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The near vision of an average person is $25 \mathrm{~cm}$. To view an object with an angular magnification of 10 , what should be the power of the microscope ?
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Verified Answer
The least distance of distinct vision of an average person, (i.e., D) is $25 \mathrm{~cm}$ In order to view an object with magnification 10.
As given that, $\quad \mathrm{v}=\mathrm{D}=25 \mathrm{~cm}$ and $\quad \mathrm{u}=-\mathrm{f}$, Now, the magnification $(\mathrm{m})$ is
$$
\begin{aligned}
&\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}} \\
&\mathrm{m}=\frac{\frac{-25}{-\mathrm{f}}}{25}=\frac{25}{\mathrm{f}}
\end{aligned}
$$
So, $\mathrm{f}=\frac{\mathrm{D}}{\mathrm{m}}=\frac{25}{\mathrm{~m}}=\frac{25}{10}=2.5=0.025 \mathrm{~m}$.
Hence, power of lens $\mathrm{P}=\frac{1}{0.025}=40 \mathrm{D}$
As given that, $\quad \mathrm{v}=\mathrm{D}=25 \mathrm{~cm}$ and $\quad \mathrm{u}=-\mathrm{f}$, Now, the magnification $(\mathrm{m})$ is
$$
\begin{aligned}
&\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}} \\
&\mathrm{m}=\frac{\frac{-25}{-\mathrm{f}}}{25}=\frac{25}{\mathrm{f}}
\end{aligned}
$$
So, $\mathrm{f}=\frac{\mathrm{D}}{\mathrm{m}}=\frac{25}{\mathrm{~m}}=\frac{25}{10}=2.5=0.025 \mathrm{~m}$.
Hence, power of lens $\mathrm{P}=\frac{1}{0.025}=40 \mathrm{D}$
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