Question

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 $40cm$. The area of the image is $9$ times that of the square. The focal length of the lens is :

• A. $36cm$
• B. $27cm$
• C. $60cm$
• D. $30cm$

Answer 1

Answer: (D)

If side of object square $=\ell$
and side of image square $={\ell }^{\prime }$
From question, $\frac{{\ell }^{\prime 2}}{\ell }=9$
or $\frac{{\ell }^{\prime }}{\ell }=3$
i.e., magnification $m=3$
$u=-40cm$
$v=3\times 40=120cm$
$f=?$
From formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{120}-\frac{1}{-40}=\frac{1}{f}$
or, $\frac{1}{f}=\frac{1}{120}+\frac{1}{40}=\frac{1+3}{120}\therefore f=30cm$