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 – A
B. 180 – 2A
C. 90 – A
D. A/2

Explanation

To find the angle of minimum deviation (Δ) for a prism with a refracting angle (A) and a refractive index (μ) given as cot(A/2), we can use the formula for the refractive index of a prism:

\[ \mu = \frac{\sin\left(\frac{A + \Delta_{min}}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

### Step-by-Step Solution:

1. **Write down the given information:**
- Refracting angle of the prism, \( A \)
- Refractive index, \( \mu = \cot\left(\frac{A}{2}\right) \)

2. **Substitute the value of μ into the prism formula:**
\[ \cot\left(\frac{A}{2}\right) = \frac{\sin\left(\frac{A + \Delta_{min}}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

3. **Rewrite cotangent in terms of sine and cosine:**
\[ \cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

4. **Set the two expressions equal:**
\[ \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + \Delta_{min}}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

5. **Cancel \(\sin\left(\frac{A}{2}\right)\) from both sides (assuming it is not zero):**
\[ \cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + \Delta_{min}}{2}\right) \]

6. **Use the identity \(\sin\left(90^\circ - x\right) = \cos(x)\):**
\[ \sin\left(90^\circ - \frac{A}{2}\right) = \sin\left(\frac{A + \Delta_{min}}{2}\right) \]

7. **Equate the angles:**
\[ 90^\circ - \frac{A}{2} = \frac{A + \Delta_{min}}{2} \]

8. **Multiply through by 2 to eliminate the fraction:**
\[ 180^\circ - A = A + \Delta_{min} \]

9. **Rearranging gives:**
\[ \Delta_{min} = 180^\circ - A - A = 180^\circ - 2A \]

10. **Final Result:**
The angle of minimum deviation is:
\[ \Delta_{min} = 180^\circ - 2A \]