The magnifying power of a compound microscope is high, if

Both objective and eyepiece have short focal length

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