Question

What will be the angle when the sun will be seen with the horizon if a person finds that the sun rays reflected by the still water in a lake are polarized? The refractive index of water is $1.327$.
A) ${57^ \circ }$
B) ${75^ \circ }$
C) ${37^ \circ }$
D) ${53^ \circ }$

Answer 1

- Hint: - From Brewster’s Law,
$\tan {\theta _B} = \mu$
where, ${\theta _B}$ is the Brewster’s angle and $\mu$ is the refractive index of water.
Now, to get the angle when the sun will be seen with the horizon, find the difference between ${90^ \circ }$ and Brewster’s angle.

Complete Step by Step Solution: -
The Brewster’s Law states that if the ray is made to fall on a surface of transparent medium such that the refracted ray makes an angle of ${90^ \circ }$ with the reflected ray, then the maximum polarization of light is achieved.
The angle which produces a ${90^ \circ }$ angle between refracted ray and reflected ray is called Brewster’s Angle. It is denoted by ${\theta _B}$.
Now, according to the question, it is given that the reflected ray is polarized then the incident angle becomes equal to the Brewster’s angle.
Therefore, according to Brewster’s Law –
$\tan {\theta _B} = \mu \cdots (1)$
Let $\mu$ be the refractive index
The value of refractive index is already given in question which is –
$\mu = 1.327$
Putting the value in equation $(1)$
$\tan {\theta _B} = 1.327 \\\ {\theta _B} = {\tan ^{ - 1}}(1.327) \\\$
Now, finding the inverse of $\tan$
$\therefore {\theta _B} = {53^ \circ }$
Because, the Brewster’s angle produces ${90^ \circ }$ angle between reflected ray and refracted ray
Therefore, the angle the sun will be seen with the horizon is –
${90^ \circ } - {\theta _B} \\\ \Rightarrow {90^ \circ } - {53^ \circ } \\\ \Rightarrow {37^ \circ } \\\$
Hence Option (C) is Correct.

Note: - During the filtration, the electric field vectors are restricted to a single plane with respect to the direction of propagation, then the light is said to be polarized. In the polarization of light all the waves vibrate in the same plane.