Waves (stationary waves, interference, diffraction, refraction) – A-level Physics exam questions with mark scheme

1Core3 marks

A progressive wave has a frequency of 50 Hz and a wavelength of 0.60 m.

(a) Calculate the phase difference, in radians, between two points on the wave that are 0.15 m apart along the direction of travel. [2]

(b) Calculate the speed of the wave. [1]

(a) phase difference = $\dfrac{0.15}{0.60} \times 2\pi$(1)
(a) $\dfrac{\pi}{2}$ rad allow 1.57 rad(1)
(b) $c = f\lambda = 50 \times 0.60 = 30$ m s$^{-1}$(1)
Answer: (a) $\pi/2$ rad (b) 30 m s$^{-1}$
Examiner tip: Phase difference = (distance ÷ wavelength) × 2π; giving the answer in degrees when radians are asked costs the mark.

2Core3 marks

(a) State what is meant by a plane-polarised wave. [1]

(b) Explain why sound waves cannot be polarised. [1]

(a) oscillations (of the field / particles) are in one plane only (containing the direction of travel) allow vibrations; allow 'oscillations in one direction only'; reject 'wave travels in one plane / one direction'(1)
(b) sound is longitudinal: the oscillations are parallel to the direction of travel (energy transfer), so there is no perpendicular oscillation to restrict 'sound is longitudinal' alone is enough for the mark(1)
(c) e.g. aligning a TV / radio aerial with the transmitted signal; Polaroid (polarising) sunglasses reducing glare; polarising filters in photography; stress analysis of plastics allow any other valid application, e.g. LCD screens, 3D cinema glasses; reject 'sunglasses' without reference to Polaroid / polarising(1)
Answer: (a) oscillations in one plane only (b) sound is longitudinal (c) e.g. aligning TV aerials
Examiner tip: Say 'oscillations in one plane', not 'waves in one direction' — direction of travel is not what polarisation restricts.

3Standard6 marks

A string is stretched between a vibration generator and a pulley. The vibrating length of the string is 0.80 m and the mass per unit length is $1.2\times10^{-3}$ kg m$^{-1}$. A mass of 0.50 kg hangs from the end of the string.

(a) Calculate the frequency of the first harmonic. [2]

(b) The frequency is increased until the third harmonic is seen. State this frequency and the number of nodes on the string, including both ends. [2]

(a) $T = 0.50 \times 9.81 = 4.9$ N and $f = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}} = \dfrac{1}{1.60}\sqrt{\dfrac{4.905}{1.2\times10^{-3}}}$(1)
(a) 40 Hz(1)
(b) 120 Hz ecf 3 × their (a)(1)
(b) 4 nodes(1)
(c) waves reflect at the ends, so two waves of the same frequency (and similar amplitude) travel in opposite directions and superpose(1)
(c) nodes form where the waves always cancel (destructive interference); antinodes where they reinforce(1)
Answer: (a) 40 Hz (b) 120 Hz, 4 nodes (c) superposition of incident and reflected waves
Examiner tip: Tension is the weight of the hanging mass (mg), not the mass itself — forgetting × 9.81 gives a frequency about 3 times too small.

4Standard5 marks

A student carries out a Young's double-slit experiment. Laser light passes through two slits 0.40 mm apart. The fringes are observed on a screen 2.5 m from the slits. The student measures the distance across 10 fringe spacings as 34 mm.

(a) Calculate the wavelength of the light. [3]

(b) Explain why the student measures across 10 fringe spacings rather than one. [1]

(a) $w = 34/10 = 3.4$ mm(1)
(a) $\lambda = \dfrac{ws}{D} = \dfrac{3.4\times10^{-3}\times0.40\times10^{-3}}{2.5}$(1)
(a) $5.4\times10^{-7}$ m allow $5.44\times10^{-7}$ m / 540 nm / 544 nm(1)
(b) the larger distance has a smaller percentage (fractional) uncertainty (same absolute uncertainty from the ruler) allow 'reduces the uncertainty in the fringe spacing' / 'the uncertainty is shared over 10 spacings'; reject 'more accurate' alone(1)
(c) e.g. never look directly into the beam / avoid reflective surfaces / wear laser safety goggles / display a warning sign(1)
Answer: (a) $5.4\times10^{-7}$ m (b) reduces percentage uncertainty (c) e.g. do not look into the beam
Examiner tip: Convert both mm values to metres; leaving s in mm gives an answer 1000 times too big.

5Standard5 marks

Light of wavelength 589 nm is incident normally on a diffraction grating with 300 lines per mm.

(a) Calculate the angle of the first-order maximum. [3]

(b) Determine the highest order maximum that can be observed. [2]

(a) $d = \dfrac{1\times10^{-3}}{300} = 3.33\times10^{-6}$ m(1)
(a) $\sin\theta = \dfrac{\lambda}{d} = \dfrac{589\times10^{-9}}{3.33\times10^{-6}} = 0.177$(1)
(a) $\theta = 10.2°$(1)
(b) maximum $n = \dfrac{d}{\lambda}$ (using $\sin\theta = 1$) $= 5.7$(1)
(b) highest order = 5 (must round down)(1)
Answer: (a) 10.2° (b) 5th order
Examiner tip: Always round the maximum order DOWN: sin θ cannot exceed 1, so 5.7 means 5 orders, not 6.

6Hard7 marks

A step-index optical fibre has a core of refractive index 1.52 and a cladding of refractive index 1.48.

(a) Calculate the critical angle at the core–cladding boundary. [2]

(b) Calculate the speed of light in the core. [1]

(d) Explain how modal dispersion arises and why using a narrower core reduces it. [2]

(a) $\sin c = \dfrac{n_2}{n_1} = \dfrac{1.48}{1.52}$(1)
(a) $c = 76.8°$ allow 77°(1)
(b) $\dfrac{3.00\times10^{8}}{1.52} = 1.97\times10^{8}$ m s$^{-1}$(1)
(c) any two from: protects the core from damage / scratches (which could let light escape); prevents signal crossing to adjacent fibres (keeps data secure); has a lower refractive index than the core so total internal reflection occurs at the boundary allow: increases the critical angle, reducing modal dispersion — first reason(1)
(c) second, different reason from the list above(1)
(d) rays entering at different angles take different paths of different lengths, so arrive at different times (pulse broadening)(1)
(d) a narrow core allows only (nearly) one path / fewer paths, so less difference in arrival time(1)
Answer: (a) 76.8° (b) $1.97\times10^{8}$ m s$^{-1}$ (c) protection; prevents crossover (d) different path lengths; narrow core → fewer paths
Examiner tip: For the critical angle at a core–cladding boundary, sin c = n₂/n₁ — not 1/n₁, which only applies to an air boundary.

7Stretch5 marks

A ray of monochromatic light is incident on one face of an equilateral glass prism at point X, as shown. The angle of incidence at X is 30°. The refractive index of the glass is 1.50.

Determine whether the ray leaves the prism at the second face (point Y). If it does, calculate the angle between the emerging ray and the normal at Y. [5]

angle of refraction at X: $\sin r_1 = \dfrac{\sin 30°}{1.50}$, $r_1 = 19.5°$(1)
angle of incidence at Y: $r_2 = 60° - 19.5° = 40.5°$ (geometry of prism)(1)
critical angle: $\sin c = \dfrac{1}{1.50}$, $c = 41.8°$ alternative for this and the next mark: $\sin\theta = 1.50\sin 40.5° \approx 0.97$(1)
$40.5° < 41.8°$ so total internal reflection does not occur and the ray emerges alternative: $\sin\theta < 1$ so refraction is possible and the ray emerges(1)
$\sin\theta = 1.50 \sin 40.5°$, $\theta = 77°$ allow 76.9°–77.1°; allow ecf from their angle of incidence at Y if less than the critical angle(1)
Answer: Ray emerges at Y (40.5° < critical angle 41.8°), at 77° to the normal
Examiner tip: The angle of incidence at the second face comes from geometry (prism angle − first refraction angle); show that step explicitly.

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