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A wire mesh consisting of very small squares is viewed at a distance of $8 \mathrm{~cm}$ through a magnifying converging lens of focal length $10 \mathrm{~cm}$, kept close to the eye. The magnification produced by the lens is
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5
$u=-8 \mathrm{~cm}, f=10 \mathrm{~cm}$
As, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$,
$$
\therefore \quad \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}+\frac{1}{-8}=\frac{1}{10}-\frac{1}{8}=\frac{4-5}{40}=\frac{-1}{40}
$$
or, $v=-40 \mathrm{~cm}$.
Magnification produced by the lens, $m=\frac{v}{u}=\frac{-40}{-8}=5$.
This is a virtual image, erect and on the same side as the object.
As, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$,
$$
\therefore \quad \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}+\frac{1}{-8}=\frac{1}{10}-\frac{1}{8}=\frac{4-5}{40}=\frac{-1}{40}
$$
or, $v=-40 \mathrm{~cm}$.
Magnification produced by the lens, $m=\frac{v}{u}=\frac{-40}{-8}=5$.
This is a virtual image, erect and on the same side as the object.
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