The exit surface of a prism with refractive index $n$ is coated with a material having refractive index $\frac{n}{2}$. When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is $\_\_\_\_$.

Let the refractive index of the prism be $n_1 = n$ and the refractive index of the coating on the exit surface be $n_2 = \frac{n}{2}$.
At the condition of minimum deviation, the angle of refraction at the first surface $r_1$ and the angle of incidence at the second surface $r_2$ are equal, i.e., $r_1 = r_2 = \frac{A}{2}$, where $A$ is the prism angle.
The problem states that at the exit surface, the light meets the condition of critical angle. Therefore, $r_2 = \theta_c$.
The critical angle $\theta_c$ for the interface between the prism and the coating is given by $\sin \theta_c = \frac{n_2}{n_1}$.
Substituting the values, $\sin \theta_c = \frac{n/2}{n} = \frac{1}{2}$.
Thus, $\theta_c = 30^\circ$.
Since $r_2 = \theta_c$, we have $r_2 = 30^\circ$.
Using the relation $r_1 = r_2 = \frac{A}{2}$ for minimum deviation, we get $\frac{A}{2} = 30^\circ$.
Therefore, $A = 60^\circ$.