Q. In compound microscope, magnifying power is 95 and the distance of object from objective lens is $\frac{1}{3.8}\ $cm. The focal length of objective lens is $\ \frac{1}{4}\ $cm. What is the magnification of eye piece?
A) 5
B) 10
C) 100
D) 200
Explanation
(a) For the objective, $\frac{1}{v_{o}} - \frac{1}{- 1/3.8} = \frac{1}{1/4}$
Or $\ \frac{1}{v_{o}} + 3.8 = 4\ or\ \frac{1}{v_{o}} = 0.2 = \frac{1}{5}$
or$\ \ v_{0} = 5\ cm$
Now, $M_{0} = \frac{5}{- \frac{1}{3.8}} = - 19$
Again, $M = M_{O} \times M_{e}$
$- 95 = - 19 \times M_{e}\ $or$\ \ M_{e} = \frac{95}{19} = 5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.