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

Explanation:
The relation between the refractive index (\( n \)), refracting angle of the prism (\( A \)), and the angle of minimum deviation (\( \delta_{\text{min}} \)) is given by the prism formula:
\( n = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)} \)

1. Applying Given Condition:
According to the question, the angle of minimum deviation is equal to the refracting angle of the prism (\( \delta_{\text{min}} = A \)). Substituting this into the formula:
\( n = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin\left(\frac{A}{2}\right)} \)

Using the trigonometric identity \( \sin A = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right) \):
\( n = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = 2\cos\left(\frac{A}{2}\right) \)

2. Determining Limiting Conditions:
For minimum deviation, the angle of incidence (\( i \)is given by:
\( i = \frac{A + \delta_{\text{min}}}{2} = \frac{A + A}{2} = A \)
For a light ray to physically enter the prism, the angle of incidence must be strictly less than the grazing angle of \( 90^\circ \):
\( i < 90^\circ \implies A < 90^\circ \)

Since \( A \) is a positive angle for a real prism, the range of \( A \) is:
\( 0^\circ < A < 90^\circ \implies 0^\circ < \frac{A}{2} < 45^\circ \)

3. Finding the Range of Refractive Index:
Taking the cosine function across this inequality (reversing the inequality signs as cosine values decrease from \( 0^\circ \) to \( 45^\circ \)):
\( \cos(45^\circ) < \cos\left(\frac{A}{2}\right) < \cos(0^\circ) \)
\( \frac{1}{\sqrt{2}} < \cos\left(\frac{A}{2}\right) < 1 \)

Multiplying the entire inequality by 2:
\( 2 \cdot \frac{1}{\sqrt{2}} < 2\cos\left(\frac{A}{2}\right) < 2 \cdot 1 \)
\( \sqrt{2} < n < 2 \)

Hence, Option D is the correct choice.