Practice question
Question
What is the angular position of the second secondary maximum in a single-slit diffraction pattern if
the slit width is \( 7.0 \, \mu\text{m} \) and the wavelength is \( 560 \, \text{nm} \)?
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. Secondary maxima occur at θ ≈ ((n + (1/2))λ/a) . For the second secondary maximum, n = 2 . λ = 5.6 × 10⁻⁷ m , a = 7.0 × 10⁻⁶ m . sin θ = ((2 + (1/2)) × 5.6 × 10⁻⁷/7.0 × 10⁻⁶) = (2.5 × 5.6 × 10⁻⁷/7.0 × 10⁻⁶) = 0.2 , θ = sin⁻¹(0.2) ≈ 11.5°
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