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 when its focal length in air is given as 10 cm, we can use the formula for the focal length of a lens in different media. Here are the steps to solve the problem:

### Step-by-Step Solution:

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

2. **Use the Lens Maker's Formula:**
The lens maker's formula for the focal length \( F \) of a lens in a medium is given by:
\[ \frac{1}{F} = \left( \frac{\mu_{lens}}{\mu_{medium}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( \mu_{lens} \) is the refractive index of the lens material (glass), \( \mu_{medium} \) is the refractive index of the surrounding medium (air or water), and \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

3. **Calculate the Focal Length in Air:**
In air, the refractive index is approximately 1, so we can write:
\[ \frac{1}{F_{air}} = \left( \frac{\mu_g}{1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This simplifies to:
\[ \frac{1}{10} = \left( \frac{3/2}{1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
\[ \frac{1}{10} = \left( \frac{1}{2} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

4. **Calculate the Focal Length in Water:**
Now, we apply the same formula for the focal length in water:
\[ \frac{1}{F_{water}} = \left( \frac{\mu_g}{\mu_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{F_{water}} = \left( \frac{3/2}{4/3} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Simplifying the fraction:
\[ \frac{3/2}{4/3} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8} \]
Thus:
\[ \frac{1}{F_{water}} = \left( \frac{9}{8} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
\[ = \left( \frac{1}{8} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

5. **Relate the Focal Lengths:**
From the previous calculations, we can relate \( F_{water} \) and \( F_{air} \):
\[ \frac{F_{water}}{F_{air}} = \frac{\left( \frac{9}{8} - 1 \right)}{\left( \frac{3}{2} - 1 \right)} = \frac{\frac{1}{8}}{\frac{1}{2}} = \frac{1}{4} \]
Therefore:
\[ F_{water} = 4 \times F_{air} \]
Substituting \( F_{air} = 10 \, \text{cm} \):
\[ F_{water} = 4 \times 10 = 40 \, \text{cm} \]

### Final Answer:
The focal length of the lens in water is \( F_{water} = 40 \, \text{cm} \).

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