Question

A parallel beam of light of wavelength $6000\dot{A}$ is incident normally on a slit of width $0.2\text{mm}$. The diffraction pattern is observed on a screen which is placed at the focal plane of a convex lens of focal length $50\text{cm}$. If the lens is placed close to the slit. The distance between the minimum on both sides of the central maximum will be

• A. $1\text{mm}$
• B. $3\text{mm}$
• C. $2\text{mm}$
• D. $4\text{mm}$

Answer 1

The correct option is B $3\text{mm}$

Given:
$\lambda =6000\dot{A}$
$b=0.2\text{mm}=0.2\times {10}^{-3}\text{m}$
$f=50\text{cm}=0.5\text{m}$

The angular position of first minima from the central maximum is given by,

$\sin \theta =\frac{\lambda }{b}=\frac{6000\times {10}^{-10}}{0.2\times {10}^{-3}}$

$\Rightarrow \sin \theta =3\times {10}^{-3}$

For small $\theta$, $\sin \theta \approx \theta$

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

If the lens is placed close to the slit, then

$\tan \theta \approx \theta =\frac{x}{f}$

Where, $x$ is the distance of first minimum from the central maximum.

Therefore, the distance between two minima on both sides of the central maximum is

$2x=2f\theta =2\times 0.5\times 3\times {10}^{-3}$

$\Rightarrow 2x=3\text{mm}$

Hence, option $\left(\text{C}\right)$ is correct.