The magnifying power of a compound microscope is high, if
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Both objective and eyepiece have short focal length
The magnifying power of a compound microscope is given by:
$M = M_{objective} \times M_{eyepiece}$
where, $M_{objective} = \frac{L}{f_o}$ and $M_{eyepiece} = \frac{D}{f_e}$
Here:
$f_o =$ focal length of objective
$f_e =$ focal length of eyepiece
$L =$ length of microscope tube
$D =$ least distance of distinct vision
For high magnifying power, $f_o$ and $f_e$ should be small.
Answer: Both objective and eyepiece have short focal length