Question

The angle of polarisation for any medium is $60^{\circ}$, what will be critical angle for this:

Answer 1

Hint: In Optics, the ratio of the angle of incidence and the angle of refraction is $90^{\circ}$ is termed as the critical angle. The ratio of light ray velocities in the air to the provided medium is a refractive index. So, the relation between the critical angle and refractive index can be proved as the Critical angle is inversely proportional to the refractive index.

Complete answer:
Relation between angle of polarisation and refractive index is given by:
$\mu = tan \theta_{p}$
Where, $\mu$ is the refractive index.
$\theta_{p}$ is the angle of polarisation.
Given: $\theta_{p} = 60^{\circ}$
$\mu = tan 60^{\circ}$
$\mu = \sqrt{3} = 1.732 $
Relation between critical angle and refractive index is given by:
$\mu = \dfrac{1}{sin \theta_{c}}$
Where, $\theta_{c}$ is the critical angle.
$ sin \theta_{c}= \dfrac{1}{\mu }$
$ \theta_{c}= sin^{-1} \dfrac{1}{\mu }$
Put the value of in the above equation.
$ \theta_{c}= sin^{-1} \dfrac{1}{\sqrt{3}}$

Note:
Polarization angle or Brewster's angle is an incidence angle at which light with a unique polarization is perfectly carried through a transparent dielectric surface, with no reflection. When an unpolarized ray is an incident at this angle, the light bounced back from the surface is perfectly polarized.