A nbsp-person with normal near point 25 cm using a compound microscope with objective nbsp-of focal length 8-0 mm and an eye piece of focal length 2-5 cm can bring an nbsp-object placed at 9-0 mm from the objective in sharp focus- The separation nbsp-between two lenses and magnification respectively are7-49 cm- 113-36 cm- 446-00 cm- 229-47 cm- 88

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Question

A person with normal near point 25 cm using a compound microscope with objective of focal length 8.0 mm and an eye piece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. The separation between two lenses and magnification respectively are

9.47 cm, 88

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3.36 cm, 44

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6.00 cm, 22

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7.49 cm, 11

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Solution

The correct option is D 9.47 cm, 88
Here, $d = 25 c m, f_{0} = 8.0 m m, f_{e} = 2.5 c m, u_{0} = - 9.0 m m = - 0.9 c m$
Now, $\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}} ∴ \frac{1}{u_{e}} = \frac{1}{v_{e}} - \frac{1}{f_{e}} = \frac{1}{- 25} - \frac{1}{2.5} = \frac{- 11}{25}$
$(∵ v_{e} = - d = - 25 c m)$
$u_{e} = \frac{- 25}{11} = 2.27 c m$
Again, $\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$
$\frac{1}{v_{0}} = \frac{1}{f_{0}} + \frac{1}{u_{0}} = \frac{1}{0.8} + \frac{1}{- 0.9} = \frac{0.9 - 0.8}{0.72} = \frac{0.1}{0.72}$
$v_{0} = \frac{0.72}{0.1} = 7.2 c m$
Therefore, separation between two lenses $= u_{e} + v_{0} = 2.27 + 7.2 = 9.47 c m$
Magnifying power, $m = \frac{v_{0}}{u_{0}} (1 + \frac{d}{f e}) = \frac{7.2}{0.9} (1 + \frac{25}{2.5}) = 88$

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