The minimum deviation produced by a glass prism of angle `40^(@)` is `36^(@)`. The refractive index of the glass will be

Options

A. 1.4
B. 1.2
C. 1.5
D. 1.8

Explanation

To find the refractive index of the glass prism, we can use the formula for the refractive index in terms of the angle of the prism and the angle of minimum deviation. The formula is given by:

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

where:
- \( \mu \) is the refractive index,
- \( A \) is the angle of the prism,
- \( D_m \) is the minimum deviation.

### Step-by-Step Solution:

1. **Identify the given values**:
- Angle of the prism, \( A = 40^\circ \)
- Minimum deviation, \( D_m = 36^\circ \)

2. **Calculate \( \frac{A + D_m}{2} \)**:
\[ \frac{A + D_m}{2} = \frac{40^\circ + 36^\circ}{2} = \frac{76^\circ}{2} = 38^\circ \]

3. **Calculate \( \frac{A}{2} \)**:
\[ \frac{A}{2} = \frac{40^\circ}{2} = 20^\circ \]

4. **Use the sine function**:
- Calculate \( \sin(38^\circ) \) and \( \sin(20^\circ) \).
- Using a calculator:
- \( \sin(38^\circ) \approx 0.6157 \)
- \( \sin(20^\circ) \approx 0.3420 \)

5. **Substitute the values into the refractive index formula**:
\[ \mu = \frac{\sin(38^\circ)}{\sin(20^\circ)} = \frac{0.6157}{0.3420} \approx 1.80 \]

6. **Final result**:
The refractive index of the glass prism is approximately \( \mu \approx 1.80 \).