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(A) Let the area of the source be $A_0$. For the two positions of the lens,the magnification $m_1$ and $m_2$ are given by $m_1^2 = \frac{A_1}{A_0}$ an

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$\sqrt{A_1 A_2}$

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(A) Let the area of the source be $A_0$. For the two positions of the lens,the magnification $m_1$ and $m_2$ are given by $m_1^2 = \frac{A_1}{A_0}$ and $m_2^2 = \frac{A_2}{A_0}$.From the property of conjugate positions of a lens,the product of the magnifications is $m_1 	imes m_2 = 1$.Therefore,$(m_1 	imes m_2)^2 = 1^2 = 1$.Substituting the expressions for $m_1^2$ and $m_2^2$,we get $\frac{A_1}{A_0} 	imes \frac{A_2}{A_0} = 1$.This simplifies to $A_0^2 = A_1 A_2$.Thus,the area of the source is $A_0 = \sqrt{A_1 A_2}$.

$A$ 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

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