Figure Illustrates an aligned system consisting of three lenses. The system is located in air. Find the distance between the first lens and a point lying on the axis to the left of the system, at which that point and its image are located symmetrically with respect to the lens system.
 

$\begin{array}{l}\mathrm{f}=\frac{100}{\mathrm{p}\left(\text{ dioptre }\right)} \mathrm{cm}\\ {\mathrm{f}}_3={\mathrm{f}}_1=\frac{100}{10}=10\mathrm{cm}\\ {\mathrm{f}}_2=\frac{100}{-10}=-10\mathrm{cm}\end{array}$
Applying lens formula for L₁
$\begin{array}{l}\frac{1}{{\mathrm{v}}_1}-\frac{1}{-\mathrm{x}}=\frac{1}{10}\\ \Rightarrow {\mathrm{v}}_1=\frac{10\mathrm{x}}{\mathrm{x}-10}\end{array}$
Applying lens formula for L₃
$\begin{array}{l}\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{u}}_3}=\frac{1}{10}\\ \Rightarrow {\mathrm{u}}_3=\frac{10\mathrm{x}}{10-\mathrm{x}}\end{array}$
Now, for L₂
$\begin{array}{r}{\mathrm{u}}_2={\mathrm{v}}_1-5=\frac{10\mathrm{x}}{\mathrm{x}-10}-5=\frac{5\mathrm{x}+50}{\mathrm{x}-10}\\ {\mathrm{v}}_2={\mathrm{u}}_3+5=\frac{10\mathrm{x}}{10-\mathrm{x}}+5=\frac{5\mathrm{x}+50}{10-\mathrm{x}}\end{array}$
Applying lens formula for L₂
$\begin{array}{l}\frac{1}{{\mathrm{v}}_2}-\frac{1}{{\mathrm{u}}_2}=\frac{1}{{\mathrm{f}}_2}\\ \Rightarrow \frac{10-\mathrm{x}}{5\mathrm{x}+50}-\frac{\mathrm{x}-10}{5\mathrm{x}+50}=\frac{1}{-10}\\ \Rightarrow \mathrm{x}=\frac{50}{3}\mathrm{cm}\end{array}$