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(A) Given: Object side $s = 3\, cm$,Object area $A_0 = s^2 = 9\, cm^2$. Object distance $u = -25\, cm$,Focal length $f = -10\, cm$.Using the mirror fo

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(A) Given: Object side $s = 3\, cm$,Object area $A_0 = s^2 = 9\, cm^2$. Object distance $u = -25\, cm$,Focal length $f = -10\, cm$.Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.$\frac{1}{v} = \frac{1}{-10} - \frac{1}{-25} = \frac{-5 + 2}{50} = \frac{-3}{50}$.So,$v = -\frac{50}{3}\, cm$.Linear magnification $m = -\frac{v}{u} = -\left(\frac{-50/3}{-25}ight) = -\frac{2}{3}$.Areal magnification $m_A = m^2 = \left(-\frac{2}{3}ight)^2 = \frac{4}{9}$.Image area $A_I = m_A 	imes A_0 = \frac{4}{9} 	imes 9 = 4\, cm^2$.

$A$ square cardboard of side $3\, cm$ is placed $25\, cm$ from a concave mirror of focal length $10\, cm$. What is the area enclosed by the image of the cardboard? (The plane of the cardboard is perpendicular to the principal axis.)

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