Question

A plane mirror is placed $22.5cm$ in front of a concave mirror of focal length $10cm$. An object can be placed between the two mirrors so that the first image in both the mirror coincides. The object distance from the spherical mirror is

• A. $15cm$
• B. $25cm$
• C. $30cm$
• D. $20cm$

Answer 1

Answer: (A)

$15cm$

It can seen from the figure that, 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 equal and erect image of object at a distance $(22.5-x)$ behind the mirror. Now according to the given problem, the image coincided with the image formed by the plane mirror, therefore for concave mirror.
$v=-[22.5+22.5-x]=-(45-x)$

and object distance
$u=-x$

given focal length is

$f=-10$

So using mirror formula,

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

$\frac{1}{-(45-x)}+\frac{1}{-x}=\frac{1}{-10}⟹\frac{45}{45x-x^2}=\frac{1}{10}$

$x^2-45x+450=0$

$(x-30)(x-15)=0$

Hence $x=30cm$ or $x=15cm$

.
But as the distance between the two mirror is $22.5$, $x=30$ is not admissible . So the object must be at a distance of $15$ cm from the concave mirror