Question

An object is placed in front of a concave mirror of focal length $f$. A virtual image is formed with a magnification of 2. To obtain a real image of same magnification, the object has to moved by a distance

• A. $f$
• B. $\frac{f}{2}$
• C. $\frac{f}{2}$
• D. $\frac{2f}{3}$

Answer 1

The correct option is A $f$

In the first case, let $x$ be the distance of the object from the mirror. Then,
$u=-x$
Since, $m=\frac{-v}{u}=2$ [given]
$v=+2x$

Using $\frac{1}{v}+\frac{1}{u}=\frac{1}{(-f)}$
or $\frac{1}{2x}-\frac{1}{x}=-\frac{1}{f}$ or $x=\frac{f}{2}$
where $f$ is the magnitude of focal length.

In the second case, let $y$ be the distance of object from the mirror. Then,
$u=-y,v=-2y$
[since magnification $m=-2$ for real image]

So, $\frac{1}{-2y}-\frac{1}{y}=\frac{1}{(-f)}$
$\therefore y=\frac{3}{2}f$
So, object will have to be moved by a distance of
$y-x=\frac{3f}{2}-\frac{f}{2}=f$