Q. In a compound microscope, the focal length of the objective and the eye lens are $2.5\ cm$ and $5\ cm$ respectively. An object is placed at 3.75 $cm$ before the objective and image is formed at the least distance of distinct vision, then the distance between two lenses will be $(i.e.$ length of the microscopic tube)
A) 11.67 $cm$
B) 12.67 $cm$
C) 13.00 $cm$
D) 12.00 $cm$
Explanation
(a) $L_{D} = v_{o} + u_{e}$ and for objective lens $\frac{1}{f_{o}} = \frac{1}{v_{o}} - \frac{1}{u_{o}}$
Putting the values with proper sign convention
$\frac{1}{+ 2.5} = \frac{1}{v_{o}} - \frac{1}{( - 3.75)} \Rightarrow v_{o} = 7.5\ cm$
For eye lens $\frac{1}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{u_{e}}$
$\Rightarrow \frac{1}{+ 5} = \frac{1}{( - 25)} - \frac{1}{u_{e}} \Rightarrow u_{e} = - 4.16\ cm$
$\Rightarrow \left| u_{e} \right| = 4.16\ cm$
Hence $L_{D} = 7.5 + 4.16 = 11.67\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.