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(B) Given: Focal length $f = -10 \ cm$,Object distance $u = -25 \ cm$,Area of object $A_{	ext{obj}} = (3 \ cm)^2 = 9 \ cm^2$.Using the mirror formula

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(B) Given: Focal length $f = -10 \ cm$,Object distance $u = -25 \ cm$,Area of object $A_{	ext{obj}} = (3 \ cm)^2 = 9 \ cm^2$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,the magnification $m$ is given by $m = \frac{f}{f-u}$.Substituting the values: $m = \frac{-10}{-10 - (-25)} = \frac{-10}{15} = -\frac{2}{3}$.The area of the image $A_{	ext{image}}$ is related to the area of the object by $A_{	ext{image}} = m^2 	imes A_{	ext{obj}}$.$A_{	ext{image}} = \left(-\frac{2}{3}ight)^2 	imes 9 = \frac{4}{9} 	imes 9 = 4 \ cm^2$.

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

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