Question

A ray of light is incident at an angle of ${45}^{\circ }$ on a square slab of a transparent material. What should be the refractive index of the material of the slab so that the emergent ray travels along the vertical face AB?

• A. $\sqrt{2}$
• B. $\sqrt{\frac{3}{2}}$
• C. $1.5$
• D. $\sqrt{\frac{5}{3}}$

Answer 1

The correct option is B

$\sqrt{\frac{3}{2}}$

Consider the refractive index of the slab to be x.
Light ray travels from medium 1 to medium 2.
By Snell's law,
$\frac{sini}{sinr}=\frac{Refractiveindexofmedium2}{Refractiveindexofmedium1}$
At the first interface
$\frac{sin{45}^o}{sinr}=\frac{x}{1}$ ..(1)
At the second interface
$cosr=\frac{1}{x}$
$\Rightarrow sinr=\frac{\sqrt{x^2-1}}{x}$ ..(2)
from (1) and (2)
$\frac{1}{\sqrt{2}\times \frac{\sqrt{x^2-1}}{x}}=x$
$\Rightarrow \frac{1}{\sqrt{2}}=\sqrt{x^2-1}$
$\Rightarrow x^2-1=\frac{1}{2}$
$\Rightarrow x^2=\frac{3}{2}$
$\Rightarrow x=\sqrt{\frac{3}{2}}$