Example: A thin converging lens of focal length $f=25cm$ forms the image of an object on a screen placed at a distance of $75cm$ from the lens. The screen is moved closer to the lens by a distance of $25cm$. Find the distance through which the object has to be shifted so that its image on the screen is sharp again.

Solution:
Focal length of the converging lens: $f=25cm$
Case 1: Image distance: $v=75cm$.
Let object distance be $u$. 
Using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, we get:
$\frac{1}{75}-\frac{1}{u}=\frac{1}{25}$
or, $u=37.5cm$
This means that the object is at a distance of $37.5cm$ from the lens initially.
Case 2: The screen is shifted by $25cm$ towards the lens.
New image distance: $v=75-25=50cm$
Let new object distance be $u$
Using: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, we get:
$\frac{1}{50}-\frac{1}{u}=\frac{1}{25}$
or, $u=50cm$
 So, the object has to be shifted by a distance: $d=50-37.5=12.5cm$ away from the lens.