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The relation between the linear magnification $m$, the object distance ${ }^u$ and the focal length $f$ is
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Verified Answer
The correct answer is:
$m=\frac{f}{f-u}$
$\because m=-\frac{v}{u}$ also $\frac{1}{f}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{u}{f}=\frac{u}{v}+1$
$\Rightarrow-\frac{u}{v}=1-\frac{u}{f} \Rightarrow \frac{-v}{u}=\frac{f}{f-u}$ so $m=\frac{f}{f-u}$.
$\Rightarrow-\frac{u}{v}=1-\frac{u}{f} \Rightarrow \frac{-v}{u}=\frac{f}{f-u}$ so $m=\frac{f}{f-u}$.
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