Question

A convex lens made of glass (refractive index = 1.5) has a focal length of 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to:

• A. 72 cm
• B. 96 cm
• C. 24 cm
• D. 48 cm

Hint:

When a lens is immersed in a medium with a different refractive index, its focal length changes. Use the lens maker’s formula to compute the new focal length by considering the ratio of refractive indices.

Answer 1

The Correct Option is B

Step 1: The lens maker's formula for focal length \( f \) in a medium with refractive index \( \mu \) and radii of curvature \( R_1 \) and \( R_2 \) is given by: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] For air, the focal length is \( f = 24 \, \text{cm} \) and the refractive index is \( \mu_{\text{air}} = 1.5 \). The refractive index of water is \( \mu_{\text{water}} = 1.33 \). Step 2: To find the focal length in water, \( f' \), we use the relationship derived from the lens maker's formula: \[ \frac{1}{f'} = \left( \frac{\mu_{\text{water}} - 1}{\mu_{\text{air}} - 1} \right) \cdot \frac{1}{f} \] Substituting the given values: \[ \frac{1}{f'} = \left( \frac{1.33 - 1}{1.5 - 1} \right) \cdot \frac{1}{24} \] \[ \frac{1}{f'} = \frac{0.33}{0.5} \cdot \frac{1}{24} \] Solving for \( f' \): \[ f' = \frac{96}{1} = 96 \, \text{cm} \] Therefore, the focal length in water is 96 cm.