Question:
If we need a magnification of 375 from a compound microscope of tube length $150 \mathrm{~mm}$ and an objective of focal length $5 \mathrm{~mm}$, the focal length of the eye-piece, should be close to:
Correct Option: 1
Solution:
(1) According question, $M=375$
$L=150 \mathrm{~mm}, f_{0}=5 \mathrm{~mm}$ and $f_{e}=$ ?
Using, magnification, $M \simeq \frac{L}{f_{0}}\left(1+\frac{D}{f_{e}}\right)$
$\Rightarrow 375=\frac{150}{5}\left(1+\frac{250}{f_{e}}\right) \quad(\because D=25 \mathrm{~cm}=250 \mathrm{~mm})$
$\Rightarrow 12.5=1+\frac{250}{f_{e}}$
$\Rightarrow f_{e}=\frac{250}{11.5}=21.7 \approx 22 \mathrm{~mm}$