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How do superposition, interference, diffraction and stationary waves arise, and what do they reveal about light?

Wave properties: the principle of superposition, two-source interference and the Young double-slit experiment, the diffraction grating, and stationary waves with nodes and antinodes.

A focused answer to the Eduqas A-Level Physics Component 3 wave properties content, covering the principle of superposition, two-source interference and the Young double-slit experiment, the diffraction grating equation, and stationary waves with their nodes and antinodes.

Generated by Claude Opus 4.813 min answer

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  1. What this dot point is asking
  2. The answer
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What this dot point is asking

Eduqas wants you to state the principle of superposition, explain two-source interference and analyse the Young double-slit experiment with y=λDay = \frac{\lambda D}{a}, use the diffraction grating equation dsin⁡θ=nλd\sin\theta = n\lambda, and describe stationary (standing) waves with their nodes and antinodes and the conditions for resonance.

The answer

The principle of superposition

Two-source interference and the double slit

The diffraction grating

Stationary waves

Examples in context

Interference and diffraction gratings are the basis of spectroscopy, used to identify elements in stars and laboratory samples from their characteristic spectral lines. Thin-film interference produces the colours of soap bubbles and oil films and underlies anti-reflection coatings on lenses. Stationary waves explain the notes of stringed and wind instruments and the resonant modes of microwave ovens and laser cavities.

Try this

Q1. State the principle of superposition. [1 mark]

  • Cue. When waves overlap, the resultant displacement is the vector sum of the individual displacements.

Q2. Light of wavelength 500 nm500\ \text{nm} falls on a grating with spacing 2.0×10−6 m2.0 \times 10^{-6}\ \text{m}. Find the first-order diffraction angle. [2 marks]

  • Cue. sin⁡θ=nλd=500×10−92.0×10−6=0.25\sin\theta = \frac{n\lambda}{d} = \frac{500 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.25, so θ=14.5∘\theta = 14.5^\circ.

Q3. State the distance between adjacent nodes on a stationary wave of wavelength λ\lambda. [1 mark]

  • Cue. Half a wavelength, λ2\frac{\lambda}{2}.

Exam-style practice questions

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

Eduqas 20194 marksIn a Young double-slit experiment, slits separated by 0.50 mm0.50\ \text{mm} are illuminated by light of wavelength 590 nm590\ \text{nm}. The fringes are observed on a screen 2.4 m2.4\ \text{m} away. Calculate the fringe spacing.
Show worked answer →

Fringe spacing from y=λDay = \dfrac{\lambda D}{a}, where aa is the slit separation and DD the screen distance.

y=(590×10−9)(2.4)0.50×10−3=1.416×10−60.50×10−3=2.83×10−3 my = \dfrac{(590 \times 10^{-9})(2.4)}{0.50 \times 10^{-3}} = \dfrac{1.416 \times 10^{-6}}{0.50 \times 10^{-3}} = 2.83 \times 10^{-3}\ \text{m}, about 2.8 mm2.8\ \text{mm}.

Markers reward y=λDay = \frac{\lambda D}{a}, correct powers of ten, and the fringe spacing about 2.8 mm2.8\ \text{mm}.

Eduqas 20224 marksMonochromatic light of wavelength 633 nm633\ \text{nm} is shone normally onto a diffraction grating with 300 lines per millimetre300\ \text{lines per millimetre}. Calculate the angle of the first-order maximum.
Show worked answer →

The grating spacing is d=1300×103 lines per metre=3.33×10−6 md = \dfrac{1}{300 \times 10^{3}\ \text{lines per metre}} = 3.33 \times 10^{-6}\ \text{m}.

First order (n=1n = 1) from dsin⁡θ=nλd\sin\theta = n\lambda: sin⁡θ=nλd=633×10−93.33×10−6=0.190\sin\theta = \dfrac{n\lambda}{d} = \dfrac{633 \times 10^{-9}}{3.33 \times 10^{-6}} = 0.190.

θ=sin⁡−1(0.190)=11.0∘\theta = \sin^{-1}(0.190) = 11.0^\circ.

Markers reward finding dd from the lines per metre, dsin⁡θ=nλd\sin\theta = n\lambda, and the first-order angle about 11∘11^\circ.

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