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

This problem requires calculating the focal lengths of two plano-convex lenses with a refractive index \( n = 1.5 \). The lens maker's formula is employed for this calculation.

Lens Maker's Formula:

The formula is stated as:

\[\frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\]

Here, \( f \) represents the focal length, \( n \) is the refractive index of the lens material, \( R_1 \) is the radius of curvature of the first surface, and \( R_2 \) is the radius of curvature of the second surface. For a plano-convex lens, \( R_2 = \infty \), simplifying the formula to:

\[\frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{\infty}\right) = (n - 1) \frac{1}{R_1}\]

Focal Length of the First Lens (\( f_1 \)):

Given \( R_1 = 2 \, \text{cm} \) and \( n = 1.5 \) (lens in air), the simplified lens maker's formula yields:

\[\frac{1}{f_1} = (1.5 - 1) \frac{1}{2}\]\[\frac{1}{f_1} = 0.5 \times \frac{1}{2} = \frac{0.5}{2} = 0.25 \, \text{cm}^{-1}\]\[f_1 = \frac{1}{0.25} = 4 \, \text{cm}\]

Focal Length of the Second Lens (\( f_2 \)):

For the second lens, \( R_1 = 3 \, \text{cm} \) and it is immersed in a liquid with a refractive index of 1.2. The effective refractive index is calculated as \( n_{\text{eff}} = \frac{1.5}{1.2} \).

\[\frac{1}{f_2} = \left(\frac{1.5}{1.2} - 1\right) \frac{1}{3}\]\[\frac{1}{f_2} = \left(\frac{0.3}{1.2}\right) \times \frac{1}{3}\]\[\frac{1}{f_2} = \frac{0.3}{3.6} = \frac{1}{12}\]\[f_2 = 12 \, \text{cm}\]

Ratio of \( f_1 \) and \( f_2 \):

The ratio between the focal lengths is calculated as:

\[\text{Ratio} = \frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3}\]

Therefore, the ratio of \( f_1 \) to \( f_2 \) is \( 1 : 3 \).

The final answer is \( 1 : 3 \).