Question

A ray of light travelling in air falls on the surface of a rectangular slab of a plastic material whose refractive index is 1.6. If the incident ray makes an angle of $53^\circ$ with the normal$\left( {\sin 53^\circ = 4/5} \right)$, find the angle made by the refracted ray with the normal?

Answer 1

To solve this problem,we are going to use Snell’s law according to which the ratio of sine of angle of incidence to the sine of angle of refraction is equal to the refractive index of a medium.

Formula used:
${n_1}\sin {\theta _i} = {n_2}\sin {\theta _f}$
Here, ${n_1}$ is the refractive index of the first medium, ${\theta _i}$ is the angle of incidence, ${n_2}$ is the refractive index of the second medium and ${\theta _f}$ is the angle of refraction.

Complete step by step answer:
When the ray of light travels from one medium to another medium, the path of light bends depending on a refractive index of the second medium. If the refractive index of the second medium is less than the refractive index of the first medium, the ray of light bends away from the normal and if the refractive index of the second medium is greater than the refractive index of the first medium, the ray of light bend towards the normal.
The refractive index of the air is 1. In this question, the ray of light is travelling from air to a rectangular slab of refractive index 1.6 which is greater than the refractive index of the air. Therefore, the refracted ray will bend towards the normal and angle made by the refracted ray will be less than the angle of incidence.
Use Snell’s law to determine the angle of refraction as follows,
${n_1}\sin {\theta _i} = {n_2}\sin {\theta _f}$
Here, ${n_1}$ is the refractive index of the air, ${\theta _i}$ is the angle of incidence, ${n_2}$ is the refractive index of rectangular slab and ${\theta _f}$ is the angle of refraction.
Substitute 1 for ${n_1}$, $53^\circ$ for ${\theta _i}$, 1.6 for ${n_2}$ in the above equation.
$\left( 1 \right)\sin \left( {53^\circ } \right) = \left( {1.6} \right)\sin {\theta _f}$
$\Rightarrow \sin {\theta _f} = \dfrac{{\sin \left( {53^\circ } \right)}}{{1.6}}$
$\Rightarrow \sin {\theta _f} = 0.50$
$\Rightarrow {\theta _f} = {\sin ^{ - 1}}\left( {0.50} \right)$
$\therefore {\theta _f} = 30^\circ$
Therefore, the angle made by the refracted ray with the normal is $30^\circ$.

Note: The angle of refraction is taken with respect to the normal and not with the horizontal surface of the rectangular slab. Also, the angle of incidence is not the angle made by the incident ray with the horizontal surface of the rectangular slab but the angle made by the ray with the normal.