A convex lens of focal length 16 cm forms a real image double the size of the object. The distance of the object from the lens is

given that m = 2
Focal length of convex lens = 16 cm
$$\begin{gathered} \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{size} \mathrm{of} \mathrm{image}}{\mathrm{size} \mathrm{of} \mathrm{object}} \\ \frac{\mathrm{v}}{\mathrm{u}}=\frac{2\left(\mathrm{size} \mathrm{of} \mathrm{object}\right)}{\mathrm{size} \mathrm{of} \mathrm{object}} \\ \frac{\mathrm{v}}{\mathrm{u}}=2,\mathrm{f}=16\mathrm{cm}\Rightarrow \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\ \mathrm{v}=+\mathrm{ve},\mathrm{u}=-\mathrm{ve} \mathrm{i}.\mathrm{e}.,\mathrm{v}=2\mathrm{u}, \mathrm{u}=-\mathrm{u}(\because \mathrm{image} \mathrm{formed} \mathrm{is} \mathrm{real}) \\ \frac{1}{\mathrm{f}}=\frac{1}{2\mathrm{u}}-\frac{1}{\left(-\mathrm{u}\right)}\Rightarrow \frac{1}{16}=\frac{1+2}{2\mathrm{u}}\Rightarrow \frac{1}{16}=\frac{3}{2\mathrm{u}} \\ 2\mathrm{u}=48 \\ \mathrm{u}=24\mathrm{cm} \end{gathered}$$