In Fraunhofer diffraction pattern, slit width is 0.3 mm and screen is at 1.5 m away from the lens. If wavelength of light used is 4500, then the distance between the first minimum on either side of the central maximum is [ θ is small and measured in radian.] (a) 1.5 mm (b) 2.25 mm (c) 3.25 mm (d) 4.5 mm

Answer: (d). Distance between first minima on either side = (2λ D)/(a) = 2 × 4500 × 10⁻¹⁰ × 1.50.3 × 10⁻³ = 4.5 × 10⁻³ m So the answer is (d) 4.5 mm.

In Fraunhofer diffraction pattern, slit width is 0.3 mm and screen is… | MHT CET 2025 (25 Apr, Shift 1)

Q18MHT CET 2025 · 25 APR · SHIFT 1Medium

Diffraction at a Single Slit

In Fraunhofer diffraction pattern, slit width is 0.3 mm and screen is at 1.5 m away from the lens. If wavelength of light used is $4500\text{ \AA}$ , 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: (d)

Distance between first minima on either side $= \dfrac{2\lambda D}{a}$
$= \dfrac{2 \times 4500 \times 10^{-10} \times 1.5}{0.3 \times 10^{-3}} = 4.5 \times 10^{-3}$ m
So the answer is (d) $4.5$ mm.

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