An object is placed 15, cm from a convex mirr | vedclass

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(A) Given: Object distance $u = -15\, cm$,Radius of curvature $R = +60\, cm$.Focal length $f = R / 2 = +60 / 2 = +30\, cm$.Using the mirror formula: $

Vedclass · Accepted Answer

$v = 7.5\, cm, m = 0.5$

Vedclass · Answer

(A) Given: Object distance $u = -15\, cm$,Radius of curvature $R = +60\, cm$.Focal length $f = R / 2 = +60 / 2 = +30\, cm$.Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Substituting the values: $\frac{1}{v} + \frac{1}{-15} = \frac{1}{30}$.$\frac{1}{v} = \frac{1}{30} + \frac{1}{15} = \frac{1 + 2}{30} = \frac{3}{30} = \frac{1}{10}$.Therefore,$v = +10\, cm$.The image is formed $10\, cm$ behind the mirror.Magnification $m = -\frac{v}{u} = -\frac{10}{-15} = \frac{2}{3} \approx 0.67$.The image is virtual,erect,and diminished.

An object is placed $15\, cm$ from a convex mirror of radius of curvature $60\, cm$. Find the position of the image and its magnification.

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