The focal length of a thin lens in vacuum is `f`. If the material of the lens has `mu = 3//2`, its focal length when immersed in water of refractive index `4//3` will be.

Options

A. f
B. `4/3`f
C. 2f
D. 4f

Answer

To find the focal length of a thin lens when it is immersed in water, we can use the lens maker's formula. The formula relates the focal length of a lens to the refractive indices of the lens material and the surrounding medium.

### Step-by-Step Solution:

1. **Understanding the Lens Maker's Formula**:
The lens maker's formula is given by:
\[ \frac{1}{F} = \left(\frac{\mu_l}{\mu_s} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
where:
- \( F \) is the focal length of the lens,
- \( \mu_l \) is the refractive index of the lens material,
- \( \mu_s \) is the refractive index of the surrounding medium,
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

2. **Case 1: Lens in Vacuum**:
For the lens in vacuum, the refractive index of the surrounding medium \( \mu_s = 1 \) and the refractive index of the lens material \( \mu_l = \frac{3}{2} \). Thus, we can write:
\[ \frac{1}{F} = \left(\frac{3/2}{1} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Simplifying this gives:
\[ \frac{1}{F} = \left(\frac{1}{2}\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

3. **Case 2: Lens Immersed in Water**:
Now, when the lens is immersed in water, the refractive index of the surrounding medium \( \mu_s = \frac{4}{3} \). The formula for the new focal length \( F' \) becomes:
\[ \frac{1}{F'} = \left(\frac{3/2}{4/3} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Simplifying \( \frac{3/2}{4/3} \):
\[ \frac{3/2}{4/3} = \frac{3 \times 3}{2 \times 4} = \frac{9}{8} \]
Thus, we have:
\[ \frac{1}{F'} = \left(\frac{9}{8} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{9}{8} - \frac{8}{8}\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{8} \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

4. **Relating Focal Lengths**:
Now, we can relate \( \frac{1}{F} \) and \( \frac{1}{F'} \):
\[ \frac{1}{F'} = \frac{1}{8} \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{4} \left(\frac{1}{2} \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\right) = \frac{1}{4} \cdot \frac{1}{F} \]
Therefore, we have:
\[ F' = 4F \]

### Final Answer:
The focal length of the lens when immersed in water is:
\[ F' = 4F \]