Ray of light travels from water into glass. What should be the angle of incidence, so that the reflected and refracted rays are perpendicular to each other? ($\mu_w = 1.33$, $\mu_g = 1.547$)

Hint: Brewster's law says that as we incident a light at a certain angle called the Brewster's angle, from one medium to another, as the light passes from one medium to the other, it starts reflecting some light back into the first medium. At this angle, the incident light is at right angle to the reflected light.
Formula used:
The Brewster's angle is given by the relation:
$\tan \theta_B = \dfrac{n_2}{n_1} $
Where $n_1$ is for medium 1 from where the light is incident on medium 2 with refractive index $n_2$.

Complete answer:
Suppose we assume medium 1 as water and medium 2 as glass. As we are given that the light is incident from water (medium 1) to glass (medium 2), it will bend towards the normal upon refraction as glass is a denser medium. At a certain angle of incidence called Brewster's angle, we also start getting some reflected light back in the medium such that reflected and refracted rays make a right angle with respect to each other.
For our purpose, we can write the formula for obtaining Brewster's angle as:
$\tan \theta_B = \dfrac{\mu_g}{\mu_w} $
Substituting the given values, we get:
$\eqalign{
  & \tan {\theta _B} = \dfrac{{1.547}}{{1.33}} = 1.631 \cr
  & \Rightarrow {\theta _B} = 49.3^\circ \cr} $
This is the value of the angle at which if we incident the light in water to glass then a right angle forms between reflected and refracted rays.

Note:
Here, the definition and the name of the angle has to be very clear. This phenomenon is also relevant in total internal reflection but the light travels from denser medium to rarer medium. There the angle of incidence for which the reflected light grazes the surface is called critical angle.