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 fe = 5 cm
$\large -30 = m_o (1 + \frac{25}{5}) $
mo = – 5
Negative sign implies that image formed by objective is inverted.