Question

In a Fraunhofer diffraction at single slit of width d with incident light of wavelength $5500A^0$, the first minimum is observed at angle ${30}^0$. The first secondary maximum is observed at an angle $\theta$ =

• A. $si{n}^{-1}\left[\frac{1}{\sqrt{2}}\right]$
• B. $si{n}^{-1}\left[\frac{1}{4}\right]$
• C. $si{n}^{-1}\left[\frac{3}{4}\right]$
• D. $si{n}^{-1}\left[\frac{\sqrt{3}}{2}\right]$

Answer 1

Answer: (C)

$si{n}^{-1}\left[\frac{3}{4}\right]$

Slit width = d
$\lambda =5500A^0=5.5\times {10}^{-7}m,{\theta }_n={30}^0$
For first secondary minima, $dsin{\theta }_n=\lambda$
$d=\frac{\lambda }{sin{\theta }_n}=\frac{5.5\times {10}^{-7}}{sin{30}^0}=11\times {10}^{-7}$m
For the first secondary maxima, $dsin{\theta }_n=\frac{3\lambda }{2}$
i.e. $sin{\theta }_n=\frac{3\lambda }{2d}=\frac{3\times 5.5\times {10}^{-7}}{2\times 11\times {10}^{-7}}=sin{\theta }_n=\frac{3}{4}or{\theta }_n=si{n}^{-1}\left(3/4\right)$