Practice question
Question
What is the angular width of the central maximum in a single-slit diffraction pattern if the slit width
is \( 3.0 \, \mu\text{m} \) and the wavelength is \( 450 \, \text{nm} \)?
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. Angular width 2θ = (2λ/a) . λ = 4.5 × 10⁻⁷ m , a = 3.0 × 10⁻⁶ m . sin θ = (λ/a) = (4.5 × 10⁻⁷/3.0 × 10⁻⁶) = 0.15 , θ = sin⁻¹(0.15) ≈ 8.6° , 2θ ≈ 17.2° . Using Δ = d sinθ, y = n
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