Question

Refractive index of a rectangular glass slab is $\mu =\sqrt{3}$. A light ray incident at an angle ${60}^{\circ }$ is displaced laterally through $2.5\text{cm}$. Distance travelled by light in the slab is

• A. $4\text{cm}$
• B. $5\text{cm}$
• C. $2.5\sqrt{3}\text{cm}$
• D. $3\text{cm}$

Answer 1

The correct option is B $5\text{cm}$


Lateral shift $x=d\sin (i-r)\dots \dots \left(1\right)$
$x=2.5\text{cm};i={60}^{\circ };\mu =\sqrt{3}$
Using snell's law
$\mu =\frac{\sin i}{\sin r};\sqrt{3}=\frac{\sin {60}^o}{\sin r}\Rightarrow \sin r=\frac{1}{2}$
$\Rightarrow r={30}^{\circ }$
Lateral shift
$x=d\sin (i-r)$
$2.5=d\sin ({60}^{\circ }-{30}^{\circ })=d\sin {30}^{\circ }$
$d=2.5\times 2$
$d=5\text{cm}$