Two thin converging lenses are placed on a common axis, so that the centre of one of them coincides with
the focus of the other. An object is placed at a distance twice the focal length from the left-hand lens. Where will its image be? What is the lateral magnification? The focal length of each lens is f.

The image formed by first lens will be at distance 2f with lateral magnification m₁ = -1
For second lens this image will behave as a virtual object.
Using the lens formula, $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$ we have, $\frac{1}{\mathrm{V}}-\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}}$
$\therefore \mathrm{v}=\frac{\mathrm{f}}{2},{\mathrm{m}}_2=\frac{{\mathrm{v}}_2}{{\mathrm{u}}_2}=\frac{(\mathrm{f}/2)}{\mathrm{f}}=\frac{1}{2}$
Therefore, final image (real) is formed at a distance f/2 right from the second lens with total lateral magnification, $\mathrm{m}={\mathrm{m}}_1\times {\mathrm{m}}_2=(-1)\times \left(\frac{1}{2}\right)=-\frac{1}{2}$