A compound microscope has a magnification of \( 30 \). The focal length of its eyepiece is \( 5 \) cm. Assuming the final image to be formed at the least distance of distinct vision (\( 25 \) cm), the magnification produced by the objective is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5
For a compound microscope the total magnification is the product of the magnifications of the objective and the eyepiece: \( m = m_o m_e \).
With the final image at the least distance of distinct vision, the eyepiece acts as a simple magnifier with
\( \displaystyle m_e = 1 + \frac{D}{f_e} = 1 + \frac{25\,\mathrm{cm}}{5\,\mathrm{cm}} = 6 \)
\( \displaystyle m_o = \frac{m}{m_e} = \frac{30}{6} = 5 \)