In figure (i), a thin lens of focal length 10 cm is shown. The lens is cut into two equal parts, and the parts are arranged as shown in figure (ii). An object AB of height 1 cm is kept at a distance of 7.5 cm from the arrangement. What will be the height of the final image?

When both the lenses are placed beside each other, then the focal length of the combination can best be found by using power formula:

$\begin{array}{l}\mathrm{p}={\mathrm{p}}_1+{\mathrm{p}}_2\\ {\mathrm{p}}_{\text{tot }}=2\mathrm{p}\\ \Rightarrow {\mathrm{f}}_{\text{tot }}=\frac{\mathrm{f}}{2}=\frac{10}{2}=5\mathrm{cm}\end{array}$

Now applying lens formula.

$\begin{array}{l}\mathrm{u}=-7.5\mathrm{cm},\mathrm{f}=5\mathrm{cm}:\\ \frac{1}{\mathrm{v}}-\frac{1}{-7.5}=\frac{1}{5}\\ \Rightarrow \mathrm{v}=15\mathrm{cm}\end{array}$

Now applying magnification formula :

$\begin{array}{l}\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{15}{7.5}=2.\\ \text{ Height of image }=\\ \mathrm{m}\times \text{ height of object }\\ \Rightarrow {\mathrm{h}}_{\mathrm{i}}=2\times 1\mathrm{cm}=2\mathrm{cm}\end{array}$