Q. The focal lengths of the objective and the eye-piece of a compound microscope are $2.0\ cm$ and $3.0\ cm$ respectively. The distance between the objective and the eye-piece is $15.0\ cm$. The final image formed by the eye-piece is at infinity. The two lenses are thin. The distances in $cm$ of the object and the image produced by the objective measured from the objective lens are respectively
A) 2.4 and 12.0
B) 2.4 and 15.0
C) 2.3 and 12.0
D) 2.3 and 3.0
Explanation
(a) When final image is formed at infinity, length of the tube $= v_{o} + f_{e}$
$\Rightarrow 15 = v_{o} + 3 \Rightarrow v_{o} = 12\ cm$
For objective lens $\frac{1}{f_{o}} = \frac{1}{v_{o}} - \frac{1}{u_{o}}$
$\Rightarrow \frac{1}{( + 2)} = \frac{1}{( + 12)} - \frac{1}{u} \Rightarrow u_{o} = - 2.4\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.