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(B) Given: Focal length $f = -20 \ cm$ (for a concave mirror),Object distance $u = -10 \ cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \f

Vedclass · Accepted Answer

Enlarged,upright,virtual

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(B) Given: Focal length $f = -20 \ cm$ (for a concave mirror),Object distance $u = -10 \ cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Substituting the values: $\frac{1}{-20} = \frac{1}{v} + \frac{1}{-10}$.$\frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{2-1}{20} = \frac{1}{20}$.So,$v = +20 \ cm$.The positive sign of $v$ indicates that the image is formed behind the mirror,which means it is virtual.The magnification $m = -\frac{v}{u} = -\frac{20}{-10} = +2$.Since $m$ is positive,the image is upright. Since $|m| > 1$,the image is enlarged.Therefore,the image is enlarged,upright,and virtual.

If an object is placed $10 \ cm$ in front of a concave mirror of focal length $20 \ cm,$ the image will be

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