An object 3 , cm high is placed at a distance | vedclass

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Question

(A) Given: Height of object $O = 3 \, cm$,Object distance $u = -20 \, cm$,Focal length $f = +12 \, cm$.Using the lens formula $\frac{1}{v} - \frac{1}{

Vedclass · Accepted Answer

Position: $30 \, cm$ behind the lens,Nature: Real and Inverted,Size: $4.5 \, cm$

Vedclass · Answer

(A) Given: Height of object $O = 3 \, cm$,Object distance $u = -20 \, cm$,Focal length $f = +12 \, cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have:$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{12} + \frac{1}{-20} = \frac{5 - 3}{60} = \frac{2}{60} = \frac{1}{30}$.Therefore,$v = +30 \, cm$. The positive sign indicates the image is formed on the other side of the lens.Using magnification $m = \frac{I}{O} = \frac{v}{u}$,we have:$\frac{I}{3} = \frac{30}{-20} = -1.5$.$I = 3 	imes (-1.5) = -4.5 \, cm$.The image is real,inverted,and $4.5 \, cm$ in size.

An object $3 \, cm$ high is placed at a distance of $20 \, cm$ in front of a convex lens of focal length $12 \, cm$. Find the position,nature,and size of the image formed.

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