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 focal length of a lens in different media is given by: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] In air: \[ f = 24 \, \text{cm}, \quad \mu_{\text{air}} = 1.5 \] In water: \[ \mu_{\text{water}} = 1.33 \]

Step 2: Using the lens maker's formula, the focal length in water \( f' \) is related to the focal length in air \( f \) by the refractive index ratio: \[ \frac{1}{f'} = \left( \frac{\mu_{\text{water}} - 1}{\mu_{\text{air}} - 1} \right) \cdot \frac{1}{f} \] Substitute the values: \[ \frac{1}{f'} = \left( \frac{1.33 - 1}{1.5 - 1} \right) \cdot \frac{1}{24} \] \[ \frac{1}{f'} = \frac{0.33}{0.5} \cdot \frac{1}{24} \] \[ f' = \frac{96}{1} = 96 \, \text{cm} \] Thus, the focal length changes to 96 cm.