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An object placed at a distance of $16 \mathrm{cm}$ from a $\begin{array}{llll}\text { convex lens } & \text { produces } & \text { an image } & \text { of }\end{array}$ magnification $m(m>1)$. If the object is moved towards the lens by $8 \mathrm{cm}$, then again an image of magnification $m$ is obtained. The numerical value of the focal length of the lens is
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$12 \mathrm{cm}$
Linear magnification, $m=\frac{f}{f+u}$
As given that magnification is same for both cases this is possible if the two different values of $u$ are of opposite sign.
$i, \theta,+\frac{f}{f-16}=\frac{-f}{f-8}$
$\Rightarrow \quad 16-f=f-8$
$\Rightarrow \quad 2 f=24$
or $\quad f=12 \mathrm{cm}$
As given that magnification is same for both cases this is possible if the two different values of $u$ are of opposite sign.
$i, \theta,+\frac{f}{f-16}=\frac{-f}{f-8}$
$\Rightarrow \quad 16-f=f-8$
$\Rightarrow \quad 2 f=24$
or $\quad f=12 \mathrm{cm}$
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