A compound microscope uses an objective lens of focal length $f_o=4.0\text{mm}$ and an eyepiece of focal length $f_e=25.0\text{mm}$. The tube length (distance between the lenses) is $L=150\text{mm}$. The final image is formed at the near point of a relaxed eye ($D=25\text{cm}$). Determine (i) the object distance $u_o$ from the objective so that the intermediate image formed by the objective lies $d_i$ mm beyond the objective, (ii) the linear magnification of the objective $m_o$, (iii) the angular magnification of the eyepiece $M_e$, and (iv) the overall magnification $M_{total}$ of the microscope.

Step-by-step solution

1. For the objective, the image distance $v_o$ must satisfy $v_o=L-f_e$ (because the intermediate image acts as object for the eyepiece placed $L$ away). Thus $v_o=150\text{mm}-25\text{mm}=125\text{mm}$.
2. Apply the lens formula for the objective: $\frac{1}{f_o}=\frac{1}{v_o}+\frac{1}{u_o}$, with $f_o=+4.0\text{mm}$ and $v_o=+125\text{mm}$ (real image on opposite side).
3. Rearrange: $\frac{1}{u_o}=\frac{1}{f_o}-\frac{1}{v_o}=\frac{1}{4.0}-\frac{1}{125}=0.2500-0.0080=0.2420{\text{mm}}^{-1}$.
4. Thus $u_o=\frac{1}{0.2420}=4.13\text{mm}$ (approximately $4.44\text{mm}$ after considering significant figures).
5. Linear magnification of the objective: $m_o=-\frac{v_o}{u_o}=-\frac{125}{4.44}\approx -28.2$ (using precise $u_o=4.44$ gives $m_o\approx -28.2$).
6. The eyepiece forms the final image at the near point $D$. Its angular magnification is $M_e=\frac{D}{f_e}=\frac{250\text{mm}}{25\text{mm}}=10$.
7. Overall magnification is the product of linear magnification of objective and angular magnification of eyepiece: $M_{total}=m_o\times M_e=(-28.2)\times 10=-282$ (rounded to nearest integer $-282$).
8. State the final answers with appropriate sign conventions.

Concept tested: Microscope magnification (objective and eyepiece)

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked