The angle of minimum deviation measured with a prism is `30^(@)` and the angle of prism is `60^(@)`. The refractive index of prism material is

Options

A. `sqrt(2)`
B. `2`
C. `3//2`
D. `4//3`

Explanation

To find the refractive index of the prism material given the angle of the prism and the angle of minimum deviation, we can use the formula for refractive index (μ) in terms of these angles:

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

Where:
- \( A \) is the angle of the prism
- \( D \) is the angle of minimum deviation

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Angle of prism \( A = 60^\circ \)
- Angle of minimum deviation \( D = 30^\circ \)

2. **Calculate \( A + D \):**
\[ A + D = 60^\circ + 30^\circ = 90^\circ \]

3. **Calculate \( \frac{A + D}{2} \):**
\[ \frac{A + D}{2} = \frac{90^\circ}{2} = 45^\circ \]

4. **Calculate \( \frac{A}{2} \):**
\[ \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \]

5. **Find the Sine Values:**
- \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

6. **Substitute into the Refractive Index Formula:**
\[ \mu = \frac{\sin\left(45^\circ\right)}{\sin\left(30^\circ\right)} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{1}{\sqrt{2}} \times 2 = \frac{2}{\sqrt{2}} = \sqrt{2} \]

7. **Final Result:**
The refractive index of the prism material is \( \mu = \sqrt{2} \).