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The magnification produced by the objective lens of a compound microscope is 25. The focal length of eye piece is 5 cm and it forms final image at least distance of distinct vision. The magnifying power of the compound microscope is

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150

$\sqrt{150}$

answer is C.

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Detailed Solution

$\begin{array}{l} m_{0} = 25, f_{e} = 5 cm; m = m_{0} m_{e} \\ m = 25 [1 + \frac{D}{f_{e}}]; m = 150 \end{array}$

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