Question

Figure below shows a lens of refractive index, $ \mu = 1.4 $. $ C_1 $ and $ C_2 $ are the centres of curvature of the two faces of the lens of radii of curvature 4 cm and 8 cm respectively. The lens behaves as a

• A. diverging lens of focal length 20 cm
• B. converging lens of focal length 20 cm
• C. converging lens of focal length 12 cm
• D. diverging lens of focal length 12 cm

Hint:

For a lens with two curved surfaces, use the lens maker's formula to calculate the focal length based on the radii of curvature and the refractive index. Ensure to use the correct signs for the radii.

Answer 1

The Correct Option is B

The lens is a plano-convex lens, where: 
- \( R_1 = 4 \, \text{cm} \) (radius of curvature for the convex face), 
- \( R_2 = -8 \, \text{cm} \) (radius of curvature for the concave face, negative because it's opposite in direction). 
Using the lens maker's formula: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Substitute the given values: \[ \frac{1}{f} = (1.4 - 1) \left( \frac{1}{4} - \frac{1}{-8} \right) \] Simplifying this: \[ \frac{1}{f} = 0.4 \left( \frac{1}{4} + \frac{1}{8} \right) = 0.4 \left( \frac{2}{8} + \frac{1}{8} \right) = 0.4 \times \frac{3}{8} \] \[ \frac{1}{f} = \frac{1.2}{8} = \frac{3}{20} \] 
Thus, the focal length \( f = 20 \, \text{cm} \).