To solve this problem, we need to use the formula for the angle of minimum deviation in a prism.$\\$Given:$\\$• The refracting angle of the prism is $A. \\$• The refractive index of the material of the prism is $\mu = \cot\left(\frac{A}{2}\right). \\$The formula for the angle of minimum deviation $\delta_m$ in a prism is:$\\ \delta_m = 2i - A \\$where $i$ is the angle of incidence when the deviation is minimum.$\\$For minimum deviation, the angle of incidence $i$ is equal to the angle of emergence $e, \\$and the relation between the refractive index and the angles is given by Snell's law:$\\ \mu = \frac{\sin i}{\sin r} \\$At minimum deviation, $i = e$ and $r = \frac{A}{2}. \\$Thus, we have:$\\ \mu = \frac{\sin i}{\sin\left(\frac{A}{2}\right)} \\$Substitute $\mu = \cot\left(\frac{A}{2}\right): \\ \cot\left(\frac{A}{2}\right) = \frac{\sin i}{\sin\left(\frac{A}{2}\right)} \\$Rearrange to find $\sin i: \\ \sin i = \cot\left(\frac{A}{2}\right) \cdot \sin\left(\frac{A}{2}\right) \\$Since $\cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)},$ we have:$\\ \sin i = \cos\left(\frac{A}{2}\right) \\$Therefore, $i = \frac{\pi}{2} - \frac{A}{2} \\$Substitute $i$ into the formula for $\delta_m: \\ \delta_m = 2\left(\frac{\pi}{2} - \frac{A}{2}\right) - A \\$Simplify:$\\ \delta_m = \pi - A - A \\ \delta_m = \pi - 2A \\$Convert $\pi$ to degrees: $\pi = 180^\circ \\ \delta_m = 180^\circ - 2A \\$Therefore, the angle of minimum deviation is $180^\circ - 2A. \\$This corresponds to Option 1.