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Magnification in Thin Lens

In a compound...

Question

In a compound microscope, the magnified virtual image is formed at a distance of $25 cm$ from the eye-piece. The final image formed at a distance lens $1 cm .$ 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 close to (up to two decimal places)

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Solution

Given, $v = 25 cm; L = 20 cm$
$f_{0} = 1 cm$ , magnification, $M = 100$

For the near point, the magnification is

$M = \frac{L}{f_{0}} (1 + \frac{D}{f_{e}})$
$\Rightarrow 100 = \frac{20}{1} (1 + \frac{25}{f_{e}})$
$\Rightarrow f_{e} = \frac{25}{4} = 6.25 cm$

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