As we know that, the length of image is the difference between the images formed by mirror of the extremities of object.
As the object distance is $u$, consider the two ends of the object, be at distance $\mathrm{u}_1=(\mathrm{u}-\mathrm{L} / 2)$ and $\mathrm{u}_2=(\mathrm{u}+\mathrm{L} / 2)$ respectively so that $\left|u_1-u_2\right|=L$
Consider the image of the two ends be formed at $v_1$ and $v_2$, respectively so that the image length on principal axis would be
$$
L^{\prime}=\left|\mathbf{v}_1-v_2\right|
$$
By using mirror formula, we get
or,
$$
\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}
$$
So,
$$
\begin{aligned}
v &=\frac{f u}{u-f} \\
v_1 &=\left(\frac{\mathrm{fu}_1}{u_1-f}\right) \\
v_2 &=\left(\frac{\mathrm{fu}_2}{u_2-f}\right)
\end{aligned}
$$
To find the value of $\mathrm{v}_1$ and $\mathrm{v}_2$, put the value of $\mathrm{u}_1$ and $\mathrm{u}_2$ in (iii) and (iv) equations respectively.
By solving, the positions of two images are
$$
\begin{aligned}
&v_1=\frac{f(u-L / 2)}{u-f-L / 2}, \\
&v_2=\frac{f(u+L / 2)}{u-f+L / 2},
\end{aligned}
$$
For image length, substituting the value of $v_1$ and $v_2$ in equation (i), we get,
$$
L^{\prime}=\left|v_1-v_2\right| \quad\left[L^{\prime}=\frac{f^2 L}{\left((u-f)^2-L^2 / 4\right)}\right] .
$$
Since, the object is short and kept away from focus, we get $\therefore \quad \mathrm{L}< < (\mathrm{u}-\mathrm{f})$ or $\left(\mathrm{L}^2 / 4\right)< < (\mathrm{u}-\mathrm{f})^2$
Thus, neglecting the term $\left(\mathrm{L}^2 / 4\right)$
So,
$$
L^{\prime}=\frac{f^2}{(u-f)^2} L
$$
This is the required expression of length of image.