AS Physics: Superposition — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Superposition revision notes

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Section A

1. State the principle of superposition. [2]

2. Explain what is meant by coherent, and why coherence is necessary for an observable interference pattern. [3]

3. Give three differences between a progressive wave and a stationary wave. [3]

4. Explain how a stationary wave forms on a stretched string. [2]

5. Explain why diffraction is most noticeable when a gap is comparable in width to the wavelength of the wave passing through it. [2]

6. Explain why a stationary wave does not transfer energy along its length, despite its particles oscillating with real kinetic and potential energy. [2]

7. Explain why two independent light sources of the same colour do not produce a stable, observable interference pattern. [2]

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Section B

8. In a double-slit experiment, light of wavelength 590 nm passes through slits 0.35 mm apart onto a screen 2.4 m away.

(a) Calculate the fringe spacing. [3]

(b) State and explain the effect on fringe spacing of (i) increasing the slit separation, (ii) using blue light instead. [4]

(c) Explain why this experiment is evidence that light behaves as a wave. [2]

9. A diffraction grating has 600 lines per millimetre. Light of wavelength 633 nm is directed normally at it.

(a) Calculate the grating spacing d. [1]

(b) Calculate the angle of the second-order maximum. [3]

(c) Determine the highest order that can be observed. [3]

(d) Give one advantage of a grating over a double slit for measuring wavelength. [1]

10. A stationary wave is set up on a string of length 1.2 m fixed at both ends, vibrating in its fundamental mode at 85 Hz.

(a) Sketch or describe the positions of nodes and antinodes. [2]

(b) Calculate the wave speed. [3]

11. Two loudspeakers, coherent and in phase, are placed 1.5 m apart and emit sound of wavelength 0.68 m. A microphone moves along a line parallel to the speakers, 4.0 m away. Because the speaker separation (1.5 m) and the distance to the line (4.0 m) are comparable in size rather than one being much larger than the other, the small-angle double-slit fringe formula x = λD/a does not apply here. Find the exact distance from the centre line to the first position of maximum loudness, by setting the path difference between the two speakers equal to one wavelength and solving for the geometry directly. [4]

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Answers

1. When two or more waves meet at a point, the resultant displacement [1] is the vector sum of the individual displacements [1].

2. A constant phase difference [1] and the same frequency [1]. Without it the phase relationship varies randomly, so the pattern of maxima and minima shifts too rapidly to be observed [1].

3. Progressive waves transfer energy, stationary waves do not [1]; for an ideal progressive wave in a uniform, non-attenuating medium, amplitude is the same at all points, whereas a stationary wave’s amplitude varies from zero at nodes to maximum at antinodes [1] (a real progressive wave can lose amplitude to attenuation, geometric spreading or absorption as it travels — the equal-amplitude comparison assumes an idealised, undamped wave); in a stationary wave all points between adjacent nodes are in phase, whereas phase varies continuously along a progressive wave [1].

4. A stationary wave forms when two progressive waves of the same frequency and amplitude, travelling in opposite directions, superpose [1] — for example a wave reflected back along a stretched string, so the incident and reflected waves overlap [1].

5. Diffraction is most noticeable when the gap width is comparable to the wavelength [1]; a gap much wider than the wavelength produces little noticeable spreading, while a narrow gap produces pronounced spreading [1].

6. Energy is stored within each section between adjacent nodes, oscillating between kinetic and potential forms as the string moves [1], rather than being transferred along the wave’s length as in a progressive wave [1].

7. Two independent sources cannot maintain a constant phase difference with each other [1], so the pattern of maxima and minima shifts too rapidly to observe — a single source split into two coherent paths, as in a double slit, is needed instead [1].

8. (a) x = λD ÷ a [1] = (590 × 10⁻⁹ × 2.4) ÷ (0.35 × 10⁻³) [1] = 4.05 × 10⁻³ m ≈ 4.0 mm [1].

(b) (i) Fringe spacing decreases [1] — it is inversely proportional to slit separation [1]. (ii) Fringe spacing decreases [1] — blue light has a shorter wavelength, and spacing is proportional to λ [1].

(c) Interference — the production of alternating maxima and minima by superposition — is a wave property [1] and cannot be explained by a classical, ray-like particle model of light [1] (at this level, “particle model” means the classical picture of light as simple particles travelling in straight lines; it does not refer to modern quantum theory, which does account for interference).

9. (a) d = 1 ÷ (600 × 10³) [1] = 1.667 × 10⁻⁶ m.

(b) d sin θ = nλ, so sin θ = (2 × 633 × 10⁻⁹) ÷ (1.667 × 10⁻⁶) [1] = 0.7595 [1] θ = 49.4° [1].

(c) Maximum order when sin θ = 1: n = d ÷ λ [1] = (1.667 × 10⁻⁶) ÷ (633 × 10⁻⁹) = 2.63 [1]. Round down: the highest observable order is n = 2 [1]. Rounding up is a standard error — order 3 would require sin θ > 1, which is impossible.

(d) The maxima are sharper/narrower, so their angular positions can be measured more precisely [1] (brightness itself depends on the illumination, slit width and how intensity is normalised, so the robust, guaranteed advantage is sharpness of the maxima, not brightness).

10. (a) Nodes at both fixed ends [1]; a single antinode at the centre [1].

(b) In the fundamental, L = λ/2, so λ = 2 × 1.2 = 2.4 m [1]. v = fλ = 85 × 2.4 [1] = 204 m s⁻¹ [1].

11. With D = 4.0 m and a = 1.5 m of comparable size, λD/a is not valid here – it assumes D >> a, which doesn’t hold [1]. Instead, place the sources at (±0.75, 0) and the point of the first maximum at (x, 4.0), a perpendicular distance x from the centre line. The two source-to-point distances are r₁ = √[(x + 0.75)² + 4.0²] and r₂ = √[(x − 0.75)² + 4.0²], and the first maximum occurs where the path difference r₁ − r₂ = λ = 0.68 m [1].

Using r₁² − r₂² = (r₁ − r₂)(r₁ + r₂) = 2 × 0.75 × x = 1.5x, and r₁ − r₂ = 0.68, gives r₁ + r₂ = 1.5x ÷ 0.68. Substituting r₁ = ½(0.68 + 1.5x ÷ 0.68) into r₁² = (x + 0.75)² + 16 and solving the resulting equation for x [1] gives x ≈ 2.06 m [1] from the centre line – notably different from the (invalid) small-angle estimate of 1.81 m. This also shows the fringes are not evenly spaced this close to the sources, so “the distance between adjacent maxima” isn’t a single well-defined number here; only the position of a specific named maximum is.

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Where marks are usually lost

• Saying coherence requires equal amplitude.
• Reversing the relationship between slit separation and fringe spacing.
• Rounding the maximum order up.
• Using λ instead of λ/2 for the fundamental on a fixed string.
• Saying stationary waves transfer energy.