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Physics Question on single slit diffraction

In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is 5 mm. The screen on which the diffraction pattern is displayed is at a distance of 80 cm from the slit. The wavelength is 6000AA6000{AA} The slit width (in mm) is about

A

0.576

B

0.348

C

0.192

D

0.096

Answer

0.192

Explanation

Solution

Slit width, w=Dλdw=\frac{D\lambda }{d} Given, D=80 cm=80×10−2m,D=80\text{ }cm=80\times {{10}^{-2}}m, λ=6000A =6000×10−10m,\lambda =6000{A}\text{ }=6000\times {{10}^{-10}}m, d=52mm=5×10−32md=\frac{5}{2}mm=\frac{5\times {{10}^{-3}}}{2}m ∴\therefore w=80×10−2×6000×10−10×25×10−3w=\frac{80\times {{10}^{-2}}\times 6000\times {{10}^{-10}}\times 2}{5\times {{10}^{-3}}} w=0.192×10−3m=0.192 mmw=0.192\times {{10}^{-3}}m=0.192\text{ }mm