Question
$A$ concave mirror forms a real image three times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of object. If the shift of object is $6\, cm$. The shift of the screen & focal length of mirror are

Answer

$\frac{v}{u}=\frac{I}{o}=3$$(i)$

and $\frac{v^{\prime}}{u^{\prime}}=\frac{I}{o}=2 \Rightarrow \frac{v^{\prime}}{u+6}=2(\mathrm{ii})$

Form $( i )$ and $(ii),$ we get

$v^{\prime}=2\left(\frac{v}{3}+6\right)$$(iii)$

$\Rightarrow v^{\prime}-\frac{2}{3} v=12$

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{1}{v^{\prime}}+\frac{1}{u^{\prime}}$

$\Rightarrow \frac{1}{v}+\frac{3}{u}=\frac{1}{v^{\prime}}+\frac{2}{u^{\prime}}$

$\Rightarrow 4\left(\frac{3}{2}-1\right)$

$\Rightarrow v=\frac{4}{3} v^{\prime}(\mathrm{iv})$

From $(iii)$ and $(iv),$

$v^{\prime}-\frac{2}{3} \times \frac{4}{3} v^{\prime}=12$

$\Rightarrow \frac{1}{9} v^{\prime}=12 \Rightarrow v^{\prime}=108 \mathrm{cm}$

$\Rightarrow v=\frac{4}{3} v^{\prime}=144 \mathrm{cm}$

Shifting will be $144-108=36 c m$ toward the image.

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