In a 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 ____________ .

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answer is 6.25.

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Detailed Solution

$\begin{array}{l} M a g n i f i c a t i o n f o r N e a r p o i n t \\ M = \frac{v_{0}}{u_{0}} (1 + \frac{D}{f_{e}}) \\ P r a c t i c a l l y, w e w r i t e, u_{0} = f_{0} a n d v_{0} = ℓ \\ M \approx \frac{ℓ}{f_{0}} (1 + \frac{D}{f_{e}}) \\ Here, \frac{ℓ}{f_{0}} (1 + \frac{D}{f_{e}}) = 100 \\ \Rightarrow \frac{20}{1} (1 + \frac{25}{f_{e}}) = 100 \\ \Rightarrow f_{e} = 6.25 cm \end{array}$

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