Physics - Geometrical Optics Question with Solution | TestHub

From the lens formula (for first lens) $$\begin{array}{rl}\frac{1}{f_1}& =\frac{1}{v_1}-\frac{1}{u_1}\Rightarrow \frac{1}{2}=\frac{1}{v_1}-\frac{1}{(-4)}\\ \Rightarrow \frac{1}{2}+\frac{1}{4}& =\frac{1}{v_1}=\frac{3}{4}\\ \Rightarrow v_1& =\frac{4}{3},u_2=3-\frac{4}{3}=5/3\end{array}$$ This image will be treated as the source for second lens, then again from lens formula, we have $\frac{1}{f_2}=\frac{1}{v_2}-\frac{1}{u_2}4$ $\Rightarrow \frac{1}{1}=\frac{1}{v_2}+5/3$ $\Rightarrow 1-5/3=\frac{1}{v_2}\Rightarrow v_2=-3/2$ [By Eq. (i)] This is the final image distance from 2 nd lens. So, the overall distance of image from the primary source(or object) Let $d=4+3-1.5=5.5$