To solve this problem, we need to use the formula for the angle of minimum deviation in a prism.$\\$Given:$\\$• Angle of incidence $i = 45^\circ \\$• Angle of prism $A = 60^\circ \\$For minimum deviation, the angle of incidence $i$ is equal to the angle of emergence e.} \\ Also, the angle of refraction $r_1 = r_2 = r. \\$Using the prism formula:$\\ A = r_1 + r_2 \\ A = 2r \\$Substitute the given angle of the prism:$\\ 60^\circ = 2r \\ r = \frac{60^\circ}{2} = 30^\circ \\$The formula for the angle of minimum deviation $\delta_m$ is:$\\ \delta_m = 2i - A \\$Substitute the known values:$\\ \delta_m = 2 \times 45^\circ - 60^\circ \\ \delta_m = 90^\circ - 60^\circ \\ \delta_m = 30^\circ \\$Now, let's find the refractive index $\mu$ of the material of the prism using Snell's law:$\\ \mu = \frac{\sin i}{\sin r} \\$Substitute the known values:$\\ \mu = \frac{\sin 45^\circ}{\sin 30^\circ} \\ \mu = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} \\ \mu = \sqrt{2} \\$Therefore, the angle of minimum deviation is $30^\circ$ and the refractive index is $\sqrt{2}. \\$This corresponds to Option 4.