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Physics Question on Wave optics

In Fraunhofer diffraction pattern, slit width is 0.2 mm0.2\, mm and screen is at 2 m2 \,m away from the lens. If wavelength of light used is 5000 A˚5000 \, \mathring A then the distance between the first minimum on either side of the central maximum is (θ\theta is small and measured in radian)

A

10−1m10^{-1} m

B

10−2m10^{-2} m

C

2×10−2m2 \times 10^{-2} m

D

2×10−1m2 \times 10^{-1} m

Answer

10−2m10^{-2} m

Explanation

Solution

Distance between the first 22 minimum on other side =2λDd=\frac{2 \lambda D }{ d }
=2×5×103×10−10×22×10−1×10−3=\frac{2 \times 5 \times 10^{3} \times 10^{-10} \times 2}{2 \times 10^{-1} \times 10^{-3}}
=10×10−710−4=10 \times \frac{10^{-7}}{10^{-4}}
=10−2 m=10^{-2} \,m