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

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

Answer 1

Answer: (D)

$30\,cm$

Answer 2

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