To solve this problem, we need to determine the angle of incidence for which the light beam retraces its path after reflecting from the silvered surface of the prism.$\\$Given:$\\$• Refractive index of the prism material $n = \sqrt{2} \\$• Angle of the prism $A = 30^\circ \\$For the light to retrace its path, the angle of incidence $i$ must be such that the light undergoes total internal reflection at the silvered surface.$\\$The condition for retracing the path is that the angle of incidence at the first surface equals the angle of emergence at the same surface.$\\$Using Snell's law at the first surface:$\\ n \sin r = \sin i \\$where $r$ is the angle of refraction.$\\$For the prism, the relation between the angles is given by:$\\ i + e = A + \delta \\$where $e$ is the angle of emergence and $\delta$ is the deviation angle. For retracing the path, $\delta = 0,$ so i = e.} \\ Thus, $i + i = A \\ 2i = A \\ i = \frac{A}{2} \\$Substitute the given angle of the prism:$\\ i = \frac{30^\circ}{2} = 15^\circ \\$However, this is the angle for minimum deviation. For retracing the path, we need the condition for total internal reflection at the silvered surface.$\\$The critical angle $\theta_c$ for total internal reflection is given by:$\\ \sin \theta_c = \frac{1}{n} \\ \theta_c = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ \\$For the light to retrace its path, the angle of incidence at the first surface should be such that the angle of refraction at the second surface is $45^\circ. \\$Using Snell's law at the second surface:$\\ n \sin r = \sin 45^\circ \\ \sqrt{2} \sin r = \frac{1}{\sqrt{2}} \\ \sin r = \frac{1}{2} \\ r = 30^\circ \\$Since $r = 30^\circ$ at the second surface, the angle of incidence $i$ at the first surface should be such that:$\\ \sin i = n \sin r \\ \sin i = \sqrt{2} \cdot \frac{1}{2} \\ \sin i = \frac{\sqrt{2}}{2} \\ i = 45^\circ \\$Therefore, the angle of incidence for which the light retraces its path is $45^\circ. \\$This corresponds to Option 2.