Question

A lens is placed between the source of light and a wall. It forms images of area ${A_1}$ and ${A_2}$ on the wall for its two different positions. The area of the source of light is
A) $\sqrt {{A_1}{A_2}}$
B) $\dfrac{{{A_1} + {A_2}}}{2}$
C) ${\left( {\dfrac{{\sqrt {{A_1}} + \sqrt {{A_2}} }}{2}} \right)^2}$
D) None of these.

Answer 1

Magnification is defined as the ratio of image height to the height of the object it can be also defined as the ratio of image distance to the object distance from the lens. Magnification is the property through which the mirror or lens enlarges the size of the object.

Formula used:
The formula of the magnification is given by,
$M = \dfrac{I}{O}$
Where $M$ is the magnification $I$ is the image size and $O$ is the object size.

Complete step by step answer:
It is given in the problem that a lens is placed between the source of light and a wall. It forms images of area ${A_1}$ and ${A_2}$ on the wall for its two different positions and we need to find the area of the source light.
The formula of the magnification is given by,
$M = \dfrac{I}{O}$
Where $M$ is the magnification $I$ is the image size and $O$ is the object size.
Let the sides of the object be $a$ and $b$ then the area of the image will be equal to,
$\Rightarrow A = ma \times mb$
Where A is the area and m is the magnification of the lens.
Let the magnification for the two different positions be${m_1}$ and${m_2}$ then the multiplication of the two will give us,
$\Rightarrow {m_1} \times {m_2} = 1$
Area of image for position 1 will be,
$\Rightarrow {A_1} = {m_1}^2 \times \left( {ab} \right)$………eq. (1)
The area of the image for position 2 will be,
$\Rightarrow {A_2} = {m_2}^2 \times \left( {ab} \right)$………eq. (2)
Multiplying equation (1) and equation (2) we get,
$\Rightarrow {A_1} \times {A_2} = \left[ {{m_1}^2 \times \left( {ab} \right)} \right] \times \left[ {{m_2}^2 \times \left( {ab} \right)} \right]$
Since ${m_1} \times {m_2} = 1$ therefore we get,
$\Rightarrow {A_1} \times {A_2} = {\left( {ab} \right)^2}$
$\Rightarrow ab = \sqrt {{A_1} \times {A_2}}$

The area of source light is$ab = \sqrt {{A_1} \times {A_2}}$. The correct option for this problem is option A.

Note:
Since the source and wall are not displaced therefore when the position of the lens is changed then the new magnification becomes reciprocal of the previous magnification or in other words we can say that the image distance and object distance will get interchange.