IB DP Physics: Superposition, double-slit interference, single-slit diffraction and gratings (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover C.3 Wave phenomena (second half) in the IB Physics guide, first assessment 2025: superposition and Young’s double slit (SL and HL), then single-slit diffraction, modulation and gratings (HL only, labelled). Support: study guide, revision notes, IB DP Physics hub and checklist.

Questions

1. A pulse of height +5.0 cm travels towards a pulse of height −3.5 cm on the same rope. State the principle that gives the rope’s shape when they overlap, and find the displacement when their centres coincide. [2]

2. (a) Define coherent sources. [1] (b) Explain why light from two separate filament lamps does not form stable fringes on a screen. [2]

3. Two ultrasound transmitters emit in phase at 40 kHz. The speed of sound is 340 m s⁻¹. A detector at P is 214.0 mm from one transmitter and 231.0 mm from the other. At Q the distances are 120.00 mm and 124.25 mm. Determine the type of interference at P and at Q. [4]

4. Sodium light of wavelength 589 nm illuminates a double slit of separation 0.45 mm. The screen is 1.35 m away. (a) Calculate the fringe separation. [2] (b) Find the distance from the central maximum to the fourth bright fringe. [1] (c) The screen is moved to half the distance and the slit separation is doubled. Find the new fringe separation. [2]

5. In a double-slit experiment d = 0.20 mm and D = 2.50 m. The distance across six adjacent fringe spacings is 4.5 cm. (a) Calculate the wavelength of the light. [3] (b) Explain why the measurement is taken across several spacings. [1]

6. A laser beam lights both slits of a double slit. (a) Explain why a single laser beam gives coherent sources. [2] (b) At a bright fringe each slit alone would give intensity I. State the intensity with both slits open. [2] (c) One slit is covered. Describe the change on the screen. [2]

7. (HL only) Light of wavelength 600 nm passes through a single slit of width 0.10 mm. The screen is 2.4 m away. Calculate the angle to the first minimum and the width of the central maximum. [4]

8. (HL only) A slit of width 0.20 mm gives a single-slit pattern. The slit is narrowed to 0.10 mm. (a) State and explain the change in the width of the central maximum. [2] (b) State the change in its intensity. [1] (c) State the effect of using longer-wavelength light with the original slit. [1]

9. (HL only) A double slit has d = 0.45 mm and b = 0.15 mm. The light has wavelength 600 nm and the screen is 1.8 m away. (a) Calculate the fringe separation. [2] (b) Find the distance from the centre to the first minimum of the single-slit pattern. [2] (c) State which double-slit order is missing. [1] (d) State how many bright fringes lie within the central maximum of the single-slit pattern. [1] (e) Describe how fringe brightness varies across the central maximum. [1]

10. (HL only) A grating has 700 lines per mm. A second-order maximum of monochromatic light appears at 40.0° to the normal. (a) Calculate the wavelength. [3] (b) State the highest order observed and the total number of maxima. [2]

11. (HL only) White light (400 nm to 700 nm) is incident normally on a grating with 500 lines per mm. (a) Describe and explain the central maximum. [2] (b) Calculate the angles of the first-order violet and red light. [2] (c) Show that the second-order and third-order spectra overlap. [3] (d) Give one advantage of a grating over a double slit for measuring wavelength. [1]

12. (HL only) Light containing wavelengths 450 nm and 640 nm falls normally on a grating with 800 lines per mm. Calculate the first-order angles for both, and determine which wavelengths have a second order. [5]

Answers

1. Principle of superposition: resultant displacement is the sum of the individual displacements [1]. +5.0 + (−3.5) = +1.5 cm [1]. [2] Examiner insight: the sign must be carried through the sum; giving 8.5 cm shows the sign was ignored.

2. (a) Sources with the same frequency and a constant phase difference [1]. (b) Each lamp emits in random bursts, so the phase difference between them keeps changing [1]. Maxima and minima shift about too quickly to see, so the screen is evenly lit [1]. [3] Examiner insight: “different colours” is not an explanation; the answer must mention the changing phase relationship.

3. λ = 340 / 40 000 = 8.5 mm [1]. P: path difference = 17.0 mm = 2λ [1], so constructive [1]. Q: path difference = 4.25 mm = ½λ, so destructive [1]. [4] Examiner insight: showing the division of path difference by λ is what supports the conclusion; a bare “constructive” is weak.

4. (a) s = λD/d = (589 × 10⁻⁹ × 1.35)/(0.45 × 10⁻³) [1] = 1.77 × 10⁻³ m [1]. (b) 4s = 7.07 mm [1]. (c) s ∝ D/d, which becomes (1/2)/2 = ¼ of the original [1], so s = 0.442 mm [1]. [5] Examiner insight: conversion of mm and nm to metres is the usual place where answers go wrong by powers of ten.

5. (a) s = 0.045/6 = 7.5 × 10⁻³ m [1]. λ = sd/D = (7.5 × 10⁻³ × 0.20 × 10⁻³)/2.50 [1] = 6.0 × 10⁻⁷ m (600 nm) [1]. (b) A fixed reading uncertainty is shared by six spacings, so the percentage uncertainty in s is smaller [1]. [4] Examiner insight: “to be more accurate” alone is vague; say what is reduced.

6. (a) Both slits receive light from the same wavefront of one source [1], so any change in the laser affects both beams alike and their phase difference is constant [1]. (b) Amplitude doubles [1]; intensity depends on amplitude squared, so intensity = 4I [1]. (c) The fringes disappear [1] because there is no second beam to interfere with; the old bright positions fall to ¼ of their brightness (and the screen is lit by one beam) [1]. [6] Examiner insight: stating “2I” is a common slip; the answer should link intensity to amplitude squared.

7. θ = λ/b = (600 × 10⁻⁹)/(0.10 × 10⁻³) [1] = 6.0 × 10⁻³ rad [1]. Distance to first minimum = Dθ = 2.4 × 6.0 × 10⁻³ = 14.4 mm. Width = 2 × 14.4 = 28.8 mm [1]. The factor of two is applied for both sides [1]. [4] Examiner insight: the angle θ = λ/b reaches only the first minimum, so the full width needs the doubling, and units (rad, mm) should be stated.

8. (a) The central maximum becomes twice as wide [1] because θ = λ/b, so halving b doubles θ [1]. (b) Intensity decreases [1] because less light passes through the slit. (c) The central maximum becomes wider [1] since θ increases with λ. [4] Examiner insight: “narrower slit, narrower pattern” is the commonly lost point; link width and spread inversely.

9. (a) s = λD/d = (600 × 10⁻⁹ × 1.8)/(0.45 × 10⁻³) [1] = 2.4 mm [1]. (b) y = λD/b = (600 × 10⁻⁹ × 1.8)/(0.15 × 10⁻³) [1] = 7.2 mm [1]. (c) d/b = 3, so the third order is missing [1]. (d) Orders 0, ±1, ±2: 5 bright fringes [1]. (e) The fringes are brightest at the centre and fade towards the edges of the central maximum, following the single-slit envelope [1]. [7] Examiner insight: the envelope must be linked to the slit width b, and the fringe spacing to d; mixing them loses the explanation.

10. (a) d = 10⁻³/700 = 1.43 × 10⁻⁶ m [1]. λ = d sin θ/n = 1.43 × 10⁻⁶ × sin 40.0°/2 [1] = 4.59 × 10⁻⁷ m (459 nm) [1]. (b) d/λ = 3.11, so the highest order is 3 [1]; maxima n = 0, ±1, ±2, ±3 give 7 [1]. [5] Examiner insight: the order is a whole number below d/λ; rounding 3.11 up to 4 breaks sin θ ≤ 1.

11. (a) The central maximum is white [1]: at θ = 0 the path difference is zero for every wavelength, so all colours are in phase [1]. (b) d = 2.00 × 10⁻⁶ m. Violet: sin θ = 400 × 10⁻⁹/2.00 × 10⁻⁶ = 0.200, θ = 11.5° [1]. Red: sin θ = 0.350, θ = 20.5° [1]. (c) Third-order violet: sin θ = 3 × 0.200 = 0.600, θ = 36.9° [1]. Second-order red: sin θ = 2 × 0.350 = 0.700, θ = 44.4° [1]. The third-order violet (36.9°) is at a smaller angle than the second-order red (44.4°) and larger than the second-order violet (23.6°), so the spectra overlap [1]. (d) The maxima are narrower and brighter, so angles can be measured more precisely [1]. [8] Examiner insight: “show that” asks for both angles to be calculated and compared; one value alone is not a demonstration.

12. d = 10⁻³/800 = 1.25 × 10⁻⁶ m [1]. First order, 450 nm: sin θ = 0.360, θ = 21.1° [1]; 640 nm: sin θ = 0.512, θ = 30.8° [1]. Second order, 450 nm: sin θ = 0.720, θ = 46.1° [1]; 640 nm: sin θ = 1.024, which exceeds 1, so no second order [1]. [5] Examiner insight: showing that sin θ exceeds 1 is how a missing order is justified; saying “too big” is not enough.

Where marks are usually lost

• Not converting mm, nm and lines per mm to metres.
• Using nλ for dark fringes.
• Swapping the slit width b and the slit separation d.
• Forgetting to double the angle or distance to get the full width of the central maximum.
• Saying intensity doubles when the amplitude doubles.
• Rounding the maximum order up instead of down.
• Not explaining coherence in terms of constant phase difference.
• Giving angles in the wrong order for violet and red.

Next steps

• Revision notes
• Study guide
• IB DP Physics hub and printable checklist
• All free 10-minute diagnostics
• Book a free trial class: /trial/

Official syllabus

Physics guide, first assessment 2025, International Baccalaureate Organization, sub-topic C.3 Wave phenomena.