A plane mirror is placed at a distance of $50\text{cm}$ from a concave mirror of focal length $16\text{cm}$. Where should a short object be placed between the mirrors and facing both the mirrors so that its virtual image in the plane mirror coincides with the real image in concave mirror?

The correct option is C $30\text{cm}$ from the plane mirror
Let the object be at a distance x from the plane mirror.

The distance of object from concave mirror $=50-x$

For the plane mirror, object and image distances are equal.
$A^{\prime }M=AM=x$


$\Rightarrow O{A}^{\prime }=OM+A^{\prime }M=50+x$
For the concave mirror image distance is $50+x$

Using sign convention
$u=-(50-x),v=-(50+x),f=-16\text{cm}$

From mirror formula,
$\frac{1}{50-x}+\frac{1}{50+x}=\frac{1}{16}$
$\Rightarrow \frac{100}{{50}^2-x^2}=\frac{1}{16}\Rightarrow x=30\text{cm}$

The object must be placed at a distance of $30\text{cm}$ from the plane mirror.

Why this Question? Caution: Not to confuse with case of multiple reflection. In this question we are finding image of an object using both plane mirror and concave mirror.