JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Geometrical Optics

Take the converging lens first with $f_1=5\text{cm}$ and the object at $u=-20\text{cm}$. From $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ we get $\frac{1}{v}=\frac{1}{5}-\frac{1}{20}=\frac{3}{20}$, so $v=\frac{20}{3}\text{cm}$ to the right of the first lens. This image serves as the object for the diverging lens placed $2\text{cm}$ away, so its object distance is $\frac{20}{3}-2=\frac{14}{3}\text{cm}$ on the far side, a virtual object. For the diverging lens with $f_2=-5\text{cm}$, $\frac{1}{v}=-\frac{1}{5}+\frac{3}{14}=\frac{1}{70}$, so $v=70\text{cm}$. The final image forms $70\text{cm}$ to the right of point $B$ and is real.