Two thin lenses of focal lengths f1 and f2 are in contact and coaxial. The combination is equivalent to a single lens of power
Answer & explanation
Correct answer: option 4
$\text{Equivalent focal length is given by } \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
$\Rightarrow F = \frac{f_1f_2}{f_1+f_2}$
Equivalent Power $ P = \frac{1}{F} = \frac{f_1+ f_2}{f_1f_2}$