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.

Answer 2

Step 1: Given data.
Refractive index of glass (\( \mu_g \)) = 1.5
Refractive index of air (\( \mu_a \)) = 1.0
Refractive index of water (\( \mu_w \)) = 1.33
Focal length in air (\( f_1 \)) = 24 cm.

Step 2: Lens maker’s formula.
The lens maker’s formula for a lens in a medium of refractive index \( \mu_m \) is:
\[ \frac{1}{f} = \left( \frac{\mu_l}{\mu_m} - 1 \right)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where \( \mu_l \) is the refractive index of the lens material and \( R_1, R_2 \) are the radii of curvature of the two surfaces.

Step 3: Ratio of focal lengths.
For the same lens (same radii of curvature), the product \( \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \) remains constant.
Hence, the focal lengths in air and water are related as:
\[ \frac{f_2}{f_1} = \frac{\left( \frac{\mu_l}{\mu_a} - 1 \right)}{\left( \frac{\mu_l}{\mu_w} - 1 \right)} \]

Step 4: Substitute the values.
\[ \frac{f_2}{24} = \frac{\left( \frac{1.5}{1.0} - 1 \right)}{\left( \frac{1.5}{1.33} - 1 \right)} = \frac{0.5}{\left( 1.1278 - 1 \right)} = \frac{0.5}{0.1278} \approx 3.91 \] \[ f_2 = 24 \times 3.91 = 93.84 \approx 96 \, \text{cm} \]

Final Answer:
\[ \boxed{96 \, \text{cm}} \]