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 D $30\text{cm}$ from the plane mirror
Let the distance between object $\text{AB}$ and plane mirror is $x\text{cm}$.

$\therefore$ distance between object $\text{AB}$ and concave mirror is $(50-x)\text{cm}$

So, from the above diagram,

$AM=A^{\prime }M=x$

Image formed in the plane mirror will be at a distance $x$ from plane mirror towards right and thus $(50+x)$ from concave mirror.

$\therefore {OA}^{\prime }=OM+A^{\prime }M=(50+x)\text{cm}$

For concave mirror:

Object distance for concave mirror, $u=-(50-x)$
Image distance for concave mirror, $v=-(50+x)$
And, focal length, $f=-16\text{cm}$

Applying mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{-(50+x)}+\frac{1}{-(50-x)}=\frac{-1}{16}$

$\Rightarrow \frac{50-x+50+x}{{50}^2-x^2}=\frac{1}{16}$

$\Rightarrow 100\times 16=2500-x^2$

$\Rightarrow x=30\text{cm}$

Hence correct option is (d).