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- 2070 Asar · 5 marks
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Intensity distribution in interference
Short answer
Interference is the modification of intensity when coherent waves superpose. For amplitudes A₁, A₂ and phase difference δ, resultant amplitude squared is A₁² + A₂² + 2A₁A₂ cos δ, so intensity I = A₁² + A₂² + 2A₁A₂ cos δ. Maxima at δ = 2nπ, minima at δ = (2n+1)π.
What is interference? Explain the intensity distribution in interference with mathematical treatment. (Show that two coherent sources of constant phase difference δ and amplitudes A1 and A2 superimpose to give intensity I = A1² + A2² + 2A1A2 cos δ.)
Interference is the modification of intensity produced when two or more coherent light waves superpose, giving maxima and minima in the resultant intensity.
Intensity distribution
S1 o--------\
\--> P path difference Δ
S2 o--------/
- Let two coherent waves of amplitudes A₁ and A₂ and phase difference δ reach point P: y₁ = A₁ sin ωt, y₂ = A₂ sin(ωt + δ)
- By the principle of superposition: y = y₁ + y₂ = A₁ sin ωt + A₂ sin(ωt + δ)
- Expand: y = sin ωt (A₁ + A₂ cos δ) + cos ωt (A₂ sin δ)
- Put A₁ + A₂ cos δ = A cos φ and A₂ sin δ = A sin φ, so y = A sin(ωt + φ).
- Squaring and adding:
A² = (A₁ + A₂ cos δ)² + (A₂ sin δ)²
= A₁² + A₂² + 2 A₁A₂ cos δ
- Intensity is proportional to the square of the amplitude, so
I = A₁² + A₂² + 2 A₁A₂ cos δ
Maxima and minima
- Maximum: cos δ = 1, δ = 2nπ (path difference nλ): I_max = (A₁ + A₂)²
- Minimum: cos δ = −1, δ = (2n+1)π (path difference (2n+1)λ/2): I_min = (A₁ − A₂)²
If A₁ = A₂ = a: I = 4a² cos²(δ/2), varying between 4a² and 0. The average intensity is 2a², equal to the sum of separate intensities, so energy is conserved: it is only redistributed.
✓ Answer checked by Claude Opus 5.5 (high effort)
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