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

Given that object is real and image is virtual.

Image distance=v, Object distance=-u.

$$\begin{gathered} \mathrm{Magnification} \left(\mathrm{m}\right) =-\frac{-\mathrm{v}}{\mathrm{u}}=2 \\ \Rightarrow \mathrm{v}=2\mathrm{u} \end{gathered}$$

By using mirror equation, 

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

Now in the second case with same magnification, 

v'= -2u'

by using mirror equation,

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

Distance that the object is moved = v'-v= f