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(A) $1$. The convex lens forms an image of an object at infinity at its focal plane. Thus,the image formed by the convex lens acts as a virtual object

Vedclass · Accepted Answer

$2.5$

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(A) $1$. The convex lens forms an image of an object at infinity at its focal plane. Thus,the image formed by the convex lens acts as a virtual object for the concave lens.$2$. The distance of this image from the convex lens is $f_1 = 30 \ cm$. Since the concave lens is placed $26 \ cm$ in front of the convex lens,the distance of this image from the concave lens is $u = 30 - 26 = 4 \ cm$. Since it is behind the concave lens,$u = +4 \ cm$.$3$. For the concave lens,$f_2 = -20 \ cm$ and $u = +4 \ cm$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} - \frac{1}{4} = \frac{1}{-20} \implies \frac{1}{v} = \frac{1}{4} - \frac{1}{20} = \frac{5-1}{20} = \frac{4}{20} = \frac{1}{5}$. So,$v = 5 \ cm$.$4$. The magnification $m = \frac{v}{u} = \frac{5}{4} = 1.25$.$5$. The size of the final image $I = m 	imes O = 1.25 	imes 2 \ cm = 2.5 \ cm$.

$A$ convex lens of focal length $30 \ cm$ forms an image of height $2 \ cm$ for an object situated at infinity. If a concave lens of focal length $20 \ cm$ is placed coaxially at a distance of $26 \ cm$ in front of the convex lens,what will be the size of the final image in $cm$?

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