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(A) The lateral magnification $m$ is defined as the ratio of the height of the image $h_i$ to the height of the object $h_o$,i.e.,$m = \frac{h_i}{h_o}

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) The lateral magnification $m$ is defined as the ratio of the height of the image $h_i$ to the height of the object $h_o$,i.e.,$m = \frac{h_i}{h_o}$.Given the lateral magnification $m = 0.4$.For a two-dimensional object placed perpendicular to the principal axis,the area magnification $m_A$ is the square of the lateral magnification.$m_A = \frac{A_i}{A_o} = m^2$.Given the area of the object $A_o = 100\,cm^2$.Substituting the values: $A_i = A_o 	imes m^2 = 100 	imes (0.4)^2$.$A_i = 100 	imes 0.16 = 16\,cm^2$.Therefore,the area of the image is $16\,cm^2$.

$A$ square object of area $100\,cm^2$ is placed perpendicular to the principal axis of a concave mirror. If the lateral magnification of the mirror,for the above object position,is $0.4$,then the area of the image will be......$cm^2$.

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