A plane mirror is placed 22.5 cm in front of a concave mirror of focal length 10cm. Find where an object can be placed between the two mirrors, so that the first image from both the mirrors coincides. (Distance from plane mirror)

As shown in figure, if the object is placed at a distance x from the concave mirror, its distance from the plane mirror will be (22.5-x). So plane mirror will form equal and erect image of object at a distance (22.5-x) behind the mirror.

Now as according to given problem the image formed by concave mirror coincides with the image formed by plane mirror, therefore for concave mirror
$v=-[22.5+(22.5-x)]=-(45-x)$ and $u=-x$  
So $\frac{1}{-(45-x)}+\frac{1}{-x}=\frac{1}{-10}$ or $\frac{45}{(45x-x^2)}=\frac{1}{10}$  
i.e., $x^2-45x+450=0$ or $(x-30)(x-15)=0$  
i.e., $x=30$ cm or x = 15 cm
But as the distance between two mirrors is 22.5 cm, x = 30 cm is not admissible. So the object must be at a distance of 15 cm from the concave mirror.