A prism has a refractive index cot(A/2), where A is the refracting angle of the prism. The minimum deviation due to this prism is

The correct option is π – 2A.

To find the minimum deviation due to the prism, we start with the prism formula relating the refractive index ($\mu$), the refracting angle of the prism ($A$), and the angle of minimum deviation (${\delta }_m$):
$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

We are given that the refractive index of the prism is:
$\mu =\cot \left(\frac{A}{2}\right)$

Using the trigonometric identity $\cot \left(\theta \right)=\frac{\cos \left(\theta \right)}{\sin \left(\theta \right)}$, we can express the refractive index as:
$\mu =\frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

Now, equating the two expressions for $\mu$:
$\frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

Assuming $\sin \left(\frac{A}{2}\right)\ne 0$, we can cancel it from the denominators on both sides:
$\cos \left(\frac{A}{2}\right)=\sin \left(\frac{A+{\delta }_m}{2}\right)$

Using the trigonometric identity relating sine and cosine, $\cos \left(\theta \right)=\sin \left(\frac{\pi }{2}-\theta \right)$, we can rewrite the left side of the equation:
$\sin \left(\frac{\pi }{2}-\frac{A}{2}\right)=\sin \left(\frac{A+{\delta }_m}{2}\right)$

Comparing the angles on both sides:
$\frac{\pi }{2}-\frac{A}{2}=\frac{A+{\delta }_m}{2}$

Multiply the entire equation by 2 to clear the fractions:
$\pi -A=A+{\delta }_m$

Solving for the angle of minimum deviation (${\delta }_m$):
${\delta }_m=\pi -2A$

Thus, the minimum deviation due to the prism is indeed π – 2A.