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How do superposition, interference and diffraction let us measure wavelength?

Superposition and coherence, two-source interference and path difference, diffraction, and the diffraction grating equation.

A focused answer to WJEC A-Level Physics Unit 2 wave properties, covering superposition and coherence, two-source interference and path difference, diffraction, and using the diffraction grating equation to measure the wavelength of light.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

WJEC wants you to apply the principle of superposition, explain coherence and interference, relate constructive and destructive interference to path difference, describe diffraction, and use the diffraction grating equation. The grating equation is a dependable calculation question, and the conceptual marks come from explaining coherence and the path-difference conditions clearly.

The answer

Superposition and coherence

Where two crests coincide the waves reinforce (constructive interference); where a crest meets a trough they cancel (destructive interference).

Interference and path difference

Diffraction

Diffraction is the spreading of waves as they pass through a gap or around an obstacle. The spreading is greatest when the gap width is comparable to the wavelength, which is why sound (long wavelength) diffracts round a doorway but light (very short wavelength) barely does.

The diffraction grating

A diffraction grating has many equally spaced slits a distance dd apart. Sharp bright maxima appear at angles where dsin⁡θ=nλd\sin\theta = n\lambda, with nn the order. Measuring the angle of a known order lets you find the wavelength precisely, a specified practical.

Examples in context

Example 1. Reading a spectrum from a star. Astronomers pass starlight through a diffraction grating to spread it into a spectrum. Because dsin⁡θ=nλd\sin\theta = n\lambda sends each wavelength to a different angle, the dark absorption lines appear at precise positions, revealing the elements in the star and, through any shift, its motion. The grating is the workhorse instrument of spectroscopy.

Example 2. The colours on a CD. A CD has data tracks spaced about 1.6 μm1.6\,\mu\text{m} apart, acting as a reflection grating. White light reflecting off it satisfies dsin⁡θ=nλd\sin\theta = n\lambda at different angles for different colours, so you see the surface flash rainbow colours. The same equation that measures a star's wavelength explains the shimmer on a disc in your hand.

Try this

Q1. Light of wavelength 600 nm600\,\text{nm} hits a grating with 300 000300\,000 lines per metre. Find the angle of the first-order maximum. [3 marks]

  • Cue. d=1300000=3.33×10−6 md = \frac{1}{300000} = 3.33\times10^{-6}\,\text{m}; sin⁡θ=nλd=600×10−93.33×10−6=0.18\sin\theta = \frac{n\lambda}{d} = \frac{600\times10^{-9}}{3.33\times10^{-6}} = 0.18, so θ≈10.4∘\theta \approx 10.4^\circ.

Q2. State the path-difference condition for constructive interference. [1 mark]

  • Cue. A whole number of wavelengths, nλn\lambda.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20206 marksMonochromatic light passes through a diffraction grating with 500500 lines per millimetre. The second-order maximum is observed at an angle of 36.0∘36.0^{\circ}. Calculate the wavelength of the light and the angle of the first-order maximum.
Show worked answer →

First find the slit spacing dd, then use dsin⁡θ=nλd\sin\theta = n\lambda.

Slit spacing: d=1500×103=2.0×10−6 md = \dfrac{1}{500 \times 10^{3}} = 2.0 \times 10^{-6}\,\text{m} (500 lines per mm is 5.0×1055.0 \times 10^{5} lines per metre).

Wavelength from the second order (n=2n = 2):

λ=dsin⁡θn=2.0×10−6×sin⁡36.0∘2=2.0×10−6×0.5882=5.88×10−7 m\lambda = \dfrac{d\sin\theta}{n} = \dfrac{2.0 \times 10^{-6} \times \sin 36.0^{\circ}}{2} = \dfrac{2.0 \times 10^{-6} \times 0.588}{2} = 5.88 \times 10^{-7}\,\text{m}.

First-order angle (n=1n = 1): sin⁡θ1=nλd=1×5.88×10−72.0×10−6=0.294\sin\theta_1 = \dfrac{n\lambda}{d} = \dfrac{1 \times 5.88 \times 10^{-7}}{2.0 \times 10^{-6}} = 0.294, so θ1=17.1∘\theta_1 = 17.1^{\circ}.

Markers reward the slit spacing from lines per metre, the wavelength near 590 nm590\,\text{nm}, and the first-order angle.

WJEC 20173 marksExplain why two separate filament lamps cannot produce an observable interference pattern, whereas a double slit illuminated by a single laser can.
Show worked answer →

A stable, observable interference pattern requires two coherent sources, meaning they have the same frequency and a constant phase difference.

Two separate filament lamps emit light with randomly and rapidly changing phase, so the phase difference between them varies continuously. The interference pattern shifts far faster than the eye can follow and averages out to uniform brightness.

A double slit lit by one laser splits a single coherent beam into two, so the two slits act as coherent sources with a fixed phase relationship and produce a stable pattern of bright and dark fringes. Markers reward the need for coherence and explaining why independent lamps fail to provide it.

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