AQA A-level Physics coverage

Waves

Section 3.3
6 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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3.3.1.1

Progressive waves

  • A progressive wave transfers energy without a net transfer of matter; particles of a material medium oscillate about fixed equilibrium positions.
  • Use v=fλv=f\lambda, with vv in m s1\text{m s}^{-1}, ff in Hz\text{Hz} and λ\lambda in m\text{m}.
  • For points separated by xx along a wave, the phase difference is Δϕ=2πx/λ\Delta\phi=2\pi x/\lambda radians, or 360x/λ360x/\lambda degrees, reduced by whole cycles when appropriate.
  • Amplitude is the maximum displacement from equilibrium. Do not confuse it with wavelength, which is the shortest distance between points in phase.

Tier 1 · Easy

1 mark
ORIGINAL

Successive crests of a progressive wave are 0.80m0.80\,\text{m} apart and pass a point every 2.5ms2.5\,\text{ms}. Calculate the wave speed.

Tier 2 · Standard

3 marks
ORIGINAL

Two sensors on a ripple tank are 0.24m0.24\,\text{m} apart along the direction of a progressive wave of wavelength 0.64m0.64\,\text{m}. Determine the magnitude of their phase difference in both degrees and radians.

Tier 3 · Hard

5 marks
ORIGINAL

Water waves of frequency 2.4Hz2.4\,\text{Hz} and wavelength 0.35m0.35\,\text{m} travel from P towards Q. Q is 0.22m0.22\,\text{m} beyond P along the direction of travel. Determine the wave speed, the phase change from P to Q, and the delay between a crest passing P and that crest passing Q.

3.3.1.2

Longitudinal and transverse waves

  • In a transverse wave, the displacement of particles or fields is perpendicular to the direction of energy transfer; electromagnetic waves and waves on a stretched string are examples.
  • In a longitudinal wave, particle displacement is parallel to the direction of energy transfer, producing compressions and rarefactions; sound in air is an example.
  • Only transverse waves can be polarised. Polarisation is therefore evidence that electromagnetic waves are transverse.
  • A receiving aerial gives the strongest signal when aligned with the transmitted electric-field oscillations. Malus's law is not required for this specification.
  • All electromagnetic waves travel at 3.00×108m s13.00\times10^8\,\text{m s}^{-1} in a vacuum; do not use the speed of sound for a radio signal.

Tier 1 · Easy

2 marks
ORIGINAL

State whether a sound wave travelling through air is longitudinal or transverse, and state the direction in which the air molecules oscillate.

Tier 2 · Standard

3 marks
ORIGINAL

A radio transmitter produces a vertically polarised wave. Explain why the signal received by a straight aerial decreases when the aerial is rotated from vertical towards horizontal.

Tier 3 · Hard

5 marks
ORIGINAL

At a distance of 1.20km1.20\,\text{km}, a detector receives a radio pulse and a sound pulse that were emitted simultaneously. Take the speed of sound as 340m s1340\,\text{m s}^{-1}. Calculate the arrival-time difference and explain why passing the radio wave through a correctly oriented polariser supports its classification as transverse.

3.3.1.3

Principle of superposition of waves and formation of stationary waves

  • Superposition means that the resultant displacement at a point is the vector sum of the displacements due to the individual waves.
  • A stationary wave is formed by two progressive waves of the same frequency travelling in opposite directions; it has nodes of zero amplitude and antinodes of maximum amplitude.
  • For a string fixed at both ends, adjacent nodes are λ/2\lambda/2 apart. The nnth harmonic has nn loops, n+1n+1 nodes and frequency fn=n(2l)1T/μf_n=n(2l)^{-1}\sqrt{T/\mu}.
  • In the first harmonic, l=λ/2l=\lambda/2 and f=(2l)1T/μf=(2l)^{-1}\sqrt{T/\mu}. A common error is to set l=λl=\lambda.
  • A stationary wave has no net energy transfer along the pattern; particles within one loop oscillate in phase, while particles in adjacent loops are in antiphase.

Tier 1 · Easy

1 mark
ORIGINAL

State the difference between a node and an antinode in a stationary wave.

Tier 2 · Standard

3 marks
ORIGINAL

A string of length 0.75m0.75\,\text{m} is fixed at both ends. Its tension is 45N45\,\text{N} and its mass per unit length is 1.8×103kg m11.8\times10^{-3}\,\text{kg m}^{-1}. Determine the frequency of its first harmonic.

Tier 3 · Hard

5 marks
ORIGINAL

A 0.90m0.90\,\text{m} string fixed at both ends has mass per unit length 2.5×103kg m12.5\times10^{-3}\,\text{kg m}^{-1}. Its third-harmonic stationary wave has frequency 120Hz120\,\text{Hz}. Determine the string tension and the numbers of nodes and antinodes in this pattern.

3.3.2.1

Interference

  • Coherent sources have a constant phase difference and the same frequency. Coherence is required for a stable interference pattern.
  • Constructive interference occurs for path difference nλn\lambda; destructive interference occurs for path difference (n+12)λ(n+\tfrac12)\lambda when the sources are in phase.
  • For Young's double slits at small angles, fringe spacing is w=λD/sw=\lambda D/s, where DD is slit-to-screen distance and ss is slit separation.
  • Measure across several fringe spacings and divide to reduce percentage uncertainty. Keep λ\lambda, DD, ss and ww in consistent units.
  • A laser supplies monochromatic coherent light, but laser beams must never be viewed directly or through an optical instrument.

Tier 1 · Easy

2 marks
ORIGINAL

State the two conditions that two sources must satisfy to be coherent.

Tier 2 · Standard

3 marks
ORIGINAL

Light of wavelength 600nm600\,\text{nm} illuminates two slits separated by 0.40mm0.40\,\text{mm}. A screen is 2.4m2.4\,\text{m} from the slits. Calculate the fringe spacing.

Tier 3 · Hard

5 marks
ORIGINAL

In a Young double-slit experiment, the distance from the first to the ninth bright fringe is 28.8mm28.8\,\text{mm}. The screen is 1.80m1.80\,\text{m} from the slits and the wavelength is 520nm520\,\text{nm}. Determine the slit separation. Explain the appearance near the centre when the laser is replaced by white light.

3.3.2.2

Diffraction

  • Diffraction is the spreading of a wave after it passes through an aperture or around an obstacle; it is most pronounced when the aperture is comparable with the wavelength.
  • For a single slit, decreasing slit width or increasing wavelength increases the angular width of the central maximum.
  • For a plane transmission grating at normal incidence, maxima satisfy dsinθ=nλd\sin\theta=n\lambda, where dd is grating spacing and nn is the order.
  • Convert a line density NN into spacing with d=1/Nd=1/N in SI units, and reject an order if the calculation requires sinθ>1\sin\theta>1.
  • A monochromatic single-slit pattern has a broad bright central maximum and weaker side maxima; white light gives a white centre with coloured edges.

Tier 1 · Easy

1 mark
ORIGINAL

State how the width of the central diffraction maximum changes when monochromatic light passes through a narrower single slit.

Tier 2 · Standard

3 marks
ORIGINAL

A diffraction grating has 400400 lines per mm\text{mm}. Light of wavelength 589nm589\,\text{nm} is incident normally. Calculate the angle of the first-order maximum.

Tier 3 · Hard

5 marks
ORIGINAL

A plane transmission grating has 600600 lines per mm\text{mm}. It is illuminated normally with light of wavelength 480nm480\,\text{nm}. Determine the highest observable order and the angle of this order. Explain why the next order cannot occur.

3.3.2.3

Refraction at a plane surface

  • The refractive index is n=c/vn=c/v. At a boundary, Snell's law is n1sinθ1=n2sinθ2n_1\sin\theta_1=n_2\sin\theta_2, with angles measured from the normal.
  • Total internal reflection occurs only when light travels from higher to lower refractive index and the incidence angle exceeds the critical angle, where sinc=n2/n1\sin c=n_2/n_1.
  • A step-index fibre has a higher-index core and lower-index cladding, allowing total internal reflection while protecting the boundary and reducing signal leakage between fibres.
  • Modal dispersion arises because rays follow paths of different lengths; material dispersion arises because refractive index and speed depend on wavelength.
  • Dispersion and absorption broaden or weaken pulses, limiting bandwidth and repeater spacing. A common error is to use angles to the surface rather than to the normal.

Tier 1 · Easy

2 marks
ORIGINAL

Light travels through a transparent material at 2.00×108m s12.00\times10^8\,\text{m s}^{-1}. Calculate its refractive index.

Tier 2 · Standard

3 marks
ORIGINAL

A ray in glass of refractive index 1.521.52 meets a glass-air boundary at an incidence angle of 44.044.0^{\circ}. Determine whether total internal reflection occurs.

Tier 3 · Hard

6 marks
ORIGINAL

A step-index optical fibre has core refractive index 1.501.50 and cladding refractive index 1.471.47. A ray in the core meets the boundary at 80.080.0^{\circ} to the normal. Show that the ray is guided. Explain how material dispersion and modal dispersion broaden a light pulse in this fibre.