A convex lens of focal length `12 cm` is made of glass of `mu=(3)/(2)`. What will be its focal length when immersed in liquid of `mu=(5)/(4)` ?

Options

A. `6 cm`
B. `12 cm`
C. `24 cm`
D. `30 cm`

Explanation

To find the new focal length of a convex lens when it is immersed in a liquid, we can use the formula for the focal length of a lens in a medium different from air. The formula is given by:

\[ \frac{f'}{f} = \frac{\mu_g - 1}{\mu_g/\mu_s - 1} \]

Where:
- \( f' \) is the new focal length in the liquid,
- \( f \) is the original focal length in air,
- \( \mu_g \) is the refractive index of the lens material (glass),
- \( \mu_s \) is the refractive index of the surrounding medium (liquid).

### Step-by-Step Solution:

1. **Identify the given values**:
- Focal length of the lens in air, \( f = 12 \, \text{cm} \)
- Refractive index of glass, \( \mu_g = \frac{3}{2} = 1.5 \)
- Refractive index of the liquid, \( \mu_s = \frac{5}{4} = 1.25 \)

2. **Substitute the values into the formula**:
\[ \frac{f'}{12} = \frac{\frac{3}{2} - 1}{\frac{3/2}{5/4} - 1} \]

3. **Calculate \( \mu_g - 1 \)**:
\[ \mu_g - 1 = \frac{3}{2} - 1 = \frac{3}{2} - \frac{2}{2} = \frac{1}{2} \]

4. **Calculate \( \frac{\mu_g}{\mu_s} \)**:
\[ \frac{\mu_g}{\mu_s} = \frac{\frac{3}{2}}{\frac{5}{4}} = \frac{3}{2} \times \frac{4}{5} = \frac{12}{10} = \frac{6}{5} \]

5. **Calculate \( \frac{\mu_g}{\mu_s} - 1 \)**:
\[ \frac{6}{5} - 1 = \frac{6}{5} - \frac{5}{5} = \frac{1}{5} \]

6. **Substitute these values back into the equation**:
\[ \frac{f'}{12} = \frac{\frac{1}{2}}{\frac{1}{5}} = \frac{1}{2} \times 5 = \frac{5}{2} \]

7. **Solve for \( f' \)**:
\[ f' = 12 \times \frac{5}{2} = 30 \, \text{cm} \]

Thus, the new focal length of the lens when immersed in the liquid is **30 cm**.