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 in sharp again is

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

Answer 1

Answer: (C)

According to the first condition

$f=25cm,v=75cm$
$u=?$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{25}=\frac{1}{75}-\frac{1}{u}$
$\frac{1}{u}=\frac{1}{75}-\frac{1}{25}$
$\frac{1}{u}=\frac{1-3}{75}$
$u=-\frac{75}{2}=-37.5cm$
According to the second condition

$v_1=50cm,f=25cm,u_1=?$
$\frac{1}{f}=\frac{1}{v_1}-\frac{1}{u_1}$
$\frac{1}{25}=\frac{1}{50}-\frac{1}{u_1}$
$\Rightarrow \frac{1}{u_1}=\frac{1}{50}-\frac{1}{25}$
$\Rightarrow \frac{1}{u_1}=\frac{1-2}{50}$
$u_1=-50cm$
So, the screen is sharp again is
$\Delta u=u_1-u$
$=50-37.5$
$=12.5cm$