A convex lens of focal length ‘f’ produces a real image 3 times the size of an object, then distance of the object from the lens is

size of image = 3 size of the object
$\Rightarrow \mathrm{m}=\frac{ \mathrm{size} \mathrm{of} \mathrm{image} }{ \mathrm{size} \mathrm{of} \mathrm{object} }=3 (\mathrm{given} \mathrm{that} \mathrm{image} \mathrm{is} \mathrm{real})$
Then m = -3
$\Rightarrow \frac{\mathrm{v}}{\mathrm{u}}=-3\mathrm{v}=-3\mathrm{u},\mathrm{u}=?\mathrm{f}=+\text{ ve i.e },+\mathrm{f}$
We have
$\begin{array}{l}\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\\ \frac{-1}{3\mathrm{u}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\\ \frac{-1-3}{3\mathrm{u}}=\frac{1}{\mathrm{f}}\\ \frac{-4}{3\mathrm{u}}=\frac{1}{\mathrm{f}}\Rightarrow \mathrm{u}=\frac{-4\mathrm{f}}{3}\end{array}$
$\therefore$Distance of object is $=\frac{4\mathrm{f}}{3}$