What is the position and nature of the image | vedclass

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(D) Step $1$: Image formation by the convex lens $(f_1 = +5 \, cm)$:Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$,where $u_1

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$70 \, cm$ from point $B$ at right; real

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(D) Step $1$: Image formation by the convex lens $(f_1 = +5 \, cm)$:Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$,where $u_1 = -20 \, cm$ and $f_1 = +5 \, cm$:$\frac{1}{v_1} - \frac{1}{-20} = \frac{1}{5} \implies \frac{1}{v_1} = \frac{1}{5} - \frac{1}{20} = \frac{4-1}{20} = \frac{3}{20}$.So,$v_1 = \frac{20}{3} \, cm$ to the right of lens $A$.Step $2$: Image formation by the concave lens $(f_2 = -5 \, cm)$:The image formed by the first lens acts as an object for the second lens. The distance between the lenses is $d = 2 \, cm$.The object distance for the second lens is $u_2 = +(v_1 - d) = +(\frac{20}{3} - 2) = +\frac{14}{3} \, cm$ (virtual object).Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:$\frac{1}{v_2} - \frac{1}{14/3} = \frac{1}{-5} \implies \frac{1}{v_2} = \frac{3}{14} - \frac{1}{5} = \frac{15 - 14}{70} = \frac{1}{70}$.Thus,$v_2 = +70 \, cm$.Since $v_2$ is positive,the final image is formed $70 \, cm$ to the right of point $B$ and is real.

What is the position and nature of the image formed by the lens combination shown in the figure? ($f_1, f_2$ are focal lengths)

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