Question

The image of a square hole in a screen illuminated by light is obtained on another screen with the help of converging lens. The distance of the hole from the lens is $40\text{cm}$. If the area of the image is nine times that of the hole, the focal length of the lens is

• A. $30\text{cm}$
• B. $50\text{cm}$
• C. $60\text{cm}$
• D. $75\text{cm}$

Answer 1

The correct option is A $30\text{cm}$


Let the area of image on the screen be $A_i$ and the area of the square hole be $A_o$. It is given that area of the image is $9$ times that of the area of the object.

$A_i=9A_o$

$\frac{A_i}{A_o}=9$

This gives us areal magnification. Since linear magnification $m$ is

$m=\sqrt{\text{areal magnification}}$

$\Rightarrow m=\sqrt{9}=3$

Since image formed is real we will take the value of the magification as negative. So, $m=-3$

$m=\frac{v}{u}=-3$

$v=-3u$

$\because u=-40\text{cm}$

$v=3\times 40=120\text{cm}$

Now using the lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\Rightarrow \frac{1}{f}=\frac{1}{120}-\frac{1}{-40}=\frac{1+3}{120}$

$f=\frac{120}{4}=30\text{cm}$