According to the question,


Given, focal length, \(f=25 \mathrm{~cm}\) and distance between image of an object and screen,
\(v=75 \mathrm{~cm}\)
Now, By lens formula,
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
[ \(\because\) Because screen is moved closer to the lens.]
\(\frac{1}{u}=\frac{1}{f}-\frac{1}{v}\)
\(\begin{aligned}
\frac{1}{u} & =\frac{1}{25}-\frac{1}{75} \\
\Rightarrow \quad u & =\frac{75 \times 25}{50}=\frac{75}{2} \mathrm{~cm}
\end{aligned}\)
When the screen shift upto by \(25 \mathrm{~cm}\), then the screen will be at \(2 f\).
\(\therefore\) For to get sharp image, object has to be at \(2 f\). So, the distances is \(v-u=f\).
\(50-\frac{75}{2}=\frac{25}{2}=12.5 \mathrm{~cm}\)
\(\left[\begin{array}{l}
\text {at } 2 f \\
v=50 \mathrm{~cm}
\end{array}\right]\)
This is the distance through which the object shifted from the lens.