Question

The waves of $600\mu m$ wave length are incident normally on a slit of $1.2mm$ width. The value of diffraction angle corresponding to the first minima will be (in radian):

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{5}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is B $\frac{\pi }{6}$

Given,

$\lambda =600\mu m$

$d=1.2mm$

condition for minima

$dsin\theta =m\lambda$

for $m=1$

$sin\theta =\frac{m\lambda }{d}=\frac{\lambda }{d}$

$sin\theta =\frac{600\times {10}^{-6}}{1.2\times {10}^{-3}}=0.5$

$\theta =si{n}^{-1}\left(0.5\right)=\frac{\pi }{6}$

The correct option is B.