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Question

In a diffraction experiment, if the slit width is \( 4.0 \, \mu\text{m} \) and the wavelength is \( 800
\, \text{nm} \), what is the angular position of the first minimum?

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Explanation

**Diffraction bending** property of light waves causes bending around corners, width of central maximum inversely proportional to slit width, intensity of secondary maxima decreases with order because less constructive interference, angular position of minima θ_n = n λ/a, n=±1,±2..., second minimum n=2, third n=3, condition for third secondary maximum approx a sinθ = (2n+1)λ/2. First minimum occurs at sin θ = (λ/a) . λ = 800 nm = 8.0 × 10⁻⁷ m , a = 4.0 μm = 4.0 × 10⁻⁶ m . sin θ = (8.0 × 10⁻⁷/4.0 × 10⁻⁶) = 0.2 , so θ = sin⁻¹(0.2) ≈ 11.5° . Using Δ = d

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