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(A) Given,radius of curvature $R = 12 \ cm$. The focal length $f = R / 2 = 6 \ cm$.For a concave mirror,$f = -6 \ cm$ and object distance $u = -18 \ c

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$1 / 2$

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(A) Given,radius of curvature $R = 12 \ cm$. The focal length $f = R / 2 = 6 \ cm$.For a concave mirror,$f = -6 \ cm$ and object distance $u = -18 \ cm$. Using the magnification formula $m = f / (f - u)$:$m_{concave} = -6 / (-6 - (-18)) = -6 / 12 = -1 / 2$. The size of the image is $|m_{concave}| = 1 / 2$.For a convex mirror,$f = +6 \ cm$ and object distance $u = -18 \ cm$. Using the magnification formula $m = f / (f - u)$:$m_{convex} = 6 / (6 - (-18)) = 6 / 24 = 1 / 4$. The size of the image is $|m_{convex}| = 1 / 4$.The ratio of the size of the image formed by the convex mirror to that formed by the concave mirror is:Ratio $= |m_{convex}| / |m_{concave}| = (1 / 4) / (1 / 2) = 1 / 2$.

Two identical objects are placed in front of a convex mirror and a concave mirror,both having a radius of curvature of $12 \ cm$,at the same distance of $18 \ cm$ from the respective mirrors. The ratio of the sizes of the images formed by the convex mirror to that formed by the concave mirror is:

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