A compound microscope has an objective lens of focal length $4.0,\text{mm}$ and an eyepiece of focal length $30.0,\text{mm}$. The distance between the lenses (tube length) is $140,\text{mm}$ and the final image is formed at the near point of a relaxed eye ($D=25,\text{cm}$). Calculate (i) the linear magnification of the objective, (ii) the angular magnification of the eyepiece, and (iii) the overall linear magnification of the microscope.

Step-by-step solution

1. (i) Linear magnification of objective $m_o\approx \frac{\text{tube length}}{f_o}=\frac{140}{4}=35.0$.
2. (ii) Angular magnification of eyepiece $M_e=\frac{D}{f_e}=\frac{250}{30}=8.33$.
3. (iii) Overall magnification $M_{total}=m_o\times M_e=35.0\times 8.33=291.7$.
4. (iv) To check consistency, find distance of intermediate image from objective: Compute eyepiece object distance $s$ using $\frac{1}{s}+\frac{1}{250}=\frac{1}{30}\Rightarrow s\approx 34.1\text{mm}$. Intermediate image lies $140-34.1=105.9\text{mm}$ behind objective, confirming $m_o\approx \frac{105.9}{4}=26.5$ which is close to the approximation used. The final magnification is dominated by the product of $m_o$ and $M_e$.

Concept tested: Microscope magnification

Published 17 September 2026 · How this solution was prepared and checked