To solve this problem, we need to determine the angle between the refracted and reflected rays.$\\$Given:$\\$• Refractive index of glass $n = \sqrt{3} \\$• Angle of incidence $i = 60^\circ \\$First, we apply Snell's Law:$\\ n_1 \sin i = n_2 \sin r \\$Here, $n_1 = 1$ (refractive index of air), and $n_2 = \sqrt{3}. \\ \sin 60^\circ = \frac{\sqrt{3}}{2} \\$Substituting the values:$\\ 1 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin r \\$Simplifying:$\\ \frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin r \\ \sin r = \frac{\sqrt{3}}{2\sqrt{3}} \\ \sin r = \frac{1}{2} \\$Therefore, $r = 30^\circ. \\$The angle of reflection $r'$ is equal to the angle of incidence:$\\ r' = 60^\circ \\$The angle between the refracted and reflected rays is:$\\ \theta = r' - r \\ \theta = 60^\circ - 30^\circ \\ \theta = 30^\circ \\$However, the question asks for the angle between the refracted and reflected rays, which is actually:$\\ \theta = 180^\circ - (r + r') \\ \theta = 180^\circ - (30^\circ + 60^\circ) \\ \theta = 180^\circ - 90^\circ \\ \theta = 90^\circ \\$Therefore, the angle between the refracted and reflected rays is $90^\circ. \\$This corresponds to Option 3.