A compound microscope has an objective lens of focal length \( f_o = 1.0\,\mathrm{cm} \), an eyepiece of focal length 2.2 cm and has tube length 22 cm. The magnification produced compound microscope is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( -250 \)
For a compound microscope in normal adjustment (final image at infinity), with the object close to the focus of the objective, the magnifying power is
\( \displaystyle m = -\frac{L}{f_o}\cdot\frac{D}{f_e} \)
where \( L \) is the tube length and \( D = 25\,\mathrm{cm} \) is the least distance of distinct vision. The negative sign shows the final image is inverted.
\( L = 22\,\mathrm{cm} \)
\( f_o = 1.0\,\mathrm{cm} \)
\( f_e = 2.2\,\mathrm{cm} \)
\( D = 25\,\mathrm{cm} \)
\( \displaystyle m = -\frac{22\,\mathrm{cm}}{1.0\,\mathrm{cm}} \times \frac{25\,\mathrm{cm}}{2.2\,\mathrm{cm}} = -\frac{550}{2.2} = -250 \)