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

Answer 1

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.

\(A_i=9A_o \)

\(\dfrac{A_i}{A_o}=9\)

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, \(m=-3\)

\(m=\dfrac{v}{u}=-3\)

\(v=-3u\)

\(\because u=-40~\text{cm}\)

\(v=3 \times 40=120~\text{cm}\)

Now using the lens formula,

\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\Rightarrow \dfrac{1}{f}=\dfrac{1}{120}-\dfrac{1}{-40}=\dfrac{1+3}{120}\)

\(f=\dfrac{120}{4}=30~\text{cm}\)