The refracting angle of a prism is A and refractive index of the material of prism is `cot(A//2)` . The angle of minimum deviation will be

Options

A. `180-3A`
B. `180-2A`
C. `90-2A`
D. `180+2A`

Explanation

To solve the problem, we need to find the angle of minimum deviation (D) for a prism given its refracting angle (A) and the refractive index (n) of the material of the prism.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The refracting angle of the prism is denoted as \( A \).
- The refractive index of the material of the prism is given as \( n = \cot\left(\frac{A}{2}\right) \).

2. **Using the Formula for Refractive Index**:
- The relationship between the refractive index \( n \), the angle of the prism \( A \), and the angle of minimum deviation \( D \) is given by the formula:
\[ n = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

3. **Substituting the Refractive Index**:
- We substitute \( n \) with \( \cot\left(\frac{A}{2}\right) \):
\[ \cot\left(\frac{A}{2}\right) = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

4. **Expressing Cotangent in Terms of Sine and Cosine**:
- Recall that \( \cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \).
- Thus, we can rewrite the equation as:
\[ \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
- This simplifies to:
\[ \cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + D}{2}\right) \]

5. **Using the Co-Function Identity**:
- We can use the identity \( \sin\left(90^\circ - x\right) = \cos(x) \):
\[ \sin\left(\frac{A + D}{2}\right) = \cos\left(\frac{A}{2}\right) \implies \frac{A + D}{2} = 90^\circ - \frac{A}{2} \]

6. **Solving for Minimum Deviation \( D \)**:
- Rearranging gives:
\[ A + D = 180^\circ - A \]
\[ D = 180^\circ - 2A \]

7. **Final Result**:
- Therefore, the angle of minimum deviation \( D \) is:
\[ D = 180^\circ - 2A \]

### Conclusion:
The angle of minimum deviation for the prism is \( 180^\circ - 2A \).