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(A) For the $1$st lens (convex,$f_1 = +10 \ cm$):$u_1 = -30 \ cm$Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:$\frac{1}{v_1}

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$30 \ cm$

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(A) For the $1$st lens (convex,$f_1 = +10 \ cm$):$u_1 = -30 \ cm$Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:$\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{10} \implies \frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$$v_1 = +15 \ cm$. This image is formed $15 \ cm$ to the right of the $1$st lens.For the $2$nd lens (concave,$f_2 = -10 \ cm$):The distance between the $1$st and $2$nd lens is $5 \ cm$. The image from the $1$st lens acts as an object for the $2$nd lens at a distance $u_2 = +(15 - 5) = +10 \ cm$ (virtual object).Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:$\frac{1}{v_2} - \frac{1}{10} = \frac{1}{-10} \implies \frac{1}{v_2} = 0 \implies v_2 = \infty$.For the $3$rd lens (convex,$f_3 = +30 \ cm$):The rays are parallel after the $2$nd lens,so the object for the $3$rd lens is at infinity $(u_3 = \infty)$.Using the lens formula $\frac{1}{v_3} - \frac{1}{u_3} = \frac{1}{f_3}$:$\frac{1}{v_3} - \frac{1}{\infty} = \frac{1}{30} \implies \frac{1}{v_3} = \frac{1}{30} \implies v_3 = +30 \ cm$.The final image is formed $30 \ cm$ to the right of the third lens.

Find the position of the final image with respect to the rightmost lens formed by the lens combination shown in the figure.

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