Question

The ratio of the focal lengths of a convex lens when kept in air and when it is immersed in a liquid is 1:2. If the refractive index of the material of the lens is 1.5, then the refractive index of the liquid is

• A. 1.20
• B. 1.30
• C. 1.25
• D. 1.35
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Hint:

Use the lens maker’s formula with relative refractive indices for lenses in different media.

Answer 1

The Correct Option is A

The lens makers formula for a lens in a medium is: \[ \frac{1}{f} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where \( f \) is the focal length, \( \mu_g \) is the refractive index of the lens (glass), \( \mu_m \) is the refractive index of the medium, and \( R_1, R_2 \) are the radii of curvature (constant for the lens). - In air (\( \mu_m = \mu_a = 1 \)): \[ \frac{1}{f_a} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] - In liquid (\( \mu_m = \mu_l \)): \[ \frac{1}{f_l} = \left( \frac{\mu_g}{\mu_l} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Given the focal length ratio \( f_a : f_l = 1 : 2 \), so \( f_l = 2 f_a \), or: \[ \frac{1}{f_l} = \frac{1}{2 f_a} \] Divide the lens maker’s equations: \[ \frac{\frac{1}{f_l}}{\frac{1}{f_a}} = \frac{\frac{\mu_g}{\mu_l} - 1}{\mu_g - 1} \] Since \( \frac{1}{f_l} = \frac{1}{2} \cdot \frac{1}{f_a} \), we have: \[ \frac{\frac{1}{2 f_a}}{\frac{1}{f_a}} = \frac{1}{2} = \frac{\frac{\mu_g}{\mu_l} - 1}{\mu_g - 1} \] Given \( \mu_g = 1.5 \): \[ \frac{\frac{1.5}{\mu_l} - 1}{1.5 - 1} = \frac{1}{2} \] Simplify: \[ \frac{\frac{1.5}{\mu_l} - 1}{0.5} = \frac{1}{2} \] Multiply through by 0.5: \[ \frac{1.5}{\mu_l} - 1 = \frac{0.5}{2} = 0.25 \] Solve for \( \mu_l \): \[ \frac{1.5}{\mu_l} = 1.25 \implies \mu_l = \frac{1.5}{1.25} = \frac{1.5 \cdot 4}{1.25 \cdot 4} = \frac{6}{5} = 1.2 \] Verify: - In air: \( \frac{1}{f_a} \propto (1.5 - 1) = 0.5 \). - In liquid: \( \frac{1}{f_l} \propto \left( \frac{1.5}{1.2} - 1 \right) = 1.25 - 1 = 0.25 \). - Ratio: \( \frac{1}{f_l} / \frac{1}{f_a} = \frac{0.25}{0.5} = \frac{1}{2} \), so \( f_a / f_l = \frac{1}{2} \), confirming \( f_l = 2 f_a \). Check options: - (1) 1.20: Matches \( \mu_l = 1.2 \). - (2) 1.30: Gives \( \frac{1.5}{1.3} - 1 \approx 0.1538 \), ratio \( \frac{0.1538}{0.5} \approx 0.308 \), incorrect. - (3) 1.25: Gives \( \frac{1.5}{1.25} - 1 = 0.2 \), ratio \( \frac{0.2}{0.5} = 0.4 \), incorrect. - (4) 1.35: Gives \( \frac{1.5}{1.35} - 1 \approx 0.111 \), ratio \( \frac{0.111}{0.5} \approx 0.222 \), incorrect. The error in the original solution (\( \mu_l = 0.75 \)) arose from incorrect manipulation; the correct \( \mu_l = 1.2 \) aligns with option (1).