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  • In a compound microscope, the focal lengths of two lenses are and <img alt="\mathrm{6

In a compound microscope, the focal lengths of two lenses are \mathrm{1.5\ cm} and \mathrm{6.25\ cm}. If an object is placed at \mathrm{2\ cm} from objective and the final image is formed at \mathrm{25\ cm} from eye lens, the distance between the two lenses is:

Option: 1

\mathrm{6.00\ cm}


Option: 2

\mathrm{7.75\ cm}


Option: 3

\mathrm{9.25\ cm}


Option: 4

\mathrm{11.0\ cm}


Answers (1)

best_answer

\mathrm{\text { Here, } \mathrm{f}_0=1.5 \mathrm{~cm}, \mathrm{f}_{\mathrm{c}}=6.25 \mathrm{~cm}, \mathrm{u}_0=-2 \mathrm{~cm}, \mathrm{v}_{\mathrm{c}}=-25 \mathrm{~cm}}
For objective, 
\begin{array}{llrl} \frac{1}{\mathrm{v}_0}-\frac{1}{\mathrm{u}_0}=\frac{1}{\mathrm{f}_0} & & \therefore & \frac{1}{\mathrm{v}_0}-\frac{1}{-2}=\frac{1}{1.5} \\\\ \frac{1}{\mathrm{v}_0}=\frac{1}{1.5}-\frac{1}{2}\ \ \ \ & \text { or } \quad \mathrm{v}_0=6 \mathrm{~cm} \end{array}
For eye piece
\begin{aligned} &\mathrm{ \frac{1}{v_e}-\frac{1}{u_c}=\frac{1}{f_c} \Rightarrow \frac{1}{-25}-\frac{1}{u_c}=\frac{1}{6.25}} \\ &\mathrm{ -\frac{1}{u_c}=\frac{1}{6.25}+\frac{1}{25} \quad \text { or } \quad u_c=-5 \mathrm{~cm}} \end{aligned}
Distance between two lenses =\left|\mathrm{v}_0\right|+\left|\mathrm{u}_{\mathrm{e}}\right|=6 \mathrm{~cm}+5 \mathrm{~cm}=11 \mathrm{~cm}

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shivangi.bhatnagar

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