First slide

The lens

Question

A lens is placed between a 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

Moderate

A

$\sqrt{A_{1} A_{2}}$
B

$\frac{A_{1} + A_{2}}{2}$
C

$\frac{A_{1} - A_{2}}{2}$
D

$\frac{1}{A_{1}} + \frac{1}{A_{2}}$

Solution

Let say the length and breadth of object is a and b .
$\begin{array}{l} A r e a o f o b j e c t = a \times b \\ I n f i r s t p o s i t i o n o f d i s p l a c e m e n t m e t h o d, \\ A r e a o f I m a g e 1, A_{1} = m_{1} a \times m_{1} b \to (1) \\ I n s e c o n d p o s i t i o n o f d i s p l a c e m e n t m e t h o d, \\ A r e a o f I m a g e 2, A_{2} = m_{2} a \times m_{2} b \to (2) \\ M u l t i p l y e q u a t i o n 1 a n d 2 \\ A_{1} A_{2} = {(m_{1} m_{2})}^{2} {(a b)}^{2} \\ A l s o, i n d i s p l c e m e n t m e t h o d, m_{1} m_{2} = 1 \\ \Rightarrow A_{1} A_{2} = {(a b)}^{2} \\ \Rightarrow a b = \sqrt{A_{1} A_{2}} \\ \Rightarrow A r e a o f o b j e c t = \sqrt{A_{1} A_{2}} \end{array}$

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