Practice question
Question
What is the intensity at a point in a double-slit experiment where the phase difference is \( 2\pi \),
if the maximum intensity is \( 4I_0 \)?
Explanation
**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. Intensity I = 4I₀ cos²(Φ/2) . For Φ = 2π , I = 4I₀ cos²(π) = 4I₀ × 1 = 4I₀ . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives 4I₀, illustrating interference, diffraction and polarization principles.
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