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(C) Let the area of the source of light be $O$. For two different positions of the lens,the magnifications are $m_1$ and $m_2$.The area of the image i

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$\sqrt{A_1 A_2}$

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(C) Let the area of the source of light be $O$. For two different positions of the lens,the magnifications are $m_1$ and $m_2$.The area of the image is given by $A = m^2 O$,so $m^2 = \frac{A}{O}$.For the two positions,we have $m_1^2 = \frac{A_1}{O}$ and $m_2^2 = \frac{A_2}{O}$.Multiplying these,we get $m_1^2 m_2^2 = \frac{A_1 A_2}{O^2}$.For the displacement method,the product of magnifications is $m_1 m_2 = 1$,which implies $m_1^2 m_2^2 = 1$.Therefore,$1 = \frac{A_1 A_2}{O^2}$,which gives $O^2 = A_1 A_2$.Thus,the area of the source of light is $O = \sqrt{A_1 A_2}$.

$A$ 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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