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(B) Given: Radius of curvature $R = 22 \; cm$.For a convex mirror,the focal length $f$ is positive.$f = \frac{R}{2} = \frac{22}{2} = 11 \; cm$.The obj

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$6.2 \; cm$ on the back side of the mirror.

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(B) Given: Radius of curvature $R = 22 \; cm$.For a convex mirror,the focal length $f$ is positive.$f = \frac{R}{2} = \frac{22}{2} = 11 \; cm$.The object distance $u$ is always taken as negative in the sign convention: $u = -14 \; cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Substituting the values: $\frac{1}{11} = \frac{1}{v} + \frac{1}{-14}$.Rearranging for $v$: $\frac{1}{v} = \frac{1}{11} + \frac{1}{14}$.$\frac{1}{v} = \frac{14 + 11}{154} = \frac{25}{154}$.$v = \frac{154}{25} = 6.16 \; cm \approx 6.2 \; cm$.Since $v$ is positive,the image is formed at $6.2 \; cm$ on the back side of the mirror.

$A$ convex mirror has a radius of curvature of $22 \; cm$. If an object is placed $14 \; cm$ away from the mirror,then its image is formed at:

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