Question

A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of \( f_1 \) in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of \( f_2 \) when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of the same glass of refractive index 1.5, the ratio of \( f_1 \) and \( f_2 \) will be:

• A. \( 3 : 5 \)
• B. \( 1 : 3 \)
• C. \( 1 : 2 \)
• D. \( 2 : 3 \)

Hint:

When solving for the focal length of a plano-convex lens in different mediums, use the lensmaker’s formula considering the refractive indices of both the lens material and the surrounding medium.

Answer 1

The Correct Option is B

For a plano-convex lens, the lensmaker’s formula is: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} \right) \] Step 1: For the first lens in air: \[ \frac{1}{f_1} = \left( \frac{\mu_1}{n_{{air}}} - 1 \right) \frac{1}{R_1} \] Given that \( \mu_1 = 1.5 \), \( n_{{air}} = 1.0 \), and \( R_1 = 2 \, {cm} \): \[ \frac{1}{f_1} = \left( \frac{1.5}{1} - 1 \right) \frac{1}{2} = \frac{0.5}{2} = \frac{1}{4} \] So, \( f_1 = 4 \, {cm} \). 
Step 2: For the second lens in a liquid: \[ \frac{1}{f_2} = \left( \frac{\mu_2}{n_{{liquid}}} - 1 \right) \frac{1}{R_2} \] Given that \( \mu_2 = 1.5 \), \( n_{{liquid}} = 1.2 \), and \( R_2 = 3 \, {cm} \): \[ \frac{1}{f_2} = \left( \frac{1.5}{1.2} - 1 \right) \frac{1}{3} = \frac{0.25}{3} = \frac{1}{12} \] So, \( f_2 = 12 \, {cm} \). 
Step 3: Finding the ratio of focal lengths: \[ \frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3} \] Thus, the correct ratio is \( 1 : 3 \).