For a prism of refractive index `1.732`, the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is

Options

A. `80^(@)`
B. `70^(@)`
C. `60^(@)`
D. `50^(@)`

Explanation

To solve the problem, we need to find the angle of the prism (A) given that the refractive index (μ) is 1.732 and the angle of minimum deviation (D) is equal to the angle of the prism (A).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
We know that for a prism, the refractive index (μ) is related to the angle of the prism (A) and the angle of minimum deviation (D) by the formula:
\[ \mu = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Since we are given that \(D = A\), we can substitute \(D\) with \(A\):
\[ \mu = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)} \]

2. **Substituting the Given Refractive Index**:
We know that \(\mu = 1.732\):
\[ 1.732 = \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)} \]

3. **Using the Double Angle Identity**:
We can use the identity \(\sin(A) = 2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)\):
\[ 1.732 = \frac{2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Simplifying this gives:
\[ 1.732 = 2 \cos\left(\frac{A}{2}\right) \]

4. **Solving for Cosine**:
Dividing both sides by 2:
\[ \cos\left(\frac{A}{2}\right) = \frac{1.732}{2} = 0.866 \]

5. **Finding the Angle**:
We know that \(\cos(30^\circ) = 0.866\), therefore:
\[ \frac{A}{2} = 30^\circ \]
Multiplying both sides by 2 gives:
\[ A = 60^\circ \]

6. **Conclusion**:
The angle of the prism is \(60^\circ\).

### Final Answer:
The angle of the prism is \(60^\circ\).