The magnification of a compound microscope in normal adjustment is 200. If the focal lengths of the objective and eye-piece are 1.0 cm and 2.0 cm, respectively, then the length of the tube is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 16 cm
In normal adjustment the final image is at infinity. The total magnification is the product of the objective and eyepiece magnifications, with L the tube length and D = 25 cm the least distance of distinct vision.
\( \displaystyle m = \frac{L}{f_o}\times\frac{D}{f_e} \)
\( \displaystyle 200 = \frac{L}{1.0\,\mathrm{cm}}\times\frac{25\,\mathrm{cm}}{2.0\,\mathrm{cm}} \)
\( 200 = 12.5\,L \quad (L \text{ in cm}) \)
\( L = 16\,\mathrm{cm} \)