The power of a lens kept in air is P. When it is immersed in water, then power becomes (`mu _(water) =4/3, mu_(Lens) = 3/2 )`

Options

A. `p/2`
B. `p/3`
C. ` p/4`
D. ` p/6`

Answer

To solve the problem, we need to find the new power of a lens when it is immersed in water, given the power of the lens in air and the refractive indices of the lens and water.

### Step-by-Step Solution:

1. **Understanding the Power of a Lens**:
The power \( P \) of a lens is given by the formula:
\[ P = \frac{1}{f} \]
where \( f \) is the focal length of the lens.

2. **Initial Conditions**:
- The power of the lens in air is \( P \).
- The refractive index of the lens \( \mu_{lens} = \frac{3}{2} \).
- The refractive index of air \( \mu_{air} \approx 1 \).

3. **Focal Length in Air**:
The formula for the focal length \( f \) of the lens in air is:
\[ \frac{1}{f} = \left( \frac{\mu_{lens}}{\mu_{air}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{f} = \left( \frac{\frac{3}{2}}{1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Therefore, the power in air is:
\[ P = \frac{1}{f} = \frac{1}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

4. **Focal Length in Water**:
When the lens is immersed in water, the new focal length \( f' \) is given by:
\[ \frac{1}{f'} = \left( \frac{\mu_{lens}}{\mu_{water}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Here, \( \mu_{water} = \frac{4}{3} \). Thus,
\[ \frac{1}{f'} = \left( \frac{\frac{3}{2}}{\frac{4}{3}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Simplifying the term:
\[ \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{3 \times 3}{2 \times 4} = \frac{9}{8} \]
Therefore,
\[ \frac{1}{f'} = \left( \frac{9}{8} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \left( \frac{9}{8} - \frac{8}{8} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{8} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

5. **Finding the New Power**:
The new power \( P' \) is given by:
\[ P' = \frac{1}{f'} = \frac{1}{8} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Now, we can relate \( P' \) to \( P \):
\[ P' = \frac{1}{4} P \]

6. **Conclusion**:
Thus, the power of the lens when immersed in water becomes:
\[ P' = \frac{P}{4} \]