When a ray of light passes from a medium (air) of refractive index $\mu_1$ to another medium of refractive index $\mu_2$ less then the $\mu_1$ with angle of incidence is equal to Brewster's angle, the transmitted light is unpolarised and reflected light is plane polarised.


Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented by arrows.
Polarisation by reflection occurs when the angle of incidence is the Brewster's angle
So, $\quad \tan \mathrm{i}_{\mathrm{B}}={ }^1 \mu_2=\frac{\mu_2}{\mu_1}$
where, $\quad \mu_2 < \mu_1$
So, when the light rays travels in such a medium, the critical angle is
$$
\sin \mathrm{i}_{\mathrm{C}}=\frac{\mu_2}{\mu_1}
$$
Where, $\quad \mu_2 < \mu_1$
As $\left|\tan \mathrm{i}_{\mathrm{B}}\right|>\left|\sin \mathrm{i}_{\mathrm{C}}\right|$ for large angle $\mathrm{i}_{\mathrm{B}} < \mathrm{i}_{\mathrm{C}}$.
Hence, the polarisation by reflection occurs definitely.