Physics

Solution:

Focal length of the converging lens: $f=25cm$

Case 1: Image distance: $v=75cm$.

Let object distance be $u$.

Using $v1 −u1 =f1 $, we get:

$751 −u1 =251 $

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: $v1 −u1 =f1 $, we get:

$501 −u1 =251 $

or, $u=50cm$

So, the object has to be shifted by a distance: $d=50−37.5=12.5cm$ away from the lens.