Question

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

• A. $2.4\times {10}^{-9}\text{sec}$
• B. $5.4\times {10}^{-9}\text{sec}$
• C. $2.4\times {10}^{-7}\text{sec}$
• D. $5.4\times {10}^{-7}\text{sec}$

Answer 1

The correct option is B $5.4\times {10}^{-9}\text{sec}$

From snells law,
$\frac{\sin i}{\sin r}=\frac{3\times {10}^8}{v}=\frac{\sin {45}^{\circ }}{\sin {30}^{\circ }}=\sqrt{2}$
$\Rightarrow v=\frac{3\times {10}^8}{\sqrt{2}}\text{m/s}$

Distance travelled in the slab, $x=\frac{1\text{m}}{\cos {30}^{\circ }}=\frac{2}{\sqrt{3}}\text{m}$

So, time taken $=\frac{2\times \sqrt{2}}{\sqrt{3}\times 3\times {10}^8}=0.54\times {10}^{-8}=5.4\times {10}^{-9}\text{sec}.$

Why this Question? Tip: When light ray travels in a optical medium (for near normal incidence), optical path is given by $\mu t$