A thin lens of refractive index \( \frac{3}{2} \) is kept inside a liquid of refractive index \( \frac{4}{3} \). If the focal length of the lens in air is 10 cm, then its focal length inside the liquid is
Show Hint
The focal length of a lens changes when placed in a medium with a different refractive index. Use the formula \( f_{\text{liquid}} = f_{\text{air}} \cdot \frac{n_{\text{lens}} - 1}{n_{\text{liquid}} - 1} \).
Step 1: Understanding the lens formula.
The focal length \( f \) of a lens is given by the lens-maker's formula:
\[
\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
\]
where \( n \) is the refractive index of the lens, and \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.
However, we use the following relationship for the focal length of a lens inside a medium:
\[
f_{\text{liquid}} = f_{\text{air}} \cdot \frac{n_{\text{lens}} - 1}{n_{\text{liquid}} - 1}
\]
where:
- \( f_{\text{air}} \) is the focal length in air,
- \( n_{\text{lens}} \) is the refractive index of the lens,
- \( n_{\text{liquid}} \) is the refractive index of the liquid.
Step 2: Applying the formula.
Given:
- \( f_{\text{air}} = 10 \, \text{cm} \),
- \( n_{\text{lens}} = \frac{3}{2} \),
- \( n_{\text{liquid}} = \frac{4}{3} \).
Substitute these values into the formula:
\[
f_{\text{liquid}} = 10 \cdot \frac{\frac{3}{2} - 1}{\frac{4}{3} - 1} = 10 \cdot \frac{\frac{1}{2}}{\frac{1}{3}} = 10 \cdot 1.5 = 30 \, \text{cm}.
\]
Step 3: Conclusion.
The focal length inside the liquid is 30 cm, so the correct answer is (B).
