Focal length of a convex lense in air is `10cm`. Find its focal
length in water. Given that `mu_g=3//2` and `mu_w=4//3`.

Explanation

To find the focal length of a convex lens in water given its focal length in air and the refractive indices, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Focal length of the lens in air, \( f_{\text{air}} = 10 \, \text{cm} \)
- Refractive index of glass, \( \mu_g = \frac{3}{2} \)
- Refractive index of water, \( \mu_w = \frac{4}{3} \)

2. **Use the Lensmaker's Formula**:
The lensmaker's formula in a medium is given by:
\[ \frac{1}{f} = \left( \frac{\mu_2}{\mu_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( \mu_1 \) is the refractive index of the surrounding medium, \( \mu_2 \) is the refractive index of the lens material, and \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

3. **Calculate Focal Length in Air**:
For the lens in air, we have:
\[ \mu_1 = 1 \quad (\text{for air}), \quad \mu_2 = \mu_g = \frac{3}{2} \]
Thus, the equation becomes:
\[ \frac{1}{f_{\text{air}}} = \left( \frac{\frac{3}{2}}{1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Simplifying this gives:
\[ \frac{1}{f_{\text{air}}} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

4. **Express \( \frac{1}{R_1} - \frac{1}{R_2} \)**:
Rearranging gives:
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{2}{f_{\text{air}}} = \frac{2}{10} = \frac{1}{5} \]

5. **Calculate Focal Length in Water**:
Now, we need to find the focal length in water. Here, \( \mu_1 = \mu_w = \frac{4}{3} \) and \( \mu_2 = \mu_g = \frac{3}{2} \):
\[ \frac{1}{f_{\text{water}}} = \left( \frac{\frac{3}{2}}{\frac{4}{3}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Simplifying the first part:
\[ \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{3}{2} \cdot \frac{3}{4} = \frac{9}{8} \]
Thus, we have:
\[ \frac{1}{f_{\text{water}}} = \left( \frac{9}{8} - 1 \right) \left( \frac{1}{5} \right) = \left( \frac{1}{8} \right) \left( \frac{1}{5} \right) = \frac{1}{40} \]

6. **Final Calculation**:
Therefore, the focal length in water is:
\[ f_{\text{water}} = 40 \, \text{cm} \]

### Conclusion:
The focal length of the convex lens in water is **40 cm**.