Practice question
Question
What is the phase difference corresponding to a path difference of \( 3\lambda/2 \) in a double-slit
experiment?
Explanation
**Phase difference** corresponding to path difference Δ, φ =2π Δ/λ, for Δ=5λ/8 φ=5π/4, for Δ=9λ/4 φ=9π/2, for Δ=λ path difference φ=2π constructive, but for destructive condition path difference λ can be destructive if one reflection introduces π phase shift, resultant amplitude zero when φ=(2n+1)π. Phase difference Φ = (2π/λ) Δ . For Δ = (3λ/2) , Φ = (2π/λ) · (3λ/2) = 3π . 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 3π, illustrating interference, diffraction and polarization principles.
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