The length of the compound microscope is 14 cm. The magnifying power for relaxed eye is 25. If the focal length of eye lens is 5 cm, then the object distance for objective will be (in cm)

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2.4

1.8

1.5

2.1

answer is A.

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Detailed Solution

$\begin{array}{l} m_{e} = \frac{D}{f_{e}} = \frac{25}{5} = 5, L = v_{0} + f_{e} \\ 14 = v_{0} + 5; v_{0} = 9 cm \\ m_{0} = \frac{m}{m_{e}} = \frac{v_{0}}{u_{0}}; \frac{25}{5} = \frac{9}{u_{0}} \Rightarrow u_{0} = \frac{9}{5} = 1 .8 cm \end{array}$

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