## In This post you will get answer of Q.

The image of an illuminated square is obtained on a screen with the help of a converging lens. The distance of the square from the lens is $40, cm$. The area of the image is $9$ times that of the square. The focal length of the lens is :

Solution:

## If side of object square $=ell$

and side of image square $=ell’$

From question, $frac{ell’^{2}}{ell}=9$

or $frac{ell’}{ell }=3$

i.e., magnification $m = 3$

$u=- 40 ,cm$

$v= 3 times 40= 120, cm$

$f=?$

From formula $frac{1}{v}-frac{1}{u}=frac{1}{f}$

$frac{1}{120}-frac{1}{-40}=frac{1}{f}$

or, $frac{1}{f}=frac{1}{120}+frac{1}{40}=frac{1+3}{120} therefore f =30,cm$