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 will happen when \[ \frac{1}{f_1} = \frac{1}{f_2} \] \[ (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu - 1)\left(\frac{2}{R}\right) \] \[ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} \] Thus, the possible combination for \( R_1 \) and \( R_2 \) is \( \frac{1}{5} \, \text{cm} \) and \( \frac{1}{7} \, \text{cm} \).

Answer 2

Concept Used:

Lens-maker’s formula for a thin lens in air:

\[ \frac{1}{f}=(\mu-1)\!\left(\frac{1}{R_1}-\frac{1}{R_2}\right). \]

For the original bi-convex lens with \(R_1=\tfrac{1}{6}\,\text{cm}\) and \(R_2=-\tfrac{1}{6}\,\text{cm}\):

\[ \frac{1}{R_1}-\frac{1}{R_2}=6-(-6)=12\ \text{cm}^{-1}. \]

To keep the same power, any new convex lens must satisfy

\[ \frac{1}{R_1}-\frac{1}{R_2}=12. \]

Quick Construction:

Let \( \frac{1}{R_1}=p \) and \( -\frac{1}{R_2}=q \) with \(p,q>0\). Then \(p+q=12\). Choose \(p=5,\ q=7\):

\[ R_1=\frac{1}{5}\ \text{cm},\qquad R_2=-\frac{1}{7}\ \text{cm}. \] \[ \frac{1}{R_1}-\frac{1}{R_2}=5-(-7)=12\ (\text{same power}). \]

Answer: \( R_1=\dfrac{1}{5}\ \text{cm},\ R_2=-\dfrac{1}{7}\ \text{cm} \) (magnitudes \( \tfrac{1}{5}\ \text{cm} \) and \( \tfrac{1}{7}\ \text{cm} \)).