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A compound microscope has an objective of focal length 2cm and an eye-piece focal length 5cm. If an object is placed at a distance of 2.4cm in front of the objective, and the final image is at infinity the magnitude of magnifying power of the instrument is

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answer is 25.

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Detailed Solution

As object is at distance of 2.4 cm, in front of field lens of focal length 2cm. Field lens will forms its image at distance v such that

For eye-piece, $\frac{1}{\infty} - \frac{1}{u_{e}} = \frac{1}{5}$

$\frac{1}{v} - \frac{1}{- 2.4} = \frac{1}{2} i . e ., v = 12 c m$

$i . e ., u_{e} = - 5 c m$

So that $m = \frac{v}{u} = \frac{12}{- 2.4} = - 5$

and $m_{θ} = (\frac{D}{F_{C}}) = (\frac{25}{5}) = 5$

So, $M P = m \times m_{θ} = - 5 \times 5 = - 25$

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