Question

A wave of $600\text{µm}$ wavelength is incident normally on a slit of $1.2\text{mm}$ width. The value of diffraction angle corresponding to the first minima will be (in radians) -

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

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

For diffraction minima,
a $\sin \theta =m\lambda$
For $1^{st}$ diffraction minima, $m=1$
$\Rightarrow \sin {\theta }_1=\frac{m\lambda }{a}=\frac{600\times {10}^{-6}}{1.2\times {10}^{-3}}=\frac{1}{2}$
$\Rightarrow {\theta }_1=\frac{\pi }{6}\text{rad}$