(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.] $\; \therefore r = \left( \; \frac{\pi}{2} - i_B \right)$ $\; \therefore \mu = \left( \frac{\sin \; i_B}{\sin \; r} = \tan i_B \right)$ Thus light gets totally polarised by reflection when it is incident at an angle $i_B$ (Brewster's angle), where $i_B = \tan^{-1} \mu$ (b) The angle of incidence, of the ray, on striking the face $AC$ is $i = 60^{\circ}$ (as from figure) Also, relative refractive index of glass, with respect to the surrounding water, is $\therefore \mu_r = \frac{\; \frac{3}{2}}{\frac{4}{3}} = \; \frac{9}{8}$ Also $\sin i = \sin 60^{\circ} = \frac{\sqrt{3}}{2} = \frac{1.732}{2} = 0.866$ For total internal reflection, the required critical angle, in this case, is given by $\; \sin i_c = \frac{1}{\mu} = \frac{8}{9} \simeq 0-89$ $\therefore \; \; i < i_c$ Hence the ray would not suffer total internal reflection on striking the face $AC$