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(C) Given: Height of object $h_1 = 5 \, cm$,Object distance $u = -100 \, cm$ $(1 \, m = 100 \, cm)$,Radius of curvature $R = -20 \, cm$.Focal length $

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$0.55$

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(C) Given: Height of object $h_1 = 5 \, cm$,Object distance $u = -100 \, cm$ $(1 \, m = 100 \, cm)$,Radius of curvature $R = -20 \, cm$.Focal length $f = R/2 = -20/2 = -10 \, cm$.Using the magnification formula for a mirror: $m = \frac{h_2}{h_1} = -\frac{v}{u}$.Also,$m = \frac{f}{f - u}$.Substituting the values: $\frac{h_2}{5} = \frac{-10}{-10 - (-100)} = \frac{-10}{-10 + 100} = \frac{-10}{90} = -\frac{1}{9}$.$h_2 = 5 	imes (-\frac{1}{9}) = -\frac{5}{9} \approx -0.55 \, cm$.The magnitude of the height of the image is $0.55 \, cm$.

An object of height $5 \, cm$ is placed at a distance of $1 \, m$ from a concave mirror having a radius of curvature of $20 \, cm$. The height of the image is ....... $cm$.

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