# If we need a magnification of 375 from a compound microscope of tube length 150 mm

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:

1. $22 \mathrm{~mm}$

2. $12 \mathrm{~mm}$

3. $2 \mathrm{~mm}$

4. $33 \mathrm{~mm}$

Correct Option: 1

Solution:

(1) According question, $M=375$

$L=150 \mathrm{~mm}, f_{0}=5 \mathrm{~mm}$ and $f_{e}=$ ?

Using, magnification, $M \simeq \frac{L}{f_{0}}\left(1+\frac{D}{f_{e}}\right)$

$\Rightarrow 375=\frac{150}{5}\left(1+\frac{250}{f_{e}}\right) \quad(\because D=25 \mathrm{~cm}=250 \mathrm{~mm})$

$\Rightarrow 12.5=1+\frac{250}{f_{e}}$

$\Rightarrow f_{e}=\frac{250}{11.5}=21.7 \approx 22 \mathrm{~mm}$