$$ \textbf{Step 1: Symmetry at Minimum Deviation} $$

In a prism, when the deviation is minimum, the path of light is symmetric. This means that the angle of incidence ($i$) is equal to the angle of emergence ($i'$), and the light inside the prism makes equal angles with the prism faces. If the prism angle is $$A$$, then the refracted angle at each interface is:

$$ r = \frac{A}{2} $$

$$ \textbf{Step 2: Relating Deviation to the Angles} $$

The formula for the deviation ($D$) in a prism is given by:

$$ D = i + i' - A $$

At minimum deviation, $$i = i'$$, so:

$$ D_{\text{min}} = 2i - A $$

The special condition given is that the minimum deviation is equal to the prism angle:

$$ D_{\text{min}} = A $$

Thus:

$$ A = 2i - A \quad \Longrightarrow \quad 2i = 2A \quad \Longrightarrow \quad i = A $$

$$ \textbf{Step 3: Applying Snell's Law} $$

At the first surface, Snell's law gives:

$$ \sin i = \mu \sin r $$

Substitute the values $$i = A$$ and $$r = \frac{A}{2}$$:

$$ \sin A = \mu \sin\frac{A}{2} $$

Given that the refractive index $$\mu = \sqrt{3}$$, we have:

$$ \sin A = \sqrt{3} \sin\frac{A}{2} $$

$$ \textbf{Step 4: Solving the Equation} $$

Recall the double-angle formula for sine:

$$ \sin A = 2 \sin\frac{A}{2} \cos\frac{A}{2} $$

Substitute this into the previous equation:

$$ 2 \sin\frac{A}{2} \cos\frac{A}{2} = \sqrt{3} \sin\frac{A}{2} $$

Assuming $$\sin\frac{A}{2} \neq 0$$, we can divide both sides by $$\sin\frac{A}{2}$$:

$$ 2 \cos\frac{A}{2} = \sqrt{3} $$

Solve for $$\cos\frac{A}{2}$$:

$$ \cos\frac{A}{2} = \frac{\sqrt{3}}{2} $$

Since:

$$ \cos 30^\circ = \frac{\sqrt{3}}{2} $$

It follows:

$$ \frac{A}{2} = 30^\circ \quad \Longrightarrow \quad A = 60^\circ $$

$$ \textbf{Final Answer:} $$

The angle of the prism is $$60^\circ$$.