Question

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 $25cm$. The distance through which the object has to be shifted so that its image on the screen is sharp again is:

• A. $37.5cm$
• B. $16.25cm$
• C. $12.5cm$
• D. $13.5cm$

Answer 1

Answer: (C)

$12.5cm$

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

Case 1 : Image distance $v=75$ cm

Let object distance be $u$

Using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\therefore$ $\frac{1}{75}-\frac{1}{u}=\frac{1}{25}$ $⟹u=-37.5$ cm

This means the object is at a distance of $37.5$ cm from the lens initially.

Case 2 : The screen is shifted by $25$ cm towards the lens.

$\therefore$ New image distance $v^{\prime }=75-25=50$ cm

Let new object distance be $u^{\prime }$

Using $\frac{1}{v^{\prime }}-\frac{1}{u^{\prime }}=\frac{1}{f}$

$\therefore$ $\frac{1}{50}-\frac{1}{u^{\prime }}=\frac{1}{25}$ $⟹u^{\prime }=-50$ cm

$\therefore$ Object has to be shifted by disance $d=50-37.5=12.5$ cm away from the lens