The distance between an object and its image produced by a converging lens is \(0.72 \mathrm{~m}\). The magnification is 2 . What will be the magnification when the object is moved by \(0.04 \mathrm{~m}\) towards the lens?

Given, \(m=2\) and the image formed is real
\(\therefore\) For a real image \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u} \quad(\because u\) is negative \()\)
\(\begin{array}{ll}
\text { Or } & 1+\frac{1}{m}=\frac{u}{f} \\
\Rightarrow & 1+\frac{1}{2}=\frac{u}{f} \\
\Rightarrow & \frac{3}{2}=\frac{0.72}{f} \\
\text { or } \quad & f=\frac{0.72 \times 2}{3}=0.16 \mathrm{~m}
\end{array}\)
As the object is moved by \(0.04 \mathrm{~m}\) towards the lens, the new
\(u_{1}=0.72-0.04=0.68 \mathrm{~m}\)
Again real image is formed
So, \(1+\frac{1}{m}=\frac{u_{1}}{f}\)
\(\Rightarrow \quad \frac{1}{m}=\frac{0.04}{0.16}\)
\(\Rightarrow \quad m=4\)