Question

A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid–glass combination will be:

• A. \( \frac{800}{11} \, \text{cm} \)
• B. \( \frac{500}{11} \, \text{cm} \)
• C. \( \frac{700}{11} \, \text{cm} \)
• D. \( \frac{600}{11} \, \text{cm} \)

Hint:

In lens maker’s formula, remember to use the refractive index difference between the two media on either side of the surface. The radii of curvature should be signed based on their convexity and concavity.

Answer 1

The Correct Option is D

Given the setup, we have: 

\( \mu_l = 1.3 \), \( R_1 = 30 \, \text{cm} \), and \( R_2 = 20 \, \text{cm} \), \( \mu_l = 1.5 \)

The focal length \( f \) is given by the formula:

\( \frac{1}{f} = \left( \frac{1.3 - 1}{1} \right) \left( \frac{1}{\infty} - \frac{1}{-30} \right) \)

Simplifying, we get:

\( = \left( 1.5 - 1 \right) \left( \frac{1}{-30} - \frac{1}{-30} \right) \)

\( = \frac{0.3}{30} + \frac{0.5}{60} + \frac{1}{120} \)

\( = \frac{6 + 5}{600} = \frac{11}{600} \)

Thus, the focal length is:

\( f = \frac{600}{11} \, \text{cm} \)