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(C) Given: Object height $O = 5 \ cm$,Object distance $u = -100 \ cm$,Radius of curvature $R = -20 \ cm$.Focal length $f = R/2 = -10 \ cm$.Using the m

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$0.55$

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(C) Given: Object height $O = 5 \ cm$,Object distance $u = -100 \ cm$,Radius of curvature $R = -20 \ cm$.Focal length $f = R/2 = -10 \ cm$.Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.$\frac{1}{v} + \frac{1}{-100} = \frac{1}{-10} \Rightarrow \frac{1}{v} = -\frac{1}{10} + \frac{1}{100} = \frac{-10+1}{100} = -\frac{9}{100}$.So,$v = -\frac{100}{9} \ cm$.Magnification $m = \frac{I}{O} = -\frac{v}{u}$.$I = O 	imes (-\frac{v}{u}) = 5 	imes (-\frac{-100/9}{-100}) = 5 	imes (-\frac{1}{9}) = -\frac{5}{9} \approx -0.55 \ cm$.The size of the image is $0.55 \ cm$.

An object $5 \ cm$ tall is placed $1 \ m$ from a concave spherical mirror which has a radius of curvature of $20 \ cm$. The size of the image is $...... \ cm$.

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