A point object 'O' is placed at a distance of 30 cm from a convex lens (f=20 cm) cut into two equal halves each of which is displaced by 0.05 cm as shown in figure.

Each piece of lens forms one image and they do not coincide. Therefore number of images = 2. In this type of cutting, focal length of each piece f = 20cm i.e. same as original lens. 

For ${\mathrm{L}}_2:$ $\mathrm{u}=-30\mathrm{cm}, \mathrm{f}=+20\mathrm{cm}$

$\frac{1}{\mathrm{V}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$

$\Rightarrow \frac{1}{\mathrm{V}}-\frac{1}{-30}=\frac{1}{20}\Rightarrow \mathrm{V}=+60\mathrm{cm}$

$\frac{{\mathrm{y}}_{\text{image }}}{{\mathrm{y}}_{\text{object }}}=\frac{\mathrm{V}}{\mathrm{u}}\Rightarrow \frac{{\mathrm{y}}_{\mathrm{i}}}{0.05}=\frac{+60}{-30}\Rightarrow {\mathrm{y}}_{\mathrm{i}}=-0.1\mathrm{cm}$

Hence L₂ forms real image I₂ at a distance of 60 cm from it on the other side, below its principal axis at 0.1 cm from the axis.
Similarly, L₁ forms I₁. 

Distance between I₁ and I₂ is $0.1+0.1+0.1=0.3\mathrm{cm}$