Question

The image of a square hole in a screen illuminated by light is obtained on another screen with the help of a converging lens. The distance of the hole from the lens is 40 cm. If the area of the image is nine times that of the hole, the focal length of the lens is
\[\begin{align}
  & a)30cm \\
 & b)50cm \\
 & c)60cm \\
 & d)75cm \\
\end{align}\]

Answer 1

Hint: First we have to calculate the areal magnification, then we will calculate the linear magnification. Since a convex lens always forms the real image of an object on the other side of the lens so linear magnification is represented by a negative sign. Object on the left side of the lens is always taken as negative in sign during calculation.

Complete step-by-step answer:
Let us assume the area of square hole is A Let us assume the area of image of square hole on the careen is A^’ Since, it is given in the question that the image of a square hole is obtained with the help of converging lens and its area is 9 times of object area. Figure shows the image formation of the square hole.

Areal Magnification = \[\dfrac{{{A}_{image}}}{{{A}_{object}}}\] Areal Magnification = \[\dfrac{{{A}^{'}}}{A}=9\]. Since Linear magnification is defined as square root of areal magnification.\[Linear\,\,magnification=\sqrt{Areal\,\,magnification}\] \[\Rightarrow Linear\,\,magnification=\sqrt{9}\]
\[\therefore Linear\,\,magnification=3\]
So, Linear magnification is 3 units.
So Linear magnification is defined as the ratio of image distance to object distance. So we can write the relation between magnification, image distance and object distance. Let us assume object distance from lens is u and image distance from lens is v and magnification is represented by symbol m . Relation between them is -
\[m=\dfrac{v}{u}\](Equation 1) It is given in the question that u=40cm. According to sign convention, an object is placed at left so it is taken with negative sign and image is placed at right so it is taken with positive sign. So value of becomes u=-40cm Since image formed is real so magnification is taken as negative. Put the value of m and u in equation 1 \[-3=\dfrac{v}{-40}\] \[\Rightarrow v=120cm\]
Now we have to calculate the focal length of the lens.
Let us assume the focal length of the lens is f.
Then lens formula is
\[\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\] Put the values of u and v in this equation we get ,
\[\dfrac{1}{f}=\dfrac{1}{120}-\dfrac{1}{-40}\]
\[\Rightarrow \dfrac{1}{f}=\dfrac{1}{120}+\dfrac{1}{40}\]
\[\Rightarrow \dfrac{1}{f}=\dfrac{1+3}{120}\]
\[\Rightarrow f=\dfrac{120}{4}\]
\[\therefore f=30cm\]
Focal length of the lens is 30 cm.
So, the correct answer is “Option A”.

Note: Sign conventions are important for optics numerical. Convex lens always forms a real image when the object is placed between focus and infinity while when it is placed between focus and optical centre then a virtual image is obtained. While Concave lens always forms a virtual image whatever be the position of the object.