Q. Two thin equiconvex lenses each of focal length 0.2 m are placed coaxially with their optic centers 0.5 m apart. Then the focal length of the combination is
A) -0.4 m
B) 0.4 m
C) -0.1 m
D) 0.1 m
Explanation
(a) Equivalent focal length ($F$) of two lenses separated by distance $d$ is given by
$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1}f_{2}}$
$= \frac{1}{0.2} + \frac{1}{0.2} - \frac{0.5}{(0.2)(0.2)}$
$= 5 + 5 - 0.5 \times 5 \times 5$
$= 10 - 12.5$
$= - 2.5$
$F = \ - \frac{1}{2.5} = - 0.4\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.