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 that relates the refractive index (n), the angle of the prism (A), and the angle of minimum deviation (D). The formula is given by:

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

### Step-by-Step Solution:

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

2. **Substitute the values into the formula:**
\[ n = \frac{\sin\left(\frac{30^\circ + 60^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \]

3. **Calculate the angles:**
- Calculate \( \frac{D + A}{2} \):
\[ \frac{30^\circ + 60^\circ}{2} = \frac{90^\circ}{2} = 45^\circ \]
- Calculate \( \frac{A}{2} \):
\[ \frac{60^\circ}{2} = 30^\circ \]

4. **Evaluate the sine values:**
- \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

5. **Substitute the sine values back into the equation:**
\[ n = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{1}{\sqrt{2}} \times 2 = \frac{2}{\sqrt{2}} = \sqrt{2} \]

6. **Final result:**
The refractive index of the liquid is:
\[ n = \sqrt{2} \approx 1.414 \]

### Conclusion:
Thus, the refractive index of the liquid is \( \sqrt{2} \) or approximately 1.414.