To solve this problem, we need to analyze the behavior of light in a prism with one surface silvered.$\\$Given:$\\$• The angle of the prism is $A. \\$• Light rays fall at an angle of incidence $2A$ on the first surface.$\\$• The light returns back through the same path after reflection at the silvered surface.$\\$Let's denote:$\\$• $i = 2A$ as the angle of incidence on the first surface.$\\$• $r_1$ as the angle of refraction at the first surface.$\\$• $r_2$ as the angle of incidence at the silvered surface.$\\$Since the light returns back through the same path, the angle of incidence at the silvered surface $r_2 \\$must be equal to the angle of reflection. Therefore, $r_2 = r_1. \\$Using Snell's law at the first surface:$\\ \sin i = \mu \sin r_1 \\$Substitute $i = 2A: \\ \sin 2A = \mu \sin r_1 \\$Using the identity for $\sin 2A: \\ 2 \sin A \cos A = \mu \sin r_1 \\$At the silvered surface, the light reflects back, so the angle of incidence equals the angle of reflection.$\\$The light retraces its path, meaning the angle of refraction at the first surface equals the angle of incidence$\\$at the silvered surface. Therefore, $r_1 = A. \\$Substitute $r_1 = A$ into the equation:$\\ 2 \sin A \cos A = \mu \sin A \\$Simplify by dividing both sides by $\sin A: \\ 2 \cos A = \mu \\$Thus, the refractive index $\mu$ of the prism is $2 \cos A. \\$Therefore, the correct option is Option 2.