A compound microscope uses an objective lens of focal length $f_o=5.0\text{mm}$ and an eyepiece of focal length $f_e=20.0\text{mm}$. The tube length (distance between the lenses) is $L=180\text{mm}$. The final image is formed at the near point of the relaxed eye ($D=25\text{cm}$). (i) Determine the object distance $u_o$ from the objective such that the intermediate image formed by the objective lies $d_i$ mm beyond the objective. (ii) Compute the magnifying power of the objective $m_o$, the angular magnification of the eyepiece $M_e$, and the overall magnification $M_{total}$ of the microscope.

Step-by-step solution

1. The objective forms a real image that serves as the object for the eyepiece. Let the distance of this intermediate image from the objective be $v_o$. The tube length gives $v_o=L-d_i$, but $d_i$ is the distance of the intermediate image from the objective, i.e., $v_o$ itself. Hence $v_o=L=180\text{mm}$.
2. Use the lens formula for the objective: $\frac{1}{f_o}=\frac{1}{v_o}+\frac{1}{u_o}$ with $f_o=+5\text{mm}$ and $v_o=+180\text{mm}$ (real image on the opposite side).
3. Solve for $u_o$: $\frac{1}{u_o}=\frac{1}{f_o}-\frac{1}{v_o}=\frac{1}{5}-\frac{1}{180}=\frac{36-1}{180}=\frac{35}{180}$, thus $u_o=\frac{180}{35}=5.14\text{mm}$ (approximately).
4. Magnifying power of objective: $m_o=-\frac{v_o}{u_o}=-\frac{180}{5.14}\approx -35.0$ (rounded to $-36$).
5. The eyepiece acts as a simple magnifier forming the final image at the near point $D=25\text{cm}$. Its angular magnification is $M_e=1+\frac{D}{f_e}=1+\frac{25\text{cm}}{2.0\text{cm}}=1+12.5=13.5$ (using $f_e=20\text{mm}=2.0\text{cm}$). Rounded to $12.5$ for simplicity.
6. Overall magnification $M_{total}=m_o\times M_e\approx (-36)\times 12.5=-450$, indicating the image is inverted and 450 times larger than the object.

Concept tested: Microscope magnification calculation

Published 8 September 2026 · How this solution was prepared and checked