Question

A parallel beam of light of wavelength $6000\mathring{A}$ gets diffracted by a single slit of width $0.3\text{mm}$. The angular position of the first minima of diffracted light is,

• A. $2\times {10}^{-3}\text{rad}$
• B. $3\times {10}^{-3}\text{rad}$
• C. $8\times {10}^{-3}\text{rad}$
• D. $6\times {10}^{-3}\text{rad}$

Answer 1

The correct option is A $2\times {10}^{-3}\text{rad}$

Here, $\lambda =6000\mathring{A}=6\times {10}^{-7}\text{m}$,
$a=0.3\text{mm}=3\times {10}^{-4}\text{m}$

For first minima, $a\sin \theta =\lambda$, where $a$ is the slit width.
$\sin \theta =\frac{\lambda }{a}=\frac{6\times {10}^{-7}}{3\times {10}^{-4}}=2\times {10}^{-3}$

As $\sin \theta$ is very small, $\Rightarrow \sin \theta \approx \theta$

$\therefore \theta \approx 2\times {10}^{-3}\text{rad}$

Hence, $\left(A\right)$ is the correct answer.