Question

A ligh t ray is incident on air liquid inter face at 45°and is refracted at 30°.What is the refractive index of the liquid For what angle of incidence will the angle between reflected ray and refracted ray be 90°?

Answer 1

Answer:

Refractive index of liquid = 1.414.

Angle of incidence for which the angle between reflected ray and refracted ray is $90^\circ = 54.73^\circ$.

Explanation:

Given:

Angle of incidence, $i = 45^\circ$.

Angle of refraction, $r = 30^\circ$.

According to Snell's law,

$n_1 \sin i = n_2\sin r$

where,

$n_1$ = refractive index of air.

$n_2$ = refractive index of liquid.

Since, $n_1 =1$.

Therefore,  

$1\ \sin (45^\circ) = n_2\sin (30^\circ)\\n_2 = \dfrac{\sin(45^\circ)}{\sin(30^\circ)}\\=\dfrac{0.707}{0.5}\\= 1.414.$

We know that the angle of incidence $i$ is equal to the angle of reflection $r_e$ of light.

$i = r_e$

The angle between between reflected ray and refracted ray is $90^\circ$

Therefore,

$r_e + r = 90^\circ\\r = 90^\circ -r_e\\r = 90^\circ - i\\\Rightarrow\sin r=\sin(90^\circ -i)=\cos i.$

Putting this value in the Snell's law,

$n_1 \sin i = n_2\sin r\\1\ \sin i = 1.414\ \cos i\\\dfrac{\sin i}{\cos i} = 1.414\\\tan i = 1.414\\i = \tan^{-1} (1.414)\\= 54.73^\circ.$

This angle for which the angle between reflected ray and refracted ray is $90^\circ$ is called Brewster's angle.