A bi-convex lens has radius of curvature of both the surfaces same as $1 / 6 \mathrm{~cm}$. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides $\left(R_{1} \neq R_{2}\right)$, without any change in lens power then possible combination of $R_{1}$ and $R_{2}$ is :
Select the correct option:
A
$\frac{1}{5} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$
B
$\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{3} \mathrm{~cm}$
C
$\frac{1}{6} \mathrm{~cm}$ and $\frac{1}{9} \mathrm{~cm}$
D
$\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
This will happen when
$\frac{1}{f_{1}}=\frac{1}{f_{2}}$
$(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{-R_{2}}\right)=(\mu-1)\left(\frac{2}{R}\right)$
$\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{2}{R}$
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