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(B) $1$. The convex lens forms an image $I_1$ at its focal point. Since the object is at infinity, the image $I_1$ is formed at $f = 30\,cm$ from the

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$2.5$

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(B) $1$. The convex lens forms an image $I_1$ at its focal point. Since the object is at infinity, the image $I_1$ is formed at $f = 30\,cm$ from the convex lens.$2$. The size of this image $I_1$ is $2\,cm$.$3$. A concave lens of focal length $f_2 = -20\,cm$ is placed at $26\,cm$ from the convex lens. The image $I_1$ acts as a virtual object for the concave lens.$4$. The distance of the virtual object from the concave lens is $u = 30\,cm - 26\,cm = 4\,cm$. Since it is on the right side, we take $u = +4\,cm$.$5$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} - \frac{1}{4} = \frac{1}{-20}$$\frac{1}{v} = \frac{1}{4} - \frac{1}{20} = \frac{5-1}{20} = \frac{4}{20} = \frac{1}{5}$$v = 5\,cm$.$6$. The magnification $m$ for the concave lens is $m = \frac{v}{u} = \frac{5}{4} = 1.25$.$7$. The new size of the image $I_2$ is $m 	imes (	ext{size of } I_1) = 1.25 	imes 2\,cm = 2.5\,cm$.

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