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A person with a normal near point $(25 \mathrm{~cm})$ using a compound microscope with objective of focal length $8.0$ $\mathrm{mm}$ and an eyepiece of focal length $2.5 \mathrm{~cm}$ can bring an object placed at $9.0 \mathrm{~mm}$ from the objective in sharp focus. What is the separation between the two lenses?
Calculate the magnifying power of the microscope.
Calculate the magnifying power of the microscope.
Solution:
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Verified Answer
Here, $\mathrm{d}=25 \mathrm{~cm}, \mathrm{f}_0=8 \mathrm{~mm}=0.8 \mathrm{~cm}$
$\mathrm{f}_{\mathrm{e}}=2.5 \mathrm{~cm}, \mathrm{u}_0=-9 \mathrm{~mm}=-0.9 \mathrm{~cm}$.
$$
\begin{aligned}
&\frac{1}{\mathrm{u}_{\mathrm{e}}}=\frac{1}{\mathrm{v}_{\mathrm{e}}}-\frac{1}{\mathrm{f}_{\mathrm{e}}}=\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}=\frac{-11}{25} \\
&\therefore \mathrm{u}_{\mathrm{e}}=-\frac{25}{11}=-2.27 \mathrm{~cm} \quad \frac{1}{\mathrm{v}_0}=\frac{1}{\mathrm{u}_0}+\frac{1}{\mathrm{f}_0}=\frac{1}{-0.9}+\frac{1}{0.8} \\
&\quad=\frac{-0.8+0.9}{0.72}=\frac{0.1}{0.72} \\
&\therefore \mathrm{v}_0=\frac{0.72}{.1}=7.2 \mathrm{~cm}
\end{aligned}
$$
Therefore, seperation between two lenses $=2.27+7.2=9.47 \mathrm{~cm}$.
Magnifying power
$$
\mathrm{m}=\frac{\mathrm{v}_0}{|\mathrm{u} .|}\left(1+\frac{\mathrm{d}}{\mathrm{fe}}\right)=\frac{7.2}{0.9}\left(1+\frac{25}{2.5}\right)=88
$$
$\mathrm{f}_{\mathrm{e}}=2.5 \mathrm{~cm}, \mathrm{u}_0=-9 \mathrm{~mm}=-0.9 \mathrm{~cm}$.
$$
\begin{aligned}
&\frac{1}{\mathrm{u}_{\mathrm{e}}}=\frac{1}{\mathrm{v}_{\mathrm{e}}}-\frac{1}{\mathrm{f}_{\mathrm{e}}}=\frac{1}{-25}-\frac{1}{2.5}=\frac{-1-10}{25}=\frac{-11}{25} \\
&\therefore \mathrm{u}_{\mathrm{e}}=-\frac{25}{11}=-2.27 \mathrm{~cm} \quad \frac{1}{\mathrm{v}_0}=\frac{1}{\mathrm{u}_0}+\frac{1}{\mathrm{f}_0}=\frac{1}{-0.9}+\frac{1}{0.8} \\
&\quad=\frac{-0.8+0.9}{0.72}=\frac{0.1}{0.72} \\
&\therefore \mathrm{v}_0=\frac{0.72}{.1}=7.2 \mathrm{~cm}
\end{aligned}
$$
Therefore, seperation between two lenses $=2.27+7.2=9.47 \mathrm{~cm}$.
Magnifying power
$$
\mathrm{m}=\frac{\mathrm{v}_0}{|\mathrm{u} .|}\left(1+\frac{\mathrm{d}}{\mathrm{fe}}\right)=\frac{7.2}{0.9}\left(1+\frac{25}{2.5}\right)=88
$$
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