OCR A Level Physics A: Superposition, interference and the double slit (H556) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 4.4.3 (Superposition) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. Use c = 3.00 × 10⁸ m s⁻¹ and a speed of sound of 340 m s⁻¹. Questions 8, 9 and 11 use d sin θ = nλ and gratings, which the specification says are assessed at A level only. Revise first with the study guide or the revision notes.

Questions

1. Two pulses travel towards each other along a rope. One is a triangular pulse 6.0 cm above the rest line of the rope. The other is a triangular pulse 4.0 cm below it.

(a) State the principle of superposition. [2] (b) State the displacement of the rope when the centres of the two pulses coincide. [1] (c) Describe what happens to the pulses after they have passed through each other. [1]

2. Two waves of the same frequency have displacements y₁ = 4.0 sin θ and y₂ = 4.0 sin(θ + 120°), in cm.

(a) Find the resultant displacement at θ = 30° and at θ = 90°. [3] (b) The largest resultant displacement is 4.0 cm. Explain why the interference is neither fully constructive nor fully destructive. [1]

3. (a) Define coherent sources. [1] (b) Explain why two filament lamps side by side do not produce interference fringes. [2] (c) Explain why two slits lit by one laser beam can act as coherent sources. [1]

4. Two loudspeakers are connected in phase to one signal generator, which produces a note of 1360 Hz. A microphone is moved to point P, which is 3.15 m from one speaker and 3.90 m from the other. It is then moved to point R, which is 3.00 m and 3.625 m from them.

(a) Calculate the wavelength of the sound. [1] (b) State and justify whether the sound at P is loud or quiet. [2] (c) Calculate the phase difference between the two waves at R. [2] (d) One connection to one speaker is now reversed. State what is heard at P. [1]

5. A microwave transmitter shines on a metal plate with two narrow gaps 15.0 cm apart. A probe receiver is moved along a line parallel to the plate, 1.20 m beyond it.

(a) Describe how you would find the distance between adjacent maxima as accurately as possible. [3] (b) Five spacings between adjacent maxima add up to 1.00 m. Calculate the wavelength. [2] (c) Calculate the frequency of the microwaves. [1]

6. A double slit has slit separation 0.45 mm. A screen is 1.85 m away.

(a) Light of wavelength 470 nm is used. Calculate the fringe spacing. [2] (b) The light is changed to wavelength 650 nm. Calculate the new fringe spacing and state how the pattern changes. [2] (c) Describe the pattern seen with white light. [2]

7. A student uses slits 0.60 mm apart and measures the fringe spacing x for different slit-to-screen distances D.

D / m	1.00	1.50	2.00	2.50	3.00
x / mm	1.03	1.56	2.06	2.59	3.09

(a) State what graph to plot and what its gradient represents. [2] (b) Use the data to find the wavelength of the light. [3]

8. (A level only) A grating has 240 lines per mm. Light of wavelength 560 nm is incident normally.

(a) Calculate the grating spacing. [1] (b) Calculate the angle of the third-order maximum. [2] (c) Find the largest order of maximum that can be seen. [2]

9. (A level only) A laser beam passes at normal incidence through a grating with 1200 lines per mm. A screen is 0.800 m from the grating. The first-order maxima are 0.664 m either side of the central maximum.

(a) Calculate the angle of the first-order maximum. [2] (b) Calculate the wavelength. [2] (c) Explain why the small-angle approximation cannot be used here. [1] (d) Suggest one way to reduce the uncertainty in y. [1]

10. Derive λ = ax/D for a double slit, where a is the slit separation, D the distance to the screen and x the fringe spacing. State the condition for the result to be valid. [6]

11. (A level only) A grating has 360 lines per mm. White light has wavelengths from 410 nm to 690 nm.

(a) Calculate the grating spacing. [1] (b) Show that the second-order spectrum overlaps the third-order spectrum. [4] (c) Explain why Young’s double-slit experiment gave a classical confirmation of the wave nature of light. [3]

Answers

1. (a) When two waves meet, the resultant displacement is the sum of the individual displacements [1], taking account of their directions (signs) [1]. (b) 6.0 − 4.0 = +2.0 cm (above the rest line) [1]. (c) Each pulse continues with its original shape and speed [1]. [4] Examiner insight: “Waves add” with no mention of direction or sign leaves out the idea that makes the 2.0 cm in part (b) possible.

2. (a) At 30°: y₁ = 4.0 sin 30° = 2.0 cm and y₂ = 4.0 sin 150° = 2.0 cm [1], so the resultant is 4.0 cm [1]. At 90°: y₁ = 4.0 cm and y₂ = 4.0 sin 210° = −2.0 cm, so the resultant is 2.0 cm [1]. (b) The phase difference is 120°, which is neither 0° (constructive) nor 180° (destructive), so the amplitude is between 0 and 8.0 cm [1]. [4] Examiner insight: Sine values at 150° and 210° are where signs go wrong, so write each term out before adding.

3. (a) Sources with a constant phase difference [1]. (b) The lamps emit light in random short bursts, so the phase difference between them changes randomly and rapidly [1]. No stable pattern of maxima and minima forms, so the screen looks uniformly lit [1]. (c) Both slits are lit by the same wavefront, so any change in the laser affects both equally and their phase difference stays constant [1]. [4] Examiner insight: “Same frequency” alone is incomplete. Say the phase difference is constant.

4. (a) λ = v/f = 340 / 1360 = 0.25 m [1]. (b) Path difference = 3.90 − 3.15 = 0.75 m [1] = 3.0λ, a whole number of wavelengths, so the waves are in phase and the sound is loud [1]. (c) Path difference = 3.625 − 3.00 = 0.625 m = 2.5λ [1]. Phase difference = 2.5 × 2π = 5π rad (15.7 rad, 900°) [1]. (d) The speakers are now in antiphase, so P, with a path difference of 3λ, gives destructive interference: quiet [1]. [6] Examiner insight: Give the path difference in wavelengths, then the conclusion.

5. (a) Use one transmitter so the two gaps are coherent [1]. Move the probe slowly along a line parallel to the plate and note where the meter reading is greatest [1]. Measure the distance across several maxima with a ruler and divide by the number of spacings, repeating to check [1]. (b) x = 1.00 / 5 = 0.200 m [1]. λ = ax/D = (0.150 × 0.200) / 1.20 = 0.0250 m [1]. (c) f = c/λ = 3.00 × 10⁸ / 0.0250 = 1.2 × 10¹⁰ Hz [1]. [6] Examiner insight: Count the gaps between maxima, not the maxima.

6. (a) x = λD/a [1] = (470 × 10⁻⁹ × 1.85) / (0.45 × 10⁻³) = 1.9 × 10⁻³ m (1.93 mm) [1]. (b) x is proportional to λ, so the fringes are further apart [1]: x = (650 × 10⁻⁹ × 1.85) / 0.45 × 10⁻³ = 2.7 × 10⁻³ m (2.67 mm) [1]. (c) The central fringe is white [1]; the other fringes are coloured spectra, with blue nearer the centre than red [1]. [6] Examiner insight: Convert mm and nm to metres before substituting. Check the answer’s size against typical fringe widths of about a millimetre.

7. (a) Plot x against D, which gives a straight line through the origin [1]. The gradient is λ/a [1]. (b) Gradient = (3.09 − 1.03) × 10⁻³ / (3.00 − 1.00) = 1.03 × 10⁻³ [1]. λ = gradient × a = 1.03 × 10⁻³ × 0.60 × 10⁻³ [1] = 6.2 × 10⁻⁷ m (618 nm) [1]. [5] Examiner insight: Read the gradient using two well-separated points on the best-fit line, not neighbouring data points.

8. (a) d = 1 / (240 × 10³) = 4.17 × 10⁻⁶ m [1]. (b) sin θ = nλ/d = (3 × 560 × 10⁻⁹) / 4.167 × 10⁻⁶ = 0.403 [1], θ = 23.8° [1]. (c) d/λ = 4.167 × 10⁻⁶ / 560 × 10⁻⁹ = 7.44 [1]; the highest order is 7 [1]. [5] Examiner insight: Rounding 7.44 up to 8 gives an order needing sin θ above 1.

9. (a) tan θ = 0.664 / 0.800 = 0.830 [1], so θ = 39.7° [1]. (b) d = 1 / (1.2 × 10⁶) = 8.33 × 10⁻⁷ m, [1], then λ = d sin θ / n = 8.33 × 10⁻⁷ × sin 39.7° = 5.3 × 10⁻⁷ m [1]. (c) At 39.7°, sin θ = 0.639 but tan θ = 0.830, so sin θ is not close to tan θ [1]. (d) Measure the separation of the two first-order spots and halve it, which avoids locating the central spot [1]. [6] Examiner insight: Using y/L in place of sin θ gives about 6.9 × 10⁻⁷ m, a clearly different value; show the angle step.

10. For a point P at angle θ to the central line, the path difference between waves from the slits is a sin θ [1]. A bright fringe needs path difference nλ, so a sin θ = nλ [1]. For small θ, sin θ ≈ tan θ = y/D, where y is the distance from the centre [1]. So y = nλD/a [1]. Neighbouring fringes (n and n + 1) differ by x = λD/a [1]. Rearranging gives λ = ax/D, valid when a ≪ D (so angles are small) [1]. [6] Examiner insight: State the approximation at the step where it is used, and give the validity condition.

11. (a) d = 1 / (360 × 10³) = 2.78 × 10⁻⁶ m [1]. (b) Second-order red: sin θ = 2 × 690 × 10⁻⁹ / 2.78 × 10⁻⁶ = 0.497 [1], θ = 29.8° [1]. Third-order violet: sin θ = 3 × 410 × 10⁻⁹ / 2.78 × 10⁻⁶ = 0.443 [1], θ = 26.3°, which is smaller than 29.8°, so part of the third order lies within the second [1]. (c) Interference with dark fringes needs light from two sources to cancel [1], which is what waves can do by superposition [1]. Young’s fringes therefore supported Huygens’s wave model and not a purely particle model such as Newton’s [1]. [8] Examiner insight: “Overlap” needs a comparison of two angles, not just a statement.

Where marks are usually lost

• Defining coherence as “same phase” rather than a constant phase difference.
• Giving a conclusion without the path difference in wavelengths.
• Forgetting to swap the conditions when the sources are in antiphase.
• Counting maxima instead of the spacings between them.
• Leaving a in mm or λ in nm in a substitution.
• Using sin θ ≈ tan θ for grating angles.
• Rounding the highest grating order up.
• Deriving λ = ax/D without stating that a ≪ D.

Next steps

• Review with the revision notes and the study guide.
• Tick off the unit on the OCR A Level Physics hub and checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, section 4.4.3 Superposition.