An object is placed at a distance of 10 , cm | vedclass

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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} +

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Magnified,erect,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,$v = +20 \, cm$.Since $v$ is positive,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 erect.Since $|m| > 1$,the image is magnified.Therefore,the image is magnified,erect,and virtual.

An object is placed at a distance of $10 \, cm$ from a concave mirror of focal length $20 \, cm$. The image formed will be .......

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