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A compound microscope has an eye piece of focal length \mathrm{10\ cm} and an objective of focal length \mathrm{4\ cm}. Calculate the magnification, if an object is kept at a distance of \mathrm{5\ cm} from the objective, so that the final image is formed at the least distance of distinct vision \mathrm{25\ cm}:

Option: 1

12


Option: 2

11


Option: 3

13


Option: 4

14


Answers (1)

best_answer

\mathrm{\text { Here, } \mathrm{f}_0=4 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=10 \mathrm{~cm}, \mathrm{u}_0=-5 \mathrm{~cm}}
For objective
\mathrm{\begin{aligned} & \mathrm{\frac{1}{\mathrm{v}_0}-\frac{1}{-5}=\frac{1}{4}} \\ & \frac{1}{\mathrm{v}_0}=\frac{1}{4}-\frac{1}{5}=\frac{1}{20} \quad \text { or } \quad \mathrm{v}_0=20 \mathrm{~cm} \end{aligned}}
Magnification when the final image is formed at the least distance of distinct vision \mathrm{D(=25\ cm)} is
\mathrm{\mathrm{M}=\frac{\mathrm{v}_0}{\left|\mathrm{u}_0\right|}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{c}}}\right)=\frac{20}{5}\left(1+\frac{25}{10}\right)=14}

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