An object of height 3.6 cm is placed normall | vedclass

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(A) Given: Height of object $h_o = 3.6 \ cm$, Radius of curvature $R = 30 \ cm$. Focal length $f = R/2 = 30/2 = 15 \ cm$. Since it is a concave mirror

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$5.4$

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(A) Given: Height of object $h_o = 3.6 \ cm$, Radius of curvature $R = 30 \ cm$. Focal length $f = R/2 = 30/2 = 15 \ cm$. Since it is a concave mirror, $f = -15 \ cm$. The object is at a distance of $10 \ cm$ from the principal focus. Position of object $u = -(f + 10) = -(15 + 10) = -25 \ cm$. Using mirror formula: $1/v + 1/u = 1/f$. $1/v = 1/f - 1/u = 1/(-15) - 1/(-25) = -1/15 + 1/25 = (-5 + 3)/75 = -2/75$. $v = -75/2 = -37.5 \ cm$. Magnification $m = -v/u = h_i/h_o$. $m = -(-37.5) / (-25) = -37.5 / 25 = -1.5$. Height of image $h_i = m 	imes h_o = -1.5 	imes 3.6 = -5.4 \ cm$. The magnitude of the height of the real image is $5.4 \ cm$.

An object of height $3.6 \ cm$ is placed normally on the principal axis of a concave mirror of radius of curvature $30 \ cm$. If the object is at a distance of $10 \ cm$ from the principal focus of the mirror, then the height of the real image formed due to the mirror is (in $cm$)

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