Question

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

• A. 20 cm from concave mirror
• B. 15 cm from the concave mirror
• C. 5 cm from plane mirror
• D. 7.5 from plane mirror

Answer 1

Answer: (B)

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 from 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 $u-x ; \frac{1}{-(45-x)}+\frac{1}{-x}=\frac{1}{-10}$
$\frac{45}{\left(45 x-x^{2}\right)}=\frac{1}{10} ; x^{2}-45 x+450=0$
$(x-30)(x-15)=0 \,\,\, x=30 \,cm$ 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.