Question

If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece should be close to:

• A. 22 mm
• B. 2 mm
• C. 12 mm
• D. 33 mm

Answer 1

The correct option is A 22 mm

Magnification of compound microscope for least distance of distinct vision setting(strained eye)
$$M=\frac{L}{f_0}(1+\frac{D}{f_e})$$
where L is the tube length
$f_0$ is the focal length of objective
D is the least distance of distinct vision $=25cm$
$$\begin{array}{cc}\text{i.e.}& 375=\frac{150\times {10}^{-3}}{5\times {10}^{-3}}(1+\frac{25\times {10}^{-2}}{f_e})\\ \text{i.e.}& 12.5=1+\frac{25\times {10}^{-2}}{f_e}\\ \text{i.e.}& \frac{25\times {10}^{-2}}{f_e}=11.5\\ \therefore & f_e\approx 21.7\times {10}^{-3}m=22mm\end{array}$$