A convergent lens forms on a screen an image of lamp magnified to twice its normal size. After the lens has been moved 36 cm closer to the screen it gives an image diminished by a factor of two. Find the focal length of the lens (in cm)

For the two positions of the lens, the image is formed on the same screen. So according to reversibility of path of light when object is at a distance u, its image will be at a distance v from the lens and vice versa
$$\begin{gathered} \therefore \left|\mathrm{v}\right|-\left|\mathrm{u}\right|=36 .......\left(\mathrm{i}\right) \\ \text{ And }\left|\frac{\mathrm{v}}{\mathrm{u}}\right|=2 ........\left(\mathrm{ii}\right) \end{gathered}$$
On solving above equations, we get | u | = 36 cm and | v | = 72cm 
Now from lens formula, $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$, we have
$\frac{1}{72}-\frac{1}{-36}=\frac{1}{\mathrm{f}}\Rightarrow \mathrm{f}=24\mathrm{cm}$