Find the position of final image from first lens. Given focal...

(i) For refraction from lst lens, $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{~V}}-\frac{1}{\mathrm{u}}$$\begin{aligned} & \Rightarrow \frac{1}{\mathrm{~V}}=\frac{1}{\mathrm{f}}+\frac{\mathrm{l}}{\mathrm{u}}=\frac{1}{10}-\frac{1}{40}=\frac{3}{40} \mathrm{~cm} \\ & \Rightarrow \mathrm{v}=\frac{40}{3} \mathrm{~cm} \end{aligned}$This image acts as object for 2nd lens. $\therefore$ Object distance for 2 nd lens $\mathrm{L}_2=30-\frac{40}{3}=\frac{50}{3} \mathrm{~cm}$ (ii) Now for refraction from $L_2, \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\begin{aligned} & \Rightarrow \frac{1}{v}=\frac{1}{u}+\frac{1}{\mathrm{f}}=\frac{1}{10}-\frac{3}{50}=\frac{2}{50} \\ & \Rightarrow \mathrm{v}=25 \mathrm{~cm} \end{aligned}$$$This image is object for 3rd lens,$\text{ ( } \mathrm{v}=30-25=5 \mathrm{~cm} \text{ ) }\Rightarrow$For$\mathrm{L}_3, \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}=\frac{1}{10}-\frac{1}{5} or \mathrm{v}=-10 \mathrm{~cm}$(location of final image$=10 \mathrm{~cm}$to left of$\mathrm{L}_3$)$\therefore$Distance from$\mathrm{L}_1=30+(30-10)=50 \mathrm{~cm}$.