A convex lens and a concave lens, each with a | vedclass

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(C) For the convex lens: Object distance $u_1 = -8 \,cm$, focal length $f_1 = +4 \,cm$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{

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$18$

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(C) For the convex lens: Object distance $u_1 = -8 \,cm$, focal length $f_1 = +4 \,cm$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$, we get $\frac{1}{v_1} = \frac{1}{4} + \frac{1}{-8} = \frac{1}{8}$, so $v_1 = +8 \,cm$. This image acts as a virtual object for the concave lens.For the concave lens: The distance between the lenses is $d = 6 \,cm$. The image formed by the convex lens is $8 \,cm$ behind it, which is $8 - 6 = 2 \,cm$ behind the concave lens. Thus, for the concave lens, $u_2 = +2 \,cm$ and $f_2 = -4 \,cm$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$, we get $\frac{1}{v_2} = \frac{1}{-4} + \frac{1}{2} = \frac{1}{4}$, so $v_2 = +4 \,cm$. This means the final image is formed $4 \,cm$ to the right of the concave lens.The object is $8 \,cm$ to the left of the convex lens. The distance between the object and the convex lens is $8 \,cm$, the distance between the lenses is $6 \,cm$, and the distance between the concave lens and the final image is $4 \,cm$. The total distance between the object and the final image is $8 + 6 + 4 = 18 \,cm$.

$A$ convex lens and a concave lens, each with a focal length of $4 \,cm$, are separated by a distance of $6 \,cm$ along their axis. An object is placed $8 \,cm$ before the convex lens. The distance between the object and its final image is (in $\,cm$)

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