What will be the height of the image (in mm) | vedclass

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(D) Given: Object height $O = 2 \, mm$,Object distance $u = -20 \, cm$,Radius of curvature $R = +40 \, cm$ (for a convex mirror).Focal length $f = R/2

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$1$

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

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

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