Question

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

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

Answer 1

Answer: (B)

5 cm

Given $\mu =\sqrt{3}$
${\theta }_a={60}^0$

Lateral shift $d=t\frac{\sin ({\theta }_a-{\theta }_b^{\prime })}{\cos {\theta }_b^{\prime }}$ ...................(1)
Let the length traveled in glass be $l$.
From geometry, $t=l\cos {\theta }_b^{\prime }$ ..........................(2)
Substituting $t$ from (2) in (1)

$d=lsin({\theta }_a-{\theta }_b^{\prime })$ .............................................(3)

From Snell's law, $\frac{\sin {\theta }_a}{\sin {\theta }_b^{\prime }}=\mu$
$\Rightarrow \sin {60}^0=\sqrt{3}\sin {\theta }_b^{\prime }$
$\Rightarrow \sin {\theta }_b^{\prime }=\frac{1}{2}$
$\Rightarrow {\theta }_b^{\prime }={30}^0$

Hence, from (3)

$2.5=l\sin ({60}^0-{30}^0)$
$\Rightarrow l=5cm$