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IB Physics Higher Level · Topic C.3

C.3 Wave phenomena

113 free, high-quality past-paper-style questions on Wave phenomena for IB Physics HL, each with a full mark scheme. Covers: Two-source interference; Refraction of wavefronts; Diffraction around bodies and apertures; and more. For when you have finished the past papers, or want to practise one topic at a time.C.3 at Standard Level →Read the C.3 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.496 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1C-1A-18Two-source interference Paper 1AEasy1 mark

Two coherent sources each produce waves of intensity I0 at a point P. The waves arrive at P in phase. The intensity of a wave is proportional to the square of its amplitude.

What is the intensity at P?

Show mark scheme
Marking pointMarkNotes
Step 1Intensity ∝ amplitude²; each wave has amplitude A with I0 ∝ A².—All 3 steps must be completed — there is no mark for a part-answer.
Step 2In phase, the amplitudes add: resultant amplitude 2A.—
Step 3Intensity ∝ (2A)² = 4A², i.e. 4I0. (At a minimum it is zero, so the average over the pattern is 2I0 — energy is conserved.)✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThis is the result for waves arriving in antiphase, with equal amplitudes. Waves arriving in phase reinforce.
  • BThis is the intensity of one source alone; it ignores the second wave.
  • CThis adds the intensities. It is the displacements (and so the amplitudes) that superpose: the amplitude doubles.
  • DCorrect: the amplitude doubles, so the intensity is 2² = 4 times as large.
2C-1A-28Refraction of wavefronts Paper 1AEasy1 mark

The diagram shows wavefronts of a wave passing from medium 1 into medium 2. The wavefront spacing is 1.5 cm in medium 1 and 1.0 cm in medium 2.

What is the ratio n2/n1 of the refractive index of medium 2 to that of medium 1?

medium 1medium 2spacing 1.5 cmspacing 1.0 cm
Plane wavefronts crossing the boundary between two media (drawn to scale); the arrow shows the direction of travel in medium 1.
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Marking pointMarkNotes
Step 1The wavefront spacing is the wavelength: λ1 = 1.5 cm and λ2 = 1.0 cm. The frequency is the same on both sides.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2v = fλ, so v2/v1 = λ2/λ1.—
Step 3n1/n2 = v2/v1, so n2/n1 = λ1/λ2 = 1.5/1.0 = 1.5.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis is λ2/λ1 = v2/v1; the refractive index is inversely proportional to the speed, so the ratio must be inverted.
  • BThis assumes that because the frequency is unchanged nothing else changes; the speed and wavelength both change, so the refractive index does too.
  • CCorrect: n2/n1 = v1/v2 = λ1/λ2 = 1.5.
  • DThis squares the wavelength ratio; the refractive index is simply inversely proportional to the speed.
3C-1A-43Diffraction around bodies and apertures Paper 1AEasy1 mark

Straight water waves of wavelength 2.0 cm approach a gap of width 10 cm in a barrier, as shown. The gap is then narrowed to 2.0 cm while the frequency of the waves is kept constant.

Which row describes the waves beyond the narrower gap, compared with those beyond the 10 cm gap?

10 cmdirection of travel2.0 cmbarrier
Ripple tank viewed from above (drawn to scale): straight wavefronts approach a gap in a barrier.
Wavelength beyond the gapSpreading of the waves
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Marking pointMarkNotes
Step 1The water depth (so the wave speed) and the frequency are unchanged, so λ = v/f is unchanged beyond the gap.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Diffraction is most noticeable when the gap width is comparable with the wavelength: at 10 cm (5λ) the waves emerge mostly straight with some bending at the edges; at 2.0 cm (≈ λ) they spread out as almost circular wavefronts.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis assumes the gap “squeezes” the waves and also reverses the effect of gap width on spreading.
  • BThe spreading is right, but diffraction does not change the wavelength — speed and frequency are both unchanged.
  • CThe wavelength is right, but a narrower gap produces more spreading, not less.
  • DCorrect: same wavelength, much more spreading when the gap ≈ λ.
4C-1A-47Coherence Paper 1AEasy1 mark

The two headlamps of a parked car illuminate the same white wall, but no interference fringes can be seen where the two beams overlap.

Which statement explains this?

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Marking pointMarkNotes
Step 1Waves always superpose where they overlap, but a stable pattern of maxima and minima needs coherent sources: the same frequency and a constant phase difference.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Each lamp emits light in short, random bursts from many atoms, so the phase difference between the two beams changes randomly millions of times per second; any fringes move so fast that they average to uniform illumination.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the sources are not coherent.
  • BUnequal amplitudes would reduce the contrast of fringes (minima not completely dark) but would not remove them.
  • CSuperposition always occurs; it is the steady pattern that is missing.
  • DSlits are one way of obtaining coherent sources from a single source, but interference itself does not require slits.
5C-1A-61Reflection, refraction and transmission at boundaries Paper 1AEasy1 mark

An upward pulse travels along each of two long horizontal ropes towards the right-hand end of the rope. Rope 1 is tied to a rigid wall. Rope 2 is tied to a light ring that can slide without friction on a smooth vertical pole, so that this end is free to move.

Which row describes the reflected pulse on each rope?

rope 1rope 2wallsmooth polerope 1: end fixed to the wallrope 2: end tied to a light ring
Two ropes, each carrying an upward pulse travelling towards its right-hand end (not to scale).
Rope 1 (fixed end)Rope 2 (free end)
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Marking pointMarkNotes
Step 1At the wall the rope cannot move, so the end is always a point of zero displacement (a node). The reflected pulse must cancel the incident pulse there, so it is inverted.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the free end the ring is pulled up by the incident pulse and overshoots, so the end is a point of maximum displacement (an antinode). The reflected pulse is not inverted.—
Step 3Rope 1: inverted; rope 2: upright.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: a fixed end reflects the pulse inverted (a phase change of π); a free end reflects it upright.
  • BThis assumes that reflection never changes the orientation of the pulse; it is true only at the free end.
  • CThis assumes that every reflection inverts the pulse; it is true only at the fixed end.
  • DThis swaps the two boundary conditions: the fixed end is a node and inverts, the free end is an antinode and does not.
6C-1A-62Young's double-slit experiment Paper 1AEasy1 mark

In a double-slit experiment with monochromatic light the fringe spacing on the screen is s. The whole apparatus, including the slits and the screen, is then immersed in water of refractive index 1.33. The light source, the slit separation and the slit-to-screen distance are unchanged.

What is the new fringe spacing?

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Marking pointMarkNotes
Step 1The frequency of the light is set by the source and does not change, but the speed falls to c/1.33, so the wavelength in the water is λ/1.33.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2s = λD/d with D and d unchanged, so s ∝ λ.—
Step 3New spacing = s/1.33 = 0.75s.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis divides by the refractive index twice (s/1.33² = 0.57s), as if the fringe spacing were proportional to the square of the wavelength.
  • BCorrect: the wavelength, and so the fringe spacing, is reduced by the factor 1.33: s/1.33 = 0.75s.
  • CThis assumes that the wavelength is fixed by the source. It is the frequency that is fixed; the wavelength changes with the speed.
  • DThis multiplies by the refractive index: the light is slower in water, so its wavelength is shorter, not longer.
7C-1A-69Two-source interference Paper 1AEasy1 mark

Two coherent sources S1 and S2 emit waves of wavelength λ in phase. At point P the path difference from the two sources is 1.5λ. At P, the wave from S1 alone would have amplitude 2A and the wave from S2 alone would have amplitude A. The intensity at P due to S1 alone is I1. The intensity of a wave is proportional to the square of its amplitude.

What is the intensity at P when both sources emit?

Show mark scheme
Marking pointMarkNotes
Step 1A path difference of 1.5λ = (1 + ½)λ means the waves arrive in antiphase: destructive interference.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Resultant amplitude = 2A − A = A.—
Step 3Intensity ∝ amplitude², so I = I1 × (A/2A)² = I1/4.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis assumes that destructive interference always gives zero. Complete cancellation needs equal amplitudes.
  • BCorrect: the amplitudes subtract to A, and the intensity is (1/2)² of I1.
  • CThis subtracts the intensities (I1 − I1/4) instead of the amplitudes.
  • DThis treats 1.5λ as a condition for constructive interference: (3A/2A)²I1.
8IBP2H-023-01-01Superposition of pulses Paper 1AEasy1 mark

Two pulses of the same width and shape travel towards each other along a stretched bungee cord. Pulse P has a maximum displacement of +3.0 cm and pulse Q has a maximum displacement of −5.0 cm. What is the displacement of the cord at the point where the two maxima coincide?

−53Oxdisplacement / cmPQ
Pulses travelling towards each other
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Marking pointMarkNotes
By the principle of superposition the resultant displacement is the vector (signed) sum of the individual displacements: (+3.0) + (−5.0) = −2.0 cm. Adding the magnitudes gives 8.0 cm, which ignores that the pulses are on opposite sides of the equilibrium position. +2.0 cm gets the sign wrong: the larger pulse is below the equilibrium position.✓ 1Answer B

Answer: B

Common error: Displacements add with their signs. Only add magnitudes when both pulses are on the same side of the equilibrium position.

9IBP2H-023-01-02Double slit with diffraction envelope Paper 1AEasy1 mark

In a Young's double-slit set-up, the slit separation is halved and the distance from the slits to the screen is doubled. The wavelength is unchanged. What happens to the fringe spacing?

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Marking pointMarkNotes
The fringe spacing is s = λD/d. Doubling D doubles s. Halving d also doubles s. The overall factor is 2 × 2 = 4. 'Unchanged' comes from thinking the two changes cancel. 'One quarter' comes from inverting the dependence on both D and d.✓ 1Answer D

Answer: D

Exam tip: Write the proportionality s ∝ D/d before substituting factors. Each change then becomes a multiplication.

10IBP2H-023-00-01Diffraction gratings Paper 1AEasy1 mark

Monochromatic light is incident normally on a diffraction grating. The grating is replaced by a wider grating that has the same line spacing, so that many more lines are illuminated. What happens to the principal maxima on a distant screen?

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Marking pointMarkNotes
The positions of the principal maxima follow from nλ = d sin θ. This depends only on the line spacing d and the wavelength, so the maxima do not move. With more lines contributing, the waves cancel more completely at angles slightly away from each maximum, so the maxima become narrower. More light passes through the grating, so they are also brighter. B and C wrongly assume the number of lines changes the spacing of the maxima. D reverses the effect of adding more lines.✓ 1Answer A

Answer: A

Common error: Changing the number of lines N does not move the maxima; only d and λ do. N controls how sharp and how bright the maxima are.

11IBP2H-023-00-06Optical fibre Paper 1AEasy1 mark

In an optical fibre, the speed of light is 2.00×108 m s−1 in the core and 2.05×108 m s−1 in the cladding. Which expression gives the critical angle c at the core–cladding boundary?

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Marking pointMarkNotes
At the critical angle the refracted ray travels along the boundary, so ncore sin c = nclad sin 90°. This gives sin c = nclad/ncore. Since n = c0/v, this ratio equals vcore/vclad = 2.00/2.05, so A is correct.
B inverts the ratio, which would give a sine greater than 1.
C uses the core–air boundary instead of the core–cladding boundary.
D uses cosine, as if the angle were measured from the surface rather than from the normal.
✓ 1Answer A

Answer: A

Exam tip: Refractive index is inversely proportional to wave speed. Total internal reflection needs light to be slower in the first medium, so sin c is the slower speed divided by the faster speed, which is less than 1.

12IBP2H-023-02-01Refraction of wavefronts Paper 1AEasy1 mark

A beam of monochromatic light passes from air into a glass block. Which row correctly describes what happens to the frequency, speed and wavelength of the light in the glass?

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Marking pointMarkNotes
The frequency is set by the source and cannot change at a boundary, because the number of wavefronts arriving each second must equal the number leaving. Glass has a higher refractive index, so v = c/n is smaller. Since v = fλ with f fixed, the wavelength falls in the same ratio as the speed. Option A wrongly attributes the change to frequency, B reverses the speed change and D keeps the speed constant.✓ 1Answer C

Answer: C

Common error: Colour is linked to frequency, not wavelength. The light does not change colour in glass even though its wavelength is shorter.

13IBP2H-023-02-08Snell's law Paper 1AEasy1 mark

The refractive indices of three materials are: diamond 2.42, crown glass 1.52 and water 1.33. For which journey of a light ray can total internal reflection occur at the boundary?

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Marking pointMarkNotes
Total internal reflection is only possible when light travels from a medium of higher refractive index to one of lower refractive index, and the angle of incidence exceeds the critical angle. Only diamond (2.42) into water (1.33) meets this condition, with sin c = 1.33/2.42. In the other three cases the light enters a medium of higher refractive index.✓ 1Answer D

Answer: D

Exam tip: Check the direction of travel first. If the second medium has the higher refractive index, total internal reflection is impossible at any angle.

14IBP3H-C3-20r25-01Double slit with diffraction envelope Paper 1AEasy1 mark

Light of wavelength 600 nm passes through two narrow slits that are 0.30 mm apart. Interference fringes are observed on a screen 2.0 m from the slits. What is the distance between adjacent bright fringes?

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Marking pointMarkNotes
Fringe spacing = λD/d = (600×10−9 × 2.0)/(0.30×10−3) = 4.0×10−3 m = 4.0 mm. 2.0 mm is the distance from the central maximum to the first minimum, which is half a fringe spacing. 8.0 mm doubles the spacing as if both sides of the central maximum were included. 0.40 mm comes from a power-of-ten slip that treats 0.30 mm as 3.0 mm.✓ 1Answer B

Answer: B

Common error: Remember that fringe spacing is measured between adjacent bright fringes, not from the central maximum to the first dark fringe.

15IBP3H-C3-20r25-02Optical fibre Paper 1AEasy1 mark

Infrared light of wavelength 1500 nm in a vacuum enters the core of an optical fibre, which has refractive index 1.50. What is the wavelength of the light in the core?

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Marking pointMarkNotes
The frequency is unchanged and the speed falls by the factor n, so the wavelength becomes λ/n = 1500/1.50 = 1000 nm. 2250 nm multiplies by n instead of dividing. 1500 nm assumes the wavelength is unchanged, which is true only for the frequency. 667 nm divides by n2.✓ 1Answer D

Answer: D

Exam tip: When light enters a denser medium, frequency stays the same while speed and wavelength both decrease by the factor n.

16C-1A-04Single-slit diffraction Paper 1AMedium1 mark

Monochromatic light passes through a single slit and forms a diffraction pattern on a distant screen. The slit width is doubled and the wavelength is halved.

What happens to the width of the central maximum?

Show mark scheme
Marking pointMarkNotes
Step 1The first minimum is at sin θ = λ/b, so the angular half-width of the central maximum is proportional to λ/b.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Halving λ multiplies the width by ½; doubling b multiplies it by another ½.—
Step 3Overall factor ½ × ½ = ¼.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThe two changes act in the same direction (both narrow the pattern), so they cannot cancel.
  • BThis accounts for only one of the two changes.
  • CCorrect: width ∝ λ/b = (½)/(2) = ¼ of the original.
  • DThis is the factor if the slit were narrowed and the wavelength lengthened.
17C-1A-05Double slit with diffraction envelope Paper 1AMedium1 mark

Monochromatic light passes through two slits and forms an interference pattern on a distant screen. The pattern of bright fringes is modulated by the single-slit diffraction pattern of each slit.

The width of each slit is now reduced while the separation of the centres of the slits is kept the same. Which row describes the change in the separation of adjacent bright fringes and in the number of bright fringes within the central maximum of the diffraction envelope?

Separation of bright fringesNumber of bright fringes in the central maximum
Show mark scheme
Marking pointMarkNotes
Step 1The fringe separation is set by the separation of the slits: s = λD/d. Neither λ nor d changes, so s is unchanged.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The width of the central maximum of the envelope is set by the width of each slit: its first minimum is at θ = λ/b. A smaller b makes the envelope wider.—
Step 3A wider envelope with unchanged fringe spacing contains more bright fringes (the number is about 2d/b − 1).✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: s depends only on d, while the envelope widens as b decreases, so it contains more fringes.
  • BThis assumes a narrower slit gives a narrower diffraction pattern; the angle of the first minimum, λ/b, increases as b decreases.
  • CThis confuses the slit width b with the slit separation d in s = λD/d, as though narrower slits spread the fringes apart.
  • DThis takes the fringe spacing to shrink with the slit width, as if s were proportional to b; the separation d, not b, sets the spacing.
18C-1A-06Diffraction gratings Paper 1AMedium1 mark

Monochromatic light is incident normally on a diffraction grating. The second-order maximum is observed at an angle of 30.0° to the straight-through direction.

What is the angle of the first-order maximum?

Show mark scheme
Marking pointMarkNotes
Step 1For the grating nλ = d sin θ, so sin θ is proportional to the order n; the angle itself is not.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2sin θ1 = sin θ2/2 = sin 30.0°/2 = 0.250.—
Step 3θ1 = sin−1 0.250 = 14.5°.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: sin θ1 = ½ sin 30.0° = 0.250, so θ1 = 14.5°.
  • BThis halves the angle instead of its sine; the grating equation is linear in sin θ, not in θ.
  • CThis halves tan θ, as though the positions of the maxima on a screen were proportional to the order: tan−1(tan 30.0°/2) = 16.1°. That is true only for small angles.
  • DThis inverts the ratio, taking sin θ1 = 2 sin 30.0° = 1.00; the first order is closer to the centre than the second, not further out.
19C-1A-07Refraction: wavelength and frequency Paper 1AMedium1 mark

A ray of light passes from medium X into medium Y. The diagram, drawn on a square grid, shows the incident ray, the refracted ray and the normal at the point of incidence.

What is the ratio (wavelength of the light in Y)/(wavelength of the light in X)?

medium Xmedium Ynormal
The incident and refracted rays drawn on a square grid; the dashed line is the normal at the point of incidence.
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Marking pointMarkNotes
Step 1From the grid the incident ray moves 4 squares parallel to the boundary for every 3 squares towards it, so it is 5 squares long and sin θX = 4/5 = 0.80 (angle measured from the normal).—All 4 steps must be completed — there is no mark for a part-answer.
Step 2The refracted ray moves 3 squares across for every 4 squares down, so sin θY = 3/5 = 0.60.—
Step 3The frequency is unchanged at the boundary, so λ ∝ v, and by Snell's law vY/vX = sin θY/sin θX.—
Step 4λY/λX = 0.60/0.80 = 0.75.✓ 1Answer C

Answer: C  ·  4 stages of work, one mark

Every option, and why

  • AThis uses tangents read from the grid, tan θY/tan θX = (3/4)/(4/3) = 0.56, instead of sines.
  • BThis is the ratio of the angles themselves, 36.9°/53.1° = 0.69; Snell's law relates the sines of the angles, not the angles.
  • CCorrect: λY/λX = sin θY/sin θX = 0.60/0.80 = 0.75.
  • DThis is sin θX/sin θY = 1.33, the ratio nY/nX. The ray bends towards the normal, so the light slows down in Y and its wavelength becomes shorter, not longer.
20C-1A-11Refraction and total internal reflection Paper 1AMedium1 mark

An equilateral glass prism PQR is completely immersed in water of refractive index 1.33. A ray of light enters face PQ along the normal and then strikes face PR, as shown.

What is the least refractive index of the glass for which the ray is totally internally reflected at PR?

PQR60°60°60°waterglass
Equilateral glass prism PQR immersed in water. The ray enters face PQ along the normal and strikes face PR.
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Marking pointMarkNotes
Step 1The ray enters PQ along the normal, so it is not deviated. It meets PR at an angle of incidence equal to the angle between the two faces, 60°.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Total internal reflection needs the angle of incidence to be at least the critical angle: sin c = nwater/nglass ≤ sin 60°.—
Step 3nglass ≥ 1.33/sin 60° = 1.33/0.866 = 1.54.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • A1/sin 60° = 1.15 is the least refractive index for a prism in air; the water outside raises the critical angle.
  • BCorrect: nglass = 1.33/sin 60° = 1.54.
  • C1/sin 30° = 2.00: this takes the angle between the ray and the surface (30°) as the angle of incidence, and also ignores the water.
  • D1.33/sin 30° = 2.66: the water is included, but the angle is measured from the surface instead of from the normal.
21C-1A-13Two-source interference Paper 1AMedium1 mark

Two loudspeakers S₁ and S₂, 3.0 m apart, emit sound of wavelength 1.0 m in phase. A microphone is placed 4.0 m from S₁ along a line perpendicular to S₁S₂.

What is detected at the microphone?

3.0 m4.0 mS₁S₂M
Diagram not to scale
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Marking pointMarkNotes
Step 1Distance from S₂: √(4.0² + 3.0²) = 5.0 m (a 3-4-5 triangle).—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Path difference = 5.0 − 4.0 = 1.0 m.—
Step 3This equals exactly one wavelength (n = 1), so the two waves arrive in phase: constructive interference, a maximum.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: the path difference is one whole wavelength, so the waves arrive in phase and reinforce.
  • BThe path difference is right but the conditions are swapped: a path difference of a whole number of wavelengths gives a maximum; a minimum needs (n + ½)λ.
  • CThis uses a minus sign in Pythagoras: √(4.0² − 3.0²) = 2.65 m, so 4.0 − 2.65 = 1.35 m. The distance from S₂ is the hypotenuse, √(4.0² + 3.0²) = 5.0 m.
  • DThis adds the two perpendicular distances (4.0 + 3.0 = 7.0 m) instead of using Pythagoras, giving a path difference of 3.0 m. The conclusion happens to be right but the path difference is wrong.
22C-1A-24Young's double-slit experiment Paper 1AMedium1 mark

In a double-slit experiment the slits are 0.50 mm apart and the screen is 2.0 m from the slits. The diagram shows the bright fringes seen on the screen and the distance between the centres of the two outermost bright fringes.

What is the wavelength of the light?

14.4 mm
Bright fringes on the screen. The arrow joins the centres of the two outermost bright fringes.
Show mark scheme
Marking pointMarkNotes
Step 1The diagram shows 7 bright fringes, so there are 6 fringe spacings between the centres of the outermost ones: s = 14.4 mm/6 = 2.4 mm.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2s = λD/d gives λ = sd/D.—
Step 3λ = 2.4 × 10−3 × 0.50 × 10−3/2.0 = 6.0 × 10−7 m = 600 nm.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis takes the distance from a bright fringe to the next dark fringe (1.2 mm) as the fringe spacing.
  • BThis divides 14.4 mm by the number of bright fringes (7) instead of the number of spacings between them (6).
  • CCorrect: s = 2.4 mm and λ = sd/D = 600 nm.
  • DThis divides 14.4 mm by 5, the number of bright fringes between the two outermost ones.
23C-1A-41Two-source interference Paper 1AMedium1 mark

Two loudspeakers are connected to the same signal generator, but the leads to one of them are reversed, so that the speakers emit sound of wavelength λ in antiphase.

At which of the following path differences from the two speakers is the sound a minimum?

Show mark scheme
Marking pointMarkNotes
Step 1With sources in phase, a minimum needs a path difference of (n + ½)λ, which produces a phase difference of π.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Here the sources already differ in phase by π. The total phase difference is π + 2π(path difference)/λ, and for cancellation it must be an odd multiple of π, so the path difference must be a whole number of wavelengths.—
Step 3Minima occur at path differences of 0, λ, 2λ, …; of the options only λ.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: with the sources in antiphase, a path difference of a whole wavelength leaves the waves in antiphase, so they cancel.
  • B3λ/2 gives a minimum for sources in phase; the extra π from the reversed leads makes it a maximum.
  • C5λ/2 is the condition for a minimum with sources in phase; here the waves arrive in phase and reinforce.
  • D7λ/2 is the condition for a minimum with sources in phase; with the reversed leads this point is a maximum.
24C-1A-42Single-slit diffraction Paper 1AMedium1 mark

Light of wavelength 500 nm passes through a single slit and forms a diffraction pattern on a screen 2.00 m from the slit. The graph shows how the intensity on the screen varies with the distance y from the centre of the pattern.

What is the width of the slit?

-15-10-5051015y / mm0.00.20.40.60.81.0intensity / arbitrary units
Intensity of the single-slit diffraction pattern against distance from the centre of the pattern (drawn to scale).
Show mark scheme
Marking pointMarkNotes
Step 1From the graph the first minima are at y = ±5.0 mm.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2For small angles θ ≈ y/D = 5.0 × 10−3/2.00 = 2.5 × 10−3 rad.—
Step 3θ = λ/b, so b = 500 × 10−9/2.5 × 10−3 = 2.0 × 10−4 m = 0.20 mm.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis uses the distance between the second minima on either side (20 mm) as y.
  • BThis uses the full width of the central maximum (10 mm) as y; the angle λ/b is measured from the centre to the first minimum.
  • CCorrect: θ = 5.0 mm/2.00 m = 2.5 × 10−3 rad and b = λ/θ = 0.20 mm.
  • DThis uses b = 2λD/y with y = 5.0 mm — the expression for the full width of the central maximum applied to its half-width.
25C-1A-44Snell's law Paper 1AMedium1 mark

A layer of cooking oil of refractive index 1.47 floats on water of refractive index 1.33 in a glass dish. The oil and water surfaces are horizontal. A ray of light in air strikes the oil surface at an angle of incidence of 50°.

What is the angle between the ray in the water and the normal?

airoil, n = 1.47water, n = 1.3350°?
Diagram not to scale
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Step 1At the air–oil surface: 1.00 × sin 50° = 1.47 sin θoil, so θoil = 31.4°.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the oil–water surface (parallel to the first): 1.47 sin 31.4° = 1.33 sin θw.—
Step 3So 1.00 × sin 50° = 1.33 sin θw: the oil layer drops out. sin θw = 0.766/1.33 = 0.576, θw = 35.2°.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis inverts the ratio at the oil–water boundary (sin θw = sin 31.4° × 1.33/1.47 = 0.472), bending the ray towards the normal on entering the slower-to-faster boundary.
  • BThis is the angle in the oil; the ray refracts again at the oil–water boundary.
  • CCorrect: n sin θ is the same in every parallel layer, so sin θw = sin 50°/1.33.
  • DThis divides the angle, not its sine, by the refractive index (50°/1.33 = 37.6°).
26C-1A-45Diffraction gratings Paper 1AMedium1 mark

Monochromatic light is incident normally on two narrow slits a distance d apart, and the pattern is observed on a distant screen. The two slits are then replaced by a diffraction grating with many slits, also a distance d apart, illuminated by the same light.

Which statements about the principal maxima of the grating pattern, compared with the maxima of the two-slit pattern, are correct?

I. They are at the same angles.

II. They are narrower.

III. They are further apart.

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Marking pointMarkNotes
Step 1Principal maxima occur where the waves from adjacent slits have a path difference of a whole number of wavelengths: nλ = d sin θ for both arrangements. With the same d and λ the angles are the same, so I is correct and III is incorrect.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2With many slits, a small change of angle away from a maximum makes the waves from the many slits cancel almost completely, so each principal maximum is much narrower (and brighter): II is correct.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AII is also correct: many slits make each principal maximum much sharper.
  • BI is also correct: the condition nλ = d sin θ depends only on d and λ.
  • CIII is incorrect: the positions of the maxima depend on d, which is unchanged; it is the width of each maximum that decreases.
  • DCorrect: same angles, much narrower maxima.
27C-1A-46Superposition of pulses Paper 1AMedium1 mark

Two pulses travel towards each other along a stretched rope, each at a speed of 20 cm s−1. The diagram shows the rope at time t = 0.

What is the displacement of point P at t = 2.0 s?

P020406080100120140x / cm+6−20
The rope at t = 0 (drawn to scale; vertical scale in cm). Both pulses travel at 20 cm s−1 in the directions shown.
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Step 1In 2.0 s each pulse moves 40 cm. The triangular pulse now occupies 40–80 cm with its peak at 60 cm; the rectangular pulse occupies 60–80 cm.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2At P (70 cm) the triangular pulse is 10 cm from its peak, half-way down its 20 cm slope: displacement = 6.0 × (1 − 10/20) = +3.0 cm. The rectangular pulse gives −2.0 cm.—
Step 3By the principle of superposition the resultant displacement is the sum: +3.0 + (−2.0) = +1.0 cm.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: +3.0 cm − 2.0 cm = +1.0 cm.
  • BThis is the displacement due to the triangular pulse alone; the rectangular pulse is also at P at this instant.
  • CThis uses the peak height of the triangular pulse (6.0 − 2.0) instead of its height at P.
  • DThis adds the magnitudes (3.0 + 2.0), ignoring that the rectangular pulse is inverted.
28C-1A-48Two-source interference Paper 1AMedium1 mark

Two loudspeakers connected to the same signal generator emit sound of frequency 680 Hz in phase. The speed of sound is 340 m s−1. A student walks along a line parallel to the line joining the speakers, starting at the central loud point, and passes through three quiet points.

What is the path difference from the two speakers at the third quiet point?

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Step 1λ = v/f = 340/680 = 0.500 m.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Quiet points (destructive interference) occur where the path difference = (n + ½)λ with n = 0, 1, 2 …; the first quiet point is n = 0.—
Step 3The third quiet point has n = 2: path difference = 2.5 × 0.500 = 1.25 m.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis is 1.5λ: it assumes successive quiet points differ in path difference by λ/2 (3 × λ/2), but they differ by a whole wavelength.
  • BCorrect: 2.5λ = 1.25 m.
  • CThis is 3λ — the condition for the third loud point (constructive, nλ), not a quiet point.
  • DThis is 3.5λ, taking n = 3 for the third minimum instead of n = 2.
29C-1A-67Diffraction gratings Paper 1AMedium1 mark

White light containing all wavelengths from 400 nm to 700 nm is incident normally on a diffraction grating with 300 lines per mm. Complete spectra are formed on both sides of the central maximum.

Which statements are correct?

I. The first-order spectrum and the second-order spectrum overlap.

II. The second-order spectrum and the third-order spectrum overlap.

III. In each spectrum, violet light is diffracted through a smaller angle than red light.

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Marking pointMarkNotes
Step 1The maxima obey nλ = d sin θ, so sin θ ∝ nλ. The overlap of two orders is decided by comparing nλ at the ends of the spectrum; d does not matter.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2First order extends to 1 × 700 = 700 nm and second order starts at 2 × 400 = 800 nm: no overlap (I incorrect).—
Step 3Second order extends to 2 × 700 = 1400 nm and third order starts at 3 × 400 = 1200 nm: they overlap (II correct). Shorter wavelengths give smaller sin θ in every order (III correct).✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AI is incorrect: 700 nm (first-order red) is less than 800 nm (second-order violet), so these spectra are separate.
  • BI is incorrect, and II is correct: third-order violet (3 × 400 = 1200 nm) lies inside the second-order spectrum, which reaches 2 × 700 = 1400 nm.
  • CCorrect: only the second and third orders overlap, and violet is always diffracted least.
  • DI is incorrect: the first-order spectrum ends (700 nm) before the second-order spectrum begins (800 nm).
30IBP2H-023-01-03Single-slit diffraction Paper 1AMedium1 mark

Laser light passes through an adjustable single slit and forms a diffraction pattern on a distant screen. The laser is replaced with one of 1.5 times the wavelength, and the slit width is halved. All diffraction angles remain small. By what factor does the angular width of the central maximum change?

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For small angles, the first minimum is at θ ≈ λ/b. The central maximum has angular width 2θ, so the width is proportional to λ/b. Multiplying λ by 1.5 and dividing b by 2 gives a factor of 1.5 × 2 = 3.0.
The distractors come from wrong proportionalities:
• 0.75 (= 1.5 × 0.5) treats the width as proportional to λb.
• 1.33 (= 2/1.5) wrongly treats the width as inversely proportional to λ.
• 0.33 (= 0.5/1.5) inverts the relation, treating the width as proportional to b/λ.
✓ 1Answer C

Answer: C

Common error: A narrower slit gives a wider central maximum. Diffraction spreading increases as the gap becomes smaller relative to the wavelength, so the width is proportional to λ/b, not b/λ.

31IBP2H-023-01-04Two-source interference Paper 1AMedium1 mark

Two radio transmitters emit in phase at the same wavelength λ. A receiver moves sideways from a point on the perpendicular bisector of the line joining the transmitters. The signal falls to a minimum for the third time. What is the path difference from the transmitters to the receiver at this position?

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For in-phase sources, minima occur where the path difference is (n + ½)λ: 0.5λ, 1.5λ, 2.5λ and so on. The third minimum from the central maximum is therefore at 2.5λ. 3.0λ is a maximum, not a minimum. 1.5λ is only the second minimum. 3.5λ counts one minimum too many.✓ 1Answer B

Answer: B

Exam tip: List the sequence 0.5λ, 1.5λ, 2.5λ … explicitly before choosing. The first minimum is at half a wavelength, not one wavelength.

32IBP2H-023-01-05Double slit with diffraction envelope Paper 1AMedium1 mark

A double slit has slit separation d and each slit has width b, where d = 4b. The slits are illuminated with monochromatic light. How many bright interference fringes can be seen within the central maximum of the single-slit diffraction envelope?

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Interference maxima occur where d sinθ = nλ. The first diffraction minimum occurs where b sinθ = λ, which is the same as d sinθ = 4λ. The orders n = ±4 therefore fall exactly on the envelope minima and are missing. The fringes inside the central envelope are n = −3 to +3, giving 7. The answer 9 wrongly includes the missing ±4 orders. The answer 8 forgets the central (n = 0) fringe. The answer 4 counts only one side.✓ 1Answer B

Answer: B

Exam tip: For d/b = N (an integer), the central envelope contains 2N − 1 bright fringes. Every Nth order is missing.

33IBP2H-023-01-06Young's double-slit experiment Paper 1AMedium1 mark

A double-slit arrangement is used to test a component for an underwater laser communication link. In air the fringe spacing is 2.66 mm. The whole apparatus, including the screen, is then submerged in water of refractive index 1.33. What is the new fringe spacing?

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The frequency is unchanged in water, but the wave speed and the wavelength are both divided by n. Since s = λD/d, the spacing becomes 2.66/1.33 = 2.00 mm. The answer 2.66 mm assumes the wavelength does not change. The answer 3.54 mm multiplies by n instead of dividing. The answer 1.50 mm divides by n twice.✓ 1Answer B

Answer: B

Common error: In a denser medium the wavelength decreases. The frequency, which is set by the source, does not change.

34IBP2S-022-00-05Double slit with diffraction envelope Paper 1AMedium1 mark

Monochromatic light passes through a single slit and forms a diffraction pattern on a distant screen. The angles involved are small. The wavelength is doubled and the slit width is halved. By what factor does the angular width of the central maximum change?

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The first minimum is at θ ≈ λ/b, so the central maximum has angular width 2λ/b.
With λ → 2λ and b → b/2, the width becomes 2(2λ)/(b/2) = 4 × (2λ/b).
The factor is therefore 4.
Option B treats the width as proportional to λb, so the two changes appear to cancel.
Option C counts only the wavelength change.
Option D counts only the slit change and wrongly assumes that a narrower slit gives a narrower maximum.
✓ 1Answer A

Answer: A

Common error: Diffraction spreading increases as the slit gets narrower: θ is proportional to λ/b, so halving b doubles θ.

35IBP2S-022-01-09Double slit with diffraction envelope Paper 1AMedium1 mark

Monochromatic light passes through a single slit, and the diffraction pattern is viewed on a distant screen. The slit width is reduced from 0.20 mm to 0.10 mm. What happens to the angle between the centre of the pattern and the first minimum?

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For a single slit, the first minimum occurs at θ ≈ λ/b. Halving b therefore doubles θ, so the central maximum becomes wider. Option A assumes that a narrower slit gives a narrower pattern. Option B ignores the dependence on b. Option D wrongly squares the factor.✓ 1Answer C

Answer: C

Exam tip: A narrower gap produces more diffraction. The angular spread is inversely proportional to the slit width.

36IBP2H-023-00-02Optical fibre Paper 1AMedium1 mark

A short pulse of light is sent along a long step-index optical fibre. The pulse emerges spread out in time because of multimode (modal) dispersion. Which single change would reduce this dispersion?

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Multimode dispersion happens because rays that zig-zag at different angles travel different path lengths along the fibre. Light is guided only if it meets the core–cladding boundary at more than the critical angle, where sin c = nclad/ncore. If nclad is closer to ncore, the critical angle is closer to 90°. Only rays nearly parallel to the axis are then guided, so the range of path lengths is smaller and the pulse spreads less. Options A and D both increase the index difference, so a wider range of ray angles is guided and the spreading increases. Option C increases the spreading, because the time difference between paths is proportional to the fibre length.✓ 1Answer B

Answer: B

Exam tip: Link dispersion to the range of path lengths, and link that range to the critical angle. That chain of reasoning is what earns credit.

37IBP2H-023-00-03Two-source interference Paper 1AMedium1 mark

Two loudspeakers emit sound of wavelength 0.34 m. They are connected so that they vibrate in antiphase. A microphone is 3.00 m from one loudspeaker and 3.85 m from the other. What does the microphone detect?

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The path difference is 3.85 − 3.00 = 0.85 m. In wavelengths, this is 0.85/0.34 = 2.5λ. A path difference of 2.5λ gives a phase difference of 5π, which is equivalent to π. The sources already differ in phase by π, so the total phase difference is 6π, a whole number of cycles. The waves therefore arrive in phase and interfere constructively, giving a maximum. A applies the in-phase rule and ignores the phase difference between the sources. C ignores the fact that a half-integer path difference gives an exact half-cycle phase difference. D is wrong because whether the waves cancel depends on position.✓ 1Answer B

Answer: B

Common error: The rules 'nλ gives a maximum' and '(n + ½)λ gives a minimum' apply only to sources in phase. For antiphase sources, the two rules swap.

38IBP2H-023-00-04Snell's law Paper 1AMedium1 mark

In an aquarium, a diver shines a torch through water (refractive index 1.33) onto a flat, parallel-sided glass wall (refractive index 1.52). The beam meets the glass at 30° to the normal and then passes into air in the visitor area. What is the angle between the beam in air and the normal?

water, n = 1.33glass1.52air30°?
Diagram not to scale
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At each boundary, n sin θ stays the same: 1.33 sin 30° = 1.52 sin θglass = 1.00 sin θair. Because the glass faces are parallel, the glass index cancels out. This gives sin θair = 1.33 × 0.500 = 0.665, so θair = 41.7°. A (25.9°) is the angle inside the glass. B assumes a parallel-sided block never changes the direction, which is only true when the same medium is on both sides. D uses 1.52 sin 30° and ignores the water.✓ 1Answer C

Answer: C

Exam tip: For parallel boundaries, n sin θ is the same in every layer. Compare the first and last media directly.

39IBP2H-023-00-05Diffraction gratings Paper 1AMedium1 mark

Light of wavelength 550 nm is incident normally on a diffraction grating with 600 lines per millimetre. What is the total number of principal maxima produced?

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The line spacing is d = 1/600 mm = 1.67×10−6 m. The order must satisfy n ≤ d/λ = 1.67×10−6/5.50×10−7 = 3.03, so the highest order is 3. There are orders 1, 2 and 3 on each side, plus the zero order, giving 2 × 3 + 1 = 7. A counts one side only. B counts the zero order and one side. C leaves out the central zero-order maximum.✓ 1Answer D

Answer: D

Common error: Count both sides and include the zero order. Always round d/λ down, never up.

40IBP2H-023-02-02Reflection, refraction and transmission at boundaries Paper 1AMedium1 mark

In medical ultrasound imaging, a pulse travels through soft tissue at 1540 m s−1 and strikes a boundary with bone, where its speed is about 4000 m s−1. The pulse meets the boundary at an angle to the normal. Which statement is correct?

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Refraction depends on wave speed, not on the physical density of the material. The pulse speeds up on entering bone, so by sin θ1/sin θ2 = v1/v2 it bends away from the normal. Total internal reflection requires the wave to travel towards a medium in which it moves faster. That is the case here, from tissue into bone, with sin c = 1540/4000 and c ≈ 23°. Option C is also wrong about wavelength, because λ = v/f increases in bone.✓ 1Answer D

Answer: D

Common error: The light rule 'TIR goes from dense to less dense' really means 'from slow to fast'. For sound and ultrasound, the faster medium is often the denser one.

41IBP2H-023-02-03Double slit with diffraction envelope Paper 1AMedium1 mark

Monochromatic light is incident on a double slit. Each slit has width b and the slit separation is 4b. How many bright interference fringes are seen within the central maximum of the single-slit diffraction envelope?

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The first diffraction minimum is at sin θ = λ/b. Interference maxima occur at sin θ = nλ/d = nλ/(4b). The n = 4 maxima therefore coincide with the first diffraction minima and are missing. Inside the central envelope the visible orders are n = 0, ±1, ±2, ±3, which gives 7 fringes. Option D wrongly counts the missing ±4 orders, option A counts only one side, and option C counts 2 × 4 without allowing for the shared central fringe or the missing orders.✓ 1Answer B

Answer: B

Exam tip: If d/b is a whole number N, the central envelope contains 2N − 1 bright fringes. Order N is the first missing order.

42IBP2H-023-02-05Reflection, refraction and transmission at boundaries Paper 1AMedium1 mark

In a gym, a light training rope is tied to the end of a much heavier rope. Waves travel more slowly in the heavy rope. An upward pulse is sent along the light rope towards the join. Which describes the reflected and transmitted pulses?

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At a boundary into a medium where the wave is slower, the join behaves partly like a fixed end. The reflected pulse undergoes a phase change of π and is inverted. The transmitted pulse is always in phase with the incident pulse, so it is upright, though slower and shorter. Option D describes travel from the heavy rope into the light rope. A transmitted pulse is never inverted.✓ 1Answer A

Answer: A

Common error: Only the reflected wave can be inverted, and only when it reflects from a 'slower' medium. The transmitted wave keeps the original phase.

43IBP2H-023-02-07Young's double-slit experiment Paper 1AMedium1 mark

A bathymetric lidar system maps the sea floor using a green laser of wavelength 532 nm in air. The refractive index of sea water is 1.33. What is the wavelength of the laser light in the water?

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The frequency is unchanged and the speed falls by a factor n, so λwater = λair/n = 532/1.33 = 400 nm. Option C assumes the wavelength does not change. Option D multiplies by n instead of dividing. Option A divides by n twice.✓ 1Answer B

Answer: B

Common error: Light slows down in water, so its wavelength must get shorter. If your answer is longer than 532 nm, you have multiplied by n instead of dividing.

44IBP3H-C3-20r25-03Two-source interference Paper 1AMedium1 mark

Two radio transmitters X and Y emit waves of frequency 150 MHz in phase with each other. A receiver is 1500.0 m from X and 1503.0 m from Y. The speed of radio waves is 3.00×108 m s−1. Which statement describes the signal at the receiver?

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λ = c/f = 3.00×108/1.50×108 = 2.00 m. The path difference is 3.0 m = 1.5λ, a half-integer number of wavelengths, so the waves arrive in antiphase and cancel. Being a whole number of metres has no bearing on interference. A path difference larger than one wavelength does not guarantee a maximum; it must be a whole number of wavelengths. A non-zero path difference gives a minimum only if it is a half-integer number of wavelengths.✓ 1Answer A

Answer: A

Common error: Compare the path difference with the wavelength, never with a length in metres; only the ratio path difference ÷ λ decides the result.

45IBP3H-C3-20r25-04Diffraction gratings Paper 1AMedium1 mark

A diffraction grating has 300 lines per millimetre. Light of wavelength 500 nm is incident normally on the grating. What is the total number of bright maxima that can be seen, including the central maximum?

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The slit spacing is d = (1/300) mm = 3.33 μm. Since sin θ = nλ/d cannot exceed 1, n ≤ d/λ = 6.67, so the highest order is 6. Orders 1 to 6 appear on each side of the central maximum, giving 2×6 + 1 = 13 maxima. 12 leaves out the central maximum. 7 counts only one side including the central maximum. 6 counts only the orders on one side.✓ 1Answer C

Answer: C

Exam tip: Round d/λ down to a whole number to get the highest order, then count both sides and the central maximum.

46IBP3H-C3-20r25-05Double slit with diffraction envelope Paper 1AMedium1 mark

In a double-slit experiment the slit separation is doubled and the distance from the slits to the screen is halved. The wavelength is unchanged. How does the fringe spacing on the screen change?

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Fringe spacing s = λD/d. Doubling d halves s and halving D halves s again, so s becomes one quarter of the original. Half of the original accounts for only one of the two changes. Staying the same wrongly assumes the two changes cancel as they would if both were doubled. Four times inverts the dependence on both quantities.✓ 1Answer A

Answer: A

Common error: Fringe spacing is proportional to D and inversely proportional to d, so doubling one and halving the other does not cancel.

47IBP3H-C3-20r25-06Optical fibre Paper 1AMedium1 mark

The core of an optical fibre has refractive index 1.48 and the cladding has refractive index 1.46. What is the critical angle at the boundary between the core and the cladding?

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Total internal reflection at the core–cladding boundary needs sin θc = ncladding/ncore = 1.46/1.48 = 0.986, so θc = 80.6°. 9.4° is the complement of the critical angle (90° − 80.6°). 42.5° uses sin θc = 1/1.48, which is the core–air boundary. 44.6° results from using tan instead of sin.✓ 1Answer D

Answer: D

Exam tip: The critical angle depends on the ratio of the two refractive indices, with the smaller index on top of the fraction.

48C-1A-49Refraction and total internal reflection Paper 1AHard1 mark

Sound travels at 340 m s−1 in air and at 1500 m s−1 in water. Total internal reflection of sound can occur at a flat air–water surface.

Which row gives the critical angle, and the medium in which the sound must be travelling towards the surface?

Critical angleSound travelling in
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Marking pointMarkNotes
Step 1Snell's law in terms of speeds: sin θ1/sin θ2 = v1/v2.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2A critical angle exists only when the wave passes into the medium where it travels faster (θ2 = 90° needs sin θ1 = v1/v2 < 1). For sound this is from air into water.—
Step 3sin c = 340/1500 = 0.227, c = 13.1°. (Sound from air at more than 13.1° to the normal is totally reflected — one reason why voices above a lake are hard to hear underwater.)✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: sin c = vair/vwater and the sound must start in the slower medium, air.
  • BThis carries over the idea from light that TIR happens “from water to air”; for sound, water is the faster medium, so a wave leaving water bends towards the normal and can always escape.
  • C76.9° is the complement of the critical angle (cos c = 0.227), i.e. an angle measured from the surface, not from the normal.
  • DThis combines the wrong angle (measured from the surface) with the wrong starting medium.
49C-1A-50Diffraction gratings Paper 1AHard1 mark

A discharge lamp emits two monochromatic lines of wavelength 440 nm and 660 nm. The light is incident normally on a diffraction grating with 250 lines per mm.

What is the smallest non-zero angle at which a maximum of one wavelength coincides with a maximum of the other?

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Marking pointMarkNotes
Step 1d = 1/250 mm = 4.00 × 10−6 m.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Maxima coincide when n1 × 440 nm = n2 × 660 nm. The smallest common value is 1320 nm: third order of 440 nm and second order of 660 nm.—
Step 3sin θ = 1320 × 10−9/4.00 × 10−6 = 0.330, θ = 19.3°.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis is the first-order maximum of the 660 nm line alone; no 440 nm maximum is there (the 440 nm orders are at 6.3° and 12.7°).
  • BCorrect: 3 × 440 nm = 2 × 660 nm = 1320 nm, sin θ = 0.330.
  • CThis takes the third order of 660 nm (sin θ = 0.495); 1980 nm is not a whole-number multiple of 440 nm, so no 440 nm maximum is there (fourth order 26.1°, fifth order 33.4°).
  • DThis is the next coincidence (sixth order of 440 nm with fourth order of 660 nm, sin θ = 0.660); a coincidence occurs at a smaller angle.
50C-1A-71Optical fibre Paper 1AHard1 mark

A straight optical fibre has a core of refractive index 1.50 and a cladding of refractive index 1.40. A ray of light in air strikes the flat end face of the core at an angle θ to the axis of the fibre, as shown, and refracts into the core.

What is the largest value of θ for which the ray is totally internally reflected at the core–cladding boundary?

axiscladding n = 1.40cladding n = 1.40core n = 1.50θair
Longitudinal section of the fibre (not to scale). The ray strikes the flat end face of the core at angle θ to the axis.
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Marking pointMarkNotes
Step 1Critical angle at the core–cladding boundary: sin C = 1.40/1.50, so C = 69.0°.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The end face is perpendicular to the boundary, so the largest angle of refraction at the end face is 90° − 69.0° = 21.0°.—
Step 3At the end face 1.00 × sin θ = 1.50 × sin 21.0°, so θ = 32.6°.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis uses sin θ = sin 21.0°/1.50, applying Snell's law the wrong way round at the end face.
  • BThis is the angle of the ray inside the core. The ray bends towards the axis (the normal to the end face) as it enters the core, so the angle in air is larger.
  • CCorrect: sin θ = 1.50 sin(90° − 69.0°) = 0.539, so θ = 32.6°.
  • DThis is the critical angle at the core–cladding boundary, measured from the normal to that boundary, not the angle at the end face.
51IBP2H-023-01-07Diffraction gratings Paper 1AHard1 mark

Red light of wavelength 650 nm is incident normally on a diffraction grating with 600 lines per millimetre. What is the total number of principal maxima that can be observed, including the central maximum?

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Marking pointMarkNotes
d = 1/600 mm = 1.67 × 10−6 m. Since sinθ ≤ 1, the highest order is n ≤ d/λ = 1.67 × 10−6/650 × 10−9 ≈ 2.56, so nmax = 2. The maxima are n = 0, ±1 and ±2, giving 5 in total. The answer 2 gives only the highest order. The answer 3 counts only one side plus the centre. The answer 7 rounds 2.56 up to 3.✓ 1Answer C

Answer: C

Exam tip: Always round d/λ down. An order requiring sinθ > 1 cannot exist.

52IBP2H-023-01-08Optical fibre Paper 1AHard1 mark

Two identical pulses travel towards each other on a taut string. One pulse is displaced upwards and the other downwards. At one instant they overlap exactly and the string is momentarily straight. Which statement describes the energy of the pulses at that instant?

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When the string is straight there is no extra stretching, so the elastic potential energy associated with the pulses is momentarily zero. The string elements in the overlap region are still moving, however, because each pulse gives them a velocity in the same direction. The energy is therefore kinetic. The pulses later re-emerge unchanged, so no energy is dissipated. The ends play no part during the overlap.✓ 1Answer C

Answer: C

Common error: Zero displacement does not mean zero energy. Superposition of displacements says nothing about the velocities of the medium, and those velocities add here.

53IBP2S-022-00-07Diffraction gratings Paper 1AHard1 mark

A laboratory spectrometer uses a diffraction grating with 500 lines per millimetre. Monochromatic light of wavelength 600 nm is incident normally on the grating. What is the total number of bright maxima produced, including the central maximum?

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Marking pointMarkNotes
The line spacing is d = 1/500 mm = 2.0×10−6 m. Since sin θ ≤ 1, the highest order satisfies n ≤ d/λ = 2.0×10−6/6.0×10−7 = 3.3. The order must be a whole number, so n = 3 is the highest order. Orders 1, 2 and 3 appear on both sides of the central maximum, giving 2 × 3 + 1 = 7 maxima.
Option A gives only the highest order number, not the count of maxima. Option B leaves out the central (zero-order) maximum. Option C rounds 3.3 up to 4, but sin θ cannot exceed 1, so a fourth order cannot exist.
✓ 1Answer D

Answer: D

Common error: Round d/λ down to find the highest order. Then count the maxima on both sides and add the zero order.

54IBP2S-022-00-10Double slit with diffraction envelope Paper 1AHard1 mark

Two narrow parallel slits are cut in a coated sensor window. Each slit has width b, and their centres are separated by 4b. Monochromatic light passes through the slits. How many bright interference fringes appear within the central maximum of the single-slit diffraction envelope?

Show mark scheme
Marking pointMarkNotes
Interference maxima occur where d sin θ = mλ, with d = 4b. The first diffraction minimum occurs where b sin θ = λ, which is the same angle as d sin θ = 4λ, that is, m = 4. The m = ±4 fringes therefore fall at zero intensity and are missing. The visible fringes within the central envelope are m = −3 to +3, which is 7 fringes. Option A counts the missing ±4 fringes. Option B is 2 × 4, which omits the central fringe. Option D counts only one side.✓ 1Answer C

Answer: C

Common error: Fringes that coincide with a diffraction minimum are missing orders. Do not count them as bright fringes.

55IBP2S-022-01-08Diffraction gratings Paper 1AHard1 mark

An astronomical spectrometer uses a diffraction grating with 500 lines per millimetre. It is illuminated normally with light of wavelength 500 nm. How many bright maxima can be observed in total?

Show mark scheme
Marking pointMarkNotes
The line spacing is d = 1/500 mm = 2.0 × 10−6 m. The condition d sin θ = nλ gives n = d sin θ/λ ≤ d/λ = 4. The order n = 4 would need θ = 90°, so it cannot be observed. The highest observable order is therefore 3. Counting orders 0, ±1, ±2 and ±3 gives 7 maxima. Option A counts only the orders on one side. Option B is the value of d/λ. Option D includes the unobservable fourth orders.✓ 1Answer C

Answer: C

Common error: When d/λ is a whole number, that order lies at exactly 90° and cannot be seen. Remember to count maxima on both sides of the centre, plus the zero order.

56IBP2H-023-00-07Double slit with diffraction envelope Paper 1AHard1 mark

Monochromatic light passes through two slits. Each slit has width b, and the slit separation is d = 3b. How many bright interference fringes are seen within the central maximum of the single-slit diffraction envelope?

Show mark scheme
Marking pointMarkNotes
The interference maxima are at sin θ = nλ/d. The first diffraction minimum is at sin θ = λ/b = 3λ/d, which is exactly where the third-order fringe would be. The ±3 orders are therefore missing. Inside the envelope are orders 0, ±1 and ±2, giving 5 fringes. D includes the missing ±3 orders. C doubles d/b and leaves out the zero order. A simply takes d/b.✓ 1Answer B

Answer: B

Common error: A fringe that falls exactly on a diffraction minimum has zero intensity. Do not count it.

57IBP2H-023-00-08Diffraction gratings Paper 1AHard1 mark

White light containing wavelengths from 400 nm to 700 nm is incident normally on a diffraction grating. Which statement about the spectra is correct?

Show mark scheme
Marking pointMarkNotes
Two spectra overlap if the longest wavelength in the lower order is diffracted further than the shortest wavelength in the higher order. Compare the values of nλ, because sin θ = nλ/d and d is the same for every order.
For orders 1 and 2: 1 × 700 = 700 nm is less than 2 × 400 = 800 nm, so they never overlap. This rules out A and D.
For orders 2 and 3: 2 × 700 = 1400 nm is greater than 3 × 400 = 1200 nm, so they overlap whenever the third order appears. Since d cancels, changing the line spacing cannot separate them, which rules out C.
B is therefore correct. This holds even when only part of either spectrum is produced (1200 nm < d < 1400 nm): second-order wavelengths above 600 nm are still diffracted further than the third-order 400 nm line.
✓ 1Answer B

Answer: B

Exam tip: To test whether orders overlap, compare nλ values directly. The line spacing cancels, so it cannot create or remove an overlap.

58IBP2H-023-02-04Optical fibre Paper 1AHard1 mark

An optical fibre in air has a core of refractive index n1 and a cladding of lower refractive index n2. A manufacturer replaces the cladding with one whose refractive index is closer to n1, though still less than n1. What happens to the critical angle at the core–cladding boundary, and to the acceptance angle? The acceptance angle is the maximum angle of incidence at the air–core end face for which light is still guided along the fibre by total internal reflection.

Show mark scheme
Marking pointMarkNotes
sin c = n2/n1. Raising n2 towards n1 therefore increases c. Inside the core, rays must now meet the wall at a larger angle to the normal, so they must travel closer to the fibre axis. The largest allowed refraction angle at the end face, 90° − c, therefore decreases. By Snell's law, the largest allowed incidence angle in air also decreases: sin θmax = n1 cos c = √(n12 − n22). Option B is correct. Option A assumes the critical angle falls as the indices get closer. Option C assumes a larger critical angle makes total internal reflection easier. Option D inverts the sin c ratio.✓ 1Answer B

Answer: B

Exam tip: Chain the two boundaries. The critical angle at the wall fixes the largest refraction angle at the end face, and Snell's law at the end face then fixes the acceptance angle.

59IBP2H-023-02-06Double slit with diffraction envelope Paper 1AHard1 mark

A double-slit interference pattern is modulated by a single-slit diffraction envelope. The width of each slit is halved. The slit separation, the wavelength and the screen distance are unchanged. Which is correct?

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Marking pointMarkNotes
The fringe spacing s = λD/d depends only on the slit separation, so it is unchanged. The width of the central diffraction envelope is 2λD/b, so halving b doubles it. A wider envelope with the same fringe spacing contains more fringes. Option D assumes a narrower slit gives a narrower envelope. Options A and B wrongly link fringe spacing to slit width.✓ 1Answer C

Answer: C

Exam tip: Keep the two length scales separate. The separation d controls the fringe spacing and the slit width b controls the envelope.

60IBP3H-C3-20r25-07Wavefronts and rays Paper 1AHard1 mark

Plane water waves travel from deep water, where their speed is 0.32 m s−1, into shallow water, where their speed is 0.20 m s−1. In the deep water the wavefronts make an angle of 50° with the boundary. What angle do the wavefronts make with the boundary in the shallow water?

deep water, 0.32 m s⁻¹shallow water, 0.20 m s⁻¹50°?
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The angle between a wavefront and the boundary equals the angle between the ray and the normal, so sin θ2 = sin 50° × (0.20/0.32) = 0.479 and θ2 = 28.6°. 23.7° comes from using cos 50° instead of sin 50°. 31.3° scales the angle itself by the speed ratio (50° × 0.625). 50° assumes the wavefronts do not turn at the boundary.✓ 1Answer B

Answer: B

Common error: Snell's law relates the sines of the angles, so the angles themselves are not in the same ratio as the speeds.

61IBP3H-C3-20r25-08Diffraction gratings Paper 1AHard1 mark

White light containing wavelengths from 400 nm to 700 nm is incident normally on a diffraction grating. Which statement about the spectra seen on a distant screen is correct?

Show mark scheme
Marking pointMarkNotes
For a given grating sin θ is proportional to nλ. The first-order red end (700) is below the second-order violet end (2×400 = 800), so these do not overlap. The second-order red end (2×700 = 1400) is above the third-order violet end (3×400 = 1200), so the second and third orders overlap. The third-order violet end (1200) is far beyond the first-order red end (700), so first and third do not overlap. Orders at higher n overlap because the spectra widen in proportion to n.✓ 1Answer C

Answer: C

Exam tip: Compare nλ values: an overlap occurs when the long-wavelength end of one order exceeds nλ for the short-wavelength end of the next.

62C-1B-10Young's double-slit experiment Paper 1BEasy7 marks

A laser beam falls on a double slit of separation 0.400 mm and an interference pattern forms on a screen. The student moves the screen to five positions and measures the distance D from the screen to the front edge of the stand that holds the slide. For each position the student measures the distance across 10 fringe spacings with a ruler; each measurement has an uncertainty of ±0.5 mm.

The graph shows the fringe spacing s against D. The line of best fit has been extended to the D axis.

D / m0.801.201.602.002.40
10s / mm12.417.423.028.133.7
s / mm1.241.742.302.813.37
-0.40.00.40.81.21.62.02.42.8D / m0.00.51.01.52.02.53.03.54.0s / mm
Fringe spacing s against D, with the line of best fit extended (drawn to scale).
(a)
(i)

Calculate the percentage uncertainty in s for D = 0.80 m.

(1)
(b)
(i)

Determine the wavelength of the laser light.

(2)
(c)
(i)

The line meets the D axis at a negative value. Explain this and state how far the slits are from the point from which D was measured.

(2)
(ii)

State the effect of this systematic error on the value of λ in (b).

(1)
(d)
(i)

Suggest how, using the same ruler, the student could reduce the percentage uncertainty in (a) without moving the screen, and determine the new percentage uncertainty.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
0.5/12.4 × 100 = 4.0 %✓ 1Accept 4 %. The same percentage applies to s and 10s.
Part (b)(i)
Gradient = λ/d: large triangle, e.g. (0, 0.16) to (2.6, 3.62), gradient = 1.33 × 10−3✓ 1Accept 1.28–1.38 × 10−3 (mm per mm).
λ = 1.33 × 10−3 × 0.400 × 10−3 = 5.3 × 10−7 m✓ 1Accept 510–550 nm. Allow ECF.
Part (c)(i)
The slits are behind the front edge of the stand, so every D is too small by the same amount (a systematic error / zero offset)✓ 1
Intercept −0.12 m: the slits are 0.12 m (≈ 12 cm) further from the screen✓ 1Accept 0.10–0.14 m. Allow ECF from the line used in (b).
Part (c)(ii)
None: a constant offset in D changes the intercept but not the gradient, and λ was found from the gradient✓ 1Allow ECF from (b).
Part (d)(i)
Measure across more fringe spacings, e.g. 20 spacings (≈ 24.8 mm): 0.5/24.8 = 2.0 % (halved)✓ 1Method and a consistent new percentage both needed. "Use a more precise instrument": [0]. Allow ECF from (a).

Answers: (a)(i) 4.0 %  ·  (b)(i) 5.3 × 10−7 m  ·  (c)(i) 0.12 m  ·  (c)(ii) no effect  ·  (d)(i) 2.0 % (the remaining parts are explanations — see the table above)

63C-1B-16Refraction and total internal reflection Paper 1BEasy7 marks

Short pulses of light from a laser diode are sent into one end of a reel of optical fibre. A detector at the other end is connected to an oscilloscope, which displays the time t between the electrical trigger signal and the arrival of the light pulse. The measurement is repeated for reels of different length L. The time includes a fixed delay in the electronics and connecting cables. Each value of t is uncertain by ±2 ns.

The graph shows t against L with the line of best fit.

L / m20406080100
t / ns112208308408504
020406080100120L / m0100200300400500600t / ns
Time t against fibre length L with the line of best fit (drawn to scale).
(a)
(i)

Determine the gradient of the line, including its unit, and hence the speed of light in the fibre core.

(2)
(b)
(i)

Determine the refractive index of the core.

(1)
(c)
(i)

Each value of t is uncertain by ±2 ns. Using this uncertainty in the first and last data points, estimate the absolute uncertainty in the gradient and hence in the refractive index of the core.

(2)
(d)
(i)

Determine the delay in the electronics, and explain why plotting a graph gives a better value for the speed than calculating L/t for a single reel.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Gradient = (554 − 13)/110 = 4.92 ns m−1✓ 1Accept 4.8–5.0 ns m−1. Unit required.
v = 1/gradient = 2.03 × 108 m s−1✓ 1Accept 2.0–2.1 × 108 m s−1. Allow ECF.
Part (b)(i)
n = c/v = 3.00 × 108/2.03 × 108 = 1.48✓ 1Accept 1.44–1.50. Allow ECF from (a).
Part (c)(i)
Steepest and shallowest lines through the limits ±2 ns of (20 m, 112 ns) and (100 m, 504 ns): (506 − 110)/80 = 4.95 ns m−1 and (502 − 114)/80 = 4.85 ns m−1, so the gradient is uncertain by ±0.05 ns m−1✓ 1Accept ±0.04 to ±0.06 ns m−1.
n = c × gradient, so Δn = 3.00 × 108 × 0.05 × 10−9 = ±0.015 ≈ ±0.02✓ 1Accept ±0.01 to ±0.02 (1 s.f.). Allow ECF from (a) and (b).
Part (d)(i)
Delay = intercept on the t axis = 13 ns✓ 1Accept 9–16 ns.
The fixed delay adds the same time to every reading, so it affects only the intercept; the gradient (and v) is free of it, whereas L/t for one reel would include the delay and give a speed that is too small✓ 1Allow ECF from (a).

Answers: (a)(i) 4.92 ns m−1; 2.03 × 108 m s−1  ·  (b)(i) 1.48  ·  (c)(i) ±0.05 ns m−1; ±0.02  ·  (d)(i) 13 ns (the remaining parts are explanations — see the table above)

64C-1B-04Snell's law Paper 1BMedium7 marks

A ray box sends a narrow ray of light into the curved face of a semicircular glass block, along a radius, so that it reaches the centre of the flat face and leaves the block into air. The block is placed on a printed polar grid so that the angle of incidence inside the glass, θ2, can be set precisely. The angle of refraction in air, θ1, is read from the grid to ±1°.

Student X suggests that θ1/θ2 is constant. Student Y suggests that sin θ1/sin θ2 is constant.

θ2 / °131925303538
θ1 / ° (±1°)203040506070
θ₂θ₁glassair
Diagram not to scale
(a)
(i)

Deduce, using three sets of data, which suggestion is supported.

(2)
(b)
(i)

Using the data for θ2 = 38°, determine the refractive index n of the glass with its absolute uncertainty.

(2)
(c)
(i)

Determine the range of possible values of the critical angle for this glass.

(2)
(d)
(i)

A ray inside this glass meets the flat face at an angle of incidence of 41°. Discuss whether total internal reflection must occur.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Three calculations of each ratio, e.g. θ1/θ2 = 1.54, 1.67, 1.84 and sin ratios = 1.52, 1.53, 1.53 (for θ2 = 13°, 30°, 38°)✓ 1At least three rows, spread over the range, for both ratios.
θ1/θ2 rises steadily (by ≈ 20 %), while sin θ1/sin θ2 stays the same within ≈ 1 %, so Y is supported✓ 1The conclusion must be justified by the trend in their numbers. Two rows only: [1 max].
Part (b)(i)
n = sin 70°/sin 38° = 1.53✓ 1Snell's law as supported in (a); allow ECF from (a) only if a consistent method is used.
Using 69° and 71° (or extreme values): n ranges from 1.516 to 1.536, so Δn = ±0.01✓ 1Accept ±0.01 (±0.010). Absolute uncertainty to 1 s.f. Applying ±1° to θ2, which was set precisely: [0] for this mark.
Part (c)(i)
sin c = 1/n = 1/1.53, c = 40.9°✓ 1Allow ECF from (b).
Range using n ± Δn: 40.6° to 41.3°✓ 1Accept 40.4°–40.7° to 41.1°–41.4°. Allow ECF from (b).
Part (d)(i)
41° lies inside the range found in (c), so it cannot be said whether 41° exceeds the critical angle: total internal reflection is possible but not certain✓ 1Allow ECF from (c): the conclusion must follow from their range.

Answers: (a)(i) Y (sine ratio constant)  ·  (b)(i) 1.53 ± 0.01  ·  (c)(i) ≈ 40.6° to 41.3°  ·  (d)(i) not certain (the remaining parts are explanations — see the table above)

65C-1B-05Single-slit diffraction Paper 1BMedium7 marks

Light from a laser passes through an adjustable slit and forms a diffraction pattern on a screen 1.500 m away. The slit width is set with a screw and read from a scale as bs. The width w of the central maximum on the screen (between the first minima on either side) is measured with a ruler.

For small angles the slit width b is related to w by w = 2λD/b. The student plots bs against 1/w. The graph shows the data and the line of best fit.

bs / mm0.1300.1800.2300.2800.330
w / mm19.812.89.87.76.6
(1/w) / mm−10.05050.07810.10200.1299
0.000.030.060.090.120.150.18(1/w) / mm⁻¹0.000.050.100.150.200.250.300.350.40bₛ / mm
Slit-scale reading bs against 1/w with the line of best fit (drawn to scale).
(a)
(i)

Complete the table.

(1)
(b)
(i)

Determine the gradient of the line, including its unit.

(2)
(c)
(i)

Determine the wavelength of the laser light.

(2)
(d)
(i)

The model predicts that the line passes through the origin. Use the graph to determine the true slit width when the scale reads 0.180 mm, and state the effect, if any, of the intercept on your answer to (c).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
1/6.6 = 0.1515 mm−1✓ 1Accept 0.152 or 0.15.
Part (b)(i)
Large triangle, e.g. (0, 0.029) to (0.17, 0.363); the point from (a) lies on the line✓ 1Allow ECF from (a).
Gradient = 1.97 mm²✓ 1Accept 1.90–2.05 mm². Unit mm² (or m²) required.
Part (c)(i)
Gradient = 2λD, so λ = 1.97 × 10−6/(2 × 1.500)✓ 1Conversion of mm² to m² must be seen or implied.
λ = 6.6 × 10−7 m (656 nm)✓ 1Accept 630–685 nm. Allow ECF from (b).
Part (d)(i)
Intercept = 0.029 mm: the scale has a zero error, reading ≈ +0.03 mm when the true width is zero; true width = 0.180 − 0.03 = 0.15 mm✓ 1Accept intercept 0.02–0.04 mm and width 0.14–0.16 mm.
A zero error shifts every bs equally, so it changes only the intercept; the gradient and λ are unaffected✓ 1Allow ECF from (c).

Answers: (a)(i) 0.1515 mm−1  ·  (b)(i) 1.97 mm²  ·  (c)(i) 6.6 × 10−7 m  ·  (d)(i) 0.15 mm; no effect (the remaining parts are explanations — see the table above)

66C-1B-18Refraction and total internal reflection Paper 1BMedium7 marks

A food scientist builds a refractometer to find the sugar concentration of fruit juices. A laser beam passes along a radius of a semicircular dish holding the solution and meets the flat face at its centre. The angle of incidence is increased until the refracted ray just skims along the flat face; this is the critical angle c, measured to ±0.2°. The refractive index is then n = 1/sin c.

Solutions of known sugar concentration C (percentage by mass) are used to plot the calibration graph of n against C.

C / %01020304050
c / °48.647.847.146.245.544.8
n1.3331.3501.3651.4021.419
0102030405060C / %1.321.341.361.381.401.421.44n
Calibration graph: refractive index n against concentration C (drawn to scale).
(a)
(i)

Complete the table and determine the absolute uncertainty in n for the solution with C = 30 %.

(2)
(b)
(i)

Determine the gradient of the calibration line, including its unit.

(2)
(c)
(i)

For a sample of orange juice the critical angle is 46.4°. Determine the sugar concentration of the juice, with its absolute uncertainty.

(2)
(d)
(i)

The juice is measured at a higher temperature than the calibration solutions. Suggest how this could affect your answer to (c).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
1/sin 46.2° = 1.386✓ 1Accept 1.385–1.386.
1/sin 46.0° = 1.390 and 1/sin 46.4° = 1.381, so Δn = ±0.005✓ 1Accept ±0.004 to ±0.005 (1 s.f.). ALT: Δn = cos c Δc/sin² c with Δc in radians.
Part (b)(i)
Large triangle, e.g. (0, 1.332) to (55, 1.428)✓ 1The point from (a) lies on the line; allow ECF from (a).
Gradient = 1.73 × 10−3 %−1✓ 1Accept 1.65–1.80 × 10−3. Unit (%−1, per percentage point) required.
Part (c)(i)
n = 1/sin 46.4° = 1.381; C = (1.381 − 1.332)/1.73 × 10−3 = 28 %✓ 1Accept 26–30 %. Allow ECF from (b).
Δn ≈ ±0.005 (as in (a)), so ΔC = 0.005/1.73 × 10−3 ≈ ±3 %✓ 1Accept ±2 to ±3 percentage points, to 1 s.f. Allow ECF from (a) and (b).
Part (d)(i)
The refractive index (speed of light) of the liquid depends on its temperature/density; a warmer, less dense juice has a smaller n, so its concentration would be underestimated (a systematic error)✓ 1A direction must be given with a reason. Accept the opposite direction only if justified by a stated change in n.

Answers: (a)(i) 1.386 ± 0.005  ·  (b)(i) 1.73 × 10−3 %−1  ·  (c)(i) 28 ± 3 % (the remaining parts are explanations — see the table above)

67C-1B-02Diffraction gratings Paper 1BHard7 marks

A student uses a diffraction grating on a spectrometer to study light from a hydrogen discharge tube. The light is made parallel and falls normally on the grating. A telescope is rotated to each maximum and its angular position is read from a circular scale. Positions to the right of the scale zero are positive and to the left negative.

For the red hydrogen line of wavelength 656.3 nm, the first three orders are found on both sides. The telescope reading for the straight-through beam was not recorded. A violet line is seen in first order only on this table.

OrderReading on left / °Reading on right / °
1−9.013.8
2−20.825.6
3−33.838.6
1 (violet line)−5.19.9
(a)
(i)

Determine the reading of the straight-through position and the angle θ of the third-order red maximum.

(2)
(b)
(i)

Determine the slit separation d of the grating.

(1)
(c)
(i)

Determine the wavelength of the violet line.

(2)
(d)
(i)

Some energy levels of the hydrogen atom are −3.40 eV, −1.51 eV, −0.850 eV, −0.544 eV and −0.378 eV. Deduce the transition that produces the violet line.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Straight-through reading = mean of a left/right pair = (38.6 + (−33.8))/2 = +2.4°✓ 1Any order gives +2.4°. Accept 2.4° to the right (the scale has a zero offset).
θ3 = (38.6 − (−33.8))/2 = 36.2°✓ 1Or 38.6 − 2.4. Using 38.6° directly (ignoring the offset) is [0] for this mark.
Part (b)(i)
d = 3 × 656.3 × 10−9/sin 36.2° = 3.33 × 10−6 m✓ 1Accept 3.32–3.35 × 10−6 m (≈ 300 lines per mm). Allow ECF from (a). Use of a lower order with its corrected angle gives the same value and is accepted.
Part (c)(i)
θ = (9.9 − (−5.1))/2 = 7.5°✓ 1Or 9.9 − 2.4, using the offset from (a). Allow ECF from (a).
λ = d sin θ = 3.33 × 10−6 × sin 7.5° = 4.35 × 10−7 m✓ 1Accept 432–438 nm. Allow ECF from (b).
Part (d)(i)
Photon energy = hc/λ = 6.63 × 10−34 × 3.00 × 108/(4.35 × 10−7 × 1.60 × 10−19) = 2.86 eV✓ 1Accept 2.84–2.88 eV. Allow ECF from (c).
−0.544 − (−3.40) = 2.86 eV: from the −0.544 eV level to the −3.40 eV level✓ 1Allow ECF: the transition whose energy difference is closest to their photon energy.

Answers: (a)(i) +2.4°; 36.2°  ·  (b)(i) 3.33 × 10−6 m  ·  (c)(i) 4.35 × 10−7 m  ·  (d)(i) 2.86 eV; −0.544 eV → −3.40 eV (the remaining parts are explanations — see the table above)

68C-1B-17Young's double-slit experiment Paper 1BHard7 marks

A student estimates the Planck constant using five light-emitting diodes (LEDs). Each LED in turn illuminates a double slit of separation 0.250 mm, and the distance across 10 fringe spacings is measured on a screen 1.800 m away. The wavelength λ is calculated from the fringe spacing. The student then finds the threshold voltage Vth, the smallest potential difference at which the LED just begins to emit light.

The model is eVth = hc/λ, so a graph of Vth against 1/λ should be a straight line through the origin. The graph shows the data (note the false origins).

LEDbluecyanamberorangered
10s / mm34.037.642.644.847.7
(1/λ) / 106 m−12.121.911.691.61
Vth / V2.512.211.991.891.75
1.41.51.61.71.81.92.02.12.2(1/λ) / 10⁶ m⁻¹1.61.82.02.22.42.6Vₜₕ / V
Threshold voltage Vth against 1/λ with the line of best fit (drawn to scale; false origins).
(a)
(i)

Show that the wavelength of the red LED is about 660 nm.

(2)
(b)
(i)

Complete the table.

(1)
(c)
(i)

Determine the gradient of the line, including its unit.

(2)
(d)
(i)

Determine a value for the Planck constant and compare it with the accepted value. Suggest one reason for the difference.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
s = 47.7/10 = 4.77 mm; λ = sd/D = 4.77 × 10−3 × 0.250 × 10−3/1.800✓ 1
λ = 6.63 × 10−7 m (663 nm)✓ 1Unrounded value or full substitution required.
Part (b)(i)
1/6.63 × 10−7 = 1.51 × 106 m−1✓ 1Accept 1.51 or 1.52 (from 660 nm). Allow ECF from (a).
Part (c)(i)
Large triangle, e.g. (1.45, 1.69) to (2.15, 2.53)✓ 1The point from (b) lies on the line; allow ECF from (b).
Gradient = 1.20 × 10−6 V m✓ 1Accept 1.15–1.26 × 10−6 V m. Unit required.
Part (d)(i)
h = e × gradient/c = 1.60 × 10−19 × 1.20 × 10−6/3.00 × 108 = 6.4 × 10−34 J s✓ 1Accept 6.1–6.7 × 10−34 J s. Allow ECF from (c).
About 3 % below 6.63 × 10−34 J s; each LED emits a band of wavelengths and starts to glow when the lowest-energy (longest-wavelength) photons are produced, so Vth is systematically below hc/eλ for the measured (peak) wavelength✓ 1Allow ECF: comparison must use their value. Accept the difficulty of judging the onset of emission by eye (too late or too early) only with a direction consistent with their result.

Answers: (b)(i) 1.51 × 106 m−1  ·  (c)(i) 1.20 × 10−6 V m  ·  (d)(i) 6.4 × 10−34 J s (the remaining parts are explanations — see the table above)

69C-1B-19Single-slit diffraction Paper 1BHard7 marks

In an electron-diffraction tube, electrons accelerated from rest through a potential difference V pass through a thin layer of graphite and form bright rings on a fluorescent screen. The radius r of the inner ring is measured with vernier calipers. For this ring the geometry of the tube gives r/L = λ/d, where L = 0.130 m and d is a spacing between planes of carbon atoms.

The graph shows r against V−1/2 with the line of best fit.

V / kV2.02.53.03.54.05.0
r / mm17.014.813.812.611.910.5
V−1/2 / 10−2 V−1/22.2362.0001.8261.6901.414
1.21.41.61.82.02.22.4V−1/2 / 10−2 V−1/281012141618r / mm
Ring radius r against V−1/2 with the line of best fit (drawn to scale; false origins).
(a)
(i)

Show that the de Broglie wavelength of the electrons accelerated through 4.0 kV is about 1.9 × 10−11 m.

(2)
(b)
(i)

Complete the table.

(1)
(c)
(i)

Determine the gradient of the line, including its unit.

(2)
(d)
(i)

Hence determine d.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Ek = eV, so p = √(2meV) = √(2 × 9.11 × 10−31 × 1.60 × 10−19 × 4.0 × 103) = 3.4 × 10−23 kg m s−1✓ 1
λ = h/p = 1.94 × 10−11 m✓ 1A value to at least 3 s.f. or the full substitution is required.
Part (b)(i)
1/√4000 = 1.581 × 10−2 V−1/2✓ 1Accept 1.58.
Part (c)(i)
Large triangle, e.g. (1.30, 9.6) to (2.35, 17.8)✓ 1The point from (b) lies on the line; allow ECF from (b).
Gradient = 7.73 mm per 10−2 V−1/2 = 0.773 m V1/2✓ 1Accept 0.73–0.81 m V1/2. Unit required.
Part (d)(i)
r = Lλ/d = Lh/(d√(2meV)), so the gradient = Lh/(d√(2me))✓ 1Allow ECF from (a) and (c).
d = 0.130 × 6.63 × 10−34/(0.773 × √(2 × 9.11 × 10−31 × 1.60 × 10−19)) = 2.1 × 10−10 m✓ 1Accept 1.9–2.2 × 10−10 m. Allow ECF from (c).

Answers: (b)(i) 1.581 × 10−2 V−1/2  ·  (c)(i) 0.773 m V1/2  ·  (d)(i) 2.1 × 10−10 m (the remaining parts are explanations — see the table above)

70C-2-24Wavefronts and rays Paper 2Easy10 marks

The graph shows how the speed of long water waves depends on the depth of the water. Ocean swell of period 8.0 s travels over a sea bed where the water depth changes suddenly from 4.0 m to 1.0 m along a straight underwater ledge.

0.00.51.01.52.02.53.03.54.04.55.0water depth / m012345678wave speed / m s⁻¹
Speed of water waves against depth (drawn to scale).
(a)
(i)

Use the graph to state the wave speed in water 4.0 m deep and in water 1.0 m deep.

(1)
(ii)

Determine the wavelength of the swell in each depth of water.

(2)
(iii)

The speed of these waves is v = √(gd), where d is the depth. Show that, for swell of a fixed period, the wavelength is proportional to √d.

(1)
(b)
(i)

The rays of the swell meet the ledge at an angle of incidence of 40°. Determine the angle of refraction.

(2)
(ii)

Swell approaching a beach at an angle reaches the shore with its wavefronts almost parallel to the beach. Explain why.

(2)
(c)
(i)

A tsunami is a long water wave. Estimate the time for a tsunami to cross 3000 km of ocean 4.0 km deep, and state one assumption you make.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
6.3 m s−1 and 3.1 m s−1✓ 1Accept 6.1–6.4 and 3.0–3.3 m s−1. Both needed.
Part (a)(ii)
λ = vT = 6.26 × 8.0 = 50 m in the deep water✓ 1Allow ECF from (a)(i).
λ = 3.13 × 8.0 = 25 m in the shallow water; the period does not change at the ledge✓ 1
Part (a)(iii)
λ = vT = T√(gd), and T and g are constant, so λ ∝ √d✓ 1
Part (b)(i)
sin θ2 = sin 40° × 3.13/6.26✓ 1Allow ECF from (a)(i).
θ2 = 19°✓ 1
Part (b)(ii)
As the water becomes shallower the waves slow down, so the part of each wavefront in shallower water travels more slowly✓ 1
The wavefronts turn so that the rays bend towards the normal (towards the direction perpendicular to the shore), until the wavefronts are almost parallel to the beach✓ 1
Part (c)(i)
v = √(9.81 × 4000) = 198 m s−1, so t = 3.0 × 106/198 = 1.5 × 104 s (about 4.2 h)✓ 1
Assumption: the depth is the same everywhere (or: v = √(gd) still applies at this depth)✓ 1

Answers: (a)(i) 6.3 m s−1; 3.1 m s−1  ·  (a)(ii) 50 m; 25 m  ·  (b)(i) 19°  ·  (c)(i) about 4.2 h (the remaining parts are explanations — see the table above)

71IBP2H-023-01-09Superposition of pulses Paper 2Easy5 marks

Two athletes training with a heavy battle rope each flick one end, sending a single pulse towards the other. The centres of the two pulses are initially 6.0 m apart. Each pulse travels at 4.0 m s−1. Pulse A has a maximum displacement of +12 cm and pulse B has a maximum displacement of −7 cm. The pulses have the same width and shape.

−712Opositiondisplacement / cm4.0 m s⁻¹4.0 m s⁻¹AB6.0 m
Pulses shown at the start; centres 6.0 m apart
(a)

State the principle of superposition.

(1)
(b)

Calculate the time taken for the centres of the pulses to meet.

(2)
(c)

Determine the displacement of the rope at the point where the centres coincide.

(1)
(d)

State the shape and speed of pulse A after the pulses have passed through each other.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
When two or more waves meet, the resultant displacement at any point is the (vector) sum of the individual displacements at that point✓ 1
Part (b)
Closing speed = 4.0 + 4.0 = 8.0 m s−1, or each pulse travels 3.0 m✓ 1
t = 6.0/8.0 = 0.75 s✓ 1
Part (c)
(+12) + (−7) = +5 cm✓ 1
Part (d)
Shape unchanged (same as before meeting) and speed still 4.0 m s−1; both needed for the mark✓ 1

Exam tip: When two pulses approach each other, use the closing speed (the sum of the two speeds) to find when they meet.

72IBP2H-023-00-09Snell's law Paper 2Easy5 marks

Green light of wavelength 520 nm in air comes from a phone's display. It travels through air and strikes the glass cover of a smartphone at an angle of incidence of 35°. The refractive index of the glass is 1.52.

(a)

Calculate the angle of refraction in the glass.

(2)
(b)

Calculate the speed of light in the glass.

(1)
(c)

Determine the wavelength of the light in the glass.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
sin r = sin 35°/1.52 = 0.377✓ 1
r = 22° (22.2°)✓ 1
Part (b)
v = 3.00×108/1.52 = 1.97×108 m s−1✓ 1
Part (c)
The frequency is unchanged on refraction, so λ is proportional to v, or λglass = λair/n✓ 1
λglass = 520/1.52 = 342 nm (ECF from (b) if v/f is used)✓ 1

Common error: The frequency does not change when light enters a new medium. Only the speed and wavelength change.

73IBP2H-023-02-11Optical fibre Paper 2Easy6 marks

A long-distance telecommunications link uses infrared light of vacuum wavelength 1550 nm in a glass fibre. The core has refractive index 1.47. Take c = 3.00×108 m s−1.

(a)

Calculate the frequency of the infrared light.

(2)
(b)

Calculate the wavelength of the light in the fibre core.

(1)
(c)

State the frequency of the light in the core, giving a reason.

(1)
(d)

Calculate the minimum time taken for a signal to travel along 80 km of this fibre.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
f = c/λ = 3.00×108/1550×10−9 . f = 1.94×1014 Hz .✓ 2
Part (b)
λcore = 1550/1.47 = 1.05×10−6 m (1054 nm) .✓ 1
Part (c)
1.94×1014 Hz, unchanged, because the frequency is determined by the source and does not change at a boundary .✓ 1
Part (d)
v = c/n = 3.00×108/1.47 = 2.04×108 m s−1 . t = 80×103/2.04×108 = 3.9×10−4 s .✓ 2

Common error: Using c instead of c/n for the signal speed gives a time that is too short by a factor of 1.47.

74IBP3H-C3-20r25-10Double slit with diffraction envelope Paper 2Easy7 marks

A student shines a green laser through a double slit with slit separation 0.20 mm. The interference pattern is formed on a screen 2.5 m from the slits. The distance between the centre of the 1st bright fringe and the centre of the 9th bright fringe is 5.32 cm.

(a)

Outline the condition for a bright fringe to form at a point on the screen.

(2)
(b)

Calculate the distance between adjacent bright fringes.

(1)
(c)

Determine the wavelength of the laser light, in nm.

(2)
(d)

The green laser is replaced by a red laser of wavelength 650 nm. State and explain the change in the fringe spacing.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The waves from the two slits arrive in phase / with a phase difference of zero or a multiple of 2π✓ 1
The path difference from the two slits to that point is a whole number of wavelengths, nλ✓ 1
Part (b)
5.32 cm / 8 = 6.65 mm✓ 1
Part (c)
Use of λ = sd/D✓ 1
λ = 6.65×10−3 × 0.20×10−3 / 2.5 = 5.32×10−7 m = 532 nm✓ 1
Part (d)
The fringe spacing increases✓ 1
Because spacing s = λD/d is proportional to wavelength, and the wavelength is larger✓ 1

Common error: Eight spacings, not nine, lie between the 1st and the 9th bright fringe; dividing by 9 gives the wrong spacing.

75IBP3H-C3-20r25-11Two-source interference Paper 2Easy7 marks

At a mountain base camp, two loudspeakers A and B are mounted on posts and emit the same steady tone of frequency 680 Hz, in phase. The speed of sound in the air is 340 m s−1. A hiker stands at a point P that is 8.30 m from A and 6.55 m from B.

8.30 m6.55 mABP
Diagram not to scale
(a)

Calculate the wavelength of the sound.

(2)
(b)

Determine whether the hiker hears a loud sound or a quiet sound at P, by calculating the path difference.

(3)
(c)

State two conditions that the two sources must satisfy for a stable interference pattern to be observed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Use of λ = v/f✓ 1
λ = 340/680 = 0.500 m✓ 1
Part (b)
Path difference = 8.30 − 6.55 = 1.75 m✓ 1
= 3.5 wavelengths (ECF from (a))✓ 1
A half-integer number of wavelengths means destructive interference, so the sound at P is quiet✓ 1
Part (c)
The sources must be coherent: constant phase difference✓ 1
The same frequency (and hence the same wavelength)✓ 1

Exam tip: State the path difference as a multiple of the wavelength before commenting on the type of interference.

76C-2-04Single-slit diffraction Paper 2Medium10 marks

A geostationary communications satellite transmits to the Earth at a frequency of 12.0 GHz from a circular dish of diameter 2.40 m. The angle between the centre of the beam and its first minimum can be estimated from the single-slit result θ = λ/b, where b is the diameter of the dish.

(a)
(i)

Show that the radius of the orbit of a geostationary satellite is about 4.2 × 107 m.

(2)
(ii)

Determine the diameter of the region of the Earth's surface, directly below the satellite, that lies between the first minima of the beam.

(3)
(b)
(i)

The operator wants this region to have half the diameter. Deduce, without numerical substitution, two separate changes to the transmitter that would each achieve this.

(2)
(ii)

Explain why a receiving dish on the ground can be fixed in one position.

(1)
(c)
(i)

Estimate the time between a signal leaving a ground station and arriving at a receiver after relay by the satellite. State one assumption you make.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The gravitational force provides the centripetal force: GMm/r² = 4π²mr/T², with T = 24 h = 8.64 × 104 s✓ 1
r = (GMT²/4π²)1/3 = 4.22 × 107 m✓ 1Answer to at least 3 s.f. or full substitution required.
Part (a)(ii)
λ = c/f = 0.0250 m, so θ = 0.0250/2.40 = 1.04 × 10−2 rad✓ 1
The distance from the satellite to the Earth's surface is r − RE = 3.59 × 107 m✓ 1Using r itself gives 8.8 × 105 m: [2 max].
Diameter = 2θ(r − RE) = 7.47 × 105 m (about 750 km)✓ 1Accept 7.4–7.5 × 105 m. Allow ECF from (a)(i).
Part (b)(i)
The diameter is proportional to λ/b = c/(fb)✓ 1
Either double the diameter of the dish or double the frequency (halve the wavelength)✓ 1Both changes needed.
Part (b)(ii)
The satellite orbits above the equator in the same direction as the Earth rotates, with a period equal to the Earth's rotation period, so it stays above the same point of the surface✓ 1Period and direction (or equatorial orbit) both needed.
Part (c)(i)
t ≈ 2(r − RE)/c = 2 × 3.59 × 107/3.00 × 108 = 0.24 s✓ 1Accept 0.24–0.28 s. Allow ECF from (a)(i).
Assumption: both ground stations are close to the point directly below the satellite (or: the satellite relays the signal without delay)✓ 1

Answers: (a)(i) 4.22 × 107 m  ·  (a)(ii) 7.47 × 105 m  ·  (c)(i) 0.24 s (the remaining parts are explanations — see the table above)

77C-2-08Double-slit interference Paper 2Medium10 marks

Light from a laser diode passes through a double slit and falls on a screen 1.60 m away. A light sensor is moved across the screen, and the graph shows the variation of the intensity I it records with position y on the screen. The centres of the slits are 0.180 mm apart.

-30-20-100102030y / mm0.00.20.40.60.81.0I / arbitrary units
Intensity against position on the screen (drawn to scale).
(a)
(i)

Use the graph to determine the separation s of adjacent bright fringes.

(2)
(ii)

Determine the wavelength of the light.

(2)
(iii)

Calculate the energy of a photon of this light. Give your answer in eV.

(2)
(b)
(i)

The bright fringes are not equally bright. Use the graph to determine the width of each slit, explaining your method.

(3)
(ii)

The laser diode emits light of power 5.0 mW. Calculate the number of photons it emits per second.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Measurement across several fringes, e.g. the third-order maxima at ±17.3 mm are 6 separations apart: 34.7 mm/6✓ 1Use of a single fringe spacing: [1 max].
s = 5.8 mm✓ 1Accept 5.6–6.0 mm.
Part (a)(ii)
λ = sd/D = 5.78 × 10−3 × 0.180 × 10−3/1.60✓ 1
λ = 6.5 × 10−7 m✓ 1Accept 630–675 nm. Allow ECF from (a)(i).
Part (a)(iii)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/6.5 × 10−7 = 3.06 × 10−19 J✓ 1
E = 1.91 eV✓ 1Unit eV required for this mark. Allow ECF from (a)(ii).
Part (b)(i)
The double-slit fringes are modulated by the single-slit diffraction pattern of each slit, whose first minimum is where the fringes vanish, at y ≈ 26 mm✓ 1Accept 25–27 mm.
θ = y/D = 0.026/1.60 = 1.62 × 10−2 rad✓ 1
b = λ/θ = 4.0 × 10−5 m✓ 1Accept 3.8–4.2 × 10−5 m. Allow ECF from (a)(ii).
Part (b)(ii)
N = P/E = 5.0 × 10−3/3.06 × 10−19 = 1.6 × 1016 s−1✓ 1Allow ECF from (a)(iii).

Answers: (a)(i) 5.8 mm  ·  (a)(ii) 6.5 × 10−7 m  ·  (a)(iii) 1.91 eV  ·  (b)(i) 4.0 × 10−5 m  ·  (b)(ii) 1.6 × 1016 s−1 (the remaining parts are explanations — see the table above)

78C-2-09Refraction and total internal reflection Paper 2Medium10 marks

A diver at depth h below the calm surface of a swimming pool looks upwards. She sees the whole sky above the pool compressed into a bright circle (a "window") on the surface directly above her. Outside this circle the surface looks like a mirror. The refractive index of the water is 1.33.

diverhrbright circle (window) of radius rwaterair
Diagram not to scale
(a)
(i)

Calculate the critical angle for light in the water at the water–air boundary.

(1)
(ii)

Explain why the light from the whole sky reaches the diver from within a cone of half-angle equal to the critical angle.

(2)
(b)
(i)

Show that the radius r of the window is h/√(n² − 1).

(2)
(ii)

Calculate the radius of the window when the diver is 3.0 m below the surface, and state its radius when she descends to 6.0 m.

(2)
(iii)

Explain why, outside the window, the surface looks like a mirror to the diver.

(1)
(c)
(i)

Light of wavelength 589 nm in air enters the water. Determine its wavelength in the water and the energy of one of its photons in the water, in eV.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
sin c = 1/1.33, so c = 48.8°✓ 1
Part (a)(ii)
Light from the sky meets the surface at angles of incidence between 0° and 90°; light arriving almost along the surface (90°) is refracted at the critical angle✓ 1
All other sky light is refracted at smaller angles, so all of it travels to the diver inside a cone of half-angle c✓ 1Reference to the reversibility of the path at the critical angle also accepted.
Part (b)(i)
r = h tan c, with sin c = 1/n✓ 1
cos c = √(1 − 1/n²), so tan c = (1/n)/√(1 − 1/n²) = 1/√(n² − 1), giving r = h/√(n² − 1)✓ 1A right-angled triangle with sides 1, √(n² − 1) and n is an acceptable route.
Part (b)(ii)
r = 3.0/√(1.33² − 1) = 3.42 m✓ 1Or 3.0 × tan 48.8°. Allow ECF from (a)(i) or (b)(i).
r ∝ h, so at 6.0 m r = 6.8 m✓ 1Allow ECF from the first value.
Part (b)(iii)
Light from the pool travelling up to the surface at angles of incidence greater than the critical angle is totally internally reflected, so the diver sees the reflected pool floor✓ 1
Part (c)(i)
λwater = 589/1.33 = 443 nm✓ 1
The frequency is unchanged at the boundary, so the photon energy is unchanged: E = hc/λair = 2.11 eV✓ 1Using 443 nm (giving 2.81 eV): [0] for this mark. Unit eV required.

Answers: (a)(i) 48.8°  ·  (b)(ii) 3.42 m; 6.8 m  ·  (c)(i) 443 nm; 2.11 eV (the remaining parts are explanations — see the table above)

79C-2-12Diffraction gratings Paper 2Medium10 marks

The data on a DVD are stored along a spiral track. Adjacent turns of the track are a distance d apart, so the disc acts as a reflection grating. A laser beam of wavelength 650 nm is directed normally onto the disc, and the first-order reflected beams make an angle of 61.4° with the incident beam. The grating equation nλ = d sin θ applies.

(a)
(i)

Calculate the track separation d.

(1)
(ii)

Deduce the total number of reflected beams.

(1)
(iii)

The DVD is replaced by a CD, for which d = 1.60 μm. Determine the angles of all the reflected beams other than the zero order.

(2)
(b)

The DVD track is read at a constant linear speed of 3.49 m s−1, starting at radius 24 mm and ending at radius 58 mm.

(i)

Determine the angular velocity of the disc at the start and at the end of reading. Give an appropriate unit for your answers.

(2)
(ii)

Estimate the maximum playing time of the disc. State one assumption you make.

(3)
(iii)

Determine the average angular acceleration of the disc while it plays.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
d = λ/sin θ = 650 × 10−9/sin 61.4° = 7.40 × 10−7 m✓ 1
Part (a)(ii)
d/λ = 1.14, so only n = 0 and n = ±1 exist: 3 beams✓ 1Allow ECF from (a)(i). "2" (zero order forgotten): [0].
Part (a)(iii)
d/λ = 2.46, so orders 1 and 2 only✓ 1
θ1 = 24.0° and θ2 = 54.3° (on each side of the incident beam)✓ 1Both angles needed.
Part (b)(i)
ω = v/r = 3.49/0.024 = 145 and 3.49/0.058 = 60.2✓ 1
Unit: rad s−1✓ 1Unit mark is independent; accept s−1.
Part (b)(ii)
Track length ≈ area/d = π(0.058² − 0.024²)/7.40 × 10−7 = 1.18 × 104 m✓ 1Allow ECF from (a)(i).
Time = length/speed = 1.18 × 104/3.49 = 3.4 × 103 s ≈ 56 min✓ 1Accept 50–60 min.
Assumption: the spiral can be treated as concentric circles a distance d apart filling the whole area (or: no gaps in the track)✓ 1
Part (b)(iii)
α = (60.2 − 145)/3.4 × 103 = −2.5 × 10−2 rad s−2 (a deceleration)✓ 1Allow ECF from (b)(i) and (b)(ii).

Answers: (a)(i) 7.40 × 10−7 m  ·  (a)(ii) 3  ·  (a)(iii) 24.0°, 54.3°  ·  (b)(i) 145 rad s−1; 60.2 rad s−1  ·  (b)(ii) about 56 min  ·  (b)(iii) −2.5 × 10−2 rad s−2 (the remaining parts are explanations — see the table above)

80C-2-15Optical fibre Paper 2Medium9 marks

A step-index optical fibre of length 2.00 km has a core of refractive index 1.480 and a cladding of refractive index 1.460. Pulses of infrared radiation of wavelength 1310 nm are sent along it. Some of the radiation in a pulse travels along the axis of the core, and some zig-zags along the core, meeting the core–cladding boundary at the critical angle.

core, n₁ = 1.480cladding, n₂ = 1.460cladding2.00 km
Diagram not to scale; ray angles exaggerated
(a)
(i)

Calculate the critical angle at the core–cladding boundary.

(1)
(ii)

Show that the difference in the times taken to travel the fibre length L by the zig-zag path and by the axial path is (n1L/c)(n1/n2 − 1), where n1 and n2 are the refractive indices of the core and the cladding.

(2)
(iii)

Calculate this time difference for the 2.00 km fibre.

(1)
(b)
(i)

Each pulse spreads out by this time difference. Estimate the maximum number of pulses per second that can be sent so that successive pulses remain separate, and suggest one change to the fibre that would increase this number.

(2)
(ii)

The average power entering the fibre is 1.0 μW when pulses are sent at the rate found in (b)(i). Determine the number of photons in each pulse.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
sin c = 1.460/1.480, so c = 80.6°✓ 1
Part (a)(ii)
The zig-zag path length is L/sin c = Ln1/n2, using sin c = n2/n1✓ 1
Both rays travel at c/n1, so Δt = (n1/c)(Ln1/n2 − L) = (n1L/c)(n1/n2 − 1)✓ 1
Part (a)(iii)
Δt = (1.480 × 2000/3.00 × 108)(1.480/1.460 − 1) = 1.35 × 10−7 s✓ 1Allow ECF from (a)(ii).
Part (b)(i)
Pulses must be at least Δt apart, so the rate ≈ 1/Δt = 7.4 × 106 s−1✓ 1Allow ECF from (a)(iii).
Make n2 closer to n1 (or use a shorter fibre)✓ 1
Part (b)(ii)
Energy per pulse = 1.0 × 10−6/7.4 × 106 = 1.35 × 10−13 J✓ 1Allow ECF from (b)(i).
Photon energy = hc/λ = 6.63 × 10−34 × 3.00 × 108/1310 × 10−9 = 1.52 × 10−19 J✓ 1
Number = 1.35 × 10−13/1.52 × 10−19 = 8.9 × 105✓ 1Accept 8.8–9.0 × 105. Allow ECF from (b)(i).

Answers: (a)(i) 80.6°  ·  (a)(iii) 1.35 × 10−7 s  ·  (b)(i) 7.4 × 106 s−1  ·  (b)(ii) 8.9 × 105 (the remaining parts are explanations — see the table above)

81C-2-25Double-slit interference Paper 2Medium11 marks

A microwave transmitter T emits waves of wavelength 3.0 cm towards a metal sheet containing two narrow slits S1 and S2 whose centres are 15.0 cm apart. A receiver is moved along a line parallel to the sheet and 1.20 m from it, as shown. O is on the central axis.

TS₁S₂15.0 cmOP1.20 m0.480 m
Plan view (not to scale). The receiver moves along the dashed line.
(a)
(i)

Calculate the frequency of the microwaves.

(1)
(ii)

Explain why S1 and S2 behave as coherent sources.

(2)
(b)
(i)

Use s = λD/d to predict the distance from O of the second maximum.

(1)
(ii)

Determine the path difference S2P − S1P in terms of the wavelength.

(3)
(iii)

Explain what your answer to (b)(ii) shows about the use of s = λD/d in this experiment.

(2)
(c)
(i)

At the positions of minimum signal the receiver still detects a weak signal. Suggest two reasons for this.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
f = c/λ = 3.00 × 108/0.030 = 1.0 × 1010 Hz✓ 1
Part (a)(ii)
Both slits are illuminated by the same wavefronts from the single transmitter✓ 1
so the waves leaving them have the same frequency and a constant phase difference (zero)✓ 1
Part (b)(i)
s = 0.030 × 1.20/0.150 = 0.24 m, so the second maximum is predicted 2s = 0.48 m from O (at P)✓ 1
Part (b)(ii)
P is 0.405 m and 0.555 m from the perpendiculars through S1 and S2✓ 1Allow ECF from (b)(i).
S1P = √(1.20² + 0.405²) = 1.267 m and S2P = √(1.20² + 0.555²) = 1.322 m✓ 1
Path difference = 0.0556 m = 1.85λ✓ 1Accept 1.8λ–1.9λ.
Part (b)(iii)
The path difference at P is not 2λ, so P is not a maximum: the prediction is wrong✓ 1Allow ECF from (b)(ii).
s = λD/d assumes small angles (λ ≪ d and positions close to O); here λ/d = 0.2 and the angle is about 22°, so the true maximum is further out (≈ 0.52 m)✓ 1The value 0.52 m is not required.
Part (c)(i)
The waves from the two slits reach the receiver with different amplitudes (different distances), so they cannot cancel completely✓ 1[2 max]. Also accept: the receiver has a finite width and averages over a region around the minimum.
Microwaves reflected from walls, the bench or people reach the receiver by other paths✓ 1

Answers: (a)(i) 1.0 × 1010 Hz  ·  (b)(i) 0.48 m  ·  (b)(ii) 1.85λ (the remaining parts are explanations — see the table above)

82C-2-26Superposition of pulses Paper 2Medium12 marks

Noise-cancelling headphones contain a microphone that samples the noise reaching the ear and a small loudspeaker that emits “anti-noise”, designed to superpose with the noise at the eardrum. The speed of sound is 340 m s−1.

The graph shows the variation with time t of the pressure change p at the eardrum caused by a short noise pulse (solid line) and by the anti-noise pulse (dashed line), each on its own.

0.00.51.01.52.02.53.0t / ms-0.6-0.4-0.20.00.20.40.6p / Pa
Pressure change at the eardrum against time for the noise pulse (solid) and the anti-noise pulse (dashed), each acting alone.
(a)
(i)

Use the graph to describe the resultant pressure change at the eardrum when both pulses arrive together, stating its maximum value.

(2)
(ii)

State the phase difference between the noise and the anti-noise that the headphones are designed to produce at the eardrum.

(1)
(b)

Because of the position of the speaker, the anti-noise travels 2.0 cm further than the designer intended.

(i)

Show that, for a steady hum of frequency 150 Hz, this extra path changes the phase of the anti-noise by less than 0.1 rad.

(2)
(ii)

Determine the lowest frequency for which, because of this extra 2.0 cm, the anti-noise arrives in phase with the noise.

(2)
(iii)

Hence explain why noise-cancelling headphones work best for low-frequency noise.

(1)
(c)
(i)

A student suggests cancelling the hum of an air-conditioner by playing, through a separate loudspeaker in the room, a recording of the same hum inverted. Explain why this would not produce steady cancellation in the room.

(2)
(ii)

To cancel the 150 Hz hum, the cone of the headphone speaker, of mass 2.0 g, oscillates with simple harmonic motion of amplitude 0.10 mm. Determine the maximum resultant force on the cone.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The displacements (pressure changes) add at each instant: the resultant is a small triangular pulse with the same timing (0.5 ms to 2.5 ms)✓ 1
Maximum +0.05 Pa (0.50 − 0.45) at 1.5 ms✓ 1
Part (a)(ii)
π rad (antiphase)✓ 1Accept 180°.
Part (b)(i)
λ = 340/150 = 2.27 m✓ 1
Phase change = 2π × 0.020/2.27 = 0.055 rad (< 0.1 rad)✓ 1Answer to at least 2 s.f. required.
Part (b)(ii)
The anti-noise is designed to be in antiphase, so it arrives in phase when the extra path is half a wavelength: λ = 0.040 m✓ 1Allow ECF from (a)(ii). Use of λ = 0.020 m (giving 17 kHz): [1 max].
f = 340/0.040 = 8.5 × 103 Hz✓ 1
Part (b)(iii)
At low frequency the wavelength is long compared with the small path error, so the phase error is small and almost complete cancellation occurs; at high frequency the same path error is a large fraction of λ✓ 1Allow ECF from (b)(i) and (b)(ii).
Part (c)(i)
The recording and the hum are not coherent: their frequencies are not exactly equal and their phase difference is not constant, so cancellation would drift into reinforcement✓ 1
The path difference from the two sources is different at different points of the room, so antiphase could hold only at some points✓ 1
Part (c)(ii)
amax = ω²x0 = (2π × 150)² × 1.0 × 10−4 = 89 m s−2✓ 1
F = ma = 2.0 × 10−3 × 89 = 0.18 N✓ 1

Answers: (a)(i) +0.05 Pa  ·  (a)(ii) π rad  ·  (b)(i) 0.055 rad  ·  (b)(ii) 8.5 × 103 Hz  ·  (c)(ii) 0.18 N (the remaining parts are explanations — see the table above)

83C-2-27Single-slit diffraction Paper 2Medium9 marks

Green laser light of wavelength 532 nm is incident normally on a single narrow slit. The graph shows how the intensity of the diffracted light, relative to its maximum value, varies with the angle θ from the straight-through direction.

-12-10-8-6-4-2024681012θ / mrad0.00.20.40.60.81.0relative intensity
Relative intensity against angle for the single slit (drawn to scale).
(a)
(i)

Use the graph to determine the width of the slit.

(2)
(ii)

State how the angular width of the central maximum compares with that of a secondary maximum.

(1)
(iii)

A screen is placed 2.00 m from the slit. Calculate the width of the central maximum on the screen.

(1)
(b)
(i)

The slit width is halved. Describe two changes to the pattern.

(2)
(c)
(i)

Electrons with a de Broglie wavelength of 532 nm pass through the same slit. Determine their speed, and state whether the diffraction pattern would have the same angular width.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
First minimum at θ = 4.4 mrad✓ 1Accept 4.2–4.6 mrad.
b = λ/θ = 532 × 10−9/4.4 × 10−3 = 1.2 × 10−4 m✓ 1Accept 1.15–1.27 × 10−4 m.
Part (a)(ii)
The central maximum is twice as wide✓ 1
Part (a)(iii)
Width = 2 × 2.00 × 4.4 × 10−3 = 1.8 × 10−2 m✓ 1Allow ECF from (a)(i). Half this value (one side only): [0].
Part (b)(i)
The central maximum becomes twice as wide (θ doubles)✓ 1
The intensity is reduced, as less light passes through the slit and it is spread over a wider angle✓ 1
Part (c)(i)
p = h/λ = 6.63 × 10−34/532 × 10−9 = 1.25 × 10−27 kg m s−1✓ 1
v = p/m = 1368 m s−1✓ 1Accept 1.4 × 103 m s−1.
Yes: the angular width depends only on λ/b, and the wavelength is the same✓ 1Allow ECF from (a)(i).

Answers: (a)(i) 1.2 × 10−4 m  ·  (a)(ii) twice  ·  (a)(iii) 1.8 × 10−2 m  ·  (c)(i) 1368 m s−1; same (the remaining parts are explanations — see the table above)

84C-2-28Diffraction gratings Paper 2Medium10 marks

A laser projector for a light show combines three lasers of wavelengths 450 nm (blue), 520 nm (green) and 638 nm (red) into a single beam. The beam is incident normally on a diffraction grating with 500 lines per millimetre, and the spots are seen on a large flat wall parallel to the grating and 1.50 m from it.

(a)
(i)

Explain why the central spot appears white.

(1)
(ii)

Determine the distance on the wall between the first-order green spot and the first-order red spot on the same side.

(3)
(b)
(i)

Determine the total number of spots on the wall, assuming it is large enough.

(3)
(ii)

The grating is replaced by one with 1000 lines per millimetre. State the number of red spots, other than the central spot, that are now seen.

(1)
(c)
(i)

The red laser is replaced by a laser of a different wavelength λ. The second-order spot of this laser falls at the same position on the wall as the third-order blue spot. Determine λ, and deduce whether a third-order spot of this laser is seen.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
At the centre the path difference between waves from all slits is zero for every wavelength, so all three colours have a maximum there and their mixture appears white✓ 1
Part (a)(ii)
d = 1/(500 × 103) = 2.00 × 10−6 m✓ 1
Green: θ = 15.1°; red: θ = 18.6°✓ 1
y = 1.50 tan θ: 0.404 m and 0.505 m, so the separation is 0.10 m✓ 1Accept 0.10–0.11 m. The small-angle approximation (y = 1.50θ with θ in rad) gives 0.089 m and is not valid at these angles: [2 max].
Part (b)(i)
n ≤ d/λ since sin θ ≤ 1, with d = 2000 nm✓ 1
Blue 2000/450 = 4.4 → 4; green 3.8 → 3; red 3.1 → 3✓ 1All three required.
1 + 2 × (4 + 3 + 3) = 21 spots✓ 1Allow ECF from the orders found. 20 (central spot omitted) or 11 (one side only): [0] for this mark.
Part (b)(ii)
d/λ = 1000/638 = 1.6, so first order only: 2 red spots✓ 1
Part (c)(i)
Same angle, so the same d sin θ: 2λ = 3 × 450 nm, giving λ = 675 nm✓ 1
d/λ = 2000/675 = 2.96 < 3, so there is no third-order spot✓ 1Allow ECF for their wavelength.

Answers: (a)(ii) 0.10 m  ·  (b)(i) 21  ·  (b)(ii) 2  ·  (c)(i) 675 nm; no (the remaining parts are explanations — see the table above)

85C-2-29Reflection, refraction and transmission at boundaries Paper 2Medium13 marks

A medical ultrasound scanner uses a transducer with a flat face 12 mm wide to send short pulses of ultrasound of frequency 3.5 MHz into the body. The scanner calculates the depth of each reflecting boundary assuming that ultrasound travels at 1540 m s−1 in all soft tissue. In fact it travels at 1450 m s−1 in fat and at 1580 m s−1 in muscle.

(a)
(i)

Calculate the wavelength of the ultrasound in muscle.

(1)
(ii)

Estimate the angle through which the beam spreads beyond its edges, using θ = λ/b, and comment on the result.

(2)
(b)
(i)

The beam meets a boundary between fat and muscle from the fat side, at an angle of incidence of 20°. Calculate the angle of refraction in the muscle.

(2)
(ii)

Explain why a structure lying beyond this boundary is displayed in the wrong position sideways.

(2)
(c)

Along a line normal to the skin, an echo from an organ boundary returns 65.0 μs after the pulse is sent.

(i)

Show that the depth displayed by the scanner is 5.0 cm.

(1)
(ii)

The pulse actually passes through 2.00 cm of fat and then through muscle. Determine the true depth of the organ boundary.

(3)
(d)
(i)

About 50 mW of ultrasound power is absorbed in 20 g of tissue of specific heat capacity 3600 J kg−1 K−1. Determine the rise in temperature of this tissue during a one-minute scan, assuming no energy is conducted away, and comment on the safety of the scan.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λ = v/f = 1580/3.5 × 106 = 4.5 × 10−4 m✓ 1
Part (a)(ii)
θ ≈ 4.5 × 10−4/0.012 = 0.038 rad (about 2.2°)✓ 1Allow ECF from (a)(i).
The aperture is about 27 wavelengths wide, so diffraction is small and the beam stays narrow, allowing fine detail to be located✓ 1
Part (b)(i)
sin θ2/sin 20° = 1580/1450✓ 1
θ2 = 21.9°✓ 1Accept 22°.
Part (b)(ii)
The scanner assumes the ultrasound travels in a straight line along the original direction✓ 1
but the beam is deviated by about 2° at the boundary, so the echo comes from a point displaced sideways from the assumed line✓ 1Allow ECF from (b)(i).
Part (c)(i)
Depth = vt/2 = 1540 × 65.0 × 10−6/2 = 0.0500 m✓ 1The factor 2 (there and back) must be seen.
Part (c)(ii)
Time in fat, down and back = 2 × 0.0200/1450 = 27.6 μs✓ 1
Time in muscle = 65.0 − 27.6 = 37.4 μs, so the depth in muscle = 1580 × 37.4 × 10−6/2 = 2.96 cm✓ 1Allow ECF from the time in fat.
True depth = 2.00 + 2.96 = 4.96 cm✓ 1Accept 4.95–4.96 cm. Allow ECF from (c)(i) for the comparison.
Part (d)(i)
ΔT = Pt/(mc) = 0.050 × 60/(0.020 × 3600) = 0.042 K✓ 1
The rise is far too small to damage the tissue, so the scan is safe from heating effects✓ 1Comment must agree with the candidate's value.

Answers: (a)(i) 4.5 × 10−4 m  ·  (a)(ii) 0.038 rad  ·  (b)(i) 21.9°  ·  (c)(i) 0.0500 m  ·  (c)(ii) 4.96 cm  ·  (d)(i) 0.042 K (the remaining parts are explanations — see the table above)

86C-2-37Refraction and total internal reflection Paper 2Medium14 marks

An optical rain sensor on a car windscreen uses an infrared light-emitting diode (LED) that emits radiation of wavelength 880 nm in air. The beam enters the glass of the windscreen through a small prism and travels along the windscreen by reflection between its outer and inner surfaces, meeting the outer surface at an angle of incidence of 45.0°. After a distance of 60 mm along the windscreen it leaves through a second prism and reaches a detector. Both prisms are fixed to the inner surface.

The refractive index of the glass is 1.50 and that of water is 1.33. The outer and inner surfaces are parallel and 5.0 mm apart.

outer surface (air or water)inner surfaceglass, n = 1.50LEDdetector45°5.0 mm60 mm
Diagram not to scale
(a)
(i)

Calculate the critical angle for the glass–air boundary.

(1)
(ii)

Show that the critical angle for a glass–water boundary is about 62°.

(1)
(iii)

Explain why the signal at the detector falls when raindrops cover parts of the outer surface.

(3)
(b)
(i)

Calculate the angle of refraction of the light in a raindrop on the outer surface.

(2)
(ii)

State the range of angles of incidence at the outer surface for which this sensor can distinguish a wet windscreen from a dry one.

(1)
(iii)

After heavy rain, a continuous layer of water whose outer surface is flat and parallel to the windscreen covers the sensing area. Suggest, with a calculation, why the signal at the detector may then hardly fall.

(2)
(c)
(i)

Determine the frequency of the radiation and its wavelength in the glass.

(2)
(ii)

Determine the number of times the beam meets the outer surface while it travels the 60 mm along the windscreen.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
sin C = 1/1.50, so C = 41.8°✓ 1
Part (a)(ii)
sin C = 1.33/1.50 = 0.887, so C = 62.5°✓ 1Answer to at least 3 s.f. or full substitution required.
Part (a)(iii)
Where the outer surface is dry, 45.0° is greater than the glass–air critical angle (41.8°), so the beam is totally internally reflected towards the detector✓ 1Allow ECF from (a)(i).
Where there is water, 45.0° is less than the glass–water critical angle (62°), so total internal reflection does not occur✓ 1Allow ECF from (a)(ii).
Most of the light is then refracted (transmitted) into the drops and only a small fraction is reflected, so less reaches the detector✓ 1Some reference to light leaving the glass is needed.
Part (b)(i)
1.50 sin 45.0° = 1.33 sin θ✓ 1
θ = 52.9°✓ 1Award [1 max] for 38.8° (indices inverted).
Part (b)(ii)
Between 42° and 62°✓ 1Allow ECF from (a)(i) and (a)(ii).
Part (b)(iii)
The ray meets the flat water–air surface at the angle of refraction in the water, 52.9°, which is greater than the water–air critical angle sin−1(1/1.33) = 48.8°✓ 1Allow ECF from (b)(i). ALT: 1.50 sin 45.0° = 1.06 > 1, so no ray can be refracted into the air.
So the light is totally internally reflected at the water–air surface and returns through the water and the glass towards the detector (drops have curved surfaces, so light can leave them)✓ 1Reference to the light still reaching the detector is needed.
Part (c)(i)
f = c/λ = 3.00 × 108/880 × 10−9 = 3.41 × 1014 Hz✓ 1
λglass = 880/1.50 = 587 nm✓ 1The frequency is unchanged in the glass.
Part (c)(ii)
Between one reflection at the outer surface and the next, the beam crosses the glass twice and moves 2 × 5.0 × tan 45.0° = 10 mm along it✓ 1The first meeting is 5 mm from the entry point.
60/10 = 6 times✓ 1

Answers: (a)(i) 41.8°  ·  (a)(ii) 62.5°  ·  (b)(i) 52.9°  ·  (b)(ii) 42° to 62°  ·  (b)(iii) 52.9° > 48.8°: total internal reflection at the water–air surface  ·  (c)(i) 3.41 × 1014 Hz; 587 nm  ·  (c)(ii) 6 (the remaining parts are explanations — see the table above)

87IBP2H-023-01-10Young's double-slit experiment Paper 2Medium8 marks

A forensic laboratory checks the wavelength of a green laser pointer using a double slit of slit separation 0.25 mm. For each slit-to-screen distance D, a technician measures the distance X across ten fringe spacings.

D / m : 1.00 | 1.50 | 2.00 | 2.50
X / mm : 21.3 | 31.9 | 42.6 | 53.2

laserDdouble slitd = 0.25 mmX (10 fringe spacings)
Diagram not to scale
(a)

Outline why the technician measures across ten fringe spacings rather than one.

(1)
(b)

Explain why a graph of X against D is expected to be a straight line through the origin.

(2)
(c)

Determine the wavelength of the laser light.

(3)
(d)

The green laser is replaced with a red laser, keeping D = 2.00 m. Predict, with a reason, how X changes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
It reduces the percentage (fractional) uncertainty in the fringe spacing✓ 1
Part (b)
s = λD/d, so X = 10λD/d✓ 1
λ and d are constant, so X ∝ D✓ 1
Part (c)
Gradient = (53.2 − 21.3)/(2.50 − 1.00) ≈ 21.3 mm m−1✓ 1
s/D = gradient/10 = 2.13 × 10−3✓ 1
λ = (s/D) × d = 2.13 × 10−3 × 0.25 × 10−3 ≈ 5.3 × 10−7 m✓ 1
Part (d)
X increases✓ 1
Red light has a longer wavelength and the fringe spacing is proportional to λ✓ 1

Exam tip: Use the gradient rather than a single row of data. Remember to divide by the number of fringe spacings before applying s = λD/d.

88IBP2H-023-01-11Single-slit diffraction Paper 2Medium9 marks

In a precision-engineering workshop, the width of a narrow gap between two machined blades is checked by shining laser light of wavelength 650 nm through it. The diffraction pattern is viewed on a screen 1.80 m from the gap. The central maximum has a total width of 9.4 mm.

laser1.80 mslit, width b9.4 mm
Diagram not to scale
(a)

Explain how the first minimum on either side of the central maximum arises.

(2)
(b)

Calculate the width b of the gap.

(3)
(c)

Thermal expansion of the blade holder increases the gap width by 5.0%. Determine the new width of the central maximum.

(2)
(d)

Describe the variation of intensity across the screen, as a sketch of intensity against position would show.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The gap is treated as two halves, with each point in one half paired with a point b/2 away in the other half✓ 1
At the first minimum the path difference between paired points is λ/2, so each pair interferes destructively and the total intensity is zero✓ 1
Part (b)
Angle to first minimum θ = (9.4 × 10−3/2)/1.80 = 2.61 × 10−3 rad✓ 1
θ = λ/b, so b = λ/θ✓ 1
b = 650 × 10−9/2.61 × 10−3 ≈ 2.5 × 10−4 m✓ 1
Part (c)
Width of central maximum ∝ 1/b, since θ = λ/b and width ≈ 2Dθ✓ 1
New width = 9.4/1.05 ≈ 9.0 mm (8.95 mm); also accept recalculation using the new gap width 1.05b✓ 1
Part (d)
Central maximum of greatest intensity, with a width twice that of each secondary maximum✓ 1
Secondary maxima on both sides, symmetric and of rapidly decreasing intensity, separated by zero-intensity minima✓ 1

Common error: The given width is the full width of the central maximum. Halve it before finding the angle to the first minimum.

89IBP2H-023-01-13Two-source interference Paper 2Medium7 marks

Two loudspeakers in a sports arena are driven in phase by the same 1200 Hz test tone. A sound engineer stands at point P, which is 8.40 m from one speaker and 9.25 m from the other. The speed of sound is 340 m s−1.

9.25 m8.40 mABP
Diagram not to scale
(a)

Calculate the wavelength of the sound.

(1)
(b)

Show that P is a position of maximum loudness.

(2)
(c)

The frequency of the tone is varied. Determine the lowest frequency for which P is a position of minimum loudness.

(2)
(d)

Explain why the sound at P would not be completely silent at a minimum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
λ = 340/1200 = 0.283 m✓ 1
Part (b)
Path difference = 9.25 − 8.40 = 0.85 m✓ 1
0.85/0.283 = 3.0, a whole number of wavelengths, so the waves arrive in phase and interfere constructively✓ 1
Part (c)
Minimum requires path difference = λ/2, so λ = 1.70 m✓ 1
f = 340/1.70 = 200 Hz✓ 1
Part (d)
The waves arrive with unequal amplitudes because P is at different distances from the two speakers✓ 1
Reflections from the arena surfaces also reach P, so cancellation is incomplete✓ 1

Exam tip: The lowest frequency for a minimum corresponds to the longest wavelength, where the path difference is half a wavelength.

90IBP2H-023-01-16Double slit with diffraction envelope Paper 2Medium7 marks

In a planetarium demonstration of Young's experiment, monochromatic light passes through two narrow, equal slits. With only one slit open, the intensity at the centre of the screen is I0. Assume each slit alone illuminates the central region of the screen uniformly, and ignore any single-slit diffraction envelope.

(a)

Determine, in terms of I0, the intensity at a bright fringe when both slits are open.

(2)
(b)

Outline why the result in (a) does not violate conservation of energy.

(1)
(c)

One slit is partly covered so that its wave amplitude at the screen is 0.60 of the other's. Calculate the ratio of the maximum intensity to the minimum intensity.

(2)
(d)

The monochromatic source is replaced with white light. Outline the appearance of the fringes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
At a bright fringe the waves arrive in phase, so the amplitudes add and the resultant amplitude is doubled✓ 1
Intensity ∝ amplitude², so the intensity is 4I0✓ 1
Part (b)
Energy is redistributed from dark fringes (zero intensity) to bright fringes, so the average intensity is 2I0, which equals the sum I0 + I0 from the two slits separately✓ 1
Part (c)
Imax ∝ (1 + 0.60)² = 2.56 and Imin ∝ (1 − 0.60)² = 0.16✓ 1
Ratio Imax/Imin = 16✓ 1
Part (d)
The central fringe is white, because the path difference is zero for all wavelengths✓ 1
The other fringes are coloured, with violet (blue) nearer the centre and red further out. Further from the centre, the fringes of different colours overlap, so the pattern loses contrast and becomes near-uniform illumination✓ 1

Exam tip: Combine amplitudes first, then square to find intensity. Never add or subtract intensities directly when coherent waves interfere.

91IBP2S-022-00-15Diffraction gratings Paper 2Medium7 marks

A planetary lander carries a grating spectrometer to analyse light from a gas discharge. Light is incident normally on the grating. The hydrogen line of wavelength 656 nm produces a second-order maximum at an angle of 41.0° to the straight-through direction.

(a)

Calculate the line spacing of the grating.

(2)
(b)

Determine the highest order of the 656 nm line that can be observed.

(2)
(c)

Hydrogen also emits light of wavelength 486 nm. Calculate the angle of its third-order maximum.

(2)
(d)

Outline one advantage of using a grating rather than a double slit for this measurement.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses d sin θ = nλ .✓ 1
d = 2 × 656×10−9/sin 41.0° = 2.00×10−6 m .✓ 1
Part (b)
n ≤ d/λ = 2.00×10−6/656×10−9 = 3.05 .✓ 1
So the highest order is the third .✓ 1
ECF from (a).—
Part (c)
sin θ = 3 × 486×10−9/2.00×10−6 = 0.729 .✓ 1
θ = 46.8° .✓ 1
ECF from (a).—
Part (d)
The grating gives much sharper and brighter maxima, so angles can be measured more precisely and close wavelengths separated .✓ 1

Exam tip: Write the highest order as a whole number found by rounding d/λ down. Show the value of d/λ to earn the method mark.

92IBP2H-023-00-13Two-source interference Paper 2Medium7 marks

Two radio antennas, 30.0 m apart, emit coherent signals in phase at a frequency of 100 MHz. A car drives along a straight road that is parallel to the line joining the antennas and 3.00 km from it. A radio in the car detects a signal that rises and falls in strength as the car moves.

T₁T₂30.0 mroad3.00 kmcar
Diagram not to scale
(a)

Calculate the wavelength of the radio waves.

(1)
(b)

At one point on the road, the path difference from the two antennas is 7.5 m. State and explain the signal strength detected at this point.

(2)
(c)

Near the point on the road that is equidistant from both antennas, calculate the distance along the road between adjacent positions of maximum signal.

(2)
(d)

The antennas are rewired so that they emit in antiphase. Explain what is now detected at the point on the road that is equidistant from both antennas.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
λ = c/f = 3.00×108/1.00×108 = 3.00 m✓ 1
Part (b)
7.5 m = 2.5λ, a half-integer number of wavelengths (ECF from (a))✓ 1
The waves arrive in antiphase and interfere destructively, so the signal is a minimum✓ 1
Part (c)
s = λD/d✓ 1
s = 3.00 × 3000/30.0 = 300 m (ECF from (a))✓ 1
Part (d)
The path difference there is zero, so the waves arrive with the source phase difference of π (half a cycle)✓ 1
They interfere destructively, so the signal is a minimum instead of a maximum✓ 1

Exam tip: s = λD/d relies on a small-angle approximation. It holds only near the central maximum, where the distance along the road is much smaller than D. Further out, the spacing between maxima increases.

93IBP2H-023-00-15Refraction and total internal reflection Paper 2Medium7 marks

A car's rain sensor sends an infrared beam through the windscreen glass, which has refractive index 1.50. The beam meets the outer surface of the glass at an angle of incidence of 45°, and a detector measures the intensity of the reflected beam. The refractive index of water is 1.33.

outer surface (air or water)inner surfaceglass, n = 1.50LEDdetector45°
Diagram not to scale
(a)

Calculate the critical angle for the glass–air boundary.

(2)
(b)

Explain why the whole beam is reflected when the windscreen is dry.

(1)
(c)

Raindrops cover the outer surface at the point where the beam strikes it. Calculate the angle of refraction of the beam into the water.

(2)
(d)

Explain how the sensor can detect rain.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
sin c = 1/1.50 = 0.667✓ 1
c = 41.8°✓ 1
Part (b)
The 45° angle of incidence is greater than the critical angle of 41.8°, so total internal reflection occurs (ECF from (a))✓ 1
Part (c)
1.50 sin 45° = 1.33 sin θ, so sin θ = 0.797✓ 1
θ = 52.9°✓ 1
Part (d)
The critical angle for the glass–water boundary is sin−1(1.33/1.50) = 62.5°, which is greater than 45°, so total internal reflection no longer occurs✓ 1
Part of the beam refracts into the water, so the reflected intensity at the detector falls. This drop triggers the wipers✓ 1

Common error: The critical angle depends on both media. Replacing air with water changes it, even though the glass is the same.

94IBP2H-023-00-16Diffraction gratings Paper 2Medium9 marks

The tracks on a DVD act as a reflection diffraction grating with a line spacing of 0.74 µm. A DVD player reads the disc with a laser of wavelength 650 nm. Assume the light is incident normally on the disc and that the grating equation applies to the reflected light.

(a)

Calculate the angle of the first-order maximum for the laser light.

(2)
(b)

Show that no second-order maximum is produced.

(2)
(c)

The disc is lit normally by white light with wavelengths from 400 nm to 700 nm. Determine the angular width of the first-order spectrum.

(3)
(d)

A Blu-ray disc has a track spacing of 0.32 µm and is read with light of wavelength 405 nm. Explain why, at normal incidence, this light produces no diffracted beams from the disc.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
sin θ = 650×10−9/0.74×10−6 = 0.878✓ 1
θ = 61.4°✓ 1
Part (b)
For n = 2, sin θ = 2 × 650×10−9/0.74×10−6 = 1.76✓ 1
sin θ cannot be greater than 1, so the second order does not exist✓ 1
Part (c)
For 400 nm, sin θ = 0.541, so θ = 32.7°✓ 1
For 700 nm, sin θ = 0.946, so θ = 71.1°✓ 1
Angular width = 71.1° − 32.7° = 38° (38.4°)✓ 1
Part (d)
For n = 1, sin θ = 405/320 = 1.27✓ 1
This is greater than 1, so no first or higher order exists; only the zero-order (direct) reflection appears✓ 1

Exam tip: If nλ/d is greater than 1, that order does not exist. Showing this calculation explicitly earns the marks.

95IBP2H-023-02-09Refraction and total internal reflection Paper 2Medium8 marks

A small lamp is fixed to the floor of a swimming pool, 2.0 m below the flat water surface. The refractive index of the pool water is 1.33. Take the refractive index of air as 1.00. Light from the lamp leaves the water only through a circular region of the surface directly above the lamp.

lamp2.0 m1.5 mwaterairemerging circle: radius ?
Diagram not to scale
(a)

State two conditions for total internal reflection to occur at the water surface.

(2)
(b)

Show that the critical angle for the water–air boundary is about 49°.

(2)
(c)

Determine the radius of the circle on the surface through which light from the lamp emerges.

(2)
(d)

A ray from the lamp reaches the surface at a point 1.5 m horizontally from the point directly above the lamp. Calculate the angle between this ray and the normal after it leaves the water.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The light travels from the medium of higher refractive index (water) towards the medium of lower refractive index (air) . The angle of incidence is greater than the critical angle .✓ 2
Part (b)
sin c = 1/1.33 = 0.752 . c = 48.8°, which is about 49° .✓ 2
Part (c)
The edge of the circle is where rays meet the surface at the critical angle, so r = 2.0 tan 48.8° . r = 2.3 m (2.28 m) . Allow error carried forward from the critical angle found in the previous part.✓ 2
Part (d)
Angle of incidence: tan θ = 1.5/2.0, so θ = 36.9° and sin θ = 0.60 . sin r = 1.33 × 0.60 = 0.798, so r = 53° .✓ 2

Exam tip: In 'circle of light' problems the edge ray is the critical ray. Draw a right-angled triangle with the depth as the adjacent side and the horizontal distance as the opposite side. Use this triangle to find the angle of incidence for any ray.

96IBP2H-023-02-13Wavefronts and rays Paper 2Medium7 marks

Coastal engineers study ocean swell approaching a beach. The waves have period 8.0 s. Their speed is 6.0 m s−1 in a region of deeper water and 4.0 m s−1 in a shallower region beyond a straight underwater ledge.

shallower: 4.0 m s⁻¹deeper: 6.0 m s⁻¹40°?
Diagram not to scale
(a)

State why the period of the waves is the same on both sides of the ledge.

(1)
(b)

Calculate the wavelength of the waves in the shallower region.

(2)
(c)

The wave direction in the deeper water makes an angle of 40° with the normal to the ledge. Calculate the angle the wave direction makes with the normal in the shallower region.

(2)
(d)

Explain why wave crests arriving at a beach are often almost parallel to the shoreline, even when the swell approaches at an angle.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Each crest arriving at the ledge produces one crest beyond it, so the number of crests per second (the frequency, and therefore the period) cannot change .✓ 1
Part (b)
Use λ = vT (or λ = v/f) with the unchanged period . λ = 4.0 × 8.0 = 32 m .✓ 2
Part (c)
Use sin θ2 = (v2/v1) sin θ1 = (4.0/6.0) × sin 40° = 0.429 . θ2 = 25° (accept 25.4°) .✓ 2
Part (d)
The water becomes gradually shallower towards the shore, so the wave speed keeps decreasing . The waves are continuously refracted towards the normal to the depth contours, so the angle between the crests and the shoreline approaches zero .✓ 2

Exam tip: For water waves, use sin θ₁/sin θ₂ = v₁/v₂. There is no refractive index to look up, because the speed ratio does the same job. The frequency is fixed by the source, so only the speed and the wavelength change.

97IBP2H-023-02-14Refraction and total internal reflection Paper 2Medium8 marks

An astronomer tests a glass block for a spectrometer. The table gives the refractive index of the glass for three wavelengths, all measured in air.

λ in air / nm: 450 | 550 | 650
refractive index: 1.532 | 1.524 | 1.519

White light is incident on a flat face of the block at an angle of incidence of 60.0°. Take c = 3.00×108 m s−1.

(a)

Calculate the speed of 650 nm light in the glass.

(1)
(b)

Calculate the wavelength in the glass of the light that has wavelength 450 nm in air.

(1)
(c)

Determine the angle between the refracted 450 nm and 650 nm rays inside the glass.

(3)
(d)

The 450 nm and 650 nm light later meet an inner glass–air face, both at the same angle of incidence. This angle of incidence is gradually increased. Explain which of these two wavelengths is the first to be totally internally reflected.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
v = 3.00×108/1.519 = 1.97×108 m s−1 .✓ 1
Part (b)
λ = 450/1.532 = 294 nm .✓ 1
Part (c)
450 nm: sin r = sin 60.0°/1.532 = 0.5653, so r = 34.42° . 650 nm: sin r = sin 60.0°/1.519 = 0.5701, so r = 34.76° . Angular separation = 0.34° (accept 0.3°–0.4°) .✓ 3
Part (d)
The critical angles are 40.7° for 450 nm (sin c = 1/1.532) and 41.2° for 650 nm (sin c = 1/1.519) . The 450 nm (blue) light has the larger refractive index and so the smaller critical angle . Once the angle of incidence passes 40.7°, the 450 nm light is totally internally reflected, while the 650 nm light is still partly transmitted until 41.2° .✓ 3

Exam tip: When the answer depends on small differences in refractive index, keep at least four significant figures in intermediate angles. Round only the final separation.

98IBP3H-C3-20r25-12Double-slit interference Paper 2Medium8 marks

A telecoms technician tests two laser diodes using the same double slit, which has slit separation 0.40 mm. The screen is 1.60 m from the slits. One diode emits light of wavelength 500 nm and the other emits light of wavelength 650 nm. The diodes are switched on together and both patterns appear on the screen with a common central maximum.

(a)

Calculate the fringe spacing for each wavelength.

(3)
(b)

Calculate the separation of the 3rd bright fringes of the two patterns, on the same side of the central maximum.

(2)
(c)

Determine the smallest distance from the central maximum at which a bright fringe of one pattern coincides exactly with a bright fringe of the other pattern.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Use of s = λD/d✓ 1
500 nm: s = 2.00 mm✓ 1
650 nm: s = 2.60 mm✓ 1
Part (b)
Use of 3 × (2.60 − 2.00) (ECF from (a))✓ 1
1.8 mm✓ 1
Part (c)
Coincidence requires n × 500 = m × 650 for whole numbers n and m✓ 1
n : m = 13 : 10, so the smallest values are n = 13 and m = 10✓ 1
Distance = 13 × 2.00 = 26 mm (= 10 × 2.60)✓ 1

Exam tip: For coincident fringes, set the two path differences equal and look for the smallest pair of whole numbers; check the answer using both patterns.

99IBP3H-C3-20r25-13Diffraction gratings Paper 2Medium9 marks

A food scientist analyses the light from a sodium-vapour lamp, of wavelength 589 nm, using a diffraction grating that has 500 lines per millimetre. The light is incident normally on the grating.

(a)

Show that the slit spacing of the grating is 2.00 μm.

(1)
(b)

Calculate the angle between the central maximum and the first-order maximum.

(2)
(c)

Explain why the bright maxima from a grating are much narrower than those from a double slit of the same slit spacing.

(2)
(d)

Determine the highest order of maximum that can be observed.

(2)
(e)

Calculate the angle of the highest-order maximum from the central maximum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = 1/(500×103) m = 2.00×10−6 m✓ 1
Part (b)
Use of d sin θ = nλ✓ 1
θ = 17.1°✓ 1
Part (c)
With many slits, at a small angle away from a maximum the contributions from successive slits have steadily increasing phase differences✓ 1
These cancel over the whole set of slits, so intensity falls to zero close to the maximum / only a very narrow range of angles gives in-phase addition✓ 1
Part (d)
sin θ cannot exceed 1, so n ≤ d/λ = 3.40✓ 1
Highest order n = 3✓ 1
Part (e)
Use of sin θ = 3λ/d✓ 1
θ = 62.1°✓ 1

Common error: The highest order is the whole number below d/λ; rounding 3.40 to 4 would require sin θ greater than 1, which is impossible.

100IBP3H-C3-20r25-14Double-slit interference Paper 2Medium8 marks

In a telecoms laboratory, a laser of wavelength 640 nm illuminates a single slit of width 0.040 mm. A diffraction pattern is formed on a screen 2.50 m from the slit.

laser2.50 mslit, width 0.040 mm?
Diagram not to scale
(a)

Explain why the first minimum occurs at the angle for which the path difference between waves from the two edges of the slit is one wavelength.

(2)
(b)

Calculate the angle of the first minimum from the central axis, in radians. Use the small-angle approximation.

(2)
(c)

Calculate the width of the central maximum on the screen.

(2)
(d)

The slit width is halved. State and explain the effect on the width of the central maximum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Pair each point in the top half of the slit with the point half a slit width below it: path difference λ/2✓ 1
These pairs cancel in antiphase, so all of the waves cancel (minimum)✓ 1
Part (b)
Use of θ = λ/a✓ 1
θ = 640×10−9/0.040×10−3 = 0.016 rad✓ 1
Part (c)
Use of width = 2θD (ECF from (b))✓ 1
= 2 × 0.016 × 2.50 = 0.080 m (8.0 cm)✓ 1
Part (d)
The width doubles✓ 1
Because θ = λ/a: halving a doubles the angle, and the width is proportional to θ✓ 1

Exam tip: The central maximum is twice as wide as the distance from the centre to the first minimum; do not forget the factor of 2.

101IBP3H-C3-20r25-16Diffraction around bodies and apertures Paper 2Medium7 marks

A climber shouts from behind a rocky ridge. The sound passes through a gap in the rocks of width 1.50 m. The speed of sound in air is 340 m s−1. Consider a low-pitched voice of frequency 250 Hz and a whistle of frequency 2000 Hz.

(a)

Calculate the wavelength of each sound.

(2)
(b)

Determine the angle of the first diffraction minimum, measured from the straight-through direction, for each sound.

(3)
(c)

Explain why a listener far off to one side of the gap hears the low-pitched voice more clearly than the whistle.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Use of λ = v/f✓ 1
1.36 m for 250 Hz and 0.170 m for 2000 Hz✓ 1
Part (b)
Use of sin θ = λ/a✓ 1
250 Hz: θ = 65.0°✓ 1
2000 Hz: θ = 6.51°✓ 1
Part (c)
The 250 Hz wavelength is comparable with the gap width, so the sound diffracts strongly and spreads over a wide angle✓ 1
The 2000 Hz wavelength is much smaller than the gap, so it spreads little and stays in a narrow beam✓ 1

Common error: A wave diffracts significantly when the wavelength is comparable to or larger than the gap, so shorter wavelengths do not spread as much.

102C-2-02Double slit with diffraction envelope Paper 2Hard9 marks

In a modern version of the double-slit experiment, electrons are accelerated from rest through a potential difference of 600 V and then pass through two slits cut in a thin membrane. Each slit has width 75 nm and the centres of the slits are 300 nm apart. The electrons are detected on a screen 1.50 m beyond the slits.

(a)
(i)

Show that the de Broglie wavelength of the electrons is about 5 × 10−11 m.

(2)
(ii)

Calculate the separation of adjacent bright fringes on the screen.

(1)
(b)
(i)

Deduce the number of bright fringes that lie within the central maximum of the single-slit diffraction pattern.

(3)
(c)
(i)

The accelerating potential difference is increased to 2400 V. Deduce, without calculating a new wavelength, the new fringe separation in terms of s, and whether your answer to (b) changes.

(2)
(d)
(i)

Outline why this experiment is evidence that electrons have wave properties.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Ek = eV = 1.60 × 10−19 × 600 = 9.60 × 10−17 J and p = √(2mEk) = 1.32 × 10−23 kg m s−1✓ 1
λ = h/p = 6.63 × 10−34/1.32 × 10−23 = 5.01 × 10−11 m✓ 1Answer to at least 2 s.f. or full substitution required.
Part (a)(ii)
s = λD/d = 5.01 × 10−11 × 1.50/(300 × 10−9) = 2.51 × 10−4 m✓ 1Accept 2.5 × 10−4 m. Allow ECF from (a)(i).
Part (b)(i)
First minimum of the envelope at θ = λ/b, i.e. y = λD/b = 1.00 × 10−3 m from the centre✓ 1Allow ECF from (a)(i).
This is d/b = 4 fringe separations from the centre, so the fourth-order interference maximum falls on the envelope minimum and is missing✓ 1Allow ECF from (a)(ii) for a consistent ratio.
Orders 0, ±1, ±2 and ±3 are seen: 7 bright fringes✓ 1Award [2 max] for 9 (missing orders not removed).
Part (c)(i)
λ ∝ 1/p ∝ 1/√V, so the wavelength halves and the fringe separation becomes s/2✓ 1Award this mark for s/2 with a reason.
The envelope width (∝ λ/b) also halves, so the ratio is still d/b = 4 and there are still 7 fringes✓ 1Allow ECF from (b).
Part (d)(i)
Interference fringes (and a diffraction envelope) are produced, which only waves can do, and their spacing agrees with λ = h/p✓ 1Reference to interference or diffraction needed; "they behave like waves" alone is not enough.

Answers: (a)(i) 5.01 × 10−11 m  ·  (a)(ii) 2.51 × 10−4 m  ·  (b)(i) 7  ·  (c)(i) s/2; unchanged (the remaining parts are explanations — see the table above)

103C-2-16Two-source interference Paper 2Hard9 marks

Two radio transmitters, 1.50 km apart, broadcast the same programme in phase at a frequency of 96.0 MHz. A straight road runs parallel to the line joining the transmitters at a perpendicular distance of 12.0 km from it. A driver on the road notices that the radio signal fades and recovers regularly.

T₁T₂1.50 kmroad12.0 kmcar
Diagram not to scale
(a)
(i)

Calculate the wavelength of the radio waves.

(1)
(ii)

Explain why the signal strength varies along the road.

(2)
(b)
(i)

Show that the distance along the road between adjacent points of strongest signal, near the point opposite the midpoint of the transmitters, is about 25 m.

(2)
(ii)

Calculate the number of times per second the signal fades when the car moves at 90 km h−1.

(1)
(iii)

The car is at a point of fading when the driver starts to brake, with a uniform deceleration of 2.5 m s−2, from 25 m s−1. Determine the time until the car reaches the next point of fading.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λ = c/f = 3.00 × 108/96.0 × 106 = 3.125 m✓ 1
Part (a)(ii)
The waves from the two transmitters are coherent; as the car moves, the path difference to the car changes✓ 1
Where it is nλ the waves arrive in phase and the signal is strong; where it is (n + ½)λ they arrive in antiphase and it fades✓ 1Both conditions needed.
Part (b)(i)
d ≪ D (and the points are near the centre line), so s = λD/d applies✓ 1
s = 3.125 × 12 000/1500 = 25.0 m✓ 1Allow ECF from (a)(i).
Part (b)(ii)
90 km h−1 = 25 m s−1, so the rate = 25/25 = 1.0 s−1✓ 1Allow ECF from (b)(i).
Part (b)(iii)
Adjacent points of fading are also 25 m apart: 25 = 25t − ½ × 2.5t²✓ 1Allow ECF from (b)(i).
t = (25 − √(25² − 2 × 2.5 × 25))/2.5✓ 1Solving the quadratic; the smaller root is required.
t = 1.06 s✓ 11.0 s (constant speed assumed): [1 max].

Answers: (a)(i) 3.125 m  ·  (b)(i) 25.0 m  ·  (b)(ii) 1.0 s−1  ·  (b)(iii) 1.06 s (the remaining parts are explanations — see the table above)

104IBP2H-023-01-12Double slit with diffraction envelope Paper 2Hard9 marks

A laser-cut double slit is tested with light of wavelength 600 nm. On a screen 2.0 m away, the interference fringes have a spacing of 1.2 mm. The fringes lie within a single-slit diffraction envelope whose central maximum has a total width of 12 mm.

(a)

Calculate the slit separation d.

(2)
(b)

Calculate the width b of each slit.

(2)
(c)

Explain why the fifth-order interference maxima are missing from the pattern.

(2)
(d)

State the number of bright fringes within the central diffraction maximum.

(1)
(e)

The width of each slit is reduced while d is unchanged. Predict the effect on the pattern.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = λD/s✓ 1
d = (600 × 10−9 × 2.0)/(1.2 × 10−3) = 1.0 × 10−3 m✓ 1
Part (b)
Half-width of 6.0 mm gives θ = 6.0 × 10−3/2.0 = 3.0 × 10−3 rad✓ 1
b = λ/θ = (600 × 10−9)/(3.0 × 10−3) = 2.0 × 10−4 m✓ 1
Part (c)
d/b = 5, so the angle for d sinθ = 5λ is the same as the angle for b sinθ = λ (allow ECF from (a) and (b))✓ 1
The fifth-order maxima coincide with the first diffraction minima, where the single-slit intensity is zero✓ 1
Part (d)
9 (orders −4 to +4) (allow ECF from (c))✓ 1
Part (e)
Fringe spacing is unchanged, because λ, D and d are unchanged✓ 1
The diffraction envelope becomes wider, so more bright fringes lie within the central maximum (and the pattern is dimmer)✓ 1

Exam tip: The fringe spacing is set by d, while the envelope is set by b. Treat the two effects separately before combining them.

105IBP2H-023-01-14Diffraction gratings Paper 2Hard9 marks

A hospital laboratory spectrometer uses a diffraction grating with 600 lines per millimetre to examine the sodium doublet, which has wavelengths of 589.0 nm and 589.6 nm. The light is incident normally on the grating. The spectrometer can separate two lines only if their angular separation is at least 0.040°.

(a)

Determine the highest order of the spectrum that can be observed.

(2)
(b)

Show that the doublet cannot be separated in the first order.

(3)
(c)

Determine whether the doublet can be separated in the second order.

(2)
(d)

Explain why a grating gives much sharper maxima than a double slit.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = 1/(6.00 × 105) = 1.67 × 10−6 m✓ 1
n ≤ d/λ = 2.83, so the highest order is 2✓ 1
Part (b)
sinθ = λ/d: 0.35340 and 0.35376✓ 1
θ = 20.695° and 20.718°✓ 1
Δθ ≈ 0.022°, which is less than 0.040°, so the lines are not separated✓ 1
Part (c)
sinθ = 2λ/d = 0.70680 and 0.70752, giving θ = 44.975° and 45.034°✓ 1
Δθ ≈ 0.058°, which is greater than 0.040°, so the lines are separated; the second order exists because the highest order is 2✓ 1
Allow error carried forward for an incorrect value of d from the first part.—
Part (d)
Many slits contribute; away from the exact angle d sinθ = nλ, the waves from the many slits have a spread of phases and cancel✓ 1
So the intensity falls to near zero very close to each principal maximum, giving narrow, bright lines✓ 1

Exam tip: Angular separation increases with order, so higher orders resolve close wavelengths better, provided the order exists.

106IBP2H-023-01-15Superposition of pulses Paper 2Hard9 marks

On a test rig for a footbridge cable, two rectangular pulses travel along a long horizontal cable. Positions are measured along the cable as x.

Pulse A has displacement +4.0 cm and length 0.60 m, and moves in the +x direction at 2.0 m s−1. At t = 0 it occupies 0.40 m ≤ x ≤ 1.00 m.

Pulse B has displacement −2.5 cm and length 0.40 m, and moves in the −x direction at 2.0 m s−1. At t = 0 it occupies 3.00 m ≤ x ≤ 3.40 m.

0.511.522.533.54−2.54Ox / mdisplacement / cm2.0 m s⁻¹2.0 m s⁻¹AB
Pulses shown at t = 0
(a)

Calculate the time at which the pulses first begin to overlap.

(2)
(b)

Describe the displacement of the cable at t = 0.60 s.

(3)
(c)

Calculate the time at which the pulses have completely separated.

(2)
(d)

Pulse A later reflects from a rigidly fixed end of the cable. State the displacement and direction of motion of the reflected pulse.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Gap between leading edges = 3.00 − 1.00 = 2.00 m, closing at 4.0 m s−1✓ 1
t = 0.50 s✓ 1
Part (b)
A occupies 1.60–2.20 m and B occupies 1.80–2.20 m✓ 1
Overlap is 1.80–2.20 m, with displacement (+4.0) + (−2.5) = +1.5 cm✓ 1
From 1.60 m to 1.80 m the displacement is +4.0 cm; elsewhere it is zero✓ 1
Part (c)
Rear of A, 0.40 + 2.0t, reaches rear of B, 3.40 − 2.0t✓ 1
t = 3.00/4.0 = 0.75 s✓ 1
Part (d)
Displacement −4.0 cm (inverted)✓ 1
Travels in the −x direction at 2.0 m s−1, with unchanged length✓ 1

Common error: Track both the leading and trailing edges of each pulse. Overlap continues until the rear edges pass each other, not the front edges.

107IBP2S-022-01-15Young's double-slit experiment Paper 2Hard8 marks

A manufacturer characterises a red laser diode for a barcode scanner. The laser illuminates a double slit with slit separation d = 0.40 mm. The fringe spacing s is measured for several slit-to-screen distances D.

D / m: 1.00, 1.50, 2.00, 2.50, 3.00
s / mm: 1.62, 2.45, 3.22, 4.06, 4.85

(a)

Explain why a graph of s against D is expected to be a straight line through the origin.

(2)
(b)

Determine the wavelength of the laser light using the data.

(3)
(c)

The double slit is replaced by a diffraction grating with 300 lines per mm. Calculate the angle of the second-order maximum.

(2)
(d)

State one advantage of using the grating instead of the double slit to measure the wavelength.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
s = λD/d, and λ and d are constant✓ 1
So s is directly proportional to D, with gradient λ/d✓ 1
Part (b)
The gradient, using widely separated points, is (4.85 − 1.62)/(3.00 − 1.00) ≈ 1.62 mm m−1 = 1.62 × 10−3✓ 1
λ = gradient × d = 1.62 × 10−3 × 0.40 × 10−3✓ 1
λ ≈ 6.5 × 10−7 m (accept 640–650 nm)✓ 1
Part (c)
d = 1/300 mm = 3.33 × 10−6 m and sin θ = 2 × 6.46 × 10−7/3.33 × 10−6 = 0.388 (ECF from (b))✓ 1
θ = 22.8°✓ 1
Part (d)
The maxima are much sharper and brighter, and more widely spaced, so the angles can be measured more precisely.✓ 1

Exam tip: In data questions, find the gradient from points far apart, or from a best-fit line, rather than from a single data pair.

108IBP2H-023-00-11Optical fibre Paper 2Hard9 marks

A step-index fibre links two buildings on a campus. The fibre is 2.0 km long. The refractive index of the core is 1.48 and that of the cladding is 1.46. Take the speed of light in a vacuum as 3.00×108 m s−1.

core, n₁ = 1.48cladding, n₂ = 1.46cladding2.0 km
Diagram not to scale; ray angles exaggerated
(a)

Show that the critical angle at the core–cladding boundary is about 81°.

(2)
(b)

Calculate the time taken for light travelling along the axis of the fibre to cross the link.

(2)
(c)

A ray meets the core–cladding boundary at exactly the critical angle. Determine the difference in arrival time between this ray and the axial ray.

(3)
(d)

Estimate the maximum rate at which separate pulses can be sent along the link without overlapping at the receiver.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
sin c = 1.46/1.48 = 0.9865✓ 1
c = 80.6° ≈ 81°✓ 1
Part (b)
v = 3.00×108/1.48 = 2.027×108 m s−1✓ 1
t = 2000/2.027×108 = 9.87×10−6 s (accept 9.85×10−6 s to 9.87×10−6 s)✓ 1
Part (c)
Path length = L/sin c = 2000/0.9865 = 2027 m (or path ratio = ncore/ncladding = 1.0137)✓ 1
Time = 2027/2.027×108 = 1.0002×10−5 s✓ 1
Δt = 1.0002×10−5 − 9.867×10−6 = 1.35×10−7 s (accept 1.3×10−7 s to 1.4×10−7 s; ECF from (a) and (b))✓ 1
Part (d)
Each pulse spreads out by Δt, so successive pulses must be at least Δt apart in time✓ 1
Rate ≈ 1/Δt = 1/1.35×10−7 ≈ 7.4×106 pulses per second (accept 7×106 s−1 to 8×106 s−1; ECF from (c))✓ 1

Exam tip: Find Δt by subtracting two nearly equal times. Carry at least four significant figures in the intermediate values, or write Δt = taxial(ncore/ncladding − 1) so that no subtraction is needed.

109IBP2H-023-00-12Diffraction gratings Paper 2Hard9 marks

A spectroscope is used to check the output of a sodium street lamp. The lamp emits two close lines of wavelengths 589.0 nm and 589.6 nm. The light is incident normally on a grating with 500 lines per millimetre.

(a)

Calculate the line spacing of the grating.

(1)
(b)

Determine the highest order in which the 589.0 nm line can be observed.

(2)
(c)

Calculate the angle of the second-order maximum for the 589.0 nm line.

(2)
(d)

Determine the angular separation of the two lines in the second order.

(2)
(e)

Explain why the two lines are easier to resolve in the third order than in the second order, for the same illuminated region of the grating.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = 1/(5.00×105 m−1) = 2.00×10−6 m✓ 1
Part (b)
n ≤ d/λ = 2.00×10−6/589.0×10−9 = 3.40 (ECF from (a))✓ 1
The highest order is 3, because n must be a whole number and sin θ cannot exceed 1✓ 1
Part (c)
sin θ = 2 × 589.0×10−9/2.00×10−6 = 0.589✓ 1
θ = 36.1° (36.09°)✓ 1
Part (d)
Either: sin θ' = 2 × 589.6×10−9/2.00×10−6 = 0.5896, so θ' = 36.131°, and subtract the unrounded θ = 36.088°—
Or: use Δθ = nΔλ/(d cos θ) = 2 × 0.6×10−9/(2.00×10−6 × cos 36.1°)✓ 1
Δθ ≈ 0.043° (7.4×10−4 rad) (ECF from (c))✓ 1
Part (e)
Resolvance R = λ/Δλ = nN, so for the same number N of illuminated lines, the resolvance is proportional to the order n. It is 1.5 times larger in third order than in second order✓ 1
Resolving these lines needs R ≈ 589/0.6 ≈ 980, so about 330 illuminated lines are needed in third order, compared with about 490 in second order. Equivalently, the separation grows from 0.043° to about 0.11°, faster than the width of each maximum✓ 1

Common error: Subtracting two rounded angles such as 36.1° − 36.1° gives zero. Keep extra figures, or use the differential form of the grating equation.

110IBP2H-023-00-14Young's double-slit experiment Paper 2Hard8 marks

In a laboratory, laser light of wavelength 633 nm passes through a double slit. For each slit-to-screen distance D, a student measures the distance X spanned by 10 fringe spacings.

D / m: 1.00, 1.50, 2.00, 2.50
X / mm: 12.7, 19.0, 25.3, 31.6

laserDdouble slitdX (10 fringe spacings)
Diagram not to scale
(a)

Outline why the student measures 10 fringe spacings rather than one.

(1)
(b)

Determine the slit separation d, using the gradient of the data.

(3)
(c)

The central maximum of the single-slit diffraction envelope contains exactly 9 bright fringes, and the 5th-order fringes are missing. Determine the width b of each slit.

(2)
(d)

The slit width is halved while the slit separation stays the same. Describe the changes to the pattern.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
This reduces the percentage uncertainty in the fringe spacing, because the ruler's fixed absolute uncertainty is spread over a larger distance✓ 1
Part (b)
Gradient of X against D = (31.6 − 12.7)/(2.50 − 1.00) = 12.6 mm m−1 (accept 12.6–12.7 from a line of best fit)✓ 1
Divide by 10 to get one fringe spacing per metre of D: 1.26 mm m−1 = 1.26×10−3, and this equals λ/d✓ 1
d = 633×10−9/1.26×10−3 = 5.0×10−4 m✓ 1
Part (c)
The 5th-order fringe coincides with the first diffraction minimum, so λ/b = 5λ/d✓ 1
b = d/5 = 1.0×10−4 m (allow error carried forward from the value of d found above)✓ 1
Part (d)
The fringe spacing is unchanged, but the diffraction envelope doubles in width, so about 19 fringes now lie within the central maximum✓ 1
The overall intensity of the pattern decreases, because less light passes through each slit✓ 1

Exam tip: Remember to divide by 10 before using λ/d: the gradient here is for 10 fringe spacings, not one.

111IBP2H-023-02-10Double slit with diffraction envelope Paper 2Hard9 marks

A green laser of wavelength 532 nm illuminates a double slit. Each slit has width 0.040 mm and the centres of the slits are 0.20 mm apart. The pattern is viewed on a screen 1.80 m from the slits. Assume small angles throughout.

laser1.80 mdouble slitd = 0.20 mmslit width b = 0.040 mm
Diagram not to scale
(a)

Calculate the separation of adjacent bright interference fringes on the screen.

(2)
(b)

Calculate the width on the screen of the central maximum of the single-slit diffraction envelope.

(2)
(c)

Determine the number of bright fringes visible within the central diffraction maximum.

(2)
(d)

Describe how the intensity of the light varies with position across the central diffraction maximum. Refer to your earlier answers.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
s = λD/d = (532×10−9 × 1.80)/(0.20×10−3) . s = 4.8×10−3 m (4.79 mm) .✓ 2
Part (b)
Distance from the centre to the first diffraction minimum = λD/b = (532×10−9 × 1.80)/(0.040×10−3) = 23.9 mm . Full width = 2 × 23.9 mm ≈ 48 mm (47.9 mm) .✓ 2
Part (c)
d/b = 5 (or 47.9/4.79 = 10 fringe spacings across the envelope), so the 5th-order interference maxima fall on the first diffraction minima and are missing . The visible orders are 0, ±1, ±2, ±3 and ±4, giving 9 bright fringes . Allow error carried forward from the earlier parts.✓ 2
Part (d)
There is a bright fringe at the centre, with further bright fringes equally spaced about 4.8 mm apart and dark regions between them . The brightest fringe is at the centre, and the peak intensities decrease smoothly on each side, following the single-slit envelope . The intensity falls to zero at about ±24 mm, where the 5th-order fringes are missing, so 9 peaks lie within the central maximum and the ±4th-order peaks are the faintest .✓ 3

Exam tip: Find the ratio d/b first. It tells you straight away which interference order is missing and how many fringes fit inside the envelope. It also checks your fringe-separation and envelope-width results.

112IBP2H-023-02-15Double slit with diffraction envelope Paper 2Hard8 marks

A forensic laboratory checks a laser-etched double slit of unknown dimensions. A laser of wavelength 650 nm illuminates the slits and the pattern is observed on a screen 2.00 m away. The distance across 10 fringe spacings is measured as 26 mm. The fourth-order bright fringes on both sides of the centre are absent.

(a)

Calculate the slit separation d.

(2)
(b)

Explain why the fourth-order bright fringes are absent.

(2)
(c)

Determine the width b of each slit.

(2)
(d)

Calculate the angular width of the central maximum of the diffraction envelope.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Fringe spacing s = 26/10 = 2.6 mm . d = λD/s = (650×10−9 × 2.00)/(2.6×10−3) = 5.0×10−4 m .✓ 2
Part (b)
The fourth-order interference maximum is at the same angle as the first minimum of the single-slit diffraction pattern of each slit . Each slit sends no light in that direction, so there is no light to interfere constructively and the fringe is missing .✓ 2
Part (c)
Equate the angles of the first diffraction minimum and the fourth interference maximum: λ/b = 4λ/d . b = d/4 = 1.25×10−4 m, or 1.3×10−4 m . Allow follow-through from the slit separation calculated in the first part.✓ 2
Part (d)
Half-angle θ = λ/b = 650×10−9/1.25×10−4 = 5.2×10−3 rad . Using b = 1.3×10−4 m gives 5.0×10−3 rad, which is also accepted. Full angular width = 2θ ≈ 1.0×10−2 rad . Accept 1.04×10−2 rad. Allow follow-through from the slit width found in the previous part.✓ 2

Common error: Divide the measured distance by the number of spacings, not the number of fringes. 10 spacings span 11 bright fringes.

113IBP3H-C3-20r25-15Double slit with diffraction envelope Paper 2Hard9 marks

In a double-slit experiment using light of wavelength 600 nm, each slit has width 0.10 mm and the distance between the centres of the slits is 0.30 mm. The screen is 1.5 m from the slits.

(a)

Outline what is meant by a missing order in a double-slit pattern.

(2)
(b)

Determine which orders are missing.

(2)
(c)

Determine how many bright fringes are seen inside the central diffraction maximum.

(3)
(d)

Calculate the width of the central diffraction maximum on the screen, in mm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A bright fringe expected from the double-slit interference is absent✓ 1
Because it falls at a position of zero intensity in the single-slit diffraction pattern (a minimum of the envelope)✓ 1
Part (b)
Missing when d sin θ = mλ and a sin θ = λ coincide, so m = d/a = 3✓ 1
Orders 3, 6, 9, ... are missing✓ 1
Part (c)
The central diffraction maximum extends to the first minimum, at m = 3✓ 1
Fringes with m = 0, ±1, ±2 lie inside it✓ 1
5 bright fringes✓ 1
Part (d)
Use of width = 2λD/a✓ 1
= 2 × 600×10−9 × 1.5 / 0.10×10−3 = 18 mm✓ 1

Exam tip: Compare the ratio d/a with whole numbers to find missing orders, then use that ratio to count the fringes inside the envelope.