Find the position of final image from first lens. Given focal length of each lens is 10 cm .

(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 $\mathrm{L}_1=30+(30-10)=50 \mathrm{~cm}$.