An object 1 , cm tall is placed 4 , cm in fro | vedclass

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(B) Given: Object height $h_o = 1 \, cm$,Image height $h_i = 3 \, cm$,Object distance $u = -4 \, cm$.Since the image is upright,the magnification $m =

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concave mirror of radius of curvature $12 \, cm$.

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(B) Given: Object height $h_o = 1 \, cm$,Image height $h_i = 3 \, cm$,Object distance $u = -4 \, cm$.Since the image is upright,the magnification $m = \frac{h_i}{h_o} = \frac{+3}{+1} = +3$.Using the magnification formula for a spherical mirror: $m = \frac{f}{f - u}$.Substituting the values: $3 = \frac{f}{f - (-4)} = \frac{f}{f + 4}$.$3(f + 4) = f \Rightarrow 3f + 12 = f \Rightarrow 2f = -12 \Rightarrow f = -6 \, cm$.$A$ negative focal length indicates a concave mirror.The radius of curvature $R = 2|f| = 2 	imes 6 = 12 \, cm$.Thus,a concave mirror of radius of curvature $12 \, cm$ is required.

An object $1 \, cm$ tall is placed $4 \, cm$ in front of a mirror. In order to produce an upright image of $3 \, cm$ height,one needs a:

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