An object 2.5 , cm high is placed at a distan | vedclass

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(D) Given: Object height $O = +2.5 \, cm$,object distance $u = -10 \, cm$,and radius of curvature $R = -30 \, cm$ (for a concave mirror).Focal length

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(D) Given: Object height $O = +2.5 \, cm$,object distance $u = -10 \, cm$,and radius of curvature $R = -30 \, cm$ (for a concave mirror).Focal length $f = R/2 = -30/2 = -15 \, cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Substituting the values: $\frac{1}{-15} = \frac{1}{v} + \frac{1}{-10}$.$\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$.So,$v = +30 \, cm$.Magnification $m = \frac{I}{O} = -\frac{v}{u}$.$\frac{I}{2.5} = -\frac{30}{-10} = 3$.$I = 3 	imes 2.5 = 7.5 \, cm$.

An object $2.5 \, cm$ high is placed at a distance of $10 \, cm$ from a concave mirror of radius of curvature $30 \, cm$. The size of the image is ........ $cm$.

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