If we need a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eye-piece, when the final image is at near point is

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22mm

12mm

2mm

33mm

answer is A.

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Detailed Solution

According question, $M = 375$
$L = 150 m m, f_{0} = 5 m m a n d f_{e} = ?$
Using, magnification, $M ≃ \frac{L}{f_{0}} (1 + \frac{D}{f_{e}})$
$\Rightarrow 375 = \frac{150}{5} (1 + \frac{250}{f_{e}}) (∵ D = 25 c m = 250 m m) \Rightarrow 12.5 = 1 + \frac{250}{f_{e}} \Rightarrow f_{e} = \frac{250}{11.5} = 21.7 \approx 22 m m$

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