Question

The radius of the central bright spot in the diffraction pattern of a circular hole of diameter $d$ with light of wavelength $\lambda$ is $r$ and the distance between the circular hole and screen is $D$, Now, the whole set up is immersed in a liquid of refractive index $\mu$ and the distance between the hole and the screen is reduced by $2$ times, the radius of central bright spot reduced by $4$ times. The value of $\mu$ is

Answer 1

Case 1 :-
Radius of central bright spot is given by

$r=\frac{1.22\lambda D}{d}...\left(1\right)$

Case 2:-
When the whole set up is immersed in a liquid of refractive indes $\mu$

$r^{\prime }=\frac{1.22{\lambda }^{\prime }{D}^{\prime }}{d}$

Given that,
$r^{\prime }=\frac{r}{4}$ and $D^{\prime }=\frac{D}{2}$

By Snell's law,
${\lambda }^{\prime }=\frac{\lambda }{\mu }$

$\Rightarrow \frac{r}{4}=\frac{1.22\lambda D}{2\mu d}$

$\Rightarrow \frac{r}{4}=\frac{r}{2\mu }\left(\text{from}\right(1\left)\right)$

$\Rightarrow \mu =2$

$\therefore$ Correct answer $=2$.