Question

A bi-convex lens has radius of curvature of both the surfaces same as $ \frac{1}{6} \, \text{cm} $. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides $ (R_1 \neq R_2) $, without any change in lens power then possible combination of $ R_1 $ and $ R_2 $ is:

• A. \( \frac{1}{3} \, \text{cm} \) and \( \frac{1}{3} \, \text{cm} \)
• B. \( \frac{1}{5} \, \text{cm} \) and \( \frac{1}{7} \, \text{cm} \)
• C. \( \frac{1}{3} \, \text{cm} \) and \( \frac{1}{7} \, \text{cm} \)
• D. \( \frac{1}{6} \, \text{cm} \) and \( \frac{1}{9} \, \text{cm} \)

Hint:

When dealing with lens formulas, always ensure that the radii of curvature for both sides of the lens match the conditions for the required lens power.

Answer 1

The Correct Option is B

This occurs when \[\frac{1}{f_1} = \frac{1}{f_2}\] \[\left(\mu - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu - 1)\left(\frac{2}{R}\right)\] \[\frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R}\] Therefore, the possible values for \( R_1 \) and \( R_2 \) are \( \frac{1}{5} \, \text{cm} \) and \( \frac{1}{7} \, \text{cm} \).