Thank you for reporting, we will resolve it shortly
Q.
The distance between an object and its real image produced by a converging lens is $0.72\, m$. The magnification is $2$. What will be the magnification when the object is moved by $0.04\, m$ towards the lens ?
Given, $m=2$ and the image formed is real
$\therefore $ For a real image $\frac{1}{f}=\frac{1}{v}+\frac{1}{u} \,\,\,\,\,(\because u$ is negative $)$
or $ 1+\frac{1}{m}=\frac{u}{f} $
$\Rightarrow 1+\frac{1}{2}=\frac{u}{f} $
$\Rightarrow \frac{3}{2}=\frac{0.72}{f}$
or$f=\frac{0.72 \times 2}{3}=0.16\, m$
As the object is moved by $0.04 \,m$ towards the lens, the new
$u_{1}=0.72-0.04=0.68 \,m$
Again real image is formed
So, $1+\frac{1}{m}=\frac{u_{1}}{f}$
$\Rightarrow \frac{1}{m}=\frac{0.04}{0.16} $
$\Rightarrow m=4 $