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In compound microscope, the magnified virtual image is formed at a distance of 25 cm from the eye – piece. The focal length of its objective lens is 1 cm. If the magnification is 100 and the tube length of the microscope is 20 cm, then the focal length of the eye – piece lens (in cm) is.......
(Round off to nearest integer)

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answer is 6.

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Detailed Solution

When the final image is formed at least distance of distinct vision (i.e, at 25cm ) from eye – piece of a compound microscope , then the magnification is given by

$m = - \frac{V_{0}}{u_{0}} (1 + \frac{D}{f e})$

$m = - \frac{L}{f_{0}} (1 + \frac{D}{f e})$

$- 100 = - \frac{20}{1} (1 + \frac{25}{f e})$

$5 = 1 + \frac{25}{f e} \Rightarrow 4 = \frac{25}{f e}$

$\Rightarrow f e = \frac{25}{4} = 6.25 = 6 c m$

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