The two lenses of a compound microscope are of focal lengths 2 cm and 5 cm. If an object is placed at a distance of 2.1 cm from the objective of focal length 2 cm the final image forms at the least distance of distinct vision of a normal eye. Find the magnifying power of the microscope is

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120

answer is C.

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Detailed Solution

$\begin{array}{l} f_{0} = 2 cm, f_{e} = 5 cm, u_{0} = - 2 .1 cm \\ \frac{1}{f_{0}} = \frac{1}{v_{0}} - \frac{1}{u_{0}}, \\ v_{0} = 42 cm \\ m_{0} = \frac{v_{0}}{u_{0}}, \\ m = m_{0} m_{e}; m = \frac{v_{0}}{u_{0}} [1 + \frac{D}{f_{e}}] = 120 \end{array}$

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