The critical angle for a certain transparent medium with respect to...

The critical angle $C$ is related to the refractive index $\mu$ of the medium by the formula:
$\sin(C) = \dfrac{1}{\mu}$

Given $C = \sin^{-1}\left(\dfrac{3}{5}\right)$, we have $\sin(C) = \dfrac{3}{5}$.
Therefore, the refractive index is $\mu = \dfrac{5}{3}$.

According to Brewster's law, the polarising angle $i_p$ is given by:
$\tan(i_p) = \mu$

Substituting the value of $\mu$:
$\tan(i_p) = \dfrac{5}{3}$
$\Rightarrow i_p = \tan^{-1}\left(\dfrac{5}{3}\right)$

Answer: $\tan^{-1}\left(\dfrac{5}{3}\right)