JEE Main · 2008 · 27 Apr 2008 · Offline · Geometrical Optics

For a convex lens the thin-lens relation $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ links object and image distances. With the real-is-positive convention for a real object and its real image it rearranges to $v=\frac{uf}{u-f}$, a rectangular hyperbola: as $u$ approaches $f$ the image distance runs off to infinity, and as $u$ grows large $v$ settles towards $f$. The curve has asymptotes $u=f$ and $v=f$ and is symmetric about $u=v$, where object and image are equidistant at $2f$. The graph showing this hyperbolic fall from large $v$ towards $v=f$ is the correct one.