A thin convex lens of refractive index 1.5cm has 20cm focal length in air. If the lens in completely immersed in a liquid of refractive index. 1.6, then its focal length will be

Options

A. `-160cm`
B. `-100cm`
C. `+10cm`
D. `+100cm`

Explanation

To solve the problem of finding the focal length of a convex lens when it is immersed in a liquid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Refractive index of the lens (n_lens) = 1.5
- Focal length of the lens in air (f_a) = 20 cm
- Refractive index of the liquid (n_liquid) = 1.6

2. **Use the Lens Maker's Formula:**
The lens maker's formula is given by:
\[ \frac{1}{f} = (n_{lens} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( f \) is the focal length, \( n_{lens} \) is the refractive index of the lens, and \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

3. **Calculate the Focal Length in Air:**
For the lens in air, we can express the formula as:
\[ \frac{1}{f_a} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Simplifying this gives:
\[ \frac{1}{20} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Therefore:
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{10} \]

4. **Calculate the Focal Length in Liquid:**
When the lens is immersed in a liquid, the lens maker's formula becomes:
\[ \frac{1}{f_l} = \left( \frac{n_{lens}}{n_{liquid}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values we have:
\[ \frac{1}{f_l} = \left( \frac{1.5}{1.6} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
We already found that \( \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{10} \). Therefore:
\[ \frac{1}{f_l} = \left( \frac{1.5}{1.6} - 1 \right) \cdot \frac{1}{10} \]

5. **Calculate \( \frac{1.5}{1.6} - 1 \):**
\[ \frac{1.5}{1.6} = 0.9375 \quad \text{thus} \quad 0.9375 - 1 = -0.0625 \]
Therefore:
\[ \frac{1}{f_l} = -0.0625 \cdot \frac{1}{10} = -0.00625 \]

6. **Find \( f_l \):**
Taking the reciprocal gives:
\[ f_l = \frac{1}{-0.00625} = -160 \text{ cm} \]

### Final Answer:
The focal length of the lens when immersed in the liquid is \( f_l = -160 \text{ cm} \).