For the first secondary maximum of a single-slit Fraunhofer diffraction pattern, which approximate pair is correct?
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Question

For the first secondary maximum of a single-slit Fraunhofer diffraction pattern, which approximate pair is correct?

G1 Physics · Accepted Answer

For a single slit, I = I₀[sin u ÷ u]². Secondary maxima satisfy tan u = u, excluding u = 0. The first positive secondary-maximum solution is approximately u ≈ 4.493. Since u = πa sinθ ÷ λ, we get a sinθ ÷ λ = u ÷ π ≈ 4.493 ÷ π ≈ 1.43. Thus, a sinθ ≈ 1.43λ. At u ≈ 4.493, I ÷ I₀ = [sin4.493 ÷ 4.493]² ≈ 0.047. Therefore the first secondary maximum occurs near a sinθ ≈ 1.43λ with intensity about 0.047I₀. Correct option: A.