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

To solve this problem, we need to determine the focal lengths of two plano-convex lenses made of glass with a refractive index \( n = 1.5 \). We will use the lens maker's formula to find the focal lengths.

Lens Maker's Formula: 

The lens maker's formula is given by:

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

where \( f \) is 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, \( R_2 = \infty \), so the formula simplifies 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 (\( f_1 \)):

The radius of curvature \( R_1 = 2 \, \text{cm} \) and the lens is in air, so \( n = 1.5 \). Using the simplified lens maker's formula:

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

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

The radius of curvature \( R_1 = 3 \, \text{cm} \) and the lens is immersed in a liquid with refractive index 1.2, so the effective refractive index \( n_{\text{eff}} = \frac{1.5}{1.2} \).

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

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

Finally, the ratio of the focal lengths is given by:

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

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

The correct answer is therefore: \( 1 : 3 \)

Answer 2

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 \( 1 : 3 \).