The size of the image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3^rd of its size. At what distance, the object has been placed from the mirror?

A diminished image is formed in both concave and convex mirrors.
For concave mirrors:
It is given the focal length f = -20cm and magnification m = -1/3.
Now, the object distance u is given by,
$\Rightarrow \mathrm{m}=-\mathrm{v}/\mathrm{u}$
$\Rightarrow -1/3=-\mathrm{v}/\mathrm{u}$
$\Rightarrow \mathrm{v}=\mathrm{u}/3$
Using the mirror formula,
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$

$\frac{1}{\mathrm{f}}=\frac{3}{\mathrm{u}}+\frac{1}{\mathrm{u}}$
$\Rightarrow \mathrm{u}=4\mathrm{f}$

$\Rightarrow \mathrm{u}=-80\mathrm{cm}$
Hence, the object distance is -80cm from the concave mirror.

For convex mirrors:
It is given that the focal length f = 20cm and magnification m = 1/3.
Now, the object distance ‘u; is given by,
$\Rightarrow \mathrm{m}=-\mathrm{v}/\mathrm{u}$

$\Rightarrow 1/3=-\mathrm{v}/\mathrm{u}$
$\Rightarrow \mathrm{v}=-\mathrm{u}/3$
Using the mirror formula,
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}$

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}-\frac{3}{\mathrm{u}}$
$\Rightarrow \mathrm{u}=-2\mathrm{f}$
$\Rightarrow \mathrm{u}=-40\mathrm{cm}$
Hence, the object distance is -40cm from the convex mirror.
Hence, The object is placed at a distance of -80cm from the concave mirror and a distance of -40cm from the convex mirror.