Question

A light ray is incident on air-liquid interface at ${45}^{\circ }$ and is refracted at ${30}^{\circ }$. What is the refractive index of the liquid? For what angle of incidence will the angle between reflected ray and refracted ray be ${90}^{\circ }$?

Answer 1

Given : $i={45}^0$ $r={30}^0$ $\left({\mu }_{air}\right)=1$

To find : ${\mu }_{water}$

Solution: From snell's law

${\mu }_1\sin i={\mu }_2\sin r$

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

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

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

Now let us take that for angle of incidence $\alpha$ the angle between reflected ray and refracted ray be ${90}^0$ then

$\alpha +r+{90}^0={180}^0$ (since i = angle of reflection from laws of reflection)

Hence $r={90}^0-\alpha$

Now from snell's law

${\mu }_1\sin \alpha ={\mu }_2\sin r$

$1\sin \alpha =\sqrt{2}\sin ({90}^0-\alpha )$

$\tan \alpha =\sqrt{2}$

$\alpha ={\tan }^{-1}\sqrt{2}$