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A compound microscope consists of an objective lens of focal length $2.0 \mathrm{~cm}$ and an eyepiece of focal length $6.25 \mathrm{~cm}$ separated by a distance of $15 \mathrm{~cm}$. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision $(25 \mathrm{~cm})$, and $(b)$ at infinity? What is the magnifying power of the microscope in each case?
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Given : $\mathrm{f}_0=2 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=6.25 \mathrm{~cm} . \mathrm{v}_{\mathrm{e}}=-25 \mathrm{~cm}, \mathrm{u}_{\mathrm{e}}=?$
(a) $\frac{1}{\mathrm{v}_{\mathrm{e}}}=\frac{1}{\mathrm{u}_{\mathrm{e}}}-\frac{1}{\mathrm{f}_{\mathrm{e}}}$,
$\frac{1}{\mathrm{u}_{\mathrm{e}}}=\frac{1}{\mathrm{v}_{\mathrm{e}}}-\frac{1}{\mathrm{f}_{\mathrm{e}}}=\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$,
$u_e=\frac{-25}{5}=-5 \mathrm{~cm}$
$\mathrm{v}_0=15-5=10 \mathrm{~cm}$.
$\frac{1}{\mathrm{u}_0}=\frac{1}{\mathrm{v}_0}-\frac{1}{\mathrm{f}_0}=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$
$u_0=\frac{-10}{4}=2.5 \mathrm{~cm}$.
magnifying power
$\mathrm{m}=\frac{\mathrm{v}_0}{\left|\mathrm{u}_0\right|}\left(1+\frac{\mathrm{d}}{\mathrm{f}_{\mathrm{e}}}\right)=\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=20$
(b) As, $\mathrm{v}_{\mathrm{e}}=\infty \quad \therefore \mathrm{u}_{\mathrm{c}}=\mathrm{f}_{\mathrm{c}}=6.25 \mathrm{~cm}$
$$
\mathrm{v}_0=15-6.25=8.75 \mathrm{~cm}
$$
$$
\begin{aligned}
&\frac{1}{\mathrm{u}_0}=\frac{1}{\mathrm{v}_0}-\frac{1}{\mathrm{f}_0}=\frac{1}{8.75}-\frac{1}{2}=\frac{2-8.75}{17.5}=\frac{-6.75}{17.5} \\
&\therefore \mathrm{u}_0=-\frac{17.5}{6.75}=-2.59 \mathrm{~cm}
\end{aligned}
$$
Magnifying power,
$$
\mathrm{m}=\frac{\mathrm{v}_0}{1 \mathrm{uu}_0 1}\left(1+\frac{\mathrm{d}}{\mathrm{fe}}\right)=\frac{8.75}{2.59}\left(1+\frac{25}{6.25}\right)=13.51
$$
(a) $\frac{1}{\mathrm{v}_{\mathrm{e}}}=\frac{1}{\mathrm{u}_{\mathrm{e}}}-\frac{1}{\mathrm{f}_{\mathrm{e}}}$,
$\frac{1}{\mathrm{u}_{\mathrm{e}}}=\frac{1}{\mathrm{v}_{\mathrm{e}}}-\frac{1}{\mathrm{f}_{\mathrm{e}}}=\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$,
$u_e=\frac{-25}{5}=-5 \mathrm{~cm}$
$\mathrm{v}_0=15-5=10 \mathrm{~cm}$.
$\frac{1}{\mathrm{u}_0}=\frac{1}{\mathrm{v}_0}-\frac{1}{\mathrm{f}_0}=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}$
$u_0=\frac{-10}{4}=2.5 \mathrm{~cm}$.
magnifying power
$\mathrm{m}=\frac{\mathrm{v}_0}{\left|\mathrm{u}_0\right|}\left(1+\frac{\mathrm{d}}{\mathrm{f}_{\mathrm{e}}}\right)=\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=20$
(b) As, $\mathrm{v}_{\mathrm{e}}=\infty \quad \therefore \mathrm{u}_{\mathrm{c}}=\mathrm{f}_{\mathrm{c}}=6.25 \mathrm{~cm}$
$$
\mathrm{v}_0=15-6.25=8.75 \mathrm{~cm}
$$
$$
\begin{aligned}
&\frac{1}{\mathrm{u}_0}=\frac{1}{\mathrm{v}_0}-\frac{1}{\mathrm{f}_0}=\frac{1}{8.75}-\frac{1}{2}=\frac{2-8.75}{17.5}=\frac{-6.75}{17.5} \\
&\therefore \mathrm{u}_0=-\frac{17.5}{6.75}=-2.59 \mathrm{~cm}
\end{aligned}
$$
Magnifying power,
$$
\mathrm{m}=\frac{\mathrm{v}_0}{1 \mathrm{uu}_0 1}\left(1+\frac{\mathrm{d}}{\mathrm{fe}}\right)=\frac{8.75}{2.59}\left(1+\frac{25}{6.25}\right)=13.51
$$
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