Question

Critical angle of a certain medium is ${\sin }^{-1}\left(\frac{3}{5}\right)$. Thge polariging angle of the medium is:

Answer 1

According to Brewster's law when unpolarised light is incident at polarizing angle$\left(i_p\right)$ on an interface separating a rarer medium from a denser medium, of refractive index $\mu$, such that

$\mu =\tan i_p$

then light is reflected in the rarer medium is completely polarized.

Also, we know critical angle $\left(i_c\right)$ is given as

$\mu =\frac{1}{\sin i_c}$

$\sin i_c=\frac{1}{\mu }$

$\Rightarrow i_c={\sin }^{-1}\left(\frac{1}{\mu }\right)$

According to the question

$i_c={\sin }^{-1}\left(\frac{3}{5}\right)$

$\Rightarrow {\sin }^{-1}\left(\frac{1}{\mu }\right)$

$={\sin }^{-1}\left(\frac{3}{5}\right)$

$\Rightarrow \mu =\frac{5}{3}$

Thus, finding polarizing angle

$\mu =\tan i_p$

$i_p={\tan }^{-1}\left(\frac{5}{3}\right)$.