A lateral object of height h_o = 0.5, cm is p | vedclass

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(C) Given: Focal length $f = +80\, cm$ (for a convex lens),object height $h_o = 0.5\, cm$,and object distance $u = -60\, cm$.Using the lens formula: $

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virtual,erect and $2\, cm$ high

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(C) Given: Focal length $f = +80\, cm$ (for a convex lens),object height $h_o = 0.5\, cm$,and object distance $u = -60\, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{80} = \frac{1}{v} - \frac{1}{-60} \implies \frac{1}{v} = \frac{1}{80} - \frac{1}{60}$.Calculating the common denominator: $\frac{1}{v} = \frac{3 - 4}{240} = -\frac{1}{240}$.Thus,$v = -240\, cm$. Since $v$ is negative,the image is virtual.Magnification $m = \frac{v}{u} = \frac{-240}{-60} = +4$.Image height $h_i = m 	imes h_o = 4 	imes 0.5\, cm = 2\, cm$.Since $m$ is positive,the image is erect. Therefore,the image is virtual,erect,and $2\, cm$ high.

$A$ lateral object of height $h_o = 0.5\, cm$ is placed on the optical axis of a bi-convex lens of focal length $f = 80\, cm$,at an object distance $u = -60\, cm$. The image formed is:

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