Question

A light ray falling at an angle of ${60}^{\circ }$ with the surface of a clean slab of ice of thickness $1.00m$ is refracted into it at an angle of ${15}^{\circ }$. Calculate the time taken by the light rays to cross the slab. Speed of light in vacuum = $3\times {10}^{8}m/s$

Answer 1

From Snell's Law, $\frac{\sin i}{\sin r}=\mu =\frac{c}{v}$

$\frac{\sin {60}^{\circ }}{\sin {15}^{\circ }}=\frac{c}{v}$

$v=\frac{c\sin {15}^{\circ }}{\sin {60}^{\circ }}=\frac{3\times {10}^8\times 0.258}{0.5}$

$v\simeq \frac{3}{2}\times {10}^{8}m/s$$l\cos {15}^{\circ }=1m$

$l=\frac{1}{\cos {15}^{\circ }}=vt$

$t=\frac{1}{\cos {15}^{\circ }\left(v\right)}\simeq \frac{2}{3}\times {10}^{-8}\sec$