Q. A combination of two thin lenses with focal lengths $f_{1}$ and $f_{2}$ respectively forms an image of distant object at distance 60 $cm$ when lenses are in contact. The position of this image shift by $30\ cm$ towards the combination when two lenses are separated by $10\ cm$. The corresponding values of $f_{1}$ and $f_{2}$ are
A) $30\ cm,\ - 60\ cm$
B) $20\ cm,\ - 30\ cm$
C) $15\ cm,\ - 20\ cm$
D) $12\ cm,\ - 15\ cm$
Explanation
(b) $\frac{1}{60} = \frac{1}{f_{1}} + \frac{1}{f_{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
And $\frac{1}{30} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{10}{f_{1}f_{2}}$ …(ii)
On solving (i) and (ii) $f_{1}f_{2} = - 600$ and $f_{1} + f_{2} = - 10$
Hence $f_{1} = 20\ cm$ and $f_{2} = - 30\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.