Question

An image of a bright square is obtained on a screen with the aid of a convergent lens. The distance between the square and the lens is $40cm$. The area of the image is nine times larger than that of the square. Select the correct statement(s).

• A. Image is formed at a distance of $120cm$ from the lens
• B. Image is formed at a distance of $360cm$ from the lens
• C. Focal length of the lens is $30cm$
• D. Focal length of the lens is $36cm$

Answer 1

Answer: (A), (C)

Focal length of the lens is $30cm$
Image is formed at a distance of $120cm$ from the lens

Given that the area of image becomes $9$ times larger than the image of square, so

$\frac{v}{u}=\frac{3}{1}\Rightarrow v=3u$

And given $u=-40$

So, $v=120$

So, using the lens formula

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{3u}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{3(-40)}-\frac{1}{(-40)}=\frac{1}{f}$

$f=30cm$.