A microscope has an objective of focal length 1.5 crn and eye piece of focal length 2.5 cm. If the distance between objective and eyepiece is 25 cm. what is the approximate value of magnification produced for relaxed eye?

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110

140

answer is C.

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Detailed Solution

$\begin{array}{l} Length of the tube is L = v_{o} + f_{e} \\ v_{o} = L - f_{e} = 25 - 2 .5 = 22 .5 cm \\ Now applying \frac{1}{v_{o}} - \frac{1}{u_{o}} = \frac{1}{f_{o}} \\ we have \frac{1}{22 .5} - \frac{1}{u_{o}} = \frac{1}{1 .5} \\ ∴ |u_{o}| \approx 1 .6 cm \\ ∴ |M| = \frac{v_{o}}{u_{o}} \times \frac{D}{f_{e}} = (\frac{22 .5}{1 .6}) (\frac{25}{2 .5}) \approx 140 \end{array}$

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