Question

A ray of light is incident on one face of a transparent slab of thickness $15\text{cm}$. The angle of incidence is ${60}^{\circ }$. If the lateral displacement of the ray on emerging from the parallel plane is $5\sqrt{3}\text{cm}$, the refractive index of the material of the slab is

• A. $1.414$
• B. $1.532$
• C. $1.732$
• D. None

Answer 1

The correct option is C $1.732$

$i={60}^{\circ }$
Displacement $=t\sec r\sin (i-r)=5\sqrt{3}$
Apply $\sin (i-r)=\sin i\cos r-\cos i\sin r$
$=15\sec r[\frac{\sqrt{3}}{2}\cos r-\frac{\sin r}{2}]=5\sqrt{3}$
$\Rightarrow \frac{\sqrt{3}}{2}-\frac{\tan r}{2}=\frac{1}{\sqrt{3}}$
$\Rightarrow r={30}^{\circ }$
Now $\mu \sin r=\sin i$
$\mu =\frac{\sin i}{\sin r}=\frac{\sqrt{3}}{2}\times \frac{2}{1}=\sqrt{3}=1.732$