A square card of side length 1, mm is being s | vedclass

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(A) Given: Object distance $u = -9\, cm$,Focal length $f = 10\, cm$.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Substituting the

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(A) Given: Object distance $u = -9\, cm$,Focal length $f = 10\, cm$.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Substituting the values: $\frac{1}{v} - \frac{1}{-9} = \frac{1}{10} \Rightarrow \frac{1}{v} + \frac{1}{9} = \frac{1}{10}$.Solving for $v$: $\frac{1}{v} = \frac{1}{10} - \frac{1}{9} = \frac{9 - 10}{90} = -\frac{1}{90}$.Thus,$v = -90\, cm$.The linear magnification $m$ is given by $m = \frac{v}{u} = \frac{-90}{-9} = 10$.The area magnification is $m^2 = (10)^2 = 100$.The original area of the card $A_0 = (1\, mm)^2 = 1\, mm^2 = 0.01\, cm^2$.The apparent area $A_i = m^2 	imes A_0 = 100 	imes 0.01\, cm^2 = 1\, cm^2$.

$A$ square card of side length $1\, mm$ is being seen through a magnifying lens of focal length $10\, cm$. The card is placed at a distance of $9\, cm$ from the lens. The axis is perpendicular to the plane of the card. The apparent area of the card through the lens is......$cm^2$

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