The angular width of the central maximum in a single-slit diffraction pattern is 60°. The slit width is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young’s fringes are observed on a screen placed 50 cm from the slits. If the observed fringe width is 1 cm, what is the slit separation (distance between the centres of the two slits)?
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Question

The angular width of the central maximum in a single-slit diffraction pattern is 60°. The slit width is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young’s fringes are observed on a screen placed 50 cm from the slits. If the observed fringe width is 1 cm, what is the slit separation (distance between the centres of the two slits)?

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G1 Physics · Accepted Answer

For a single slit, the angular width of the central maximum is 2θ, where the first minimum satisfies a sinθ = λ. Given 2θ = 60°, so θ = 30°. Hence, λ = a sin30° = (1 μm)(1/2) = 0.5 μm. In Young’s double-slit experiment, β = λD/d. Here β = 1 cm = 10⁻² m, D = 50 cm = 0.5 m, and λ = 0.5 μm = 0.5×10⁻⁶ m. Therefore, d = λD/β = (0.5×10⁻⁶ × 0.5) / 10⁻² = 25×10⁻⁶ m = 25 μm. Therefore the correct answer is D.