When an object is placed in front of a concave mirror of focal length f, a virtual image is produced with a magnification of 2. To obtain a real image with a magnification of 2, the object has to be moved by a distance equal to nf, where n is

A concave mirror can form virtual, erect and enlarged images when the object is kept at between focus and the pole of the mirror. When the object is moved between the center of curvature and the focus, the images formed are real, inverted and enlarged.

Mirror formula is given by,

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

Let the distance of the object from the pole be $-u$ .Let the distance of the image be $\text{v}$ .

Let the focal length of the mirror be f.

Magnification,

$m=-\frac{-v}{u}$

$\Rightarrow 2=\frac{v}{u}$

Hence,

$v=2u$

Applying the mirror formula here,

$\frac{1}{-f}=\frac{1}{2u}\frac{1}{-u}$

$\Rightarrow -\frac{1}{f}=\frac{1}{2u}-\frac{1}{u}$

Therefore,

$u=\frac{f}{2}$

Let the new position of the image be

$v^{\text{'}}$

and that of object be

$u^{\text{'}}$

The magnification is the same.

$m=-\frac{v^{\text{'}}}{u^{\text{'}}}$

$v^{\text{'}}=-2u^{\text{'}}$

Applying the mirror formula here,

$\frac{1}{-f}=\frac{1}{-2u^{\text{'}}}\frac{1}{-u^{\text{'}}}$

$\Rightarrow u^{\text{'}}=\frac{3f}{2}$

The shift in the position of the object is given by,

$u^{\text{'}}-u=\frac{3f}{2}-\frac{f}{2}$

$u^{\text{'}}-u=f$