The refractive index of a prism with apex angle $$A$$ is $$\cot\frac{A}{2}$$. The angle of minimum deviation is :

A prism has apex angle $$A$$ and refractive index $$\mu = \cot\frac{A}{2}$$. We need to find the angle of minimum deviation $$\delta_m$$.

We begin by recalling the prism formula for minimum deviation.

For a prism, the refractive index is related to the apex angle and minimum deviation by:

$$ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} $$

Next, we substitute $$\mu = \cot\frac{A}{2}$$ into this formula:

$$ \cot\frac{A}{2} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} $$

Since $$\cot\frac{A}{2} = \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}$$, we can write:

$$ \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} $$

Multiplying both sides by $$\sin\frac{A}{2}$$ yields:

$$ \cos\frac{A}{2} = \sin\left(\frac{A + \delta_m}{2}\right) $$

Then, using the identity $$\cos\theta = \sin\left(90° - \theta\right)$$, we have:

$$ \sin\left(90° - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right) $$

Equating the arguments leads to:

$$ 90° - \frac{A}{2} = \frac{A + \delta_m}{2} $$

Multiplying both sides by 2 gives:

$$ 180° - A = A + \delta_m $$

Solving for $$\delta_m$$, we find:

$$ \delta_m = 180° - 2A $$

The correct answer is Option (4): $$\delta_m = 180° - 2A$$.