Optics Question 814
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 :
Options:
A) 36 cm
B) 27 cm
C) 60 cm
D) 30 cm
Show Answer
Answer:
Correct Answer: D
Solution:
[d] If side of object square = $ \ell $ and side of image square = $ \ell $ ?
From question, $ \frac{\ell {{’}^{2}}}{\ell }=9\text{ or }\frac{\ell ‘}{\ell }=3 $ i.e., magnification $ m=3~\text{ }u=-40cm $
$ v=3\times 40=120cm\text{ f=?} $ From formula, $ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} $
$ \frac{1}{120}-\frac{1}{-40}=\frac{1}{f}\text{ or }\frac{1}{f}=\frac{1}{120}+\frac{1}{40}=\frac{1+3}{120} $
$ \therefore 30cm $
BETA
