A concave mirror forms on a screen a real image of thrice the linear dimensions of the object. Object and screen are moved until the image is twice the size of the object. If the shift of the object is $6 cm$, find the shift of the screen and the focal length of the mirror (both in $cm$)

Since the image formed is real, so for the first case, we have
$$\begin{gathered} m=-3 \\ \Rightarrow -\frac{v_1}{u_1}=-3 \\ \Rightarrow v_1=3u_1 \end{gathered}$$
Since $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$$\begin{gathered} \Rightarrow \frac{1}{\left(-3u_1\right)}+\frac{1}{\left(-u_1\right)}=\frac{1}{-f} \\ \Rightarrow f=\frac{3u_1}{4} ...\left(1\right) \end{gathered}$$
Similarly, for the second case, we have 
 $$\begin{gathered} m=-2 \\ \Rightarrow -\frac{v_2}{u_2}=-2 \\ \Rightarrow v_2=2u_2 \end{gathered}$$
Since $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$$\begin{gathered} \Rightarrow \frac{1}{-\left(2u_2\right)}+\frac{1}{\left(-u_2\right)}=\frac{1}{-f} \\ \Rightarrow f=\frac{2u_2}{3} ...\left(2\right) \end{gathered}$$
From $\left(1\right)$ and $\left(2\right)$, we get
    $\frac{3u_1}{4}=\frac{2u_2}{3}$
$\Rightarrow u_2=\frac{9u_1}{8} ...\left(3\right)$
But according to the problem, we are given that the shift of the object is 6 cm, so we get
$$\begin{gathered} u_2-u_1=6 cm \\ \Rightarrow \frac{9}{8}{u}_1-u_1=6 \\ \Rightarrow u_1=48 cm \end{gathered}$$
Since, $v_1=3u_1$
$\Rightarrow v_1=144 cm$
So, $f=\frac{3u_1}{4}=\frac{3}{4}\left(48\right)=36 cm$
Now $u_2=\frac{9u_1}{8}=\frac{9}{8}\left(48\right)=54 cm$
and  $v_2=2u_1=2\left(54\right)=108 cm$
So, shift of the screen is $\left(v_1-v_2\right)$
$\Rightarrow \left(v_1-v_2\right)=144-108=36 cm$