Question

The velocity of light in air is $3 \times {10^8}m{s^{ - 1}}$ and that in water is $2.2 \times {10^8}m{s^{ - 1}}$. The polarising angle of incidence is:
(A) ${45^ \circ }$
(B) ${50^ \circ }$
(C) ${53.4^ \circ }$
(D) ${63^ \circ }$

Answer 1

Hint: We need to define refractive index in terms of speed of light in different media. We can relate the angle of polarization to the refractive index.

Formula Used: The formulae used in the solution are given here.
${\text{Refractive index }}\eta {\text{ = }}\dfrac{{{v_{air}}}}{{{v_{medium}}}}$ where the velocity of light in air is ${v_{air}}$ and the velocity of light in another medium is ${v_{medium}}$.
$\eta = \tan {i_p}$ where, $\eta $ is the refractive index of the medium and ${i_p}$ is the polarization angle.

Complete Step by Step Solution
The ratio between the speed of light in medium to speed in a vacuum is the refractive index. When light travels in a medium other than the vacuum, the atoms of that medium continually absorb and re-emit the particles of light, slowing down the speed of light.
Mathematically, ${\text{Refractive index }}\eta {\text{ = }}\dfrac{{{v_{air}}}}{{{v_{medium}}}}$ where the velocity of light in air is ${v_{air}}$ and the velocity of light in another medium is ${v_{medium}}$.
It has been said that, the velocity of light in air is $3 \times {10^8}m{s^{ - 1}}$ and that in water is $2.2 \times {10^8}m{s^{ - 1}}$.
Thus, the refractive index of water with respect to air is given by,
${\text{Refractive index }}\eta {\text{ = }}\dfrac{{{v_{air}}}}{{{v_{water}}}} = \dfrac{{3 \times {{10}^8}}}{{2.2 \times {{10}^8}}} = 1.36$. Thus, the refractive index is $1.36$.
According to Brewster’s law, when an unpolarized light of known wavelength is incident on a transparent substance surface, it experiences maximum plan polarization at the angle of incidence whose tangent is the refractive index of the substance for the wavelength.The maximum polarization a ray of light may be achieved by letting the ray fall on a surface of a transparent medium in such a way that the refracted ray makes an angle of ${90^ \circ }$ with the reflected ray.
Simply put, Brewster was able to determine that the refractive index of the medium is numerically equal to the tangent angle of polarization.
Mathematically, $\eta = \tan {i_p}$ where, $\eta $ is the refractive index of the medium and ${i_p}$ is the polarization angle.
In this case we have, $\eta = 1.36 = \tan {i_p}$.
The angle of polarization is given by,
${i_p} = {\tan ^{ - 1}}\left( {1.36} \right)$
$ \Rightarrow {i_p} = {53.673^ \circ }$
The answer is, ${53.673^ \circ }$.
So the correct option is C.

Note:
Refractive index is also referred to as refraction index or index of refraction. The speed of light in a medium depends on the properties of the medium. In electromagnetic waves, the speed is dependent on the optical density of the medium. Optical density is the tendency of the atoms in a material to restore the absorbed electromagnetic energy. The more optically dense material is, the slower the speed of light. One such indicator of the optical density of a medium is the refractive index.