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(D) Given: Object height $h_o = 2 \ mm = 2 	imes 10^{-3} \ m$. Radius of curvature $R = +40 \ cm$ (for convex mirror). Focal length $f = R/2 = +20 \

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(D) Given: Object height $h_o = 2 \ mm = 2 	imes 10^{-3} \ m$. Radius of curvature $R = +40 \ cm$ (for convex mirror). Focal length $f = R/2 = +20 \ cm$. Object distance $u = -20 \ cm$. Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{20} = \frac{1}{20}$. $\frac{1}{v} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \Rightarrow v = +10 \ cm$. Magnification $m = -\frac{v}{u} = \frac{h_i}{h_o}$. $m = -\frac{10}{-20} = +0.5$. $h_i = m 	imes h_o = 0.5 	imes 2 \ mm = 1 \ mm$.

What will be the height of the image when an object of height $2 \ mm$ is placed on the principal axis of a convex mirror (of radius of curvature $40 \ cm$) at a distance of $20 \ cm$ from the mirror?

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