Q: A compound microscope has a magnifying power 30. The focal length of its eyepiece is 5 cm. Assuming the final image to be at the least distance of distinct vision (25 cm), calculate the magnification produced by objective.

Sol: In case of compound microscope,

$\large M = m_o \times m_e $ …(i)

And in case of final image at least distance of distinct vision,

$\large m_e = (1 + \frac{D}{f_e}) $ …(ii)

From (i) & (ii)

$\large M = m_o (1 + \frac{D}{f_e}) $

Here M = -30 ; D = 25 cm and f_{e} = 5 cm

$\large -30 = m_o (1 + \frac{25}{5}) $

m_{o} = – 5

Negative sign implies that image formed by objective is inverted.