Question

A light ray falling at an angle of ${45}^{\circ }$ with the surface of a clean slabof ice of thickness 1.00 m is refracted into it at an angle of ${30}^{\circ }$. Calculate the time taken by the light rays to cross the slab. Speed of light in vacuum $=3\times {10}^{8}m{s}^{-1}$

Answer 1

We know, $\frac{sini}{sinr}=\frac{3\times {10}^8}{v}$

$=\frac{sin{45}^{\circ }}{sin{30}^{\circ }}=\frac{\left(\frac{1}{\sqrt{2}}\right)}{\left(\frac{1}{2}\right)}$

$=\frac{2}{\sqrt{2}}=\sqrt{2}$

$\Rightarrow v=\frac{3\times {10}^8}{\sqrt{2}}m/sec$

Distance travelled by light in the slab is,

$x=\frac{1m}{cos{30}^{\circ }}=\frac{2}{\sqrt{3}}m$

So, time taken $=\frac{2}{\sqrt{3}}\times \frac{\sqrt{2}}{3\times {10}^8}$

$=0.54\times {10}^{-8}$

$=5.4\times {10}^{-9}\text{sec}$