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How do coherent waves interfere, and how does a diffraction grating split light into its colours?

Coherence and path difference, constructive and destructive interference, and the diffraction grating equation relating wavelength, slit spacing and the angle to a maximum.

An SQA Higher Physics answer on interference and diffraction, covering coherence and path difference, constructive and destructive interference, and the diffraction grating equation relating wavelength, slit spacing and the angle to a maximum.

Generated by Claude Opus 4.811 min answer

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  1. What this key area is asking
  2. Coherence and path difference
  3. Constructive and destructive interference
  4. The diffraction grating
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to define coherence and path difference, state the conditions for constructive and destructive interference, and apply the diffraction grating equation dsin⁡θ=mλd\sin\theta = m\lambda to find wavelengths, angles or slit spacings.

Coherence and path difference

A steady interference pattern needs coherent sources, which is why interference experiments use a single light source split into two (for example a laser through two slits) rather than two independent lamps. The pattern that appears depends only on the path difference at each point on the screen.

Constructive and destructive interference

When two coherent waves arrive in phase (crest meeting crest) they reinforce, doubling the amplitude. When they arrive exactly out of phase (crest meeting trough) they cancel. The alternating bright and dark fringes on the screen are simply the places where the path difference satisfies each condition.

The diffraction grating

A diffraction grating has many equally spaced slits. Light passing through is diffracted, and the beams from all the slits interfere, producing sharp bright maxima at specific angles.

The zero-order maximum (m=0m = 0) is straight ahead and contains all colours. Higher orders appear at larger angles, and because the angle depends on wavelength, the grating spreads white light into a spectrum, with red (longer wavelength) deflected more than blue. A grating with more lines per millimetre has a smaller dd and so spreads the spectrum out more.

Examples in context

A CD or DVD acts as a reflection grating: the closely spaced tracks split white light into the rainbow colours you see on the disc. A spectrometer in a chemistry lab uses a diffraction grating to spread light from a source into its spectral lines, identifying the elements present (the link to emission spectra). Noise-cancelling headphones use destructive interference of sound waves to cancel unwanted noise. Anti-reflection coatings on lenses and solar cells use a thin film whose thickness makes reflected waves interfere destructively, reducing glare and increasing transmitted light.

Try this

Q1. State the path-difference condition for constructive interference. [1 mark]

  • Cue. The path difference equals a whole number of wavelengths, mλm\lambda.

Q2. A grating has 400400 lines per millimetre. Calculate its slit spacing. [2 marks]

  • Cue. d=1×10−3400=2.5×10−6 md = \frac{1 \times 10^{-3}}{400} = 2.5 \times 10^{-6}\ \text{m}.

Q3. Light of wavelength 6.0×10−7 m6.0 \times 10^{-7}\ \text{m} passes through a grating with d=3.0×10−6 md = 3.0 \times 10^{-6}\ \text{m}. Calculate the angle to the first-order maximum. [3 marks]

  • Cue. sin⁡θ=mλd=1×6.0×10−73.0×10−6=0.20\sin\theta = \frac{m\lambda}{d} = \frac{1 \times 6.0 \times 10^{-7}}{3.0 \times 10^{-6}} = 0.20, so θ=11.5∘\theta = 11.5^{\circ}.

Exam-style practice questions

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

SQA Higher 20184 marksMonochromatic light of wavelength 5.9 times ten to the power minus seven m is shone through a diffraction grating with 300 lines per millimetre. Calculate the slit spacing and the angle to the first-order maximum.
Show worked answer →

Slit spacing: 300 lines per millimetre means d=1×10−3300=3.33×10−6d = \frac{1 \times 10^{-3}}{300} = 3.33 \times 10^{-6} m.

First-order maximum uses the grating equation. Relationship: dsin⁡θ=mλd\sin\theta = m\lambda with m=1m = 1.

Substitution: sin⁡θ=1×5.9×10−73.33×10−6=0.177\sin\theta = \frac{1 \times 5.9 \times 10^{-7}}{3.33 \times 10^{-6}} = 0.177.

Answer: θ=sin⁡−1(0.177)=10.2\theta = \sin^{-1}(0.177) = 10.2 degrees.

Markers reward converting lines per millimetre to a spacing, using m=1m = 1, and the angle. Forgetting to invert the line count to get the spacing is the usual error.

SQA Higher 20223 marksExplain what is meant by coherent waves, and state the condition on the path difference for constructive interference and for destructive interference.
Show worked answer →

Coherent waves have a constant phase relationship and the same frequency (and so the same wavelength). This is why a single laser, or light from one source split into two paths, is used to produce a steady interference pattern.

Constructive interference (a bright fringe or maximum) occurs where the path difference is a whole number of wavelengths: path difference=mλ\text{path difference} = m\lambda for integer mm.

Destructive interference (a dark fringe or minimum) occurs where the path difference is an odd number of half-wavelengths: path difference=(m+12)λ\text{path difference} = (m + \tfrac{1}{2})\lambda.

Markers reward the definition of coherence, the whole-wavelength condition for maxima, and the odd-half-wavelength condition for minima.

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