In Fraunhofer diffraction pattern, slitwidth is 0.5 mm and screen is at 2 m away from the lens. If wavelength of light used is 5500, then the distance between the first minimum on either side of the central maximum is ( θ is small and measured in radian) (a) 1.1 mm (b) 2.2 mm (c) 4.4 mm (d) 5.5 mm

Answer: (c). Distance between first minima on either side: W = (2Dλ)/(a). W = 2 × 2 × 5500×10⁻¹⁰0.5×10⁻³ = 4.4×10⁻³ m So the answer is (c) 4.4 mm.

In Fraunhofer diffraction pattern, slitwidth is 0.5 mm and screen is… | MHT CET 2024 (11 May, Shift 1)

Q29MHT CET 2024 · 11 MAY · SHIFT 1Easy

Diffraction at a Single Slit

In Fraunhofer diffraction pattern, slitwidth is 0.5 mm and screen is at 2 m away from the lens. If wavelength of light used is $5500\text{ \AA}$ , then the distance between the first minimum on either side of the central maximum is ( $\theta$ is small and measured in radian)

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Answer: (c)

Distance between first minima on either side: $W = \dfrac{2D\lambda}{a}$ .
$W = \dfrac{2 \times 2 \times 5500\times10^{-10}}{0.5\times10^{-3}} = 4.4\times10^{-3}\text{ m}$
So the answer is (c) $4.4\text{ mm}$ .

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