In a single-slit diffraction experiment, light of wavelength 628 nm falls on a slit of width 0.2 mm. If the angular width of the central maximum is α×10⁻² degrees, find α to the nearest integer.
Physics · Wave Optics · diffraction of light
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Question

In a single-slit diffraction experiment, light of wavelength 628 nm falls on a slit of width 0.2 mm. If the angular width of the central maximum is α×10⁻² degrees, find α to the nearest integer.

G1 Physics · Accepted Answer

For a single slit, the angular width of the central maximum is approximately Δθ = 2λ ÷ a in radians. Given, λ = 628 nm = 628×10⁻⁹ m, a = 0.2 mm = 0.2×10⁻³ m. Thus, Δθ = 2×628×10⁻⁹ ÷ (0.2×10⁻³) = 6.28×10⁻³ rad. Convert to degrees: Δθ = 6.28×10⁻³×180 ÷ π ≈ 0.360°. Therefore, Δθ ≈ 36×10⁻² degrees. Hence, α = 36.