The magnifying power of a compound microscope is high, if

The correct answer is Option (3) → Both objective and eyepiece have short focal length

The magnifying power of a compound microscope is the product of the magnification of the objective and that of the eyepiece.

    \( \displaystyle m = m_o \times m_e \approx \frac{L}{f_o} \times \frac{D}{f_e} \)

Here \( L \) is the distance between the lenses (tube length) and \( D \) is the least distance of distinct vision. The magnifying power is inversely proportional to both focal lengths, so it is large when both are small.

NCERT (2026-27):
The Microscope – “… Thus, the total magnification [(according to Eq. (9.33)], when the image is formed at infinity, is

    \( \displaystyle m = m_o m_e = \left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right) \qquad (9.45) \)

Clearly, to achieve a large magnification of a small object (hence the name microscope), the objective and eyepiece should have small focal lengths.”