Question

A light ray falling at an angle of ${45}^0$ which the surface of a clean slab of ice of thickness 1.00 m is refracted into it at an angle of 30${}^0$. alculate the time taken by the light rays to cross the slab. Speed of light in the vacuum =3$\times {10}^8$ m/s${}^{-1}$

Answer 1

We have,

$Thickness=1m$

Incident angle$={45}^0$

Refracted angle $={30}^0$

Speed of light $=3\times {10}^{8}m/s$

So,

${\mu }_{air}sin{45}^0={\mu }_{glass}sin{30}^0$

$\frac{1}{\sqrt{2}}={\mu }_g\frac{1}{2}$

${\mu }_g=\sqrt{2}$

Now,

$t=\frac{AC}{velocity}$ $(\mu =\frac{c}{v})$

$=\frac{\sqrt{2}}{3\times {10}^8\times \sqrt{2}}s$

Thus, the time taken by the light rays to cross the slab is:

$=\frac{1}{3\times {10}^8}s$