Interference — Paper 2 · A Level Physics 9702

5 (a) State the relationship between the intensity and the amplitude of a wave. … … [1] (b) Microwaves of the same amplitude and wavelength are emitted in phase from two sources P and Q. The sources are arranged as shown in Fig. 5.1. P 1.840 m X 2.020 m path of detector Q Fig. 5.1 A microwave detector is moved along a path that is parallel to the line joining P and Q. A series of intensity maxima and intensity minima are detected. When the detector is at a point X, the distance PX is 1.840 m and the distance QX is 2.020 m. The microwaves have a wavelength of 6.0 cm. (i) Calculate the frequency of the microwaves. frequency = … Hz [2] (ii) Describe and explain the intensity of the microwaves detected at X. … … … … … [3] (iii) Describe the effect on the interference pattern along the path of the detector due to each of the following separate changes. 1. The wavelength of the microwaves decreases. … … 2. The phase difference between the microwaves emitted from the sources changes to 180°. … … [2] [Total: 8]

8 marks

Mark scheme: 5(a) B1 5(b)(i) v = fλ or c = fλ C1 f = 3.00 × 108 / 0.060 = 5.0 × 109 Hz A1 5(b)(ii) (at X path) difference = 3λ M1 (at X phase) difference = 0 or 1080° M1 so intensity is at a maximum/it is an intensity maximum A1 5(b)(iii) 1. decrease in the distance between (adjacent intensity) maxima/minima B1 2. (intensity) maxima and minima exchange places B1

This question in 9702/23 May/June 2018

4 (a) State the principle of superposition. … … … [2] (b) An arrangement for demonstrating the interference of light is shown in Fig. 4.1. B P D Q B D light central wavelength a 22 mm B bright 610 nm fringe D B D B 2.7 m screen double slit Fig. 4.1 (not to scale) The wavelength of the light is 610 nm. The distance between the double slit and the screen is 2.7 m. An interference pattern of bright fringes and dark fringes is observed on the screen. The centres of the bright fringes are labelled B and centres of the dark fringes are labelled D. Point P is the centre of a particular dark fringe and point Q is the centre of a particular bright fringe, as shown in Fig. 4.1. The distance across five bright fringes is 22 mm. (i) The light waves leaving the two slits are coherent. State what is meant by coherent. … … [1] (ii) 1. State the phase difference between the waves meeting at Q. phase difference = … ° 2. Calculate the path difference, in nm, of the waves meeting at P. path difference = … nm [2] (iii) Determine the distance a between the two slits. a = … m [3] (iv) A higher frequency of visible light is now used. State and explain the change to the separation of the fringes. … … [1] (v) The intensity of the light incident on the double slit is now increased without altering its frequency. Compare the appearance of the fringes after this change with their appearance before this change. … … … … [2] [Total: 11]

11 marks

Mark scheme: 4(a) when (two or more) waves meet (at a point) B1 (resultant) displacement is the sum of the individual displacements B1 4(b)(i) constant phase difference (between the waves) B1 4(b)(ii) 1. phase difference = 360° or 0 B1 2. path difference = 1.5λ = 1.5 × 610 = 920 nm A1 4(b)(iii) λ = ax / D C1 x = 22 / 4 (= 5.5 mm) or 22 × 10–3 / 4 (= 5.5 × 10–3 m) C1 a = (610 × 10–9 × 2.7) / (5.5 × 10–3) = 3.0 × 10–4 m A1 4(b)(iv) shorter wavelength and (so) separation decreases B1 4(b)(v) • no change to fringe separation/fringe width/number of fringes • bright fringes are brighter • dark fringes are unchanged Any two of the above three points, 1 mark each. B2

This question in 9702/21 Oct/Nov 2018

5 (a) By reference to two waves, state: (i) the principle of superposition … … … … [2] (ii) what is meant by coherence. … … [1] (b) Two coherent waves P and Q meet at a point in phase and superpose. Wave P has an amplitude of 1.5 cm and intensity I. The resultant intensity at the point where the waves meet is 3I. Calculate the amplitude of wave Q. amplitude = … cm [2] (c) The apparatus shown in Fig. 5.1 is used to produce an interference pattern on a screen. laser light a wavelength 680 nm D double-slit screen Fig. 5.1 (not to scale) Light of wavelength 680 nm is incident on a double-slit. The slit separation is a. The separation between adjacent fringes is x. Fringes are viewed on a screen at distance D from the double-slit. Distance D is varied from 2.0 m to 3.5 m. The variation with D of x is shown in Fig. 5.2. 10.0 8.0 x / mm 6.0 4.0 2.0 0 2.0 2.5 3.0 3.5 D / m Fig. 5.2 (i) Use Fig. 5.2 to determine the slit separation a. a = … m [3] (ii) The laser is now replaced by another laser that emits light of a shorter wavelength. On Fig. 5.2, sketch a possible line to show the variation with D of x for the fringes that are now produced. [2] [Total: 10]

10 marks

Mark scheme: 5(a)(i) (two) waves meet/overlap (at a point) B1 (resultant) displacement is sum of the displacement of each wave B1 5(a)(ii) constant phase difference (between the waves) B1 5(b) I ∝ A 2 3I / I = (A + 1.5)2 / 1.52 C1 A = 1.1 cm A1 5(c)(i) λ = ax / D C1 e.g. a = 680 × 10–9 × 2.0 / 4.0 × 10–3 C1 a = 3.4 × 10–4 m A1 5(c)(ii) straight line from positive value on x-axis and always below ‘old’ line B1 straight line with a smaller positive gradient than ‘old’ line B1

This question in 9702/22 Feb/March 2019

4 (a) For a progressive water wave, state what is meant by: (i) displacement … … [1] (ii) amplitude. … … [1] (b) Two coherent waves X and Y meet at a point and superpose. The phase difference between the waves at the point is 180°. Wave X has an amplitude of 1.2 cm and intensity I. Wave Y has an amplitude of 3.6 cm. Calculate, in terms of I, the resultant intensity at the meeting point. intensity = … [2] (c) (i) Monochromatic light is incident on a diffraction grating. Describe the diffraction of the light waves as they pass through the grating. … … … [2] (ii) A parallel beam of light consists of two wavelengths 540 nm and 630 nm. The light is incident normally on a diffraction grating. Third-order diffraction maxima are produced for each of the two wavelengths. No higher orders are produced for either wavelength. Determine the smallest possible line spacing d of the diffraction grating. d = … m [3] (iii) The beam of light in (c)(ii) is replaced by a beam of blue light incident on the same diffraction grating. State and explain whether a third-order diffraction maximum is produced for this blue light. … … … [2] [Total: 11]

11 marks

Mark scheme: 4(a)(i) distance (in a specified direction of particle/point on wave) from the equilibrium position B1 4(a)(ii) the maximum distance (of particle/point on wave) from the equilibrium position or the maximum displacement (of particle/point on wave) B1 4(b) I ∝ A2 C1 IR / I = (3.6 – 1.2)2 / (1.2)2 resultant intensity = 4.0I A1 4(c)(i) as wave(s) pass through the slit(s) B1 wave(s) spread (into geometric shadow) B1 4(c)(ii) nλ = d sin θ C1 3λ = d sin 90° or 3λ = d C1 d = 3 × 630 × 10–9 = 1.9 × 10–6 m A1 4(c)(iii) wavelength of blue light is shorter (than 540 nm/630 nm/wavelengths of original light) M1 (so) third order diffraction maximum is produced A1

This question in 9702/22 May/June 2019

5 A ripple tank is used to demonstrate the interference of water waves. Two dippers D1 and D2 produce coherent waves that have circular wavefronts, as illustrated in Fig. 5.1. D1 D2 X Fig. 5.1 The lines in the diagram represent crests. The waves have a wavelength of 6.0 cm. (a) One condition that is required for an observable interference pattern is that the waves must be coherent. (i) Describe how the apparatus is arranged to ensure that the waves from the dippers are coherent. … … [1] (ii) State one other condition that must be satisfied by the waves in order for the interference pattern to be observable. … … [1] (b) Light from a lamp above the ripple tank shines through the water onto a screen below the tank. Describe one way of seeing the illuminated pattern more clearly. … … [1] (c) The speed of the waves is 0.40 m s–1. Calculate the period of the waves. period = … s [2] (d) Fig. 5.1 shows a point X that lies on a crest of the wave from D1 and midway between two adjacent crests of the wave from D2. For the waves at point X, state: (i) the path difference, in cm path difference = … cm [1] (ii) the phase difference. phase difference = … ° [1] (e) On Fig. 5.1, draw one line, at least 4 cm long, which joins points where only maxima of the interference pattern are observed. [1] [Total: 8]

8 marks

Mark scheme: 5(a)(i) the dippers are connected to the same vibrator/motor B1 5(a)(ii) (the overlapping waves have) similar/same amplitude B1 5(b) any means of ‘freezing’ the pattern e.g. use a stroboscope/strobe B1 5(c) vT = λ or v = fλ and f = 1 / T C1 T = 0.060 / 0.40 = 0.15 s A1 5(d)(i) path difference = 3.0 cm A1 5(d)(ii) phase difference = 180° A1 5(e) line drawn joining points where only maxima are observed (i.e. through points where wavefronts intersect) of length at least 4 cm B1

This question in 9702/21 Oct/Nov 2019

5 (a) State what is meant by the wavelength of a progressive wave. … … [1] (b) A cathode-ray oscilloscope (CRO) is used to analyse a sound wave. The screen of the CRO is shown in Fig. 5.1. 1 cm 1 cm Fig. 5.1 The time-base setting of the CRO is 2.5 ms cm–1. Determine the frequency of the sound wave. frequency = … Hz [2] (c) The source emitting the sound in (b) is at point A. Waves travel from the source to point C along two different paths, AC and ABC, as shown in Fig. 5.2. 20.8 m C A 8.0 m reflecting B surface Fig. 5.2 (not to scale) Distance AB is 8.0 m and distance AC is 20.8 m. Angle ABC is 90°. Assume that there is no phase change of the sound wave due to the reflection at point B. The wavelength of the waves is 1.6 m. (i) Show that the waves meeting at C have a path difference of 6.4 m. [1] (ii) Explain why an intensity maximum is detected at point C. … … … [2] (iii) Determine the difference between the times taken for the sound to travel from the source to point C along the two different paths. time difference = … s [2] (iv) The wavelength of the sound is gradually increased. Calculate the wavelength of the sound when an intensity maximum is next detected at point C. wavelength = … m [1] [Total: 9]

9 marks

Mark scheme: 5(a) distance moved by wavefront/energy during one cycle/oscillation/period (of source) or minimum distance between two wavefronts or distance between two adjacent wavefronts B1 5(b) (T =) 2.0 × 2.5 (= 5.0 ms) or 2.0 × 2.5 × 10–3 (= 5.0 × 10–3 s) C1 f = 1 / (5.0 × 10–3) = 200 Hz A1 5(c)(i) (path difference =) 8.0 + (20.82 – 8.02)0.5 – 20.8 = 6.4 (m) A1 5(c)(ii) • path difference = 4λ • waves (meet at C) in phase • constructive interference (of waves) any two points, one mark each B2 5(c)(iii) v = 200 × 1.6 v = 320 (m s–1) C1 ∆t = 6.4 / 320 or 27.2 / 320 – 20.8 / 320 = 0.020 s A1 5(c)(iv) 3λ = 6.4 3λ = 2.1 m A1

This question in 9702/22 Oct/Nov 2019

5 (a) Light waves emerging from the slits of a diffraction grating are coherent and produce an interference pattern. Explain what is meant by: (i) coherence … … [1] (ii) interference. … … [1] (b) A narrow beam of light from a laser is incident normally on a diffraction grating, as shown in Fig. 5.1. second order maximum spot 51° zero order 51° maximum spot laser light diffraction grating second order maximum spot screen Fig. 5.1 (not to scale) Spots of light are seen on a screen positioned parallel to the grating. The angle corresponding to each of the second order maxima is 51°. The number of lines per unit length on the diffraction grating is 6.7 × 105 m–1. (i) Determine the wavelength of the light. wavelength = … m [2] (ii) State and explain the change, if any, to the distance between the second order maximum spots on the screen when the light from the laser is replaced by light of a shorter wavelength. … … … [1] [Total: 5]

5 marks

Mark scheme: 5(a)(i) (coherence means) constant phase difference (between waves) B1 5(a)(ii) (interference is) the sum/addition/combination of the displacements of overlapping/meeting waves B1 5(b)(i) nλ = d sinθ C1 λ = sin 51° / (2 × 6.7 × 105) = 5.8 × 10–7 m A1 5(b)(ii) smaller angle (corresponding to second order maxima and so) shorter distance (between second order maxima spots) B1

This question in 9702/23 Oct/Nov 2019

4 (a) (i) By reference to the direction of propagation of energy, state what is meant by a longitudinal wave. … … [1] (ii) State the principle of superposition. … … … [2] (b) The wavelength of light from a laser is determined using the apparatus shown in Fig. 4.1. double slit screen light 3.7 × 10–4 m 2.3 m Fig. 4.1 (not to scale) The light from the laser is incident normally on the plane of the double slit. The separation of the two slits is 3.7 × 10–4 m. The screen is parallel to the plane of the double slit. The distance between the screen and the double slit is 2.3 m. A pattern of bright fringes and dark fringes is seen on the screen. The separation of adjacent bright fringes on the screen is 4.3 × 10–3 m. (i) Calculate the wavelength, in nm, of the light. wavelength = … nm [3] (ii) The intensity of the light passing through each slit was initially the same. The intensity of the light through one of the slits is now reduced. Compare the appearance of the fringes before and after the change of intensity. … … … … [2] [Total: 8]

8 marks

Mark scheme: 4(a)(i) vibrations (of particles) are parallel to direction of energy propagation B1 4(a)(ii) waves meet/overlap (at a point) B1 (resultant) displacement is sum of individual displacements B1 4(b)(i) λ = ax / D C1 = (3.7 × 10–4 × 4.3 × 10–3) / 2.3 C1 = 6.9 × 10–7 (m) = 690 nm A1 4(b)(ii) • no change to fringe separation/fringe width/number of fringes • bright fringes are darker • dark fringes are brighter Any two marking points, 1 mark each B2

This question in 9702/21 May/June 2020

4 Two progressive sound waves Y and Z meet at a fixed point P. The variation with time t of the displacement x of each wave at point P is shown in Fig. 4.1. 6 4 wave Y x / μm 2 0 0 1.0 2.0 3.0 tt // msms 4.0 –2 wave Z –4 –6 Fig. 4.1 (a) Use Fig. 4.1 to state one quantity of waves Y and Z that is: (i) the same … [1] (ii) different. … [1] (b) State and explain whether waves Y and Z are coherent. … … [1] (c) Determine the phase difference between the waves. phase difference = … ° [1] (d) The two waves superpose at P. Use Fig. 4.1 to determine the resultant displacement at time t = 0.75 ms. resultant displacement = … μm [1] (e) The intensity of wave Y at point P is I. Determine, in terms of I, the intensity of wave Z. intensity = … [2] (f) The speed of wave Z is 330 m s–1. Determine the wavelength of wave Z. wavelength = … m [3] [Total: 10]

10 marks

Mark scheme: 4(a)(i) frequency or period B1 4(a)(ii) amplitude B1 4(b) constant phase difference so coherent B1 4(c) 120° B1 4(d) resultant displacement = 4.0 μm – 1.0 μm = 3.0 μm B1 4(e) I ∝ A2 C1 intensity of Z = (22 / 42) I = 0.25 I A1 4(f) v = λ / T or v = fλ and f = 1 /T C1 330 = λ / 3.0 × 10–3 C1 λ = 0.99 m A1

This question in 9702/23 May/June 2020

4 (a) State the principle of superposition. … … … [2] (b) A transmitter produces microwaves that travel in air towards a metal plate, as shown in Fig. 4.1. microwave metal transmitter microwave plate receiver X Fig. 4.1 The microwaves have a wavelength of 0.040 m. A stationary wave is formed between the transmitter and the plate. (i) Explain the function of the metal plate. … … [1] (ii) Calculate the frequency, in GHz, of the microwaves. frequency = … GHz [3] (iii) A microwave receiver is initially placed at position X where it detects an intensity minimum. The receiver is then slowly moved away from X directly towards the plate. 1. Determine the shortest distance from X of the receiver when it detects another intensity minimum. distance = … m 2. Determine the number of intensity maxima that are detected by the receiver as it moves from X to a position that is 9.1 cm away from X. number = … [2] [Total: 8]

8 marks

Mark scheme: 4(a) (two or more) waves meet (at a point) B1 (resultant) displacement is the sum of the individual displacements B1 4(b)(i) it is a (wave) reflector / it reflects (the wave) B1 4(b)(ii) v = fλ or c = fλ C1 f = 3.0 × 108 / 0.040 = 7.5 × 109 (Hz) = 7.5 × 109 / 109 (GHz) C1 = 7.5 GHz A1 4(b)(iii) 1 distance = 0.020 m A1 2 number = 5 A1

This question in 9702/22 Feb/March 2021

4 (a) For a progressive wave, state what is meant by its period. … … [1] (b) State the principle of superposition. … … … [2] (c) Electromagnetic waves of wavelength 0.040 m are emitted in phase from two sources X and Y and travel in a vacuum. The arrangement of the sources is shown in Fig. 4.1. X path of detector 1.380 m Z 1.240 m Y Fig. 4.1 (not to scale) A detector moves along a path that is parallel to the line XY. A pattern of intensity maxima and minima is detected. Distance XZ is 1.380 m and distance YZ is 1.240 m. (i) State the name of the region of the electromagnetic spectrum that contains the waves from X and Y. … [1] (ii) Calculate the period, in ps, of the waves. period = … ps [3] (iii) Show that the path difference at point Z between the waves from X and Y is 3.5 λ, where λ is the wavelength of the waves. [1] (iv) Calculate the phase difference between the waves at point Z. phase difference = … ° [1] (v) The waves from X alone have the same amplitude at point Z as the waves from Y alone. State the intensity of the waves at point Z. … [1] (vi) The frequencies of the waves from X and Y are both decreased to the same lower value. The waves stay within the same region of the electromagnetic spectrum. Describe the effect of this change on the pattern of intensity maxima and minima along the path of the detector. … … [1] [Total: 11]

11 marks

Mark scheme: 4(a) time for one oscillation/vibration/cycle or time between adjacent wavefronts (passing the same point) or shortest time between two wavefronts (passing the same point) B1 4(b) (when two or more) waves meet/overlap (at a point) B1 (resultant) displacement is sum of the individual displacements B1 4(c)(i) microwave(s) B1 4(c)(ii) v = λ / T or v = fλ and f = 1/T C1 T = 0.040 / 3.00 × 108 C1 = 1.33 × 10–10 (s) = 1.33 × 10–10 / 10–12 (ps) = 130 ps A1 4(c)(iii) (1.380 – 1.240) / 0.040 = 3.5 or 1.380 / 0.040 – 1.240 / 0.040 = 3.5 A1 4(c)(iv) phase difference = 1260° or 180° A1 4(c)(v) (always) zero A1 4(c)(vi) increase in distance between (adjacent intensity) maxima/minima A1

This question in 9702/22 May/June 2021

4 (a) State the principle of superposition. … … … [2] (b) Two waves, with intensities I and 4I, superpose. The waves have the same frequency. Determine, in terms of I, the maximum possible intensity of the resulting wave. maximum intensity = … I [2] (c) Coherent light of wavelength 550 nm is incident normally on a double slit of slit separation 0.35 mm. A series of bright and dark fringes forms on a screen placed a distance of 1.2 m from the double slit, as shown in Fig. 4.1. The screen is parallel to the double slit. screen 1.2 m light 0.35 mm wavelength 550 nm double slit Fig. 4.1 (not to scale) (i) Determine the distance between the centres of adjacent bright fringes on the screen. distance = … m [3] (ii) The light of wavelength 550 nm is replaced with red light of a single frequency. State and explain the change, if any, in the distance between the centres of adjacent bright fringes. … … … [1] [Total: 8]

8 marks

Mark scheme: 4(a) (when two or more) waves meet/overlap (at a point) B1 (resultant) displacement is sum of the individual displacements B1 4(b) intensity ∝ amplitude2 C1 maximum intensity = 9I A1 4(c)(i) x = λD / a C1 = (550 × 10–9 × 1.2) / (0.35 × 10–3) C1 = 1.9 × 10–3 m A1 4(c)(ii) red light has longer wavelength (than 550 nm) so distance (between fringes) increases B1 Question Answer Marks

This question in 9702/23 May/June 2021

5 (a) A microphone and cathode-ray oscilloscope (CRO) are used to analyse a sound wave of frequency 5000 Hz. The trace that is displayed on the screen of the CRO is shown in Fig. 5.1. 1.0 cm 1.0 cm Fig. 5.1 (i) Determine the time-base setting, in s cm–1, of the CRO. time-base setting = … s cm–1 [2] (ii) The intensity of the sound detected by the microphone is now increased from its initial value of I to a new value of 3I. The frequency of the sound is unchanged. Assume that the amplitude of the trace on the CRO screen is proportional to the amplitude of the sound wave. On Fig. 5.1, sketch the new trace shown on the screen of the CRO. [3] (b) An arrangement for demonstrating interference using light is shown in Fig. 5.2. 3.6 × 10–4 m P light from laser, wavelength 630 nm D double slit screen Fig. 5.2 (not to scale) The wavelength of the light from the laser is 630 nm. The light is incident normally on the double slit. The separation of the two slits is 3.6 × 10–4 m. The perpendicular distance between the double slit and the screen is D. Coherent light waves from the slits form an interference pattern of bright and dark fringes on the screen. The distance between the centres of two adjacent bright fringes is 4.0 × 10–3 m. The central bright fringe is formed at point P. (i) Explain why a bright fringe is produced by the waves meeting at point P. … … [1] (ii) Calculate distance D. D = … m [3] (c) The wavelength λ of the light in (b) is now varied. This causes a variation in the distance x between the centres of two adjacent bright fringes on the screen. The distance D and the separation of the two slits are unchanged. On Fig. 5.3, sketch a graph to show the variation of x with λ from λ = 400 nm to λ = 700 nm. Numerical values of x are not required. x 0 400 700 λ/ nm Fig. 5.3 [1] [Total: 10]

10 marks

Mark scheme: 5(a)(i) period or T = 1 / 5000 (= 2  10–4 s) C1 time-base setting = 1.5  2  10–4 / 6.0 or 2  10–4 / 4.0 A1 = 5  10–5 s cm–1 5(a)(ii) new trace drawn with same period as original trace B1 new trace drawn with amplitude greater than 1.0 cm M1 new trace drawn with amplitude of 1.7 cm A1 5(b)(i) path difference (from slits to P) is zero or phase difference (between waves at P) is zero (so constructive interference) B1 5(b)(ii) = ax / D C1 D = (3.6  10–4  4.0  10–3) / 630  10–9 C1 = 2.3 m A1 5(c) upward sloping straight line starting from a non-zero value of x at = 400 nm B1

This question in 9702/22 Feb/March 2023

4 (a) State the principle of superposition. … … … [2] (b) Coherent light is incident normally on two identical slits X and Y. The diffracted light emerging from the slits superposes to produce an interference pattern on a screen positioned at a distance of 1.9 m from the slits. Fig. 4.1 shows the arrangement and the central part of the interference pattern of bright and dark fringes formed on the screen. 1.9 m dark fringe X coherent 0.65 mm bright fringe light 1.7 mm Y screen Fig. 4.1 (not to scale) The separation of the slits is 0.65 mm. The distance between the centres of adjacent bright fringes is 1.7 mm. Calculate the wavelength λ of the light. λ = … m [3] (c) Light waves from slits X and Y in (b) arrive at a point between adjacent bright fringes on the screen. Fig. 4.2 shows the variation of displacement with time for the waves arriving at the point where they meet. wave from X wave from Y displacement 0 time Fig. 4.2 A student makes two statements about the waves at this point: Statement 1: ‘The phase difference between the waves is 90°.’ Statement 2: ‘The amplitude of the resultant wave is zero.’ (i) Explain how statement 1 is correct. … … … [1] (ii) State and explain whether statement 2 is correct. … … … [1] (d) The width of each slit in (b) is decreased by the same amount. There is no change to the separation of the slits. Describe and explain the effect, if any, of this change on the appearance of the interference pattern. … … … [2] [Total: 9]

9 marks

Mark scheme: 4(a) (when two or more) waves meet / overlap (at a point) B1 (resultant) displacement is the sum of the individual displacements B1 4(b) = ax / D C1 = [(0.65  10–3)  (1.7  10–3)] / 1.9 C1 = 5.8  10–7 m A1 4(c)(i) waves are out of phase by a quarter of a cycle / period B1 or when one wave has maximum/minimum displacement, the other wave has zero displacement 4(c)(ii) (statement 2 is) not correct (because) B1 waves do not have phase difference of 180° / not in antiphase or one wave has some displacement when other has no displacement or the displacements of the waves are not always equal and opposite 4(d) more diffraction (of light/waves by slits) B1 or light/waves are more spread (by slits) or light/waves (from slits) have less intensity more (bright/dark) fringes B1 or bright fringes are less bright / are dimmer / have lower intensity or no change to fringe spacing/separation/width

This question in 9702/21 Oct/Nov 2023

6 (a) Coherent visible light of a single frequency is incident normally on a double slit. This produces a pattern of bright and dark interference fringes on a screen, as illustrated in Fig. 6.1. fringe pattern on screen screen double slit bright fringe X light 1.2 mm 10.2 mm dark fringe 3.1 m bright fringe Y Fig. 6.1 (not to scale) There are seven bright fringes. (i) Explain how the pattern of bright and dark interference fringes is formed. … … … … … … [3] (ii) The distance between the centres of bright fringe X and bright fringe Y in the pattern is 10.2 mm. The slit spacing is 1.2 mm. The distance from the slits to the screen is 3.1 m. Calculate the wavelength of the light incident on the slits. wavelength = … m [3] (iii) The light is replaced by different visible light with a shorter wavelength. State how the new fringe separation will compare to the original fringe separation. … [1] (b) A stationary wave is formed on a stretched string AB, as shown in Fig. 6.2. string P Q mean position of string A B R Fig. 6.2 P, Q and R are points on the string. (i) On Fig. 6. 2, draw a cross (×) to show the position of a node. [1] (ii) State the phase difference between P and Q. phase difference = … ° [1] (iii) State the phase difference between P and R. phase difference = … ° [1] [Total: 10]

10 marks

Mark scheme: 6(a)(i) Any three from: B3 • Light diffracts at the (two) slits. • Light (from each slit) meets / superposes (at the screen). • When the phase difference is 0 (degrees) a bright fringe / (intensity) maximum is formed. • When the phase difference is 180 (degrees) a dark fringe / (intensity) minimum is formed. 6(a)(ii)  = ax / D C1  = 1.2  10–3  (10.2  10–3 / 6) / 3.1 C1  = 6.6  10–7 m A1 6(a)(iii) (new fringe separation will be) smaller B1 6(b)(i) A cross at the intersection of the string and the mean position line. B1 6(b)(ii) 0 A1 6(b)(iii) 180˚ A1

This question in 9702/22 Feb/March 2024

4 A progressive transverse wave travelling from left to right is shown at an instant in time in Fig. 4.1. R wave direction of travel T Fig. 4.1 R and T are points on the wave. (a) State the phase difference between the points R and T. phase difference = … ° [1] (b) On Fig. 4.1, draw an arrow at point T to show the direction of movement of point T at the instant shown. [1] (c) The horizontal distance between R and T is 0.62 cm, as shown in Fig. 4.2. 0.62 cm R T Fig. 4.2 (not to scale) The speed of the wave is 0.27 m s–1. Calculate the frequency of the wave. frequency = … Hz [3] (d) The wave is a water wave produced by a dipper S1 attached to a vibrator in a ripple tank. An identical dipper S2 is attached to the same vibrator. The two dippers produce an interference pattern on the water in the tank, as shown in Fig. 4.3. water P wave crests wave troughs S1 S2 Fig. 4.3 (not to scale) The wave crests from each source are represented by solid lines on Fig. 4.3 and the wave troughs are represented by dashed lines. At point P in Fig. 4.3, the wave from S1 has the same amplitude A as the wave from S2. Describe and explain the amplitude of the resultant wave at point P. … … … … [3] [Total: 8]

8 marks

Mark scheme: 4(a) 270° A1 4(b) arrow pointing vertically downwards at T A1 4(c) v = f or v =  / T and f = 1 / T C1 wavelength = 0.62  10–2  (4 / 3) ( = 0.83  10–2 m) C1 f = 0.27 / (0.83  10–2) = 33 Hz A1 4(d) resultant displacement is the sum of the displacements of the waves (from S1 and S2) or waves (from S1 and S2) superpose (at P) B1 Any one point from:  (at P) the waves (from the two sources) (always) destructively interfere  (at P) the waves have a path difference that is (always) an odd number of half-wavelengths / their path difference is one and a half wavelengths  (at P) the waves have a phase difference that is (always) 180° / they are in antiphase / crest of one wave meets trough of other wave B1 amplitude (of the resultant wave) is zero (at all times) B1

This question in 9702/23 May/June 2024

3 Two progressive water waves X and Y travel along a straight line from point A to point B. The variation of displacement of the waves with distance from A at an instant in time is shown in Fig. 3.1. 20 displacement / cm 10 wave X 0 0 0.2 0.4 0.6 0.8 1.0 distance from A / m –10 wave Y –20 Fig. 3.1 (a) State the amplitude of wave X. amplitude = … cm [1] (b) Both waves have frequency 16 Hz. (i) Determine the speed of wave X. speed = … m s–1 [2] (ii) State and explain whether X and Y are coherent. … … … [1] (c) Wave X and wave Y superpose to form a resultant wave. On Fig. 3.2, sketch the variation of displacement of the resultant wave with distance from A at the instant of time shown in Fig. 3.1. 20 displacement / cm 10 0 0 0.2 0.4 0.6 0.8 1.0 distance from A / m –10 –20 Fig. 3.2 [2] (d) The intensity of wave X is IX. The intensity of wave Y is IY. IX Use Fig. 3.1 to determine the ratio . IY ratio = … [2] [Total: 8]

8 marks

Mark scheme: 3(a) 10.0 cm A1 3(b)(i) v = f C1 = 16  0.40 = 6.4 m s–1 A1 3(b)(ii) (X and Y have a) constant phase difference (of 180°) so (they are) coherent B1 3(c) A single wave of amplitude 10.0 cm B1 A single negative sine wave of wavelength 0.40 m B1 3(d) I  A2 C1 I X = 102 / 202 IY ratio = 0.25 A1

This question in 9702/22 May/June 2025

4 A laser emits visible light of a single frequency in a vacuum. The light is incident normally on a double slit and then forms a pattern of bright and dark fringes on a screen, as shown in Fig. 4.1. screen fringe pattern on screen double slit dark fringe 3.3 mm light 1.0 × 10–3 m bright fringe 4.8 m Fig. 4.1 (not to scale) The separation of the slits is 1.0 × 10–3 m. The distance from the slits to the screen is 4.8 m. The distance between the centres of adjacent dark fringes on the screen is 3.3 mm. (a) Explain how the pattern of bright and dark fringes is formed. … … … … … [3] (b) Calculate the frequency of the light emitted by the laser. frequency = … Hz [4] (c) The double slit is removed. A second laser is placed beside the first laser. The second laser produces visible light of a different frequency from that of the first laser. The beams of light from the two lasers overlap on the screen. Explain why a steady pattern of bright and dark fringes is not formed on the screen. … … [1] [Total: 8]

8 marks

Mark scheme: 4(a) light diffracts / spreads (at slit(s)) B1 light (from each slit) superposes / interferes (at screen) B1 when phase difference is 0 / path difference is n(where n is an integer) a bright fringe is formed B1 or when phase difference is 180(°) / path difference is (n + 1) / 2 (where n is an integer) a dark fringe is formed 4(b)  = ax / D C1  = 1.0  10–3  3.3  10–3 / 4.8 C1 ( = 6.875  10–7 m) f = v /  C1 = 3.0  108 / 6.875  10–7 A1 = 4.4  1014 Hz 4(c) the light / beams (have different frequencies so) are not coherent / do not have constant phase difference B1

This question in 9702/22 Oct/Nov 2025

4 (a) State the principle of superposition. … … … [2] (b) An electromagnetic wave of wavelength 0.026 m in free space is incident normally on an aluminium sheet, as shown in Fig. 4.1. transmitter aluminium sheet electromagnetic wave Fig. 4.1 The wave reflects at the aluminium sheet and a stationary wave is formed in the region between the transmitter and the sheet. (i) Explain how the stationary wave, including its nodes and antinodes, is formed. … … … … … [3] (ii) Calculate the frequency of the electromagnetic wave. frequency = … Hz [2] (iii) State the principal region of the electromagnetic spectrum to which the wave belongs. … [1] (iv) Determine the distance between a node and an adjacent antinode. distance = … m [1] [Total: 9]

9 marks

Mark scheme: 4(a) when two (or more) waves meet M1 (resultant) displacement equals the sum of the displacements of the (two separate) waves A1 4(b)(i) the incident and reflected waves superpose B1 (the waves superpose so that the resultant) amplitude is maximum at an antinode B1 (the waves superpose so that the resultant) amplitude is minimum / zero at a node B1 4(b)(ii) c = f C1 f = (3.00 × 108) / 0.026 A1 = 1.2 × 1010 Hz 4(b)(iii) microwave B1 4(b)(iv) distance = 0.026 / 4 A1 = 6.5 × 10–3 m

This question in 9702/24 Oct/Nov 2025