(a) Applying Snell’s law for the refraction from air to glass. Refractive index of glass w.r.t. air $\,{}^{a}\mu_g$ = $\frac{\sin 60^{\circ}}{\sin 35^{\circ}}$ = $\frac{0.8660}{0.5736}$ = 1.51 (b) Now Snell’s law for the refraction from air to water $\,{}^{a}\mu_{\omega}$ = $\frac{\sin 60^{\circ}}{\sin 47^{\circ}}$ = $\frac{0.8660}{0.6560}$ = 1.32 (c) Now the light beam is incident at an angle 45° from water to glass $\,{}^{\omega}\mu_g$ = $\frac{\sin 45^{\circ}}{\sin r}$ ⇒ $\frac{1.51}{1.32}$ = $\frac{\sin 45^{\circ}}{\sin r}$ = $\frac{0.7071}{\sin r}$ sin r = $\frac{13.2 \times 0.7071}{1.51}$ = 0.6181 ∴ r = 38.2°