In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2m away from the lens. If wavelength of light used is 5000 then the distance between the first minimum on either side of the central maximum is ( θ is small and measured in radian) (a) 2 × 10 ⁻² m (b) 10 ⁻¹ m (c) 10 ⁻² m (d) 10 ⁻³ m

Answer: (d). Distance between first minima on either side: W = (2Dλ)/(a). W = 2 × 2 × 5000×10⁻¹⁰0.2×10⁻³ = 2×10⁻⁶2×10⁻⁴ = 10⁻² m So the working gives (c) 10⁻² m ( 1 cm).

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is… | MHT CET 2021 (20 Sep, Shift 1)

Q57MHT CET 2021 · 20 SEP · SHIFT 1Easy

Diffraction at a Single Slit

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2m away from the lens. If wavelength of light used is 5000 $\text{\AA}$ 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: $W = \dfrac{2D\lambda}{a}$ .
$W = \dfrac{2 \times 2 \times 5000\times10^{-10}}{0.2\times10^{-3}} = \dfrac{2\times10^{-6}}{2\times10^{-4}} = 10^{-2}\text{ m}$
So the working gives (c) $10^{-2}\text{ m}$ ( $1\text{ cm}$ ).

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