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Q.

A compound microscope has an eyepiece of focal length 10 cm and an objective of focal length 5 cm. If an object is kept at a distance of 6 cm from the objective then magnification of the microscope for the final image at least distance of distinct vision of 25 cm is

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a

15

b

17.5

c

22.5

d

25

answer is B.

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Detailed Solution

$\frac{1}{v_{0}} = \frac{1}{5} - \frac{1}{6} = \frac{1}{30} \Rightarrow v_{0} = 30 cm$

In distinct vision, $MP = \frac{v_{0}}{u_{0}} (1 + \frac{D}{f_{e}}) = \frac{30}{6} (1 + \frac{25}{10}) = 17 .5$

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A compound microscope has an eyepiece of focal length 10 cm and an objective of focal length 5 cm. If an object is kept at a distance of 6 cm from the objective then magnification of the microscope for the final image at least distance of distinct vision of 25 cm is