Light of wavelength 5000 is incident normally on a slit. The first minimum of the diffraction pattern is observed to lie at a distance of 5 mm from the central maximum on a screen placed at a distance of 2 m from the slit. The width of the slit is (a) 2 cm (b) 0.2 cm (c) 0.02 cm (d) 0.01 cm

Answer: (c). First minimum: asin θ = λ, with sin θ ≈ y/D. a = (λ D)/(y) = 5×10⁻⁷× 25×10⁻³ = 2×10⁻⁴ m. a = 0.02 cm. So the answer is (c).

Light of wavelength 5000 is incident normally on a slit. The first… | MHT CET 2023 (13 May, Shift 2)

Q42MHT CET 2023 · 13 MAY · SHIFT 2Easy

Diffraction at a Single Slit

Light of wavelength $5000\text{ \AA}$ is incident normally on a slit. The first minimum of the diffraction pattern is observed to lie at a distance of $5\text{ mm}$ from the central maximum on a screen placed at a distance of $2\text{ m}$ from the slit. The width of the slit is

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Answer: (c)

First minimum: $a\sin\theta = \lambda$ , with $\sin\theta \approx \frac{y}{D}$ .
$a = \frac{\lambda D}{y} = \frac{5\times10^{-7}\times 2}{5\times10^{-3}} = 2\times10^{-4}\ \text{m}$ .
$a = 0.02\ \text{cm}$ . So the answer is (c).

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Diffraction at a Single Slit question from MHT CET 2023 (13 May, Shift 2)

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