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The following figure shows a beam of light converging at point $P$. When a concave lens of focal length $16 \mathrm{~cm}$ is introduced in the path of the beam at a place shown by dotted line such that $O P$ becomes the axis of the lens, the beam converges at a distance $x$ from the lens. The value of $x$ will be equal to
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$48 \mathrm{~cm}$
As the beam of light is converging at point $P$, so it will act as a virtual object for concave lens.
$\therefore$ Objective distance, $u=12 \mathrm{~cm}$
Focal length of concave lens, $f=-16 \mathrm{~cm}$
Using lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\begin{array}{ll}\Rightarrow & \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{16}+\frac{1}{12}=\frac{-3+4}{48}=\frac{1}{48} \\ \text { or } & v=48 \mathrm{~cm} \\ \text { Given, } & v=x \mathrm{~cm} \\ \therefore & x=48 \mathrm{~cm}\end{array}$
$\therefore$ Objective distance, $u=12 \mathrm{~cm}$
Focal length of concave lens, $f=-16 \mathrm{~cm}$
Using lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\begin{array}{ll}\Rightarrow & \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{16}+\frac{1}{12}=\frac{-3+4}{48}=\frac{1}{48} \\ \text { or } & v=48 \mathrm{~cm} \\ \text { Given, } & v=x \mathrm{~cm} \\ \therefore & x=48 \mathrm{~cm}\end{array}$
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