A square of side 3;cm is placed at a distance | vedclass

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(A) Given: Side of square $s = 3\;cm$,so area of object $A_o = s^2 = 9\;cm^2$.Object distance $u = -25\;cm$,focal length $f = -10\;cm$.The linear magn

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$4$

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(A) Given: Side of square $s = 3\;cm$,so area of object $A_o = s^2 = 9\;cm^2$.Object distance $u = -25\;cm$,focal length $f = -10\;cm$.The linear magnification $m$ is given by $m = \frac{f}{u - f}$.Substituting the values: $m = \frac{-10}{-25 - (-10)} = \frac{-10}{-25 + 10} = \frac{-10}{-15} = \frac{2}{3}$.The area magnification $m_A$ is the square of the linear magnification: $m_A = m^2 = (\frac{2}{3})^2 = \frac{4}{9}$.The area of the image $A_i$ is given by $A_i = m_A 	imes A_o$.$A_i = \frac{4}{9} 	imes 9 = 4\;cm^2$.

$A$ square of side $3\;cm$ is placed at a distance of $25\;cm$ from a concave mirror of focal length $10\;cm$. The centre of the square is at the axis of the mirror and the plane is normal to the axis. The area enclosed by the image of the square is $......\;cm^2$.

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