Question

A beam of unpolarised light is incident on the boundary between two transparent media if the reflected light is completely plane polarised, how is its direction related to the direction of corresponding refracted light?

Answer 1

Hint:A transverse wave property that determines the geometrical direction of the oscillations is called polarization. The oscillation direction of a transverse wave is perpendicular to the wave's rotation direction. Polarisation is a key feature of light that impacts even optical devices that do not directly evaluate it.

Complete step by step answer:
Unpolarized light is described as a light wave that vibrates in more than one plane.Unpolarized light can be converted to polarised light. As unpolarized light is incident on a transparent refracting medium, the reflected light and refracted light are perpendicular to each other at a certain angle of incidence; this angle of incidence is known as a polarising angle.

Due to the polarization of reflection, the light which gets reflected gets plane-polarized completely and this makes an angle of \[\dfrac{\pi }{2}\] along with the refracting light.To find the relation between the direction of the unpolarised light with the direction of corresponding refracted light can be found out using Brewster’s law and Snell’s law.

Brewster's law is a relationship with light waves that states that the maximal polarisation of a ray of light can be obtained by allowing the ray to land on the surface of a transparent material in such a manner that the refracted ray creates an ${90^ \circ }$ angle with the reflected ray. According to the above law,
$ip + r = {90^ \circ }$
Here, $r$ is the angle of reflection and $ip$ is the polarising angle.
$r = {90^ \circ } - ip$

According to Snell's Law, the ratio of the sine of the angles of incidence and transmission equals the ratio of the refractive index of the materials at the interface. This law can be represented as:
${\mu _1}\sin i = {\mu _2}\sin r$
Here, $r$ is the angle of reflection and $i$ is the angle of incidence.
Here according to the question, ${\mu _1} = 1$ and ${\mu _2} = \mu $
So, Substituting in the Snell’s law,
$\dfrac{{\sin i}}{{\sin r}} = mu$ $ - - - - - - - - $ ( $1$ )

From the question we know that $i = {i_p}$ and from Brewster’s law we know that $r = {90^ \circ } - ip$. So, substituting in ( $1$ ) we get,
$\dfrac{{\sin {i_p}}}{{\sin ({{90}^ \circ } - {i_p})}} = mu$
$ \Rightarrow \dfrac{{\sin {i_p}}}{{\cos {i_p}}} = mu$
$ \therefore \tan {i_p} = mu$
Thus, the relation between the direction of the unpolarised light and the direction of corresponding refracted light is $\tan {i_p} = mu$ . This relation is also called the Brewster’s angle.

Note: Brewster's angle is the incidence angle at which light with a certain polarisation is precisely reflected across a transparent dielectric surface with no reflection. As unpolarized light strikes the surface at this angle, the light that is absorbed is completely polarised.