(I) Since the light ray enters perpendicular to the face AB, the angle of incidence onface AC will be45° .
\(\sin\theta_C = \frac{1}{n}\)
\(\sin\ 45^\circ = \frac{1}{n} = \frac{1}{\sqrt{2}}\ So,\ n = \sqrt{2}\)
(II) In fig.2, the face AC of the prism is surrounded by a liquid so n = \(\frac{ng}{i} = \frac{\sqrt{2}}{\left(\frac{2}{3}\right)} =\frac{\sqrt{3}}{\sqrt{2}}\)
\(sin\theta_C = \frac{1}{n} = \frac{\sqrt{2}}{\sqrt{3}}\theta _C = \sin^{-1}(\frac{\sqrt{2}}{\sqrt{3}}) = 54.6^\circ\)
Since the angle of incidence on the surface AC is 45° , which is less than the critical angle for the pair of media (glass and the liquid), the ray neither undergoes grazing along surface AC, nor does it suffer total internal reflection
Instead it passes through the surface AC and undergoes refraction into the liquid.