A converging lens of focal length 20, cm and | vedclass

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(B) Given: Focal length $f = +20\, cm$,Object distance $u = -30\, cm$.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we get $\frac{

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$1\, cm$ below $XY$

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(B) Given: Focal length $f = +20\, cm$,Object distance $u = -30\, cm$.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v} = \frac{1}{20} + \frac{1}{-30} = \frac{3-2}{60} = \frac{1}{60}$.So,$v = +60\, cm$.The magnification $m = \frac{v}{u} = \frac{60}{-30} = -2$.Let the object be at a distance $h_O = 0.5\, cm$ above the principal axis (line $XY$).The height of the image $h_I = m 	imes h_O = -2 	imes 0.5\, cm = -1\, cm$.The negative sign indicates that the image is formed below the principal axis $XY$.Thus,the image is formed $1\, cm$ below the line $XY$.

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