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  • A compound microscope has an eye piece of focal length and an objective of focal length

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{20\ cm}:

Option: 1

12


Option: 2

11


Option: 3

10


Option: 4

13


Answers (1)

best_answer

Here, \mathrm{\mathrm{u}_0=-5 \mathrm{~cm}, \mathrm{f}_0=4 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=10 \mathrm{~cm}, \mathrm{D}=20 \mathrm{~cm}}
According to lens formula
\mathrm{\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0} \quad \text { or } \quad \frac{1}{v_0}=\frac{1}{f_0}+\frac{1}{u_0}}
Substituting the given values, we get
\mathrm{\begin{aligned} & \mathrm{\frac{1}{v_0}=\frac{1}{4}+\frac{1}{-5}=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}} \\ &\mathrm{ v_0=20 \mathrm{~cm}} \end{aligned}}
Magnification, \mathrm{\mathrm{M}=\frac{\mathrm{v}_0}{\left|\mathrm{u}_0\right|}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right)}
\mathrm{=\frac{20}{5}\left(1+\frac{20}{10}\right)=12}

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