A converging lens has a focal length of `20 cm` in air. It is made of a material of refractive index `1.6`. If it is immersed in a liquid of refractive index `1.3`, what will be its new foacl length ?

Explanation

To find the new focal length of the converging lens when it is immersed in a liquid, we can use the lens maker's formula, which relates the focal length of a lens to its refractive index and the radii of curvature of its surfaces.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Focal length of the lens in air, \( f = 20 \, \text{cm} \)
- Refractive index of the lens material, \( \mu = 1.6 \)
- Refractive index of the liquid, \( \mu_{liquid} = 1.3 \)

2. **Calculate the Relative Refractive Index:**
The relative refractive index of the lens with respect to the liquid is given by:
\[ \mu_{relative} = \frac{\mu}{\mu_{liquid}} = \frac{1.6}{1.3} \]

3. **Calculate \( \mu_{relative} - 1 \):**
\[ \mu_{relative} - 1 = \frac{1.6}{1.3} - 1 = \frac{1.6 - 1.3}{1.3} = \frac{0.3}{1.3} \]

4. **Using the Lens Maker's Formula:**
The lens maker's formula in terms of the focal length is:
\[ \frac{1}{f} = (\mu_{relative} - 1) \cdot \alpha \]
where \( \alpha = \frac{1}{r_1} - \frac{1}{r_2} \).

5. **Find \( \alpha \) from the original focal length:**
From the original focal length in air:
\[ \frac{1}{20} = (1.6 - 1) \cdot \alpha = 0.6 \cdot \alpha \]
Rearranging gives:
\[ \alpha = \frac{1/20}{0.6} = \frac{1}{12} \]

6. **Calculate New Focal Length \( f_1 \):**
Now, substituting \( \mu_{relative} - 1 \) and \( \alpha \) into the lens maker's formula for the new focal length:
\[ \frac{1}{f_1} = \left(\frac{0.3}{1.3}\right) \cdot \frac{1}{12} \]
Simplifying gives:
\[ \frac{1}{f_1} = \frac{0.3}{15.6} = \frac{3}{156} = \frac{1}{52} \]
Therefore, the new focal length \( f_1 \) is:
\[ f_1 = 52 \, \text{cm} \]

### Final Answer:
The new focal length of the lens when immersed in the liquid is \( 52 \, \text{cm} \).

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