A concave lens and a convex lens are arranged | vedclass

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(C) For the concave lens:$u = -30 \ cm$,$f = -20 \ cm$Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} - \frac{1}{-30} =

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$29.2 \ cm$ to the right of concave lens

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(C) For the concave lens:$u = -30 \ cm$,$f = -20 \ cm$Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} - \frac{1}{-30} = \frac{1}{-20}$$\frac{1}{v} = -\frac{1}{20} - \frac{1}{30} = \frac{-3-2}{60} = -\frac{5}{60} = -\frac{1}{12}$$v = -12 \ cm$ (The image is formed $12 \ cm$ to the left of the concave lens).For the convex lens:The image formed by the concave lens acts as an object for the convex lens.The distance between the lenses is $5 \ cm$.$u = -(12 + 5) = -17 \ cm$,$f = +10 \ cm$Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} - \frac{1}{-17} = \frac{1}{10}$$\frac{1}{v} = \frac{1}{10} - \frac{1}{17} = \frac{17 - 10}{170} = \frac{7}{170}$$v = \frac{170}{7} \approx 24.29 \ cm$ to the right of the convex lens.Since the convex lens is $5 \ cm$ to the right of the concave lens,the final image position relative to the concave lens is $24.29 + 5 = 29.29 \ cm$ to the right of the concave lens.

$A$ concave lens and a convex lens are arranged as shown in the figure. Find the position of the final image.

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