Interference — Paper 2 · A Level Physics 9702

5 Fig. 5.1 shows the variation with time t of the displacements xA and xB at a point P of two For sound waves A and B. Examiner’s Use wave A 3 xA / 10 – 4 cm 2 1 0 0 1 2 3 4 5 6 t / ms –1 –2 –3 wave B 2 xB / 10 – 4 cm 1 0 0 1 2 3 4 5 6 t / ms –1 –2 Fig. 5.1 (a) By reference to Fig. 5.1, state one similarity and one difference between these two waves. similarity: … difference: … [2] (b) State, with a reason, whether the two waves are coherent. … … [1] (c) The intensity of wave A alone at point P is I. For Examiner’s (i) Show that the intensity of wave B alone at point P is 4 I. Use 9 [2] (ii) Calculate the resultant intensity, in terms of I, of the two waves at point P. resultant intensity = ……………………………… I [2] (d) Determine the resultant displacement for the two waves at point P (i) at time t = 3.0 ms, resultant displacement = ……………………………… cm [1] (ii) at time t = 4.0 ms. resultant displacement = ……………………………… cm [2]

10 marks

Mark scheme: 5 (a) similarity: e.g. same wavelength/frequency/period, constant phase difference B1 difference: e.g. different amplitude/phase B1 [2] (do not allow a reference to phase for both similarity and difference) (b) constant phase difference so coherent B1 [1] (c) (i) intensity ∝ amplitude2 C1 I ∝ 32 and IB ∝ 22 leading to M1 4 IB = I A0 [2] 9 (ii) resultant amplitude = 1.0 × 10-4 cm C1 1 resultant intensity = I A1 [2] 9 (d) (i) displacement = 0 B1 [1] (ii) xA = -2.6 × 10-4 cm and xB = +1.7 × 10-4 cm C1 allow ± 0.5 × 10-4 cm) resultant displacement = (-) 0.9 × 10-4 cm A1 [2]

This question in 9702/21 Oct/Nov 2005

5 (a) Fig. 5.1 shows the variation with time t of the displacement y of a wave W as it passes a point P. The wave has intensity I. y 0 t wave W Fig. 5.1 A second wave X of the same frequency as wave W also passes point P. This wave has intensity I. The phase difference between the two waves is 60°. On Fig. 5.1, sketch the variation with time t of the displacement y of wave X. [3] (b) In a double-slit interference experiment using light of wavelength 540 nm, the separation of the slits is 0.700 mm. The fringes are viewed on a screen at a distance of 2.75 m from the double slit, as illustrated in Fig. 5.2 (not to scale). coherent light 0.700 mm wavelength 540 nm screen 2.75 m Fig. 5.2 Examiner’s Use Calculate the separation of the fringes observed on the screen. separation = … mm [3] (c) State the effect, if any, on the appearance of the fringes observed on the screen when the following changes are made, separately, to the double-slit arrangement in (b). (i) The width of each slit is increased but the separation remains constant. … … … … [3] (ii) The separation of the slits is increased. … … … [2]

11 marks

Mark scheme: 5 (a) amplitude between 6.5 squares and 7.5 squares on 3 peaks B2 (allow 1 mark if outside this range but between 6.0 and 8.0 squares) correct phase (ignore lead/lag, look at x-axis only and allow ±½ square B1 [3] (b) λ = ax / D C1 540 × 10–9 = (0.700 × 10–3 x) / 2.75 C1 x = 2.12 mm A1 [3] (c) (i) same separation B1 bright areas brighter (1) dark areas, no change (1) (allow ‘contrast greater’ for 1 mark if dark/light areas not discussed) fewer fringes observed (1) any two, 1 each B2 [3] (ii) smaller separation of fringes B1 no change in brightness B1 [2]

This question in 9702/21 Oct/Nov 2007

5 Two sources S1 and S2 of sound are situated 80 cm apart in air, as shown in Fig. 5.1. For Examiner’s Use 100 cm S1 M 80 cm S2 Fig. 5.1 The frequency of vibration can be varied. The two sources always vibrate in phase but have different amplitudes of vibration. A microphone M is situated a distance 100 cm from S1 along a line that is normal to S1S2. As the frequency of S1 and S2 is gradually increased, the microphone M detects maxima and minima of intensity of sound. (a) State the two conditions that must be satisfied for the intensity of sound at M to be zero. 1. … … 2. … … [2] (b) The speed of sound in air is 330 m s–1. The frequency of the sound from S1 and S2 is increased. Determine the number of minima that will be detected at M as the frequency is increased from 1.0 kHz to 4.0 kHz. number = … [4]

6 marks

Mark scheme: 5 (a) either phase difference is π rad / 180° or path difference (between waves from S1 and S2) is ½λ / (n + ½)λ . B1 either same amplitude / intensity at M or ratio of amplitudes is 1.28 / ratio of intensities is 1.282 … B1 [2] (b) path difference between waves from S1 and S2 = 28 cm … B1 wavelength changes from 33 cm to 8.25 cm … B1 minimum when λ = (56 cm,) 18.7 cm, 11.2 cm, (8.0 cm) … B1 so two minima … B1 [4]

This question in 9702/21 May/June 2009

6 (a) State the principle of superposition. For Examiner’s … Use … … [2] (b) Coherent light of wavelength 590 nm is incident normally on a double slit, as shown in Fig. 6.1. double screen slit A coherent light 1.4 mm P wavelength 590 nm B 2.6 m Fig. 6.1 (not to scale) The separation of the slits A and B is 1.4 mm. Interference fringes are observed on a screen placed parallel to the plane of the double slit. The distance between the screen and the double slit is 2.6 m. At point P on the screen, the path difference is zero for light arriving at P from the slits A and B. (i) Determine the separation of bright fringes on the screen near to point P. separation = … mm [3] (ii) The variation with time of the displacement x of the light wave arriving at point P on the For screen from slit A and from slit B is shown in Fig. 6.2a and Fig. 6.2b respectively. Examiner’s Use x / 3 arbitrary units 2 1 0 0 time –1 –2 wave from slit A –3 Fig. 6.2a x / 2 arbitrary units 1 0 0 time –1 wave from slit B –2 Fig. 6.2b 1. State the phase difference between waves forming the dark fringe on the screen that is next to point P. phase difference = … ° [1] 2. Determine the ratio intensity of light at a bright fringe . intensity of light at a dark fringe ratio = … [3]

9 marks

Mark scheme: 6 (a) when two (or more) waves meet (at a point) B1 (resultant) displacement is (vector) sum of individual displacements B1 [2] (b) (i) λ = ax / D (if no formula given and substitution is incorrect then 0/3) C1 590 × 10–9 = (1.4 × 10–3 × x) / 2.6 C1 x = 1.1 mm A1 [3] (ii) 1. 180° (allow π if rad stated) A1 [1] 2. at maximum, amplitude is 3.4 units and at minimum, 0.6 units C1 intensity ~ amplitude2 allow I ~ a2 C1 ratio = 3.42 / 0.62 = 32 A1 [3]

This question in 9702/23 Oct/Nov 2010

7 (a) Explain the term interference. For Examiner’s … Use … … [1] (b) A ripple tank is used to demonstrate interference between water waves. Describe (i) the apparatus used to produce two sources of coherent waves that have circular wavefronts, … … … … [2] (ii) how the pattern of interfering waves may be observed. … … … … [2] (c) A wave pattern produced in (b) is shown in Fig. 7.1. For Examiner’s Use Fig. 7.1 Solid lines on Fig. 7.1 represent crests. On Fig. 7.1, (i) draw two lines to show where maxima would be seen (label each of these lines with the letter X), [1] (ii) draw one line to show where minima would be seen (label this line with the letter N). [1]

7 marks

Mark scheme: 7 (a) when waves overlap / meet, (resultant) displacement is the sum of the individual displacements B1 [1] (b) (i) two (ball-type) dippers (M1) connected to the same vibrating source /motor (A1) or one wave source described (M1) with two slits (A1) [2] (ii) lamp with viewing screen on opposite side of tank B1 means of freezing picture e.g. strobe B1 [2] (c) (i) two correct lines labelled X B1 [1] (ii) correct line labelled N B1 [1]

This question in 9702/21 May/June 2011

6 (a) Apparatus used to produce interference fringes is shown in Fig. 6.1. The apparatus is For not drawn to scale. Examiner’s Use two slits B bright fringe P dark fringe LASER C bright fringe screen Fig. 6.1 (not to scale) Laser light is incident on two slits. The laser provides light of a single wavelength. The light from the two slits produces a fringe pattern on the screen. A bright fringe is produced at C and the next bright fringe is at B. A dark fringe is produced at P. (i) Explain why one laser and two slits are used, instead of two lasers, to produce a visible fringe pattern on the screen. … … [1] (ii) State the phase difference between the waves that meet at 1. B … [1] 2. P … [1] (iii) 1. State the principle of superposition. … … … [2] 2. Use the principle of superposition to explain the dark fringe at P. … … [1] (b) In Fig. 6.1 the distance from the two slits to the screen is 1.8 m. The distance CP is For 2.3 mm and the distance between the slits is 0.25 mm. Examiner’s Calculate the wavelength of the light provided by the laser. Use wavelength = … nm [3]

9 marks

Mark scheme: 6 (a) (i) to produce coherent sources or constant phase difference B1 [1] (ii) 1. 360° / 2π rad allow n × 360° or n × 2π (unit missing –1) B1 [1] 2. 180° / π rad allow (n × 360°) – 180° or (n × 2π) – π B1 [1] (iii) 1. waves overlap / meet B1 (resultant) displacement is sum of displacements of each wave B1 [2] 2. at P crest on trough (OWTTE) B1 [1] (b) λ = ax / D C1 = 2 × 2.3 × 10–3 × 0.25 ×10–3 / 1.8 C1 = 639 nm A1 [3]

This question in 9702/22 May/June 2011

6 (a) A laser is used to produce an interference pattern on a screen, as shown in Fig. 6.1. For Examiner’s Use P2 P1 laser light 0.450 mm wavelength 630 nm screen double slit 1.50 m Fig. 6.1 (not to scale) The laser emits light of wavelength 630 nm. The slit separation is 0.450 mm. The distance between the slits and the screen is 1.50 m. A maximum is formed at P1 and a minimum is formed at P2. Interference fringes are observed only when the light from the slits is coherent. (i) Explain what is meant by coherence. … … … [2] (ii) Explain how an interference maximum is formed at P1. … … [1] (iii) Explain how an interference minimum is formed at P2. … … [1] (iv) Calculate the fringe separation. fringe separation = … m [3] (b) State the effects, if any, on the fringes when the amplitude of the waves incident on the For double slits is increased. Examiner’s Use … … … … [3]

10 marks

Mark scheme: 6 (a) (i) coherence: constant phase difference M1 between (two) waves A1 [2] (ii) path difference is either λ or nλ or phase difference is 360° or n × 360° or n2π rad B1 [1] GCE AS/A LEVEL – May/June 2012 9702 21 (iii) path difference is either λ/2 or (n + ½) λ or phase difference is odd multiple of either 180° or π rad B1 [1] (iv) w = λD / a C1 = [630 × 10–9 × 1.5] / 0.45 × 10–3 C1 = 2.1 × 10–3 m A1 [3] (b) no change to dark fringes B1 no change to separation/fringe width B1 bright fringes are brighter/lighter/more intense B1 [3]

This question in 9702/21 May/June 2012

6 (a) Monochromatic light is diffracted by a diffraction grating. By reference to this, explain For what is meant by Examiner’s Use (i) diffraction, … … … [2] (ii) coherence, … … … [1] (iii) superposition. … … … [1] (b) A parallel beam of red light of wavelength 630 nm is incident normally on a diffraction grating of 450 lines per millimetre. Calculate the number of diffraction orders produced. number of orders = … [3] (c) The red light in (b) is replaced with blue light. State and explain the effect on the diffraction pattern. … … … … … [3]

10 marks

Mark scheme: 6 (a) (i) diffraction bending/spreading of light at edge/slit B1 this occurs at each slit B1 [2] (ii) constant phase difference between each of the waves B1 [1] (iii) (when the waves meet) the resultant displacement is the sum of the displacements of each wave B1 [1] (b) d sinθ = nλ n = d / λ = 1 / 450 × 103 × 630 × 10–9 C1 n = 3.52 M1 hence number of orders = 3 A1 [3] (c) λ blue is less than λ red M1 more orders seen A1 each order is at a smaller angle than for the equivalent red A1 [3]

This question in 9702/23 May/June 2012

4 (a) Describe the diffraction of monochromatic light as it passes through a diffraction grating. For Examiner’s … Use … … [2] (b) White light is incident on a diffraction grating, as shown in Fig. 4.1. spectrum (first order) white light white (zero order) diffraction spectrum (first order) grating screen Fig. 4.1 (not to scale) The diffraction pattern formed on the screen has white light, called zero order, and coloured spectra in other orders. (i) Describe how the principle of superposition is used to explain 1. white light at the zero order, … … … [2] 2. the difference in position of red and blue light in the first-order spectrum. … … … [2] (ii) Light of wavelength 625 nm produces a second-order maximum at an angle of 61.0° For to the incident direction. Examiner’s Determine the number of lines per metre of the diffraction grating. Use number of lines = … m–1 [2] (iii) Calculate the wavelength of another part of the visible spectrum that gives a maximum for a different order at the same angle as in (ii). wavelength = ……………………..…….. nm [2]

10 marks

Mark scheme: 4 (a) waves pass through the elements / gaps / slits in the grating M1 spread into geometric shadow A1 [2] (b) (i) 1. displacements add to give resultant displacement B1 each wavelength travels the same path difference or are in phase B1 hence produce a maximum A0 [2] 2. to obtain a maximum the path difference must be λ or phase difference 360° / 2π rad B1 λ of red and blue are different B1 hence maxima at different angles / positions A0 [2] (ii) nλ = d sin θ C1 N = sin 61° / (2 × 625 × 10–9) = 7.0 × 105 A1 [2] (iii) nλ = 2 × 625 is a constant (1250) C1 n = 1 → λ = 1250 outside visible n = 3 → λ = 417 in visible n = 4 → λ = 312.5 outside visible λ = 420 nm A1 [2]

This question in 9702/21 Oct/Nov 2012

5 (a) State three conditions required for maxima to be formed in an interference pattern For produced by two sources of microwaves. Examiner’s Use 1. … … 2. … … 3. … … [3] (b) A microwave source M emits microwaves of frequency 12 GHz. Show that the wavelength of the microwaves is 0.025 m. [3] (c) Two slits S1 and S2 are placed in front of the microwave source M described in (b), as shown in Fig 5.1. P m 0.75 m S1 0.90 O M microwave detector S2 Fig. 5.1 (not to scale) The distances S1O and S2O are equal. A microwave detector is moved from O to P. The distance S1P is 0.75 m and the distance S2P is 0.90 m. The microwave detector gives a maximum reading at O. For Examiner’s State the variation in the readings on the microwave detector as it is moved slowly along Use the line from O to P. … … … … … [3] (d) The microwave source M is replaced by a source of coherent light. State two changes that must be made to the slits in Fig. 5.1 in order to observe an interference pattern. 1. … 2. … [2]

11 marks

Mark scheme: 5 (a) waves overlap / meet / superpose (B1) coherence / constant phase difference (not constant λ or frequency) (B1) path difference = 0, λ, 2λ or phase difference = 0, 2π, 4π (B1) same direction of polarisation/unpolarised (B1) max. 3 [3] (b) λ = v / f C1 f = 12 × 109 Hz C1 λ = 3 × 108 / 12 × 109 (any subject) M1 = 0.025 m A0 [3] (c) maximum at P B1 several minima or maxima between O and P B1 5 maxima / 6 minima between O and P or 7 maxima / 6 minima including O and P B1 [3] (d) slits made narrower B1 slits put closer together B1 [2] (not just ‘make slits smaller’) Allow tilting the slits M1 and explanation of axes of rotation A1 GCE AS/A LEVEL – May/June 2013 9702 21

This question in 9702/21 May/June 2013

7 A laser is placed in front of a double slit, as shown in Fig. 7.1. P double slit 12 mm laser Q bright fringes 2.8 m screen Fig. 7.1 (not to scale) The laser emits light of frequency 670 THz. Interference fringes are observed on the screen. (a) Explain how the interference fringes are formed. … … … … … … [3] (b) Show that the wavelength of the light is 450 nm. [2] (c) The separation of the maxima P and Q observed on the screen is 12 mm. The distance between the double slit and the screen is 2.8 m. Calculate the separation of the two slits. separation = … m [3] (d) The laser is replaced by a laser emitting red light. State and explain the effect on the interference fringes seen on the screen. … … … … [2]

10 marks

Mark scheme: 7 (a) waves from the double slit are coherent / constant phase difference B1 waves (from each slit) overlap / superpose / meet (not interfere) B1 maximum / bright fringe where path difference is nλ or phase difference is n360U / 2πn rad 1 or minimum / dark fringe where path difference is (n + )λ 2 or phase difference is (2n + 1) 180U / (2n + 1)π rad B1 [3] (b) v = fλ C1 λ = (3 × 108) / 670 × 1012 = 448 (or 450) (nm) M1 [2] (c) w = 12 / 9 C1 a (= Dλ / w) = (2.8 × 450 × 10–9) / (12 / 9 × 10–3) [allow nm, mm] C1 = 9.5 × 10–4 m [9.4 × 10–4 m using λ = 448 nm] A1 [3] (d) (red light has) larger / higher / longer wavelength (must be comparison) M1 fringes further apart / larger separation A1 [2]

This question in 9702/22 May/June 2014

6 (a) State one difference and one similarity between longitudinal and transverse waves. difference: … … similarity: … … [2] (b) A laser is placed in front of two slits as shown in Fig. 6.1. slits laser 0.35 mm 2.5 m screen Fig. 6.1 (not to scale) The laser emits light of wavelength 6.3 × 10–7 m. The distance from the slits to the screen is 2.5 m. The separation of the slits is 0.35 mm. An interference pattern of maxima and minima is observed on the screen. (i) Explain why an interference pattern is observed on the screen. … … … … [2] (ii) Calculate the distance between adjacent maxima. distance = … m [2] (c) State and explain the effect, if any, on the distance between adjacent maxima when the laser is replaced by another laser emitting ultra-violet radiation. … … [1]

7 marks

Mark scheme: 6 (a) difference: vibration / oscillation (of particles) / displacement of particles is parallel to energy transfer / wavefronts in longitudinal and perpendicular for transverse B1 or transverse can be polarised, longitudinal cannot be polarised similarity: both transfer / propagate energy B1 [2] (b) (i) waves from slits are coherent / constant phase relationship (B1) waves overlap (at screen) with a phase difference or have a path difference (B1) maxima where phase difference is integer ×360° (or ×2π rad) or path difference is integer ×λ or equivalent explanation of minima e.g. (n+½)×360° (B1) max. 2 [2] (ii) maxima spacing = λD / a C1 = (6.3 × 10–7 × 2.5) / 0.35 × 10–3 = 4.5 × 10–3 m A1 [2] (c) (ultra-violet has) shorter wavelength, hence smaller separation / distance A1 [1]

This question in 9702/22 Oct/Nov 2014

7 An arrangement that is used to demonstrate interference with waves on the surface of water is shown in Fig. 7.1. A wooden bar dipper D1 to d.c. power B supply motor dipper water D2 Fig. 7.1 (view from above) (a) Two dippers D1 and D2 are connected to a motor and a d.c. power supply. Initially only D1 vibrates on the water surface to produce waves. The variation with distance x from D1 of the displacement y of the water at one instant of time is shown in Fig. 7.2. 4.0 y / mm 2.0 0 0 10 20 30 40 50 x / mm –2.0 –4.0 Fig. 7.2 Using Fig. 7.2, determine (i) the amplitude of the wave, amplitude = … mm [1] (ii) the wavelength of the wave. wavelength = … mm [1] (b) The two dippers D1 and D2 are made to vibrate and waves are produced by both dippers on the water surface. (i) State and explain whether these waves are stationary or progressive. … … [1] (ii) Explain why D1 and D2 are connected to the same motor. … … [1] (c) The points A and B on Fig. 7.1 are at the distances from D1 and D2 shown in Fig. 7.3. D1A D2A D1B D2B 5.0 cm 7.0 cm 5.0 cm 6.0 cm Fig. 7.3 State and explain the variation with time of the displacement of the water on the surface at (i) A, … … … … [2] (ii) B. … … … [1]

7 marks

Mark scheme: 7 (a) (i) 3.2 mm A1 [1] (ii) 20 mm A1 [1] (b) (i) energy is transferred/propagated (through the water) or wave profile/wavefronts move (outwards from dipper) so progressive B1 [1] (ii) to produce waves with constant/zero phase difference/coherent waves B1 [1] (c) (i) path difference is λ B1 water vibrates/oscillates with amplitude about 2 × 3.2 mm B1 [2] (ii) path difference is λ / 2 so little/no motion/displacement/amplitude B1 [1]

This question in 9702/22 Oct/Nov 2015

4 (a) (i) By reference to the direction of propagation of energy, state what is meant by a transverse wave. … … [1] (ii) State the principle of superposition. … … … [2] (b) Circular water waves may be produced by vibrating dippers at points P and Q, as illustrated in Fig. 4.1. wavefront P 44 cm R 29 cm Q Fig. 4.1 (not to scale) The waves from P alone have the same amplitude at point R as the waves from Q alone. Distance PR is 44 cm and distance QR is 29 cm. The dippers vibrate in phase with a period of 1.5 s to produce waves of speed 4.0 cm s−1. (i) Determine the wavelength of the waves. wavelength = … cm [2] (ii) By reference to the distances PR and QR, explain why the water particles are at rest at point R. … … … … … [3] (c) A wave is produced on the surface of a different liquid. At one particular time, the variation of the vertical displacement y with distance x along the surface of the liquid is shown in Fig. 4.2. 1.0 y / cm 0.5 0 00 2 4 6 8 10 x / cm –0.5 –1.0 Fig. 4.2 (i) The wave has intensity I1 at distance x = 2.0 cm and intensity I2 at x = 10.0 cm. Determine the ratio intensity I2 . intensity I1 ratio = … [2] (ii) State the phase difference, with its unit, between the oscillations of the liquid particles at distances x = 3.0 cm and x = 4.0 cm. phase difference = … [1] [Total: 11]

11 marks

Mark scheme: 4 (a) (i) Displacement of particles perpendicular to direction of energy propagation B1 (ii) waves meet / overlap (at a point) B1 (resultant) displacement is sum of the individual displacements B1 (b) (i) λ = vT or λ = v / f and f = 1 / T C1 λ = 4.0 × 1.5 λ = 6.0 (cm) A1 (ii) path difference [= (44 cm – 29 cm) / 6 cm] = 2.5λ M1 either waves have path difference = (n + ½)λ or waves have phase difference = 180° M1 so destructive interference A1 (c) (i) intensity ∝ (amplitude)2 C1 ratio = (0.602 / 0.902) = 0.44 A1 (ii) phase difference = 90° A1

This question in 9702/22 Feb/March 2016

5 The variation with time t of the displacement y of a wave X, as it passes a point P, is shown in Fig. 5.1. 4.0 3.0 \ / cm ZDYH ; 2.0 1.0 0 0 1.0 2.0 3.0 4.0 5.0 W / ms –1.0 –2.0 –3.0 –4.0 Fig. 5.1 The intensity of wave X is I. (a) Use Fig. 5.1 to determine the frequency of wave X. frequency = … Hz [2] (b) A second wave Z with the same frequency as wave X also passes point P. Wave Z has intensity 2I. The phase difference between the two waves is 90°. On Fig. 5.1, sketch the variation with time t of the displacement y of wave Z. Show your working. [3] (c) A double-slit interference experiment is used to determine the wavelength of light emitted from a laser, as shown in Fig. 5.2. 0.45 mm laser light ' double slit screen Fig. 5.2 (not to scale) The separation of the slits is 0.45 mm. The fringes are viewed on a screen at a distance D from the double slit. The fringe width x is measured for different distances D. The variation with D of x is shown in Fig. 5.3. 5.0 4.0 [ / mm 3.0 2.0 1.0 0 1.5 2.0 2.5 3.0 3.5 ' / m Fig. 5.3 (i) Use the gradient of the line in Fig. 5.3 to determine the wavelength, in nm, of the laser light. wavelength = … nm [4] (ii) The separation of the slits is increased. State and explain the effects, if any, on the graph of Fig. 5.3. … … … [2] [Total: 11]

11 marks

Mark scheme: 5 (a) T = 4 (ms) or 4 × 10–3 (s) C1 f = 1 / T = 1 / 0.004 = 250 Hz A1 [2] (b) intensity ∝ (amplitude)2 and amplitude = 2.8 (2.83) (cm) B1 curve with same period and with amplitude 2.8 cm B1 curve shifted 1.0 ms to left or to right of wave X B1 [3] (c) (i) gradient = (4.5 – 2.4) × 10–3 / (3.25 – 1.75) [= 1.4 × 10–3] B1 wavelength = 0.45 × 10–3 × 1.4 × 10–3 C1 = 6.30 × 10–7 (m) C1 = 630 nm A1 [4] (ii) (gradient is equal to λ / a therefore) gradient of line is reduced B1 value of x will be reduced for all values of D or new line is completely below old line or intercept is less B1 [2]

This question in 9702/21 May/June 2016

5 (a) Light of a single wavelength is incident on a diffraction grating. Explain the part played by diffraction and interference in the production of the first order maximum by the diffraction grating. diffraction: … … interference: … … … … [3] (b) The diffraction grating illustrated in Fig. 5.1 is used with light of wavelength 486 nm. second order first order light wavelength 486 nm 59.4° zero order diffraction grating first order second order screen Fig. 5.1 (not to scale) The orders of the maxima produced are shown on the screen in Fig. 5.1. The angle between the two second order maxima is 59.4°. Calculate the number of lines per millimetre of the grating. number of lines per millimetre = … mm–1 [3] [Total: 6]

6 marks

Mark scheme: 5 (a) diffraction: spreading/diverging of waves/light (takes place) at (each) slit/ element/gap/aperture B1 interference: overlapping of waves (from coherent sources at each element) B1 path difference λ/phase difference of 360(°)/2π (produces the first order) B1 [3] (b) d sinθ = nλ or sinθ = Nnλ C1 d = (2 × 486 × 10–9) / sin 29.7° (= 1.962 × 10–6) C1 number of lines = 510 (509.7) mm–1 A1 [3]

This question in 9702/22 May/June 2016

4 (a) State what is meant by the diffraction of a wave. … … [2] (b) An arrangement for demonstrating the interference of light is shown in Fig. 4.1. laser light Y dark fringe 2.0 mm 0.41 mm X central bright fringe wavelength 580 nm Z dark fringe ' double slit screen Fig. 4.1 (not to scale) The wavelength of the light from the laser is 580 nm. The separation of the slits is 0.41 mm. The perpendicular distance between the double slit and the screen is D. Coherent light emerges from the slits and an interference pattern is observed on the screen. The central bright fringe is produced at point X. The closest dark fringes to point X are produced at points Y and Z. The distance XY is 2.0 mm. (i) Explain why a bright fringe is produced at point X. … … … … [2] (ii) State the difference in the distances, in nm, from each slit to point Y. distance = … nm [1] (iii) Calculate the distance D. D = … m [3] (iv) The intensity of the light passing through the two slits 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: 10]

10 marks

Mark scheme: 4 (a) wave incident on/passes by or through an aperture/edge B1 wave spreads (into geometrical shadow) B1 [2] (b) (i) waves (from slits) overlap (at point X) B1 path difference (from slits to X) is zero/ phase difference (between the two waves) is zero (so constructive interference gives bright fringe) B1 [2] (ii) difference in distances = λ/ 2 = 580 / 2 = 290 nm A1 [1] (iii) λ= ax / D C1 D = [0.41 × 10–3 × (2 × 2.0 × 10–3)] / 580 × 10–9 C1 = 2.8 m A1 [3] (iv) same separation/fringe width/number of fringes bright fringe(s)/central bright fringe/(fringe at) X less bright dark fringe(s)/(fringe at) Y/(fringe at) Z brighter contrast between fringes decreases Any two of the above four points, 1 mark each B2 [2]

This question in 9702/22 Oct/Nov 2016

6 (a) Interference fringes may be observed using a light-emitting laser to illuminate a double slit. The double slit acts as two sources of light. Explain (i) the part played by diffraction in the production of the fringes, … … … [2] (ii) the reason why a double slit is used rather than two separate sources of light. … … … [1] (b) A laser emitting light of a single wavelength is used to illuminate slits S1 and S2, as shown in Fig. 6.1. A S1 laser 0.48 mm light S2 screen 2.4 m B Fig. 6.1 (not to scale) An interference pattern is observed on the screen AB. The separation of the slits is 0.48 mm. The slits are 2.4 m from AB. The distance on the screen across 16 fringes is 36 mm, as illustrated in Fig. 6.2. 16 fringes 36 mm Fig. 6.2 Calculate the wavelength of the light emitted by the laser. wavelength = … m [3] (c) Two dippers D1 and D2 are used to produce identical waves on the surface of water, as illustrated in Fig. 6.3. P 7.27.2 cmcm D1 water 11.2 cm D2 Fig. 6.3 (not to scale) Point P is 7.2 cm from D1 and 11.2 cm from D2. The wavelength of the waves is 1.6 cm. The phase difference between the waves produced at D1 and D2 is zero. (i) State and explain what is observed at P. … … … [2] (ii) State and explain the effect on the answer to (c)(i) if the apparatus is changed so that, separately, 1. the phase difference between the waves at D1 and at D2 is 180°, … … … 2. the intensity of the wave from D1 is less than the intensity of that from D2. … … … [2] [Total: 10]

10 marks

Mark scheme: 6(a)(i) waves at (each) slit/aperture spread B1 (into the geometric shadow) wave(s) overlap/superpose/sum/meet/intersect B1 6(a)(ii) there is not a constant phase difference/coherence (for two separate light source(s)) or waves/light from the double slit are coherent/have a constant phase difference B1 6(b) x = λD / a C1 λ = (36 × 10–3 × 0.48 × 10–3) / (16 × 2.4) C1 = 4.5 × 10–7 m A1 6(c)(i) no movement of the water/water is flat/no ripples/disturbance B1 the path difference is 2.5λ or the phase difference is 900° or 5π rad B1 6(c)(ii) 1. surface/water/P vibrates/ripples and as (waves from the two dippers) arrive in phase B1 2. surface/water/P vibrates/ripples and as amplitudes/displacements are no longer equal/do not cancel B1

This question in 9702/22 May/June 2017

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

5 Microwaves with the same wavelength and amplitude are emitted in phase from two sources X and Y, as shown in Fig. 5.1. path of detector X A position of central maximum B position of adjacent minimum Y Fig. 5.1 (not to scale) A microwave detector is moved along a path parallel to the line joining X and Y. An interference pattern is detected. A central intensity maximum is located at point A and there is an adjacent intensity minimum at point B. The microwaves have a wavelength of 0.040 m. (a) Calculate the frequency, in GHz, of the microwaves. frequency = … GHz [3] (b) For the waves arriving at point B, determine: (i) the path difference path difference = … m [1] (ii) the phase difference. phase difference = … ° [1] (c) The amplitudes of the waves from the sources are changed. This causes a change in the amplitude of the waves arriving at point A. At this point, the amplitude of the wave arriving from source X is doubled and the amplitude of the wave arriving from source Y is also doubled. Describe the effect, if any, on the intensity of the central maximum at point A. … … … [2] (d) Describe the effect, if any, on the positions of the central intensity maximum and the adjacent intensity minimum due to the following separate changes. (i) The separation of the sources X and Y is increased. … … [1] (ii) The phase difference between the microwaves emitted by the sources X and Y changes to 180°. … … [1] [Total: 9]

9 marks

Mark scheme: 5(a) v = fλ or c = fλ C1 f = 3.0 × 108 / 0.040 C1 = 7.5 × 109 (Hz) = 7.5 GHz A1 5(b)(i) path difference = 0.020 m A1 5(b)(ii) phase difference = 180° A1 5(c) (intensity) increases C1 (intensity) increases by a factor of 4 A1 5(d)(i) minimum moves closer to the maximum or decrease in separation of maximum and minimum B1 5(d)(ii) the maximum and minimum exchange places or the maximum becomes a minimum and the minimum becomes a maximum B1

This question in 9702/22 Oct/Nov 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) For a progressive wave on a stretched string, state what is meant by amplitude. … … [1] (b) Light from a laser has a wavelength of 690 nm in a vacuum. Calculate the period of the light wave. period = … s [3] (c) A two-source interference experiment uses the arrangement shown in Fig. 5.1. a D light from laser, double slit screen wavelength λ Fig. 5.1 (not to scale) Light from a laser is incident normally on a double slit. A screen is parallel to the double slit. Interference fringes are seen on the screen at distance D from the double slit. The separation of the centres of the slits is a. The light has wavelength λ. The separation x of the centres of adjacent bright fringes is measured for different values of distance D. The variation with D of x is shown in Fig. 5.2. x 0 0 D Fig. 5.2 The gradient of the graph is G. (i) Determine an expression, in terms of G and λ, for the separation a of the slits. a = … [2] (ii) The experiment is repeated with slits of separation 2a. The wavelength of the light is unchanged. On Fig. 5.2, sketch a graph to show the results of this experiment. [2] [Total: 8]

8 marks

Mark scheme: 5(a) maximum displacement (of a point/particle on string/wave) B1 5(b) v = λ / T or v = fλ and f = 1 / T C1 T = 690 × 10–9 / 3.00 × 108 C1 = 2.3 × 10–15 s A1 5(c)(i) λ = ax / D C1 G = x / D (so) a = λ / G A1 5(c)(ii) straight line from origin always below printed line M1 line is half the height of printed line at maximum D A1

This question in 9702/23 Oct/Nov 2021

5 Light from a laser is used to produce an interference pattern on a screen, as shown in Fig. 5.1. 0.44 mm O central bright fringe P dark fringe Q bright fringe light of R dark fringe wavelength 660 nm 1.8 m double screen slit Fig. 5.1 (not to scale) The light of wavelength 660 nm is incident normally on two slits that have a separation of 0.44 mm. The double slit is parallel to the screen. The perpendicular distance between the double slit and the screen is 1.8 m. The central bright fringe on the screen is formed at point O. The next dark fringe below point O is formed at point P. The next bright fringe and the next dark fringe below point P are formed at points Q and R respectively. (a) The light waves from the two slits are coherent. State what is meant by coherent. … … [1] (b) For the two light waves superposing at R, calculate: (i) the difference in their path lengths, in nm, from the slits path difference = … nm [1] (ii) their phase difference. phase difference = … ° [1] (c) Calculate the distance OQ. distance OQ = … m [3] (d) The intensity of the light incident on the double slit is increased without changing the frequency. Describe how the appearance of the fringes after this change is different from, and similar to, their appearance before the change. … … … … … [3] (e) The light of wavelength 660 nm is now replaced by blue light from a laser. State and explain the change, if any, that must be made to the separation of the two slits so that the fringe separation on the screen is the same as it was for light of wavelength 660 nm. … … … … [2] [Total: 11]

11 marks

Mark scheme: 5(a) constant phase difference (between the waves) B1 5(b)(i) path difference = 1.5  660 = 990 nm A1 5(b)(ii) phase difference = 360°  1.5 = 540° A1 5(c)  = ax / D C1 x = (660  10–9  1.8) / 0.44  10–3 C1 = 2.7  10–3 m A1 5(d) bright fringes are brighter B1 no change to dark fringes B1 no change to (fringe) separation / (fringe) spacing B1 5(e) (blue light has) shorter wavelength M1 (so) decrease (slit) separation A1

This question in 9702/22 May/June 2022

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

5 (a) An electromagnetic wave in a vacuum has a wavelength of 8.4 × 10–6 m. (i) State the name of the principal region of the electromagnetic spectrum for the wave. … [1] (ii) Calculate the frequency, in THz, of the wave. frequency = … THz [2] (b) An arrangement that uses a double slit to demonstrate the interference of light from a laser is shown in Fig. 5.1. screen double slit light, wavelength 6.2 × 10–7 m a 2.8 m Fig. 5.1 (not to scale) The light from the laser has a wavelength of 6.2 × 10–7 m and is incident normally on the slits. The separation of the two slits is a. The slits and screen are parallel and separated by a distance of 2.8 m. An interference pattern of bright fringes and dark fringes is formed on the screen. The distance on the screen across 8 bright fringes is 22 mm, as illustrated in Fig. 5.2. P Q R dark 22 mm fringe bright fringe Fig. 5.2 (i) The light waves emerging from the two slits are coherent. State what is meant by coherent. … … [1] (ii) Calculate the separation a of the slits. a = … m [3] (c) Fringe P is the central bright fringe of the interference pattern in (b). Fringe Q and fringe R are the nearest dark fringe and the nearest bright fringe respectively to the right of fringe P, as shown in Fig. 5.2. (i) Calculate the difference in the distances (the path difference) from each slit to the centre of fringe Q. difference in the distances = … m [1] (ii) State the phase difference between the light waves meeting at the centre of fringe R. phase difference = … ° [1] [Total: 9]

9 marks

Mark scheme: 5(a)(i) infrared B1 5(a)(ii) v = f or c = f C1 f = 3.0  108 / 8.4  10–6 = 3.6  1013 (Hz) = 36 THz A1 5(b)(i) constant phase difference (between the waves) (with time) B1 5(b)(ii)  = ax / D C1 x = 22 / 8 or 2.75 (mm) or 22  10–3 / 8 or 2.75  10–3 (m) C1 a = (6.2  10–7  2.8) / (22  10–3 / 8) = 6.3  10–4 m A1 5(c)(i) difference in distances = 6.2  10–7 / 2 = 3.1  10–7 m A1 5(c)(ii) phase difference = 360° A1

This question in 9702/21 May/June 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

5 Two point sources, A and B, produce coherent electromagnetic waves. The waves from A and B are emitted in phase and have wavelength λ, as shown in Fig. 5.1. λ A B Fig. 5.1 (not to scale) The lines on Fig. 5.1 represent wavefronts. All the points on a wavefront are in phase. (a) On Fig. 5.1, mark with a cross (×): (i) the position of an interference maximum (label this cross Y) [1] (ii) the position of an interference minimum (label this cross Z). [1] (b) The waves in air have a wavelength of 2.9 × 10–5 m. An interference pattern is detected along a line parallel to AB and at a perpendicular distance of 140 m from AB. The spacing between adjacent interference maxima is 1.2 cm. (i) Calculate the separation a of the sources A and B. a = … m [3] (ii) State the principal region of the electromagnetic spectrum to which the waves belong. … [1] [Total: 6]

6 marks

Mark scheme: 5(a)(i) cross labelled Y drawn: B1 at any position where wavefronts cross or centrally in a ‘diamond’ shape formed between any adjacent wavefronts from A and B 5(a)(ii) cross labelled Z drawn on a wavefront from one source at a point midway between adjacent wavefronts from the other B1 source 5(b)(i) = ax / D C1 a = (2.9  10–5  140) / (1.2  10–2) C1 = 0.34 m A1 5(b)(ii) infrared A1

This question in 9702/23 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

6 Two coherent sources X and Y of microwaves of frequency 2.5 × 1010 Hz are a distance of 0.18 m apart in a vacuum, as shown in Fig. 6.1. P 2.3 m X A 0.18 m O Y Q Fig. 6.1 (not to scale) There is a phase difference of 90° between the waves emitted at the two sources. A microwave detector moves along the line PQ, which is parallel to the line joining the two sources and 2.3 m away from it. Point O is on the line PQ at a position that is equidistant from the two sources. Point A is the position on line PQ where the intensity of the microwaves is the greatest. (a) (i) Explain why the position of greatest intensity is not at point O. … … … … [2] (ii) On Fig. 6.1, draw a cross (×) to show the position of the point on line PQ where the intensity minimum that is the closest to point O occurs. Label this point B. [2] (b) (i) Show that the wavelength of the microwaves is 0.012 m. [2] (ii) For point A on line PQ, determine the difference in the distances Δx travelled by the microwaves from X and the microwaves from Y. Δx = … m [1] (iii) Use the formula for the double-slit interference of light to calculate the distance between adjacent intensity maxima on line PQ. distance = … m [2] [Total: 9]

9 marks

Mark scheme: 6(a)(i) for maximum intensity, waves must be in phase (at detection) B1 (phase difference at source means) waves are not in phase / have a 90° phase difference at O B1 6(a)(ii) point B labelled on the line PQ below O M1 distance OB = distance OA A1 6(b)(i) v = f C1 = (3.00  108) / (2.5  1010) = 0.012 m A1 6(b)(ii) x = 0.012 / 4 A1 = 0.0030 m 6(b)(iii) = ax / D C1 distance = D / a = 0.012  2.3 / 0.18 A1 = 0.15 m

This question in 9702/21 Oct/Nov 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