Question

Sol. [Key Concept: Equivalent focal length of a lens combination]

For convex lens $L_{1}:u = - 30\text{ }cm$ and $f = + 10\text{ }cm$

$${\begin{matrix} & \text{~}\text{Using}\text{~}\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} + \frac{1}{30} = \frac{1}{10} \Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} \\ & \ \therefore v = + \frac{30}{2} = + 15\text{ }cm \end{matrix} }{ For\ concave\ lens\ L_{2}:u^{'} = + 12\text{ }cm}$$ and $f^{'} = - 10\text{ }cm$

$$\begin{matrix} & \text{~}\text{Using}\text{~}\frac{1}{v^{'}} - \frac{1}{u^{'}} = \frac{1}{f^{'}} \Rightarrow \frac{1}{v^{'}} - \frac{1}{12} = \frac{- 1}{10} \Rightarrow \frac{1}{v^{'}} = \frac{1}{12} - \frac{1}{10} = \frac{- 1}{60} \\ & \ \therefore v^{'} = - 60\text{ }cm \\ & \ \Rightarrow \text{~}\text{Final image formed at}\text{~}60\text{ }cm\text{~}\text{to the left of concave lens.}\text{~} \end{matrix}$$