The given figure shows a convergent lens placed inside a cell filled with liquid. The lens has focal length +20 cm when in air , and its metrial has refractive index 1.50 . If the liquid has refractive index 1.60 , then focal length of the lens in the cell is

Options

A. 80 cm
B. `-80 cm `
C. `-24 cm `
D. `-160 cm `

Answer

To find the focal length of a convergent lens placed in a liquid, we can use the lens maker's formula, which relates the focal length of a lens to the refractive indices of the lens material and the surrounding medium, as well as the radii of curvature of the lens surfaces.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Focal length of the lens in air, \( f_0 = +20 \, \text{cm} \)
- Refractive index of the lens material, \( \mu_s = 1.50 \)
- Refractive index of the liquid, \( \mu_e = 1.60 \)

2. **Use the Lens Maker's Formula**:
The lens maker's formula in air is given by:
\[ \frac{1}{f_0} = \left( \mu_s - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

3. **Rearranging for \( \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)**:
From the formula, we can express \( \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \):
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{f_0} \div \left( \mu_s - 1 \right) \]
Substituting the known values:
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{20} \div (1.50 - 1) = \frac{1}{20} \div 0.50 = \frac{1}{20} \times 2 = \frac{1}{10} \]

4. **Calculate the Focal Length in the Liquid**:
When the lens is placed in the liquid, the lens maker's formula becomes:
\[ \frac{1}{f} = \left( \frac{\mu_s}{\mu_e} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{f} = \left( \frac{1.50}{1.60} - 1 \right) \left( \frac{1}{10} \right) \]
Calculate \( \frac{1.50}{1.60} \):
\[ \frac{1.50}{1.60} = 0.9375 \quad \Rightarrow \quad 0.9375 - 1 = -0.0625 \]
Now substituting back:
\[ \frac{1}{f} = -0.0625 \times \frac{1}{10} = -0.00625 \]
Therefore, the focal length \( f \) is:
\[ f = \frac{1}{-0.00625} = -160 \, \text{cm} \]

5. **Conclusion**:
The focal length of the lens in the liquid is \( -160 \, \text{cm} \).

### Final Answer:
The focal length of the lens in the cell is \( -160 \, \text{cm} \).