A-Level — physicsMedium4 min read

Superposition and Interference

Principle of superposition; constructive and destructive interference; path difference

Tutor AI Team
Tutor AI Team

# Superposition and Interference — A-Level Physics

When two waves meet, they combine according to the principle of superposition. This leads to interference patterns — a key wave phenomenon that distinguishes waves from particles.

1. Principle of Superposition

When two or more waves meet, the resultant displacement is the sum of the individual displacements.

This applies at each point and each instant.

2. Interference

Constructive Interference

Waves arrive in phase (crests meet crests). Displacements add: Aresultant=A1+A2A_{\text{resultant}} = A_1 + A_2

Condition: path difference = nλn\lambda (n=0,1,2,...n = 0, 1, 2, ...) Phase difference = 0,2π,4π,...0, 2\pi, 4\pi, ...

Destructive Interference

Waves arrive in antiphase (crests meet troughs). Displacements cancel: Aresultant=∣A1−A2∣A_{\text{resultant}} = |A_1 - A_2|

If A1=A2A_1 = A_2: complete cancellation (zero amplitude).

Condition: path difference = (n+12)λ(n + \frac{1}{2})\lambda Phase difference = π,3π,5π,...\pi, 3\pi, 5\pi, ...

3. Coherence

For a stable, observable interference pattern, the sources must be coherent:

  • Same frequency
  • Constant phase relationship

Two independent light sources are typically NOT coherent (random phase changes).

Coherent sources can be produced by:

  • Splitting one source (e.g., double slit)
  • Using lasers (coherent by nature)

4. Young's Double Slit Experiment

Light passes through two narrow slits separated by distance dd. An interference pattern of bright and dark fringes appears on a screen at distance DD.

Fringe Spacing

w=λDd\boxed{w = \frac{\lambda D}{d}}

Where:

  • ww = fringe spacing (m)
  • λ\lambda = wavelength (m)
  • DD = slit-to-screen distance (m)
  • dd = slit separation (m)

Bright fringes (maxima)

Path difference = nλn\lambda → dsin⁡θ=nλd\sin\theta = n\lambda

Dark fringes (minima)

Path difference = (n+12)λ(n + \frac{1}{2})\lambda

Significance

Young's experiment was the first strong evidence that light behaves as a wave (particles cannot produce interference).

5. Interference with White Light

Using white light instead of monochromatic:

  • Central maximum is white (all wavelengths constructively interfere at zero path difference)
  • Higher-order fringes show spectra (different wavelengths have maxima at slightly different positions)
  • Blue is closest to centre (shorter λ\lambda), red furthest
  • Pattern becomes increasingly blurred at higher orders

6. Thin Film Interference

Thin films (soap bubbles, oil on water) show coloured patterns due to interference between light reflected from the top and bottom surfaces.

Path difference = 2nt2nt (where nn = refractive index, tt = thickness)

Phase change of π\pi occurs on reflection from a denser medium.

Worked Example: Double Slit

Problem

Light of wavelength 600 nm passes through slits 0.5 mm apart. Screen is 2 m away. Find the fringe spacing.

w=λD/d=(600×10−9×2)/(0.5×10−3)=1.2×10−6/5×10−4=2.4×10−3w = \lambda D/d = (600 \times 10^{-9} \times 2)/(0.5 \times 10^{-3}) = 1.2 \times 10^{-6}/5 \times 10^{-4} = 2.4 \times 10^{-3} m = 2.4 mm

Solution

Worked Example: Path Difference

Problem

Two speakers emit sound of wavelength 0.5 m. At a point P, the distance from speaker A is 3.25 m and from speaker B is 4.00 m. Is the interference constructive or destructive?

Path difference = 4.00−3.25=0.754.00 - 3.25 = 0.75 m = 1.5λ1.5\lambda This is (1+0.5)λ(1 + 0.5)\lambda → destructive interference.

Solution

Worked Example: Finding Wavelength

Problem

In a double slit experiment, the fringe spacing is 3.5 mm, D=1.5D = 1.5 m, d=0.25d = 0.25 mm. Find the wavelength.

λ=wd/D=(3.5×10−3×0.25×10−3)/1.5=8.75×10−7/1.5=5.83×10−7\lambda = wd/D = (3.5 \times 10^{-3} \times 0.25 \times 10^{-3})/1.5 = 8.75 \times 10^{-7}/1.5 = 5.83 \times 10^{-7} m = 583 nm (yellow/orange)

Solution

8. Practice Questions

    1. State the conditions required for observable interference. (2 marks)
    1. In a double slit experiment with 500 nm light, d=0.4d = 0.4 mm and D=1.8D = 1.8 m. Calculate the fringe spacing. (2 marks)
    1. Two coherent sources emit waves of wavelength 3 cm. At a point, the path difference is 4.5 cm. State whether the interference is constructive or destructive. (2 marks)
    1. Explain why two separate light bulbs do not produce an interference pattern. (2 marks)

    Answers

    1. Sources must be coherent (same frequency, constant phase relationship). Waves must have similar amplitudes for clear fringes.

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Summary

  • Superposition: resultant = sum of individual displacements
  • Constructive: path difference = nλn\lambda; Destructive: path difference = (n+½)λ(n+½)\lambda
  • Coherent sources needed for stable interference
  • Young's double slit: w=λD/dw = \lambda D/d
  • White light gives coloured fringes; central fringe is white
Tags:
#a-level#physics#waves#superposition#interference#path-difference

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