IB — physicsHard2 min read

Superposition and Interference

Principle of superposition; constructive and destructive interference; Young's double slit; path difference

Tutor AI Team
Tutor AI Team

# Superposition and Interference — IB Physics

When two or more waves overlap, they combine according to the principle of superposition. This leads to interference — constructive or destructive — which is fundamental evidence for wave behaviour.

1. Principle of Superposition

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

2. Constructive Interference

Path difference = nλn\lambda (n=0,1,2,...n = 0, 1, 2, ...) Waves arrive in phase → amplitudes add → maximum intensity.

3. Destructive Interference

Path difference = (n+12)λ(n + \frac{1}{2})\lambda Waves arrive in antiphase → amplitudes cancel → minimum/zero intensity.

4. Coherent Sources

For a stable interference pattern, sources must be:

  • Same frequency
  • Constant phase difference

5. Young's Double Slit

s=λDd\boxed{s = \frac{\lambda D}{d}}

Where ss = fringe spacing, DD = slit-to-screen distance, dd = slit separation.

Maxima at: dsin⁡θ=nλd\sin\theta = n\lambda

Significance: proved light is a wave (1801).

6. Thin-Film Interference

Light reflected from top and bottom surfaces of a thin film interfere. Phase change of π\pi (half wavelength) occurs when reflecting from a denser medium.

Condition for constructive (in reflected light with one phase change): 2nt=(m+12)λ2nt = (m + \frac{1}{2})\lambda

Worked Example: Example 1

Problem

Double slit: λ=600\lambda = 600 nm, d=0.5d = 0.5 mm, D=2D = 2 m. s=λD/d=600×10−9×2/0.5×10−3=2.4s = \lambda D/d = 600 \times 10^{-9} \times 2/0.5 \times 10^{-3} = 2.4 mm

Solution

Worked Example: Example 2

Problem

Two speakers 3 m apart emit sound of λ=0.5\lambda = 0.5 m. At point P, distances are 4.0 m and 4.75 m. Constructive or destructive?

Path difference = 0.75 m = 1.5λ1.5\lambda = (1+0.5)λ(1 + 0.5)\lambda → destructive.

Solution

8. Practice Questions

    1. State the principle of superposition. (1 mark)
    1. In a double-slit experiment with 500 nm light, d=0.3d = 0.3 mm, D=1.5D = 1.5 m. Find the fringe spacing. (2 marks)
    1. Explain why two separate light bulbs do not produce visible interference. (2 marks)
    1. Calculate the path difference that gives the 3rd order bright fringe. (1 mark)

    Answers

    1. When two or more waves overlap, the resultant displacement at any point is the sum of the individual displacements.

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Summary

  • Superposition: resultant = sum of displacements
  • Constructive: path diff = nλn\lambda; Destructive: path diff = (n+½)λ(n+½)\lambda
  • s=λD/ds = \lambda D/d (double slit fringe spacing)
  • Coherent sources needed for stable interference pattern
Tags:
#ib#physics#waves#superposition#interference#double-slit#path-difference

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