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(B) Given: Focal length $(f)$ = $-12 \ cm$ (for a concave mirror),Object distance $(u)$ = $-18 \ cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}

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Beyond the center of curvature

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(B) Given: Focal length $(f)$ = $-12 \ cm$ (for a concave mirror),Object distance $(u)$ = $-18 \ cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Substituting the values: $\frac{1}{-12} = \frac{1}{v} + \frac{1}{-18}$.Rearranging: $\frac{1}{v} = \frac{1}{18} - \frac{1}{12}$.Calculating the common denominator $(36)$: $\frac{1}{v} = \frac{2 - 3}{36} = -\frac{1}{36}$.Therefore,$v = -36 \ cm$.The center of curvature $(C)$ is at $2f = 2 	imes 12 = 24 \ cm$.Since the image distance $(36 \ cm)$ is greater than the center of curvature $(24 \ cm)$,the image is formed beyond the center of curvature.

An object is placed at a distance of $18 \ cm$ in front of a concave mirror of focal length $12 \ cm$. Where will the image be formed?

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