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 nearly equal to

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

Answer 1

The correct option is C $1.732$

Angle of incidence $i={60}^{\circ }$
Lateral displacement = $\text{t sec r sin (i -r)}=5\sqrt{3}$
$\text{= 15 sec r}[\frac{\sqrt{3}}{2}\text{cos r}-\frac{\text{sin r}}{2}]=5\sqrt{3}$
$\Rightarrow \frac{\sqrt{3}}{2}-\frac{\text{tan r}}{2}=\frac{1}{\sqrt{3}}$
$\Rightarrow \text{tan r}=\frac{1}{\sqrt{3}}$
$\Rightarrow r={30}^{\circ }$
Now $\mu \text{sin r = sin i}$
$\mu =\frac{\text{sin i}}{\text{sin r}}=\frac{\sqrt{3}}{2}\times \frac{2}{1}=\sqrt{3}$