Question

Focal length of a convex lens of refractive index 1.5 is 3 cm. When the lens is immersed in water of refractive index \( \frac{4}{3} \), its focal length will be:

• A. 3 cm
• B. 10 cm
• C. 12 cm
• D. 1.5 cm
• E. 6 cm

Hint:

When a lens is immersed in a different medium, the focal length changes due to the change in the refractive index of the surrounding medium.

Answer 1

The Correct Option is C

Step 1: Lens Maker's Formula The focal length \( f \) of a convex lens in air is given by the lens maker’s formula: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where \( \mu \) is the refractive index of the lens material, and \( R_1, R_2 \) are the radii of curvature of the lens surfaces. 
Step 2: Modified Lens Maker’s Formula in a Medium When the lens is immersed in a medium of refractive index \( \mu_m \), the modified formula becomes: \[ \frac{1}{f_m} = \left( \frac{\mu_{{lens}}}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Dividing both equations: \[ \frac{f}{f_m} = \frac{\mu - 1}{\frac{\mu}{\mu_m} - 1} \] 
Step 3: Substituting Given Values 
Given: 
- Refractive index of the lens: \( \mu = 1.5 \) 
- Refractive index of water: \( \mu_m = \frac{4}{3} \) 
- Focal length in air: \( f = 3 \) cm \[ \frac{f}{f_m} = \frac{1.5 - 1}{\frac{1.5}{\frac{4}{3}} - 1} \] \[ \frac{f}{f_m} = \frac{0.5}{\frac{1.5 \times 3}{4} - 1} = \frac{0.5}{\frac{4.5}{4} - 1} \] \[ \frac{f}{f_m} = \frac{0.5}{\frac{4.5 - 4}{4}} = \frac{0.5}{\frac{0.5}{4}} = \frac{0.5 \times 4}{0.5} = 4 \] \[ f_m = 4f = 4 \times 3 = 12 { cm} \] 
Thus, the new focal length of the lens in water is 12 cm.