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. `60^(@)`
B. `70^(@)`
C. `50^(@)`
D. none of these

Answer

To solve the problem, we need to find the angle of the prism (denoted as \( a \)) given that the refractive index \( \mu \) is \( 1.732 \) and that the angle of minimum deviation \( \delta_m \) is equal to the angle of the prism \( a \).

### Step-by-step Solution:

1. **Understanding the relationship**:
We know from optics that the refractive index \( \mu \) of a prism is related to the angle of the prism \( a \) and the angle of minimum deviation \( \delta_m \) by the formula:
\[ \mu = \frac{\sin\left(\frac{a + \delta_m}{2}\right)}{\sin\left(\frac{a}{2}\right)} \]
Given that \( \delta_m = a \), we can substitute this into the equation.

2. **Substituting the values**:
Substituting \( \delta_m = a \) into the equation gives:
\[ \mu = \frac{\sin\left(\frac{a + a}{2}\right)}{\sin\left(\frac{a}{2}\right)} = \frac{\sin(a)}{\sin\left(\frac{a}{2}\right)} \]
We know \( \mu = 1.732 \), so we have:
\[ 1.732 = \frac{\sin(a)}{\sin\left(\frac{a}{2}\right)} \]

3. **Using the double angle formula**:
We can use the trigonometric identity for sine:
\[ \sin(a) = 2 \sin\left(\frac{a}{2}\right) \cos\left(\frac{a}{2}\right) \]
Substituting this into our equation gives:
\[ 1.732 = \frac{2 \sin\left(\frac{a}{2}\right) \cos\left(\frac{a}{2}\right)}{\sin\left(\frac{a}{2}\right)} \]
Simplifying this, we find:
\[ 1.732 = 2 \cos\left(\frac{a}{2}\right) \]

4. **Solving for \( \cos\left(\frac{a}{2}\right) \)**:
Rearranging gives:
\[ \cos\left(\frac{a}{2}\right) = \frac{1.732}{2} = 0.866 \]

5. **Finding the angle**:
The value \( 0.866 \) corresponds to \( \cos(30^\circ) \). Therefore:
\[ \frac{a}{2} = 30^\circ \]
Multiplying both sides by 2 gives:
\[ a = 60^\circ \]

### Conclusion:
The angle of the prism \( a \) is \( 60^\circ \).