibration . [1] Page 72 of 75

When the fork’s frequency is equal to the natural frequency of the air column, resonance occurs; there is maximum energy transfer and so maximum amplitude of vibrations occurs , leading to maximum loudness. [1] When the fork’s frequency is not equal to the natural frequency of the air column, no resonance occurs and loudness drops. [1] iii. Sketch: Antinode at top, node at surface of water, 1 antinode and 1 node in between [1] iv. We have:

  L1  c 4 3  L2  c 4

(1) (2)

(2) – (1): 1   L2  L1  32.4 cm [1] 2    64.8 cm [1] v  f    512 Hz   0.648 m  332 m s1 (3 s.f.) [1] Then,

v. We have:

1  1    L1  c  15.7 cm c  c    64.8 cm   15.7 cm  0.50 cm 4  4 


Therefore, antinode is 0.50 cm above the top of the tube / 16.2 cm above water surface. [1] [Total: 20] 32.

[YJC/H2/Prelims 2010/P3/6 (part)]

d a. Slit separation on the diffraction grating: Using

dsin  n

Wavelength,  / nm 400 750

1m  1.6667 106 m 5 6 10

, we have:

1 /  Angle,

2 /  (n = 1)

13.89 26.74   26.74  13.89  12.85


(n = 2)

28.69 64.16   64.16  28.69  35.47

Page 73 of 75

  Thus,

[4]  3 400 109 m   46.05 6  1.6667 10 m

sin1  b. At n = 3, the 400-nm light is at angle of


It is before the second-order fringe of the 750-nm light at an angle of 64.16°. Hence overlapping occurs since the minimum angle of the third-order spectrum is larger than the maximum angle of the second-order spectrum . [3] c. No overlapping occurs. [1] The spectrum is brighter. [1] d. A white, bright spot/fringe. [1] [Total: 10] 33.

[VJC/H2/Prelims 2010/P3/2]

a. i. Incident waves travelling along the wire to the ends are reflected. The incident and reflected waves travelling in opposite directions have the same frequency and amplitude. They superpose to form standing waves. [2] v  f   50 Hz  0.80 m  40 m s1 (2 s.f.) [1] ii. v  ff   iii. First overtone:

v 40 m s1   100 Hz (2 s.f.)  0.40 m

0.40 m

v  ff   Second overtone:

v 40 ms1   150 Hz (2 s.f.)  2 0.40 m 3

Page 74 of 75


0.40 m

b. i. When the support is shifted, the natural frequency of the wire is changed and no longer matches the driving frequency of the periodic force produced by the alternating current in the wire and magnetic field. The system no longer resonates and hence its amplitude decreases . The frequency of the a.c. source must therefore be the natural frequency of the wire which is 50 Hz. [3] ii. The length l can be reduced to 0.20 m to double the fundamental frequency . The weights attached to the wire can be increased to increase the speed of the waves to 80 m s-1. [2] [Total: 11]

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