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 step by step, we need to find the new power of the lens when it is immersed in water, given the refractive indices of the lens and water.

### Step 1: Understand the Power of a Lens
The power \( P \) of a lens in air is given by the formula:
\[ P = \frac{\mu_L}{\mu_{air}} - 1 \cdot \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:
- \( \mu_L \) is the refractive index of the lens,
- \( \mu_{air} \) is the refractive index of air (approximately 1),
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 2: Write the New Power Formula
When the lens is immersed in water, the new power \( P' \) can be expressed as:
\[ P' = \left( \frac{\mu_L}{\mu_{water}} - 1 \right) \cdot \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( \mu_{water} \) is the refractive index of water.

### Step 3: Substitute the Values
Given:
- \( \mu_{water} = \frac{4}{3} \)
- \( \mu_L = \frac{3}{2} \)

Now, we can calculate the relative refractive index \( \mu_r \):
\[ \mu_r = \frac{\mu_L}{\mu_{water}} = \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{3}{2} \cdot \frac{3}{4} = \frac{9}{8} \]

### Step 4: Calculate New Power \( P' \)
Now substituting \( \mu_r \) into the new power formula:
\[ P' = \left( \frac{9}{8} - 1 \right) \cdot \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Calculating \( \frac{9}{8} - 1 \):
\[ \frac{9}{8} - 1 = \frac{9}{8} - \frac{8}{8} = \frac{1}{8} \]

### Step 5: Relate New Power to Old Power
We can relate the new power \( P' \) to the old power \( P \):
\[ P' = \frac{1}{8} \cdot \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Since \( P = \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \), we can write:
\[ P' = \frac{1}{8} P \]

### Conclusion
Thus, the new power of the lens when immersed in water is:
\[ P' = \frac{P}{4} \]