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(C) For lens $1$: Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $f = +8 \ cm$ and $u = -12 \ cm$.$\frac{1}{8} = \frac{1}{v_1}

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$A$ real enlarged image

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(C) For lens $1$: Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $f = +8 \ cm$ and $u = -12 \ cm$.$\frac{1}{8} = \frac{1}{v_1} - \frac{1}{-12} \Rightarrow \frac{1}{v_1} = \frac{1}{8} - \frac{1}{12} = \frac{3-2}{24} = \frac{1}{24}$.So,$v_1 = 24 \ cm$. The image $A'B'$ is formed $24 \ cm$ to the right of the first lens.The distance of this image from the second lens is $d = 30 \ cm - 24 \ cm = 6 \ cm$.Since the focal length of the second lens is $f_2 = 6 \ cm$,the image $A'B'$ acts as an object placed exactly at the focus of the second lens.Therefore,the rays emerging from the second lens will be parallel,and the final image will be formed at infinity. In optical terms,an image at infinity is considered real and highly enlarged.

An object is placed $12 \ cm$ to the left of a converging lens of focal length $8 \ cm$. Another converging lens of $6 \ cm$ focal length is placed at a distance of $30 \ cm$ to the right of the first lens. The second lens will produce

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