A square wire frame of side 3 , cm is placed | vedclass

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(A) Given: Focal length $f = -10 \, cm$,Object distance $u = -25 \, cm$,Side of square object $s_o = 3 \, cm$.Area of the object $A_o = (s_o)^2 = (3)^

Vedclass · Accepted Answer

$4$

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(A) Given: Focal length $f = -10 \, cm$,Object distance $u = -25 \, cm$,Side of square object $s_o = 3 \, cm$.Area of the object $A_o = (s_o)^2 = (3)^2 = 9 \, cm^2$.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 relationship between the area of the image $A_i$ and the area of the object $A_o$ is $A_i = m^2 	imes A_o$.$A_i = (-\frac{2}{3})^2 	imes 9 = \frac{4}{9} 	imes 9 = 4 \, cm^2$.

$A$ square wire frame of side $3 \, cm$ is placed at a distance of $25 \, cm$ from a concave mirror of focal length $10 \, cm$. The area of the image is ........ $cm^2$.

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