(a) Diagram
Polarisation by reflection
(b) Justification
Writing yes/no
(a) The diagram, showing polarisation by reflection is as shown.
[Here the reflected and refracted rays are at right angle to each other.]
$\therefore \mathit{r}=\left(\frac{\mathit{\pi }}{2}-{\mathit{i}}_{\mathit{B}}\right)$
$\therefore \mathit{\mu }=\left(\frac{\mathit{s}\mathit{i}\mathit{n}{\mathit{i}}_{\mathit{B}}}{\mathit{s}\mathit{i}\mathit{n}\mathit{r}}=\tan {\mathit{i}}_{\mathit{B}}\right)$
Thus light gets totally polarised by reflection when it is incident at an angle ${\mathit{i}}_{\mathit{B}}$ (Brewster's angle), where ${\mathit{i}}_{\mathit{B}}={\tan }^{-1}\mathit{\mu }$
(b) The angle of incidence, of the ray, on striking the face $\mathrm{AC}$ is $\mathit{i}={60}^{\circ }$ (as from figure)
Also, relative refractive index of glass, with respect to the surrounding water, is
$\therefore {\mathit{\mu }}_{\mathit{r}}=\frac{\frac{3}{2}}{\frac{4}{3}}=\frac{9}{8}$
Also $\mathit{s}\mathit{i}\mathit{n}\mathit{i}=\mathit{s}\mathit{i}\mathit{n}{60}^{\circ }=\frac{\sqrt{3}}{2}=\frac{1.732}{2}=0.866$

For total internal reflection, the required critical angle, in this case, is given by
$\mathit{s}\mathit{i}\mathit{n}{\mathit{i}}_{\mathit{c}}=\frac{1}{\mathit{\mu }}=\frac{8}{9}\simeq 0-89$

Hence the ray would not suffer total internal reflection on striking the face $\mathrm{AC}$