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Question

What is the angular position of the first secondary maximum in a single-slit diffraction pattern if the
slit width is \( 6.0 \, \mu\text{m} \) and the wavelength is \( 600 \, \text{nm} \)?

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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. First secondary maximum occurs at θ ≈ (3λ/2a) . λ = 6.0 × 10⁻⁷ m , a = 6.0 × 10⁻⁶ m . sin θ = (3 × 6.0 × 10⁻⁷/2 × 6.0 × 10⁻⁶) = 0.15 , θ = sin⁻¹(0.15) ≈ 8.6° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I =

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