Physics - Geometrical Optics Question with Solution | TestHub

(i) For refraction from lst lens, $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{V}}-\frac{1}{\mathrm{u}}$ $\begin{array}{rl}& \Rightarrow \frac{1}{\mathrm{V}}=\frac{1}{\mathrm{f}}+\frac{\mathrm{l}}{\mathrm{u}}=\frac{1}{10}-\frac{1}{40}=\frac{3}{40}\mathrm{cm}\\ & \Rightarrow \mathrm{v}=\frac{40}{3}\mathrm{cm}\end{array}$ 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 $L_2,\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ $\begin{array}{rl}& \Rightarrow \frac{1}{v}=\frac{1}{u}+\frac{1}{\mathrm{f}}=\frac{1}{10}-\frac{3}{50}=\frac{2}{50}\\ & \Rightarrow \mathrm{v}=25\mathrm{cm}\end{array}$ 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 ${\mathrm{L}}_1=30+(30-10)=50\mathrm{cm}$.