A 5 , cm tall object is placed perpendicular | vedclass

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Question

(A) Given: Object height $O = 5 \, cm$,Object distance $u = -30 \, cm$,Focal length $f = +20 \, cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u}

Vedclass · Accepted Answer

$v = 60 \, cm$,Real and Inverted,$10 \, cm$

Vedclass · Answer

(A) Given: Object height $O = 5 \, cm$,Object distance $u = -30 \, cm$,Focal length $f = +20 \, cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-30} = \frac{3-2}{60} = \frac{1}{60}$.Therefore,$v = 60 \, cm$. The positive sign indicates the image is formed on the other side of the lens.Using magnification formula $m = \frac{I}{O} = \frac{v}{u}$:$m = \frac{I}{5} = \frac{60}{-30} = -2$.$I = 5 	imes (-2) = -10 \, cm$.$(i)$ Position: $60 \, cm$ from the lens on the opposite side.(ii) Nature: Real and inverted (since $v$ is positive and $I$ is negative).(iii) Size: $10 \, cm$ tall.

$A$ $5 \, cm$ tall object is placed perpendicular to the principal axis of a convex lens of focal length $20 \, cm$. The distance of the object from the lens is $30 \, cm$. Find the $(i)$ position,$(ii)$ nature,and $(iii)$ size of the image formed.

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