Question

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 $ 6000{AA} $ The slit width (in mm) is about

• A. 0.576
• B. 0.348
• C. 0.192
• D. 0.096

Answer 1

The Correct Option is C

Slit width, $ w=\frac{D\lambda }{d} $ Given, $ D=80\text{ }cm=80\times {{10}^{-2}}m, $ $ \lambda =6000{A}\text{ }=6000\times {{10}^{-10}}m, $ $ d=\frac{5}{2}mm=\frac{5\times {{10}^{-3}}}{2}m $ $ \therefore $ $ w=\frac{80\times {{10}^{-2}}\times 6000\times {{10}^{-10}}\times 2}{5\times {{10}^{-3}}} $ $ w=0.192\times {{10}^{-3}}m=0.192\text{ }mm $