The Principal of reversibility states that the position of object and image are interchangeable. Then, by the versibility of $u$ and $v$, as seen from the lens formula,
$$
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}
$$
It shows that there are two positions for which there shall be an image.
Consider the first position be when the lens is at $\mathrm{O}$.
As given that,
$$
\begin{aligned}
& &-\mathrm{u}+\mathrm{v} &=\mathrm{D} \\
\Rightarrow & & \mathrm{u} &=-(\mathrm{D}-\mathrm{v})
\end{aligned}
$$
By putting the value of $u$ in equation (i),
so, $\quad \frac{1}{(D-v)}+\frac{1}{v}=\frac{1}{f}$
or $\quad \frac{v+D-v}{(D-v)_v}=\frac{1}{f}$
$$
v^2-D v+D f=0
$$
To find the coordinate of this equation then using
$$
\begin{aligned}
&x=\left(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\right) \\
&v=\frac{D}{2} \pm \frac{\sqrt{D^2-4 D f}}{2}
\end{aligned}
$$
By putting the value of $v$ in equation (ii) then,
$$
u=-\left[D-\left[\frac{D}{2} \pm \frac{\sqrt{D^2-4 D f}}{2}\right]\right]
$$
So,
From(iii)
When, the position of object distance is
$$
\mathrm{u}_2=\frac{\mathrm{D}}{2}+\frac{\sqrt{\mathrm{D}^2-4 \mathrm{Df}}}{2}
$$
The image forms at
$$
v_2=\frac{D}{2}-\frac{\sqrt{D^2-4 D f}}{2}
$$
Similarly, when the position of the object distance is
$$
\mathrm{u}_1=\frac{\mathrm{D}}{2}-\frac{\sqrt{\mathrm{D}^2-4 \mathrm{Df}}}{2}
$$
The image forms at
$$
v_1=\frac{D}{2}+\frac{\sqrt{D^2-4 D f}}{2}
$$
The distance between the poles for these two positions of lens $(d)=v_1-v_2$
$$
\mathrm{d}=\left[\frac{\mathrm{D}}{2}+\frac{\sqrt{\mathrm{D}^2-4 \mathrm{Df}}}{2}\right] \cdot\left[\frac{\mathrm{D}}{2}-\frac{\sqrt{\mathrm{D}^2-4 \mathrm{Df}}}{2}\right]
$$
So, $\quad d=\sqrt{D^2-4 D f}$
In first case :
If $\mathrm{u}_2=\left[\frac{\mathrm{D}}{2}+\frac{\mathrm{d}}{2}\right]$, then the image is at
$$
\mathrm{v}_2=\left[\frac{\mathrm{D}}{2}-\frac{\mathrm{d}}{2}\right]
$$
So, the magnification $\mathrm{m}_2=\frac{\mathrm{v}_2}{\mathrm{u}_2}=\frac{\frac{\mathrm{D}}{2}-\frac{\mathrm{d}}{2}}{\frac{\mathrm{D}}{2}+\frac{\mathrm{d}}{2}}$
$$
m_2=\frac{D-d}{D+d}
$$
In second case :
If $\left[\mathrm{u}_1=\frac{\mathrm{D}-\mathrm{d}}{2}\right]$, then the image is
$$
\mathrm{v}_1=\left[\frac{\mathrm{D}+\mathrm{d}}{2}\right]
$$
$\therefore \quad$ The magnification $\left(m_1\right)$ is
$$
\begin{aligned}
\mathrm{m}_1 &=\left\lfloor\frac{\mathrm{v}_1}{\mathrm{u}_1}\right\rfloor=\frac{(\mathrm{D}+\mathrm{d}) \times 2}{2 \times(\mathrm{D}-\mathrm{d})} \\
\mathrm{m}_1 &=\frac{\mathrm{D}+\mathrm{d}}{\mathrm{D}-\mathrm{d}} \\
\frac{\mathrm{m}_1}{\mathrm{~m}_2} &=\left(\frac{\mathrm{D}+\mathrm{d}}{\mathrm{D}-\mathrm{d}}\right)^2
\end{aligned}
$$