A thin lens $L_1$ of focal length 20 cm placed in combination with another thin lens $L_2$ acts as a lens of focal length 12.5 cm. The power of $L_2$ is

3 D

The correct answer is Option (1) → 3 D

For two thin lenses in contact:

$ \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} $

Given: $ f = 12.5 \, \text{cm}, \, f_{1} = 20 \, \text{cm} $

$ \frac{1}{12.5} = \frac{1}{20} + \frac{1}{f_{2}} $

$ \frac{1}{f_{2}} = \frac{1}{12.5} - \frac{1}{20} $

$ \frac{1}{f_{2}} = \frac{20 - 12.5}{250} = \frac{7.5}{250} = \frac{3}{100} $

$ f_{2} = \frac{100}{3} \, \text{cm} = \frac{1}{30} \, \text{m} $

Power: $ P_{2} = \frac{100}{f_{2}(\text{cm})} = \frac{100}{\frac{100}{3}} = 3 \, D $