IB Physics HL · first assessment 2025 · Theme C

C.3 Wave phenomena: IB Physics HL exam-style questions

Wave phenomena covers what happens when waves meet a boundary, an obstacle or each other: reflection, refraction with Snell's law and the critical angle, superposition and two-source interference with the path-difference conditions, and Young's double-slit experiment.

HL adds single-slit diffraction, with the first minimum at θ = λ/b, the double-slit pattern inside a diffraction envelope, and multiple slits and diffraction gratings, nλ = d sin θ. This page has the largest set of HL questions in Theme C.

  • 66 questions
  • 322 marks
  • Paper 1A: 36
  • Paper 1B: 11
  • Paper 2: 19
  • Full mark schemes

Showing 66 of 66 questions · 322 marks

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40 practice questions on C.3 Wave phenomena

1C-1A-04
Single-slit diffraction·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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.

Syllabus understandingC.3 (HL) — single-slit diffraction including intensity patterns as given by θ = λ/b where b is the slit width Command term: Determine

2C-1A-05
Double slit with diffraction envelope·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksPredict

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.

Syllabus understandingC.3 (HL) — single-slit diffraction as given by θ = λ/b; the modulation of the two-slit interference pattern by the one-slit diffraction effect; C.3 — Young's double-slit interference as given by s = λD/d Command term: Predict

3C-1A-06
Diffraction gratings·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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.

Syllabus understandingC.3 (HL) — interference patterns from multiple slits and diffraction gratings as given by nλ = d sin θ Command term: Determine

4C-1A-07
Refraction: wavelength and frequency·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark4 steps to full marksDetermine

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.
Show mark scheme
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.

Syllabus understandingC.3 — Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1; wavefronts and rays; C.2 — the wave equation v = fλ with the frequency unchanged at a boundary Command term: Determine

5C-1A-11
Refraction and total internal reflection·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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.
Show mark scheme
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.

Syllabus understandingC.3 — Snell's law as given by n1/n2 = sin θ2/sin θ1; the critical angle and total internal reflection Command term: Determine

6C-1A-13
Two-source interference·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce

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?

Show mark scheme
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.

Syllabus understandingC.3 — the conditions for constructive and destructive interference in terms of path difference: nλ and (n + ½)λ Command term: Deduce

7C-1A-18
Two-source interference·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce

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.

Syllabus understandingC.3 — superposition of waves and wave pulses; the condition for constructive interference as given by path difference = nλ Command term: Deduce

8C-1A-24
Young's double-slit experiment·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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.

Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d where s is the separation of fringes Command term: Determine

9C-1A-28
Refraction of wavefronts·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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.
Show mark scheme
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.

Syllabus understandingC.3 — wavefronts and rays; Snell's law as given by n1/n2 = v2/v1; C.2 — the wave equation v = fλ Command term: Determine

10C-1A-41
Two-source interference·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce

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.

Syllabus understandingC.3 — the condition for constructive interference as given by path difference = nλ; the condition for destructive interference as given by path difference = (n + ½)λ; double-source interference requires coherent sources Command term: Deduce

11C-1A-42
Single-slit diffraction·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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.

Syllabus understandingC.3 (HL) — single-slit diffraction including intensity patterns as given by θ = λ/b where b is the slit width Command term: Determine

12C-1A-43
Diffraction around bodies and apertures·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksPredict

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
Show mark scheme
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.

Syllabus understandingC.3 — wave diffraction around a body and through an aperture; wavefront-ray diagrams showing refraction and diffraction Command term: Predict

13C-1A-44
Snell's law·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme
Marking pointMarkNotes
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°).

Syllabus understandingC.3 — Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1 where n is the refractive index and θ is the angle between the normal and the ray Command term: Determine

14C-1A-45
Diffraction gratings·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce

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.

Show mark scheme
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.

Syllabus understandingC.3 (HL) — interference patterns from multiple slits and diffraction gratings as given by nλ = d sin θ Command term: Deduce

15C-1A-46
Superposition of pulses·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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.
Show mark scheme
Marking pointMarkNotes
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.

Syllabus understandingC.3 — superposition of waves and wave pulses Command term: Determine

16C-1A-47
Coherence·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksExplain

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?

Show mark scheme
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.

Syllabus understandingC.3 — that double-source interference requires coherent sources Command term: Explain

17C-1A-48
Two-source interference·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme
Marking pointMarkNotes
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.

Syllabus understandingC.3 — the condition for destructive interference as given by path difference = (n + ½)λ; the condition for constructive interference as given by path difference = nλ Command term: Determine

18C-1A-49
Refraction and total internal reflection·C.3 Wave phenomena
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce

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
Show mark scheme
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.

Syllabus understandingC.3 — Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1; critical angle and total internal reflection Command term: Deduce

19C-1A-50
Diffraction gratings·C.3 Wave phenomena
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme
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.

Syllabus understandingC.3 (HL) — interference patterns from multiple slits and diffraction gratings as given by nλ = d sin θ; patterns produced from a range of monochromatic wavelengths Command term: Determine

20C-1A-61
Reflection, refraction and transmission at boundaries·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce

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)
Show mark scheme
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.

Syllabus understandingC.3 — wave behaviour at boundaries in terms of reflection, refraction and transmission; C.4 — boundary conditions for strings: fixed and free boundaries (nodes and antinodes) Command term: Deduce

21C-1A-62
Young's double-slit experiment·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme
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.

Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d; Snell's law n1/n2 = v2/v1; C.2 — the wave equation v = fλ Command term: Determine

22C-1A-67
Diffraction gratings·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce

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.

Show mark scheme
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).

Syllabus understandingC.3 (HL) — interference patterns from multiple slits and diffraction gratings as given by nλ = d sin θ; diffraction grating patterns produced from white light Command term: Deduce

23C-1A-69
Two-source interference·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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.

Syllabus understandingC.3 — superposition of waves; the condition for destructive interference as given by path difference = (n + ½)λ; intensity ∝ amplitude² Command term: Determine

24C-1A-71
Optical fibre·C.3 Wave phenomena
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

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.
Show mark scheme
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.

Syllabus understandingC.3 — Snell's law as given by n1/n2 = sin θ2/sin θ1; critical angle and total internal reflection Command term: Determine

25C-1A-99
Diffraction around a body·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce

Straight water waves of wavelength λ travel across a lake and pass a small rocky island whose width is similar to λ.

Which statements about the waves beyond the island are correct?

I. The waves spread into the region directly behind the island.

II. The wavelength of the waves directly behind the island is less than λ.

III. If the island were the same size but λ were larger, the waves would spread further into the region behind it.

Show mark scheme
Marking pointMarkNotes
Step 1Diffraction: when waves pass the edges of a body, they spread into the region behind it (I is correct).—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Diffraction does not change the speed or the frequency of the waves, so the wavelength is unchanged (II is wrong).—
Step 3The spreading is most noticeable when the wavelength is similar to or larger than the size of the body; for a fixed body, a longer wavelength spreads further behind it (III is correct).✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis recognises the spreading but overlooks that the amount of diffraction depends on the ratio of the wavelength to the size of the body.
  • BThis treats diffraction as if it were refraction, assuming that a change of direction must come with a change of wavelength. The medium is unchanged, so the speed and the wavelength are unchanged.
  • CCorrect: the waves bend into the region behind the island with unchanged wavelength, and the effect grows as λ becomes larger compared with the size of the island.
  • DThis accepts III but wrongly includes II: the waves behind the island are still in the same water, with the same speed and frequency, so λ is unchanged.

Syllabus understandingC.3 — wave diffraction around a body and through an aperture; that waves travelling in two and three dimensions can be described through the concepts of wavefronts and rays Command term: Deduce

26C-1A-100
Wavefront-ray diagram of refraction·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify

Straight water waves travel from shallow water into deep water, where they travel faster. The boundary between the two regions is straight.

Which diagram shows the wavefronts and the ray correctly?

shallowdeepAshallowdeepBshallowdeepCshallowdeepD
Plan views of wavefronts and a ray crossing the boundary from shallow into deep water. The dashed line is the normal.
Show mark scheme
Marking pointMarkNotes
Step 1The frequency is set by the source and does not change at the boundary, so λ = v/f increases in the deep water, where v is larger: the wavefronts are further apart.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Each wavefront must be continuous across the boundary. The end of a wavefront in the deep water advances further in the same time, so the wavefronts swing round and the ray bends away from the normal (sin θ2/sin θ1 = v2/v1 > 1).—
Step 3Only diagram A shows both the larger spacing and the bending away from the normal.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: the waves speed up, so the wavelength increases and the ray bends away from the normal, with each wavefront joined across the boundary.
  • BThis shows the change of wavelength but no change of direction. With a larger spacing at the same angle the wavefronts cannot join at the boundary, so this is impossible for waves arriving at an angle.
  • CThis is the diagram for waves slowing down (deep into shallow): the wavelength decreases and the ray bends towards the normal. The waves here speed up.
  • DThis assumes that neither the wavelength nor the direction changes; the speed changes at the boundary, so the wavelength must change.

Syllabus understandingC.3 — wavefront-ray diagrams showing refraction and diffraction; wave behaviour at boundaries in terms of reflection, refraction and transmission; Guidance: sketching and interpreting wavefronts and rays (incident, reflected and transmitted waves) Command term: Identify

27C-1A-101
Transmission and reflection at a boundary·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce

A light rope is joined to a heavy rope, and the ropes are stretched in a straight line. Waves travel more slowly on the heavy rope than on the light rope. An upward pulse travels along the light rope towards the join.

Which row describes the pulse reflected at the join and the pulse transmitted into the heavy rope?

Reflected pulseTransmitted pulse
Show mark scheme
Marking pointMarkNotes
Step 1The heavy rope resists the motion of the join, so the join behaves partly like a fixed end: the reflected pulse is inverted.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The transmitted pulse is driven by the upward movement of the join, so it is upright.—
Step 3The pulse takes the same time to pass the join on both ropes; it travels more slowly on the heavy rope, so its length (speed × duration) is shorter.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThis inverts the wrong pulse. The heavy rope is pulled upwards by the join, so the transmitted pulse has the same sign as the incident pulse.
  • BThis treats the join as a free end, where the reflected pulse is upright. That happens when the second medium is lighter (faster), not heavier.
  • CThis takes a slower wave to give a longer pulse. The duration of the pulse is unchanged, so a lower speed gives a shorter pulse.
  • DCorrect: the reflected pulse is inverted, as at a fixed end, and the transmitted pulse is upright and shorter because it travels more slowly for the same duration.

Syllabus understandingC.3 — wave behaviour at boundaries in terms of reflection, refraction and transmission; superposition of waves and wave pulses Command term: Deduce

28C-1A-102
Superposition of pulses·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce

Two pulses of identical shape travel towards each other along a stretched rope. One pulse is upright and the other is inverted. At one instant the pulses overlap exactly and the whole rope is straight.

Which statement about the rope at this instant is correct?

Show mark scheme
Marking pointMarkNotes
Step 1By the principle of superposition the displacements add: the upright and inverted displacements cancel, so the rope is straight in the overlap region.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The velocities of the points do not cancel. In each pulse the leading edge is moving away from equilibrium and the trailing edge back towards it; for an upright pulse moving one way and an inverted pulse moving the other, the velocities at each point add.—
Step 3So the points in the overlap region move with their greatest speed: the energy is stored as kinetic energy, and the pulses then pass through each other unchanged.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis takes zero displacement to mean zero energy. Energy is conserved; at this instant it is all kinetic.
  • BCorrect: the displacements cancel but the velocities add, so the energy of the pulses is kinetic energy of the rope at this instant.
  • CThis confuses zero displacement with zero velocity. The rope passes through the straight position at high speed.
  • DThis assumes destructive superposition destroys the pulses. Superposition is temporary: each pulse continues unchanged after the overlap.

Syllabus understandingC.3 — superposition of waves and wave pulses Command term: Deduce

29C-1A-103
Slit width and the intensity pattern·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDeduce

Monochromatic light passes through a single narrow slit and forms a diffraction pattern on a distant screen. The width of the slit is doubled; the light source is unchanged.

Which statements about the new pattern are correct?

I. The central maximum is narrower.

II. The intensity at the centre of the pattern is greater.

III. The secondary maxima are further apart.

Show mark scheme
Marking pointMarkNotes
Step 1The first minimum is at θ = λ/b, so doubling b halves the angular width of the central maximum (I is correct) and halves the spacing of the other minima, so the secondary maxima are closer together (III is wrong).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2A wider slit lets more light through, and this light is concentrated into a narrower central maximum, so the intensity at the centre increases (II is correct).✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the angles of all the minima are halved, and more light falls into a narrower central maximum, so the centre is brighter.
  • BThis accepts that the central maximum narrows but has the secondary maxima moving apart. All the minima are at multiples of λ/b, so they all move closer together.
  • CThis has the pattern spreading out as the slit widens, as if θ were proportional to b, while still accepting the brighter centre.
  • DThis includes III: the spacing of all features of the pattern is proportional to λ/b, so the secondary maxima move closer together, not further apart.

Syllabus understandingC.3 (HL) — single-slit diffraction including intensity patterns as given by θ = λ/b; Guidance: the effect of slit width on the intensity of the single-slit diffraction pattern (qualitative) Command term: Deduce

30C-1A-104
Double-slit pattern and the single-slit envelope·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

Monochromatic light passes through two identical slits, each of width b, whose centres are a distance d apart. The graph shows how the intensity of the light on a distant screen varies with the position y on the screen.

What is d/b?

-12-10-8-6-4-2024681012y / mm0.00.20.40.60.81.0relative intensity
Relative intensity on the screen against position y (drawn to scale).
Show mark scheme
Marking pointMarkNotes
Step 1The bright fringes are λD/d apart: from the graph, 2.0 mm.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The intensity envelope is the single-slit pattern of one slit; its first minimum is at λD/b from the centre: from the graph, 8.0 mm.—
Step 3Dividing: (λD/b)/(λD/d) = d/b = 8.0/2.0 = 4.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis inverts the ratio, dividing the fringe spacing by the distance to the first minimum of the envelope.
  • BCorrect: the envelope falls to zero at four fringe spacings from the centre, so d/b = 4 (the fourth bright fringes are missing).
  • CThis counts the bright fringes inside the central maximum of the envelope (7) instead of comparing distances.
  • DThis uses the full width of the central maximum of the envelope (16 mm) instead of the distance from its centre to its first minimum.

Syllabus understandingC.3 (HL) — that the single-slit pattern modulates the double slit interference pattern; single-slit diffraction including intensity patterns as given by θ = λ/b; C.3 — Young's double-slit interference as given by s = λD/d Command term: Determine

31C-1A-105
Grating: angular separation of two lines·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

Light from a mercury lamp contains two lines of wavelength 546.1 nm and 577.0 nm. The light is incident normally on a diffraction grating with 400 lines per mm.

What is the angle between the two lines in the second-order spectrum?

Show mark scheme
Marking pointMarkNotes
Step 1d = 1/(400 × 103) = 2.50 × 10−6 m.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2sin θ = 2λ/d: for 546.1 nm, sin θ = 0.4369, θ = 25.90°; for 577.0 nm, sin θ = 0.4616, θ = 27.49°.—
Step 3Angle between the lines = 27.49° − 25.90° = 1.59°.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThis is the separation in the first order (13.34° − 12.62°): the order n = 2 has been left out.
  • BThis treats sin θ as equal to θ (in radians) at all angles: 2 × (30.9 × 10−9)/(2.50 × 10−6) = 0.0247 rad. At about 27° the small-angle approximation underestimates the separation.
  • CThis doubles the first-order separation, assuming that the angles themselves are proportional to the order. It is sin θ that is proportional to n.
  • DCorrect: the two second-order angles are 25.90° and 27.49°, which differ by 1.59°.

Syllabus understandingC.3 (HL) — interference patterns from multiple slits and diffraction gratings as given by nλ = d sin θ; Guidance: multiple slit and diffraction grating patterns produced from white light and a range of monochromatic light wavelengths Command term: Determine

32C-1A-106
Young's fringes with white light·C.3 Wave phenomena
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

White light containing all wavelengths from 400 nm to 700 nm is incident on a double slit. Point P on the screen is the centre of the third bright fringe from the central fringe for light of wavelength 650 nm.

Which wavelengths in the white light have a minimum of intensity at P?

Show mark scheme
Marking pointMarkNotes
Step 1At the third bright fringe the path difference is 3 × 650 nm = 1950 nm.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2A minimum needs 1950 nm = (n + ½)λ, so λ = 1950/(n + ½) nm: 3900, 1300, 780, 557, 433, 355 … nm.—
Step 3Only 557 nm (n = 3) and 433 nm (n = 4) lie between 400 nm and 700 nm.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: the path difference 1950 nm is an odd number of half-wavelengths for 557 nm (3.5λ) and 433 nm (4.5λ), and for no other wavelength in the range.
  • BThis uses the condition for a maximum, 1950 nm = nλ, giving 650 nm and 488 nm; these wavelengths are brightest at P.
  • CThis counts the central fringe as the first bright fringe, taking the path difference as 2 × 650 = 1300 nm; 1300/2.5 = 520 nm is then the only value in the range.
  • DThis includes 780 nm (2.5λ = 1950 nm), which satisfies the condition but is not in the white light, which stops at 700 nm.

Syllabus understandingC.3 — the condition for destructive interference as given by path difference = (n + ½)λ; Young's double-slit interference as given by s = λD/d; Guidance: multiple slit and diffraction grating patterns produced from white light and a range of monochromatic light wavelengths Command term: Determine

33C-1A-107
Two sources with a phase difference·C.3 Wave phenomena
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce

Two loudspeakers S1 and S2 are connected to the same signal generator and emit sound of the same frequency. A listener walks along a line parallel to S1S2 and some distance from it, and hears loud and quiet points. An electronic delay is then added so that the sound from S2 lags the sound from S1 by one quarter of a period.

Which row describes the loud point nearest to the line through the midpoint of S1S2, and the separation of adjacent loud points, after the delay is added?

Loud point nearest the midpoint lineSeparation of loud points
Show mark scheme
Marking pointMarkNotes
Step 1On the line through the midpoint the paths are equal, so the waves now arrive a quarter of a period out of step: this is no longer a loud point.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2For the waves to arrive in phase, the wave from S2 must travel a quarter-wavelength less than the wave from S1 to make up for its late start, so the loud point is closer to S2.—
Step 3The whole pattern shifts by a quarter of the fringe separation. The separation depends on λ and on the geometry, neither of which changes, so it is unchanged.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis ignores the delay: with a constant phase difference between the sources, equal paths no longer give waves in phase.
  • BThis has the direction reversed. The wave from S2 starts late, so it must have the shorter path, which moves the loud point towards S2.
  • CCorrect: the loud point moves towards the lagging source S2 by a quarter of the fringe separation, and the separation is unchanged.
  • DThis shifts the pattern correctly but also changes the separation. A constant phase difference shifts the pattern without changing λ or the geometry, so the separation is unchanged.

Syllabus understandingC.3 — that double-source interference requires coherent sources; the condition for constructive interference as given by path difference = nλ; superposition of waves and wave pulses Command term: Deduce

34C-1A-108
Snell's law in terms of wavelength·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

Straight water waves cross a straight boundary between two regions of a ripple tank. The wavelength is λ1 in the first region and λ2 in the second. The ray in the first region makes an angle θ1 with the normal to the boundary.

What is the angle θ2 between the ray in the second region and the normal?

Show mark scheme
Marking pointMarkNotes
Step 1The frequency is the same on both sides of the boundary, so v ∝ λ and v2/v1 = λ2/λ1.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Snell's law: sin θ2/sin θ1 = v2/v1 = λ2/λ1.—
Step 3So sin θ2 = (λ2/λ1) sin θ1.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThis inverts the ratio, as if the wave bent away from the normal when it slowed down.
  • BThis applies the ratio to the angles instead of to their sines.
  • CThis uses the cosines, which would be correct only if the angles were measured from the boundary instead of from the normal.
  • DCorrect: sin θ2/sin θ1 = v2/v1 = λ2/λ1 because the frequency is unchanged.

Syllabus understandingC.3 — Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1; that waves travelling in two and three dimensions can be described through the concepts of wavefronts and rays Command term: Determine

35C-1A-109
Fringe spacing and screen distance·C.3 Wave phenomena
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine

In a double-slit experiment with slit separation d, the fringe spacing on the screen is s. The distance from the slits to the screen is not known. The screen is then moved a further distance x away from the slits, and the fringe spacing becomes s′.

What is the wavelength of the light?

Show mark scheme
Marking pointMarkNotes
Step 1s = λD/d and s′ = λ(D + x)/d.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Subtracting removes the unknown D: s′ − s = λx/d.—
Step 3λ = d(s′ − s)/x.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis adds the two equations instead of subtracting them, so the unknown D does not cancel.
  • BCorrect: the change in fringe spacing is λx/d, so λ = d(s′ − s)/x.
  • CThis subtracts correctly but then rearranges s′ − s = λx/d wrongly, multiplying by x and dividing by d instead of multiplying by d and dividing by x.
  • DThis treats x as the whole slit-to-screen distance for the second position, ignoring the original unknown distance.

Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d Command term: Determine

36C-1A-110
Superposition of two waves·C.3 Wave phenomena
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDetermine

Two waves travel in the same direction along a long rope. The graph shows the displacement each wave would produce on its own, at one instant, against position x along the rope.

What is the displacement of the rope at point P at this instant?

0.00.10.20.30.40.50.60.70.80.91.0x / m-4-3-2-101234displacement / mmPwave 1wave 2
Displacement against position for the two waves at one instant (drawn to scale).
Show mark scheme
Marking pointMarkNotes
Step 1From the graph, at P wave 1 has displacement −3.0 mm and wave 2 has displacement +1.0 mm.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2By the principle of superposition the displacements add: −3.0 + 1.0 = −2.0 mm.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThis adds the two amplitudes (3.0 + 2.0) and gives the result the sign of wave 1; the instantaneous displacements, not the amplitudes, add.
  • BThis adds the sizes of the two displacements (3.0 + 1.0), ignoring that wave 2 is above the axis at P.
  • CCorrect: −3.0 mm + 1.0 mm = −2.0 mm.
  • DThis subtracts the amplitudes (3.0 − 2.0) instead of adding the displacements at P.

Syllabus understandingC.3 — superposition of waves and wave pulses Command term: Determine

37C-1B-02
Diffraction gratings·C.3 Wave phenomena
Paper 1BHard7 marks
Data-based question5 steps to full marksDetermine

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)

Syllabus understandingC.3 (HL) — the diffraction grating equation nλ = d sin θ; E.1 — emission spectra and the photon energy of transitions between discrete energy levels; Tools — identifying and correcting a zero (systematic) error Command term: Determine

38C-1B-04
Snell's law·C.3 Wave phenomena
Paper 1BMedium7 marks
Data-based question5 steps to full marksDeduce

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
(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)

Syllabus understandingC.3 — Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1; critical angle and total internal reflection; Tools — testing a hypothesis with several data points, propagating uncertainty through a sine Command term: Deduce

39C-1B-05
Single-slit diffraction·C.3 Wave phenomena
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine

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)

Syllabus understandingC.3 (HL) — single-slit diffraction as given by θ = λ/b; Tools — linearising a relationship, gradient with its unit, a zero error identified from an intercept Command term: Determine

40C-1B-10
Young's double-slit experiment·C.3 Wave phenomena
Paper 1BEasy7 marks
Data-based question6 steps to full marksDetermine

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)

Syllabus understandingC.3 — double-source interference as given by s = λD/d; Tools — gradient, extrapolation to an intercept, systematic error, reducing a percentage uncertainty Command term: Determine

41C-1B-16
Refraction and total internal reflection·C.3 Wave phenomena
Paper 1BEasy7 marks
Data-based question5 steps to full marksDetermine

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)

Syllabus understandingC.3 — Snell's law n1/n2 = sin θ2/sin θ1 = v2/v1 and refractive index; Tools — gradient with its unit, uncertainty of a gradient from error bars, a systematic offset read from an intercept Command term: Determine

42C-1B-17
Young's double-slit experiment·C.3 Wave phenomena
Paper 1BHard7 marks
Data-based question5 steps to full marksDetermine

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)

Syllabus understandingC.3 — double-source interference s = λD/d; E.1 — photon energy E = hf; B.5/D.2 — energy transferred per unit charge, W = qV; Tools — linearising, gradient with its unit, comparison with an accepted value Command term: Determine

43C-1B-18
Refraction and total internal reflection·C.3 Wave phenomena
Paper 1BMedium7 marks
Data-based question5 steps to full marksDetermine

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)

Syllabus understandingC.3 — Snell's law, critical angle and total internal reflection; Tools — calibration curves, propagating uncertainty through a non-linear function, gradient with its unit Command term: Determine

44C-1B-19
Single-slit diffraction·C.3 Wave phenomena
Paper 1BHard7 marks
Data-based question5 steps to full marksDetermine

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)

Syllabus understandingC.3 — diffraction through an aperture; E.2 (HL) — matter waves, the de Broglie wavelength λ = h/p; D.2 — work done on a charge W = qV; Tools — linearising, gradient with its unit Command term: Determine

45C-1B-32
Fringe spacing and slit separation·C.3 Wave phenomena
Paper 1BEasy6 marks
Data-based question6 steps to full marksDetermine

A student investigates how the fringe spacing s in a double-slit interference pattern depends on the slit separation d. A slide carrying six double slits of known separation is placed in the beam of a red laser, and the pattern is viewed on a screen at a fixed distance D = 1.500 m from the slide. For each double slit the student measures the distance across ten fringe spacings, 10s, with a ruler. Each value of 10s has an uncertainty of ±0.5 mm.

The graph shows s against 1/d with the line of best fit.

d / mm0.1500.2000.2500.3000.4000.500
10s / mm64.147.138.531.424.118.8
(1/d) / mm−16.675.004.003.332.502.00
01234567(1/d) / mm⁻¹01234567s / mm
Fringe spacing s against 1/d with the line of best fit (drawn to scale).
(a)
(i)

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

(1)
(ii)

Calculate the percentage uncertainty in s for d = 0.500 mm.

(1)
(b)
(i)

Determine the gradient of the line of best fit. State its unit.

(2)
(c)
(i)

Determine the wavelength of the laser light.

(1)
(ii)

Explain whether the data support the relationship s = λD/d.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The absolute uncertainty (±0.5 mm) is the same, but it is a smaller fraction of a longer distance, so the percentage uncertainty in s is reduced (by a factor of ten)✓ 1Accept "more accurate/precise" only if linked to a smaller percentage or fractional uncertainty. "To get an average" alone: [0].
Part (a)(ii)
0.5/18.8 × 100 = 2.7 %✓ 1Accept 2.6–2.7 %. Using 0.5/1.88: [0].
Part (b)(i)
Gradient from a large triangle on the line, e.g. (6.70 − 0.93)/(7.0 − 1.0) = 0.96✓ 1Accept 0.93–0.99. Two adjacent data points only: [0] for this mark.
Unit: mm2✓ 1Accept m2 with the value converted (9.6 × 10−7 m2).
Part (c)(i)
λ = gradient/D = 0.96 × 10−6/1.500 = 6.4 × 10−7 m (641 nm)✓ 1Allow ECF from (b). Accept 620–660 nm. Gradient not converted from mm2: [0].
Part (c)(ii)
The points lie on a straight line that passes through the origin (within the uncertainties), so s ∝ 1/d, as the relationship predicts✓ 1Both straight line and through the origin needed. Allow ECF from the line of best fit in (b).

Answers: (a)(ii) 2.7 %  ·  (b)(i) 0.96 mm2  ·  (c)(i) 641 nm (the remaining parts are explanations — see the table above)

Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d; Tools — percentage uncertainty, gradient of a graph with units, testing a proportional relationship Command term: Determine

46C-1B-33
Speed of sound from two-source interference·C.3 Wave phenomena
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine

Two loudspeakers, d = 2.00 m apart, are connected to the same signal generator and emit sound in phase. A microphone is moved along a line parallel to the line joining the speakers and D = 4.00 m from it. For each frequency f of the generator, a student measures the distance w between the second quiet point on one side of the central loud point and the second quiet point on the other side. Each value of w has an uncertainty of ±0.02 m.

The fringe separation is given by s = vD/(fd), where v is the speed of sound. The graph shows w against 1/f with error bars and the line of best fit.

f / kHz2.02.53.03.54.05.0
w / m1.050.820.700.580.530.41
(1/f) / 10−4 s5.004.003.332.502.00
0123456(1/f) / 10⁻⁴ s0.00.20.40.60.81.01.2w / m
Distance w against 1/f, with error bars and the line of best fit (drawn to scale).
(a)
(i)

Show that w = 3s.

(1)
(ii)

Calculate the missing value of 1/f.

(1)
(b)
(i)

Determine the gradient of the line of best fit. State its unit.

(2)
(c)
(i)

Determine the speed of sound.

(2)
(ii)

A teacher suggests that the sound at the quiet points is never quite zero. Suggest one reason for this.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Quiet points are at ½s, 1½s, … from the centre, so the second quiet points are 1.5s from the centre on each side and w = 2 × 1.5s = 3s✓ 1Award for a clear statement or sketch locating the second quiet points at ±1.5s.
Part (a)(ii)
1/(3.5 × 103) = 2.86 × 10−4 s✓ 1Accept 2.86 or 2.857.
Part (b)(i)
Gradient from a large triangle = 0.2105 m per 10−4 s = 2105 m s (2.11 × 103 m s)✓ 1Accept 0.200–0.220 m per 10−4 s (2000–2200 m s).
Unit: m s✓ 1Accept m Hz−1.
Part (c)(i)
Gradient = 3vD/d, so v = gradient × d/(3D)✓ 1Allow ECF from (a)(i). Omitting the factor 3 (taking the gradient as vD/d): [0] for this mark.
v = 2105 × 2.00/(3 × 4.00) = 351 m s−1✓ 1Allow ECF from (b). Accept 333–367 m s−1.
Part (c)(ii)
Any one of: sound reflected from the walls, floor or ceiling also reaches the microphone; the two waves have different amplitudes at a quiet point because the distances to the speakers differ, so they do not cancel completely; background noise✓ 1Award one valid reason. "Experimental error" alone: [0].

Answers: (a)(ii) 2.86 × 10−4 s  ·  (b)(i) 2.11 × 103 m s  ·  (c)(i) 351 m s−1 (the remaining parts are explanations — see the table above)

Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d; the condition for destructive interference as given by path difference = (n + ½)λ; that double-source interference requires coherent sources; C.2 — the nature of sound waves; Tools — gradient of a graph with units, error bars, identifying sources of error Command term: Determine

47C-1B-34
Grating spacing from spots on a wall·C.3 Wave phenomena
Paper 1BHard7 marks
Data-based question5 steps to full marksDetermine

A green laser of wavelength 532 nm is directed normally at a diffraction grating. The spots of the diffraction pattern fall on a flat wall parallel to the grating, at a distance L = 0.400 m from it. Because the position of the central spot is hard to mark, for each order n the student measures the distance 2x between the two spots of that order on opposite sides of the centre.

The student calculates sin θ for each order, where θ is the angle between the straight-through direction and the beam of order n. The graph shows sin θ against n.

n2x / msin θ
10.1310.162
20.2660.316
30.438
40.6670.640
51.0550.797
0123456n0.00.20.40.60.81.0sin θ
sin θ against order n, with the line of best fit (drawn to scale).
(a)
(i)

Determine the missing value of sin θ for n = 3.

(2)
(ii)

Explain why the student plots sin θ, not x, against n.

(1)
(b)
(i)

Determine the number of lines per millimetre of the grating.

(2)
(c)
(i)

L has an uncertainty of ±0.005 m and each value of 2x has an uncertainty of ±0.005 m. Using the fifth-order data only, determine the absolute uncertainty in the number of lines per millimetre.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
x = 0.438/2 = 0.219 m✓ 1Using 2x = 0.438 m as the distance from the centre: [0] for this mark (gives 0.738).
sin θ = x/√(x2 + L2) = 0.219/√(0.2192 + 0.4002) = 0.480✓ 1Allow ECF from MP1. Accept 0.480.
Part (a)(ii)
nλ = d sin θ, so sin θ is proportional to n (a straight line through the origin); x = L tan θ, and tan θ is not proportional to sin θ at these large angles, so x against n is not linear✓ 1Must refer to the grating equation and to the non-linear relation between x and θ.
Part (b)(i)
Gradient = λ/d = 0.160 (line of best fit through the origin: 0.1596)✓ 1Accept 0.157–0.162.
Lines per metre = gradient/λ = 0.1596/(532 × 10−9) = 3.00 × 105 m−1, so 300 lines per mm✓ 1Allow ECF from the gradient (0.160 gives 301). Accept 295–305 lines per mm. Answer per metre not converted: [1 max].
Part (c)(i)
Largest sin θ = 0.5300/√(0.53002 + 0.3952) = 0.8018 and smallest sin θ = 0.5250/√(0.52502 + 0.4052) = 0.7918✓ 1Award for both extremes with x halved: x = 0.5275 ± 0.0025 m. Allow a fractional-uncertainty method that combines the uncertainties in x and L.
Lines per mm = sin θ/(5λ): from 298 to 301, so uncertainty ≈ ±2 lines per mm✓ 1Allow ECF from their extremes. Accept ±1 to ±3 lines per mm. Allow ECF from (b).

Answers: (a)(i) 0.480  ·  (b)(i) 300 lines per mm  ·  (c)(i) ±2 lines per mm (the remaining parts are explanations — see the table above)

Syllabus understandingC.3 (HL) — interference patterns from multiple slits and diffraction gratings as given by nλ = d sin θ; Tools — linearising a relationship, gradient of a graph, propagating uncertainty Command term: Determine

48C-2-02
Double slit with diffraction envelope·C.3 Wave phenomena
Paper 2Hard9 marks
Short answer & extended response6 steps to full marksDeduce

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)

Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d; C.3 (HL) — single-slit diffraction as given by θ = λ/b; that the single-slit pattern modulates the double-slit interference pattern; E.2 (HL) — matter waves and the de Broglie wavelength λ = h/p; D.2 — work done on a charge accelerated through a potential difference Command term: Deduce

49C-2-04
Single-slit diffraction·C.3 Wave phenomena
Paper 2Medium10 marks
Short answer & extended response6 steps to full marksDetermine

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)

Syllabus understandingC.3 (HL) — single-slit diffraction as given by θ = λ/b; wave diffraction through an aperture; D.1 — orbital motion and Newton's law of gravitation; C.2 — the nature of electromagnetic waves Command term: Determine

50C-2-08
Double-slit interference·C.3 Wave phenomena
Paper 2Medium10 marks
Short answer & extended response6 steps to full marksDetermine

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)

Syllabus understandingC.3 — Young's double-slit interference as given by s = λD/d; C.3 (HL) — single-slit diffraction as given by θ = λ/b and the modulation of the double-slit pattern; E.1 — the photon energy E = hf Command term: Determine

51C-2-09
Refraction and total internal reflection·C.3 Wave phenomena
Paper 2Medium10 marks
Short answer & extended response7 steps to full marksExplain

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.

(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)

Syllabus understandingC.3 — Snell's law, critical angle and total internal reflection; Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1; C.2 — the nature of electromagnetic waves; E.1 — photon energy E = hf Command term: Explain

52C-2-12
Diffraction gratings·C.3 Wave phenomena
Paper 2Medium10 marks
Short answer & extended response7 steps to full marksDetermine

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)

Syllabus understandingC.3 (HL) — interference patterns from multiple slits and diffraction gratings as given by nλ = d sin θ; A.4 (HL) — angular velocity, angular acceleration and the relationship v = ωr; Tools — estimation Command term: Determine

53C-2-15
Optical fibre·C.3 Wave phenomena
Paper 2Medium9 marks
Short answer & extended response6 steps to full marksDetermine

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.

(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)

Syllabus understandingC.3 — Snell's law, critical angle and total internal reflection; refractive index n = c/v; E.1 — photon energy E = hf Command term: Determine

54C-2-16
Two-source interference·C.3 Wave phenomena
Paper 2Hard9 marks
Short answer & extended response6 steps to full marksDetermine

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.

(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)

Syllabus understandingC.3 — superposition of waves; that double-source interference requires coherent sources; the conditions for constructive and destructive interference; s = λD/d; A.1 — equations of motion for uniform acceleration Command term: Determine

55C-2-24
Wavefronts and rays·C.3 Wave phenomena
Paper 2Easy10 marks
Short answer & extended response7 steps to full marksDetermine

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)

Syllabus understandingC.3 — waves travelling in two and three dimensions described through wavefronts and rays; wave behaviour at boundaries in terms of refraction; Snell's law as given by sin θ2/sin θ1 = v2/v1; C.2 — v = λ/T; A.1 — speed; Tools — estimation Command term: Determine

56C-2-25
Double-slit interference·C.3 Wave phenomena
Paper 2Medium11 marks
Short answer & extended response7 steps to full marksDetermine

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)

Syllabus understandingC.3 — that double-source interference requires coherent sources; the condition for constructive interference as given by path difference = nλ; Young's double-slit interference as given by s = λD/d; C.2 — the nature of electromagnetic waves Command term: Determine

57C-2-26
Superposition of pulses·C.3 Wave phenomena
Paper 2Medium12 marks
Short answer & extended response8 steps to full marksExplain

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)

Syllabus understandingC.3 — superposition of waves and wave pulses; that double-source interference requires coherent sources; the condition for destructive interference as given by path difference = (n + ½)λ; C.2 — v = fλ; C.1 — a = −ω²x; A.2 — Newton's second law Command term: Explain

58C-2-27
Single-slit diffraction·C.3 Wave phenomena
Paper 2Medium9 marks
Short answer & extended response6 steps to full marksDetermine

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)

Syllabus understandingC.3 (HL) — single-slit diffraction including intensity patterns as given by θ = λ/b; the effect of slit width on the intensity pattern (qualitative); E.2 (HL) — the de Broglie wavelength λ = h/p Command term: Determine

59C-2-28
Diffraction gratings·C.3 Wave phenomena
Paper 2Medium10 marks
Short answer & extended response6 steps to full marksDetermine

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)

Syllabus understandingC.3 (HL) — interference patterns from multiple slits and diffraction gratings as given by nλ = d sin θ; multiple slit and diffraction grating patterns produced from white light and a range of monochromatic light wavelengths Command term: Determine

60C-2-29
Reflection, refraction and transmission at boundaries·C.3 Wave phenomena
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine

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)

Syllabus understandingC.3 — wave behaviour at boundaries in terms of reflection, refraction and transmission; wave diffraction through an aperture; Snell's law as given by sin θ2/sin θ1 = v2/v1; C.3 (HL) — θ = λ/b; C.2 — v = fλ; B.1 — specific heat capacity Command term: Determine

61C-2-37
Refraction and total internal reflection·C.3 Wave phenomena
Paper 2Medium14 marks
Short answer & extended response9 steps to full marksExplain

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.

(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)

Syllabus understandingC.3 — wave behaviour at boundaries in terms of reflection, refraction and transmission; Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1; critical angle and total internal reflection; C.2 — c = fλ, the nature of electromagnetic waves Command term: Explain

62C-2-50
Diffraction of sound through a doorway·C.3 Wave phenomena
Paper 2Easy12 marks
Short answer & extended response8 steps to full marksExplain

A trumpet player T practises in a room whose door, of width 0.85 m, is open. A student S stands in the corridor outside, where she cannot see the player through the doorway, as shown. The note being played has a frequency of 233 Hz. The sound of the trumpet also contains a strong component of frequency 3.50 kHz.

The speed of sound in air is 343 m s−1.

T0.85 mSroomcorridor
Plan view of the room, the doorway and the corridor (not to scale). The lines show wavefronts of the 3.50 kHz component approaching the doorway.
(a)
(i)

Calculate the wavelength of the 233 Hz note and of the 3.50 kHz component.

(2)
(b)
(i)

The student hears the 233 Hz note clearly, but the 3.50 kHz component only faintly. Explain this observation.

(3)
(c)
(i)

Estimate, using θ = λ/b, the angle between the straight-through direction and the first minimum of the 3.50 kHz component beyond the doorway.

(2)
(ii)

Suggest why θ = λ/b cannot be applied to the 233 Hz note here.

(1)
(d)
(i)

On the diagram, draw three wavefronts of the 3.50 kHz component after they have passed through the doorway.

(2)
(e)
(i)

A concrete pillar 0.30 m wide stands in the middle of the corridor. Outline why a person standing close behind the pillar can hear the 233 Hz note.

(1)
(f)
(i)

Light from the room passes through the same doorway. Explain why the light casts a sharp-edged patch on the far wall of the corridor.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λ = 343/233 = 1.47 m✓ 1
λ = 343/3500 = 0.098 m✓ 1Accept 0.098 m.
Part (b)(i)
Sound reaches S only by diffraction (spreading) at the doorway, because S is not in the straight-through beam✓ 1
For 233 Hz the wavelength (1.47 m) is larger than the width of the doorway, so the sound spreads widely into the corridor✓ 1Allow ECF from (a).
For 3.50 kHz the wavelength (0.098 m) is much smaller than the doorway, so this sound continues mostly straight on and little reaches S✓ 1Comparison of wavelength with the doorway width required for MP2 and MP3.
Part (c)(i)
θ = 0.098/0.85✓ 1Allow ECF from (a).
θ = 0.115 rad (6.6°)✓ 1Accept 0.11–0.12 rad or 6–7°.
Part (c)(ii)
λ/b = 1.7 > 1, so there is no first minimum: the sound spreads out through all angles beyond the doorway✓ 1Allow ECF from (a). Accept "the wavelength is greater than the width of the doorway, so the formula would give an impossible angle".
Part (d)(i)
Wavefronts with the same spacing as the incident wavefronts✓ 1Judge by eye.
Wavefronts straight in the central region and curving only at the edges, spreading slightly into the regions beside the beam✓ 1Fully semicircular wavefronts: [0] for this mark.
Part (e)(i)
The wavelength (about 1.5 m) is much larger than the width of the pillar, so the sound diffracts around the pillar into the region behind it (no sound shadow)✓ 1Allow ECF from (a). Must refer to diffraction around an obstacle.
Part (f)(i)
The wavelength of light (about 5 × 10−7 m) is very much smaller than 0.85 m, so λ/b ≈ 10−6 rad: diffraction is negligible and the light travels in straight lines✓ 1Accept an order-of-magnitude estimate of λ/b. "Light is not diffracted" alone: [0].

Answers: (a)(i) 1.47 m; 0.098 m  ·  (c)(i) 0.115 rad (the remaining parts are explanations — see the table above)

Syllabus understandingC.3 — wave diffraction around a body and through an aperture; wavefront-ray diagrams showing refraction and diffraction; that waves travelling in two and three dimensions can be described through the concepts of wavefronts and rays; C.3 (HL) — single-slit diffraction including intensity patterns as given by θ = λ/b; C.2 — the wave equation v = fλ Command term: Explain

63C-2-51
Two-source interference in a ripple tank·C.3 Wave phenomena
Paper 2Easy13 marks
Short answer & extended response8 steps to full marksDetermine

Two small dippers S1 and S2, 6.0 cm apart, are fixed to the same vibrating bar in a ripple tank. They produce circular water waves of frequency 14 Hz. The diagram shows the crests and troughs of the waves from each source at one instant, drawn to scale.

S1S2PQ2.0 cm
Ripple tank viewed from above, at one instant (drawn to scale). Solid lines are crests and dashed lines are troughs; S1 waves in dark blue, S2 waves in gold.
(a)
(i)

Outline why S1 and S2 act as coherent sources.

(1)
(b)
(i)

Use the diagram to determine the wavelength and the speed of the waves.

(2)
(ii)

Determine the path difference S1P − S2P in terms of λ, and describe the motion of the water at P.

(2)
(iii)

Explain why the water at Q stays almost still.

(2)
(c)
(i)

Determine the number of lines of minimum disturbance that cross the straight line S1S2 between the sources.

(2)
(ii)

The frequency of the bar is doubled; the speed of the waves is unchanged. Deduce whether P and Q are now points of maximum or minimum disturbance.

(3)
(d)
(i)

At the original frequency, S2 is raised slightly so that the waves it produces have a smaller amplitude than those from S1. State the effect on the water at Q.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
They are driven by the same bar, so they have the same frequency and a constant phase difference (in phase)✓ 1Both "same frequency" and "constant phase difference" needed.
Part (b)(i)
λ = 1.5 cm (from the spacing of the crests and the scale bar)✓ 1Accept 1.4–1.6 cm.
v = fλ = 14 × 1.5 = 21 cm s−1✓ 1Allow ECF from their wavelength.
Part (b)(ii)
S1P − S2P = 5λ − 3λ = 2λ✓ 1Allow ECF from (b)(i) if path differences are measured from the diagram.
A whole number of wavelengths, so the waves arrive in phase: the water oscillates up and down with large amplitude (twice the amplitude of one wave)✓ 1Accept "constructive interference: crest meets crest and trough meets trough".
Part (b)(iii)
S1Q − S2Q = 4λ − 2.5λ = 1.5λ, a path difference of (n + ½)λ✓ 1Allow ECF from (b)(i).
The waves arrive in antiphase (a crest meets a trough), so their displacements cancel at all times (destructive interference)✓ 1Award MP2 only with an idea of antiphase or crest meeting trough.
Part (c)(i)
Between the sources the path difference ranges from −6.0 cm to +6.0 cm, i.e. from −4λ to +4λ✓ 1Allow ECF from (b)(i).
Minima at ±0.5λ, ±1.5λ, ±2.5λ, ±3.5λ: 8 lines✓ 1Counting one side only (4): [1 max].
Part (c)(ii)
New wavelength = 0.75 cm✓ 1Allow ECF from (b)(i).
P: path difference 3.0 cm = 4λ, so P is still a maximum✓ 1
Q: path difference 2.25 cm = 3λ, a whole number of wavelengths, so Q is now a maximum✓ 1Allow ECF from their wavelength; the conclusion must match their path differences.
Part (d)(i)
The water at Q now oscillates with a small amplitude (the difference of the two amplitudes); it is no longer still✓ 1Allow ECF from (b)(iii).

Answers: (b)(i) 1.5 cm; 21 cm s−1  ·  (b)(ii) 2λ  ·  (b)(iii) 1.5λ  ·  (c)(i) 8  ·  (c)(ii) P: maximum (4λ); Q: maximum (3λ) (the remaining parts are explanations — see the table above)

Syllabus understandingC.3 — that double-source interference requires coherent sources; the condition for constructive interference as given by path difference = nλ; the condition for destructive interference as given by path difference = (n + ½)λ; superposition of waves and wave pulses; C.2 — the wave equation v = fλ Command term: Determine

64C-2-52
Seismic reflection at a rock boundary·C.3 Wave phenomena
Paper 2Medium11 marks
Short answer & extended response7 steps to full marksDetermine

In a seismic survey, a heavy weight is dropped on the ground to produce a pulse of longitudinal waves. The waves travel down through a layer of sediment and are partly reflected from the flat, horizontal top surface of the bedrock, at depth h below the ground. Geophones (detectors) on the surface at distances x from the source record the time t at which the reflected pulse arrives. The speed of the waves in the sediment is v1.

The graph shows t2 against x2 for the reflected pulse, with the line of best fit.

0246810121416x² / 10³ m²0100020003000400050006000t² / ms²
Square of the arrival time of the reflected pulse against the square of the distance from the source (drawn to scale).
(a)
(i)

Show that t2 = (x2 + 4h2)/v12.

(2)
(ii)

Use the graph to determine v1 and h.

(3)
(b)

The speed of the waves in the bedrock is 4.50 km s−1.

(i)

Calculate the critical angle for the waves in the sediment at the top surface of the bedrock.

(1)
(ii)

Determine the smallest distance x from the source at which a geophone receives a pulse that has been totally internally reflected at the bedrock.

(2)
(iii)

Suggest why the reflected pulse recorded by geophones beyond this distance is much stronger than for geophones closer to the source.

(1)
(c)
(i)

The pulse contains waves of frequency 90 Hz. Calculate the wavelength of these waves in the sediment and in the bedrock.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The reflection is half-way between source and geophone, so each half of the path has length √((x/2)2 + h2)✓ 1A sketch showing the symmetric reflected path with the half-distance x/2 is acceptable.
Total path = 2√((x/2)2 + h2) = √(x2 + 4h2) = v1t, so t2 = (x2 + 4h2)/v12✓ 1
Part (a)(ii)
Gradient = 1/v12 = 309 ms2 per 103 m2 = 3.09 × 10−7 s2 m−2✓ 1Accept 295–320 ms2 per 103 m2.
v1 = 1/√(gradient) = 1.80 km s−1✓ 1Allow ECF from MP1. Accept 1.75–1.85 km s−1.
Intercept = 4h2/v12 = 1115 ms2, so h = ½ × 1799 × √(1115 × 10−6) = 30 m✓ 1Allow ECF from MP2 and (a)(i). Accept intercept 1050–1180 ms2 and h = 29–31 m.
Part (b)(i)
sin c = v1/v2 = 1.80/4.50, so c = 23.6°✓ 1Allow ECF from (a)(ii). Accept 23–24°.
Part (b)(ii)
The angle of incidence at the bedrock is given by tan i = (x/2)/h; TIR when this angle ≥ c✓ 1Allow ECF from (a)(ii) and (b)(i).
x = 2h tan c = 2 × 30 × tan 23.6° = 26 m✓ 1Accept 25–27 m.
Part (b)(iii)
Beyond this distance all the energy is reflected and none is transmitted into the bedrock; closer in, much of the energy is transmitted (refracted) into the bedrock✓ 1Must refer to transmitted energy at the boundary.
Part (c)(i)
Sediment: λ = 1799/90 = 20 m✓ 1Allow ECF from (a)(ii).
Bedrock: λ = 4500/90 = 50 m✓ 1The frequency is the same in both layers.

Answers: (a)(ii) 1.80 km s−1; 30 m  ·  (b)(i) 23.6°  ·  (b)(ii) 26 m  ·  (c)(i) 20 m; 50 m (the remaining parts are explanations — see the table above)

Syllabus understandingC.3 — wave behaviour at boundaries in terms of reflection, refraction and transmission; Snell's law, critical angle and total internal reflection; Snell's law as given by n1/n2 = sin θ2/sin θ1 = v2/v1; that waves travelling in two and three dimensions can be described through the concepts of wavefronts and rays; Tools — linearising a relationship, gradient and intercept Command term: Determine

65C-2-53
Grating spectrometer and a stellar line shift·C.3 Wave phenomena
Paper 2Medium14 marks
Short answer & extended response9 steps to full marksDetermine

A small spectrometer on a telescope uses a diffraction grating with 600 lines per mm. Light from a star is made into a parallel beam that is incident normally on the grating, and a camera records the angle of each part of the spectrum. The camera can record light only at angles from 0 to 50° on one side of the straight-through direction.

(a)
(i)

Calculate the separation of adjacent lines of the grating.

(1)
(ii)

Determine the angular width of the first-order spectrum of visible light, from 400 nm to 700 nm.

(2)
(iii)

The hydrogen line of wavelength 656.28 nm is studied. Deduce the orders of this line that the camera records.

(2)
(b)

In the spectrum of the star the hydrogen line is observed at 656.48 nm.

(i)

Determine the speed of the star along the line of sight and state whether it is moving towards or away from the Earth.

(2)
(ii)

Determine, in radians, the angle between the observed line and the position of the 656.28 nm line in the first-order spectrum.

(3)
(iii)

The angle in (b)(ii) is too small for the camera to detect. A student suggests using the second-order spectrum instead. Explain why this would not work with this spectrometer.

(1)
(c)
(i)

Calculate, in eV, the energy of a photon of wavelength 656.28 nm.

(2)
(ii)

In the spectrum of the star this line appears dark. Outline why.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
d = 1/(600 × 103) = 1.67 × 10−6 m✓ 1
Part (a)(ii)
sin θ = λ/d: θ = 13.9° for 400 nm and 24.8° for 700 nm✓ 1Allow ECF from (a)(i).
Angular width = 10.9°✓ 1Accept 10.9–11.0°.
Part (a)(iii)
First order: sin θ = 0.394, θ = 23.2°; second order: sin θ = 0.788, θ = 52.0°✓ 1Allow ECF from (a)(i).
52° > 50°, so only the first order is recorded (with the zero order containing all wavelengths)✓ 1Conclusion must follow from their angles.
Part (b)(i)
v = c Δλ/λ = 3.00 × 108 × 0.20/656.28 = 9.1 × 104 m s−1✓ 1
Away from the Earth (the wavelength is increased: a redshift)✓ 1
Part (b)(ii)
sin θ0 = 656.28 × 10−9 × 6.00 × 105 = 0.39377✓ 1Allow ECF from (a)(i). Values must be kept to at least 5 s.f.
sin θ = 656.48 × 10−9 × 6.00 × 105 = 0.39389, then Δθ = sin−1(0.39389) − sin−1(0.39377)✓ 1
Δθ = 1.3 × 10−4 rad✓ 1Accept 1.2–1.4 × 10−4 rad. Rounding the angles to 3 s.f. before subtracting: [2 max].
Part (b)(iii)
The shift in angle would be larger (about three times, 4 × 10−4 rad) because sin θ ∝ n, but the second-order line is at about 52°, beyond the 50° range of the camera✓ 1Allow ECF from (a)(iii). Both ideas needed.
Part (c)(i)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/656.28 × 10−9 = 3.03 × 10−19 J✓ 1
= 1.89 eV✓ 1Accept 1.89 eV.
Part (c)(ii)
Hydrogen atoms in the cooler outer layers of the star absorb photons of exactly this energy, moving to a higher energy level, so this wavelength is missing from the light that reaches us✓ 1Must refer to absorption by atoms (in the outer layers). Allow ECF from (c)(i) for the energy quoted.

Answers: (a)(i) 1.67 × 10−6 m  ·  (a)(ii) 10.9°  ·  (a)(iii) first order only  ·  (b)(i) 9.1 × 104 m s−1, away  ·  (b)(ii) 1.3 × 10−4 rad  ·  (c)(i) 1.89 eV (the remaining parts are explanations — see the table above)

Syllabus understandingC.3 (HL) — interference patterns from multiple slits and diffraction gratings as given by nλ = d sin θ; Guidance: multiple slit and diffraction grating patterns produced from white light and a range of monochromatic light wavelengths; C.5 — Δλ/λ ≈ v/c for light; spectral line shifts and the motion of stars; E.1 — photon energy E = hf; absorption spectra Command term: Determine

66C-2-54
Two-dish radio interferometer·C.3 Wave phenomena
Paper 2Hard14 marks
Short answer & extended response8 steps to full marksDetermine

Two identical radio dishes, each of diameter b, stand on an east–west line a distance d = 32.0 m apart. Their signals are added in a receiver tuned to the hydrogen line, of frequency 1420.41 MHz. A radio source on the celestial equator passes overhead; because of the Earth's rotation, the angle θ between the direction of the source and the vertical changes at a constant rate ω, the angular velocity of the Earth.

The graph shows how the combined signal varies with time t, where t = 0 when the source is directly overhead; the dashed line is the envelope of the maxima. The sensitivity of each dish can be modelled as the single-slit pattern of a slit of width b.

-600-500-400-300-200-1000100200300400500600t / s0.00.20.40.60.81.0relative signal
Combined signal from the two dishes against time as the source passes overhead (drawn to scale).
(a)
(i)

Show that the wavelength of the radio waves is about 0.21 m.

(1)
(ii)

Explain why the combined signal is a maximum when d sin θ = nλ.

(2)
(b)
(i)

Show that, for small angles, the time between adjacent maxima is λ/(dω).

(2)
(ii)

Use the graph to determine the angular velocity of the Earth, and hence the time the Earth takes to rotate once.

(3)
(c)
(i)

Use the graph to estimate the diameter b of each dish.

(2)
(ii)

The dishes are moved to a separation of 64.0 m. State and explain the effect on the time between adjacent maxima and on the time at which the envelope first falls to zero.

(2)
(d)
(i)

A cloud of hydrogen gas in the Galaxy is observed at 1420.03 MHz. Determine its speed along the line of sight and its direction of motion.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λ = c/f = 3.00 × 108/1.42041 × 109 = 0.2112 m✓ 1Answer to at least 3 s.f. or full substitution required.
Part (a)(ii)
The waves from the distant source reach the two dishes with a path difference of d sin θ (one dish is further from the source along the wavefront direction)✓ 1A sketch of the extra path is acceptable.
When the path difference is a whole number of wavelengths, the signals arrive in phase and add constructively✓ 1Award MP2 only with reference to phase.
Part (b)(i)
Adjacent maxima: n changes by 1, so θ changes by λ/d (sin θ ≈ θ)✓ 1
θ changes at rate ω, so the time is (λ/d)/ω = λ/(dω)✓ 1
Part (b)(ii)
Time between adjacent maxima from several cycles, e.g. 6 intervals from the maximum at −272 s to the maximum at +272 s: 543/6 = 90.5 s✓ 1Accept 88–93 s.
ω = λ/(d × time) = 0.2112/(32.0 × 90.5) = 7.29 × 10−5 rad s−1✓ 1Allow ECF from (a)(i) and (b)(i).
Time for one rotation = 2π/ω = 8.6 × 104 s (about 24 h)✓ 1Allow ECF from MP2. Accept 8.4–8.8 × 104 s.
Part (c)(i)
Envelope first falls to zero at t ≈ 483 s, so θ = 7.29 × 10−5 × 483 = 0.0352 rad✓ 1Accept 470–500 s. Allow ECF from (b)(ii).
b = λ/θ = 0.2112/0.0352 = 6.0 m✓ 1Allow ECF. Accept 5.7–6.3 m.
Part (c)(ii)
Time between maxima is halved (to about 45 s), because it is proportional to 1/d✓ 1Allow ECF from (b)(i) and (b)(ii).
The envelope is unchanged (zero still at about 480 s), because it depends only on λ/b for each dish✓ 1Allow ECF from (c)(i).
Part (d)(i)
v = c Δf/f = 3.00 × 108 × 0.38/1420.41 = 8.0 × 104 m s−1✓ 1Accept 80 km s−1.
The frequency is lower than emitted, so the cloud is moving away (receding)✓ 1

Answers: (a)(i) 0.2112 m  ·  (b)(ii) 7.29 × 10−5 rad s−1; 8.6 × 104 s  ·  (c)(i) 6.0 m  ·  (d)(i) 8.0 × 104 m s−1, away (the remaining parts are explanations — see the table above)

Syllabus understandingC.3 (HL) — that the single-slit pattern modulates the double slit interference pattern; single-slit diffraction including intensity patterns as given by θ = λ/b; C.3 — the condition for constructive interference as given by path difference = nλ; that double-source interference requires coherent sources; A.2 — angular velocity ω = 2π/T; C.5 — Δf/f ≈ v/c; C.2 — c = fλ Command term: Determine

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