Physics - Geometrical Optics Question with Solution | TestHub

For the first condition$$\frac{1}{f_1}=\frac{1}{v}-\frac{1}{u}\Rightarrow \frac{1}{10}=\frac{1}{v}+\frac{1}{30}$$ $\frac{1}{v}=\frac{1}{10}-\frac{1}{30}$ $v=\frac{30}{2}=15\mathrm{cm}$ $$\begin{array}{rl}\text{ where }{f}_1=10\mathrm{cm}& \begin{array}{lc}u=-30\mathrm{cm}& \\ v& =?\end{array}\end{array}$$ For the first conditionFor the second condition when concave lens is placed $$\begin{array}{l}{v}^{\prime }=(15+45)\mathrm{cm}=60\mathrm{cm}\\ \frac{1}{F}=\frac{1}{v^{\prime }}-\frac{1}{u}\end{array}$$ (where $F=$ focal length of combination) $\therefore$ $$\begin{array}{l}\frac{1}{F}=\frac{1}{60}+\frac{1}{30}\\ F=\frac{60}{3}\mathrm{cm}=20\mathrm{cm}\end{array}$$ The magnitude of focal length of concave lens $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}\Rightarrow \frac{1}{20}=\frac{1}{10}+\frac{1}{f_2}\Rightarrow f_2=-20\mathrm{cm}$ (Negative sign for concave lens)