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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$(i)$ Gravitational force with time
$(ii)$ Viscous force with time
$(iii)$ Net force acting on the ball with time
The initial shape of the wavefront of the beam is