A liquid is placed in a hollow prism of angle `60^(@)`. If angle of the minimum deviation is 30 what is the refractive index of the liquid?

Options

A. 1.41
B. 1.5
C. 1.65
D. 1.95

Answer

To find the refractive index of the liquid placed in a hollow prism, we can use the formula related to the angle of minimum deviation and the angle of the prism. Here are the steps to solve the problem:

### Step-by-Step Solution:

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

2. **Use the Formula for Refractive Index**:
The formula for the refractive index (n) of a prism in terms of the angle of minimum deviation (D) and the angle of the prism (A) is given by:
\[ n = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

3. **Substitute the Values into the Formula**:
- Calculate \( \frac{A + D}{2} \):
\[ \frac{A + D}{2} = \frac{60° + 30°}{2} = \frac{90°}{2} = 45° \]
- Calculate \( \frac{A}{2} \):
\[ \frac{A}{2} = \frac{60°}{2} = 30° \]

4. **Calculate the Sine Values**:
- \( \sin(45°) = \frac{1}{\sqrt{2}} \)
- \( \sin(30°) = \frac{1}{2} \)

5. **Substitute the Sine Values into the Formula**:
\[ n = \frac{\sin(45°)}{\sin(30°)} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{1}{\sqrt{2}} \times 2 = \frac{2}{\sqrt{2}} = \sqrt{2} \]

6. **Calculate the Numerical Value**:
- The value of \( \sqrt{2} \) is approximately 1.414.

### Final Result:
The refractive index of the liquid is approximately \( n \approx 1.414 \).