JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Geometrical Optics

Start with the convex lens: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with $u=-30\text{cm}$ and $f=20\text{cm}$ gives $\frac{1}{v}=\frac{1}{20}-\frac{1}{30}$, so $v=+60\text{cm}$, the image forming $60\text{cm}$ to the right of the lens. The telling clue is that removing the mirror leaves the image fixed, which happens only when the light strikes the mirror normally and retraces its path, that is when the lens image falls at the mirror's centre of curvature. The mirror is $80\text{cm}$ from the lens, so its centre of curvature is at $80-60=20\text{cm}$, giving $R=20\text{cm}$ and mirror focal length $f_m=\frac{R}{2}=10\text{cm}$. A concave mirror forms a virtual, erect image only when the object lies between pole and focus, so the greatest object distance for which it still gives a virtual image is its focal length, $10\text{cm}$.