Optic axis of a thin equiconvex lens is the x-axis. The co-ordinates of a point object and its image are (-40 cm, I cm) and (50 cm, -2cm) respectively. Lens is located at

$\begin{array}{l}\mathrm{Linear}\mathrm{magnification}\mathrm{m}=\frac{-2\mathrm{cm}}{1\mathrm{cm}}=-2\\ \mathrm{i}.\mathrm{e}.,\mathrm{image}\mathrm{is}\mathrm{real}\mathrm{and}\mathrm{inverted}.\end{array}$

$\begin{array}{l}\left|\mathrm{v}\right|=2\left|\mathrm{u}\right|\\ \mathrm{Let}\left|\mathrm{u}\right|=\mathrm{x}\mathrm{then}\left|\mathrm{v}\right|=2\mathrm{x}\\ \mathrm{now}\left|\mathrm{u}\right|+\left|\mathrm{v}\right|=50-\left(-40\right)=90\\ \therefore \mathrm{x}+2\mathrm{x}=90\\ \mathrm{or}\mathrm{x}=30\mathrm{cm}\end{array}$

Hence, distance of object from the lens is 30 cm and of object is 60 cm. therefore, lens should be located at -t: -10 cm as shown in figure.