For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index n of the prism satisfies.

For a prism, the angle of minimum deviation $$\delta_{\text{min}}$$ is related to the refracting angle $$A$$ and the refractive index $$n$$ by the formula:

$$ n = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)} $$

Given that $$\delta_{\text{min}} = A$$, substitute this into the formula:

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

Using the double-angle identity $$\sin A = 2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)$$:

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

Thus, $$n = 2 \cos\left(\frac{A}{2}\right)$$.

For a prism, the refracting angle $$A$$ is typically between $$0^\circ$$ and $$90^\circ$$ (acute). As $$A$$ increases from $$0^\circ$$ to $$90^\circ$$, $$\frac{A}{2}$$ increases from $$0^\circ$$ to $$45^\circ$$. The cosine function decreases from $$\cos 0^\circ = 1$$ to $$\cos 45^\circ = \frac{\sqrt{2}}{2}$$. Therefore:

When $$A \to 0^\circ$$, $$n \to 2 \times 1 = 2$$.
When $$A \to 90^\circ$$, $$n \to 2 \times \frac{\sqrt{2}}{2} = \sqrt{2}$$.

However, $$A = 0^\circ$$ is not possible for a physical prism, so $$n < 2$$. Also, $$A = 90^\circ$$ is not typical for prisms, and the ray emergence condition requires $$A < 90^\circ$$ to avoid grazing emergence, so $$n > \sqrt{2}$$. Thus, $$n$$ lies in the open interval $$(\sqrt{2}, 2)$$.

Now, comparing with the options:

A. $$n \geq 2$$ → Incorrect, as $$n < 2$$.
B. $$\sqrt{2} < n < 2\sqrt{2}$$ → Incorrect, as $$2\sqrt{2} \approx 2.828$$ is larger than 2, and $$n < 2$$.
C. $$1 < n < 2$$ → Incorrect, as it includes values less than $$\sqrt{2} \approx 1.414$$.
D. $$\sqrt{2} < n < 2$$ → Correct, as it matches the interval $$\sqrt{2} < n < 2$$.

Therefore, the refractive index $$n$$ satisfies $$\sqrt{2} < n < 2$$.