Question

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

• A. 10 cm
• B. 30 cm
• C. 40 cm
• D. 50 cm

Hint:

The focal length of a lens changes when placed in a medium with a different refractive index. Use the formula \( f_{\text{liquid}} = f_{\text{air}} \cdot \frac{n_{\text{lens}} - 1}{n_{\text{liquid}} - 1} \).

Answer 1

The Correct Option is C

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).