JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Geometrical Optics

For a diverging lens the image of any real object is virtual, erect and nearer the lens than the object, so $|v|<|u|$ throughout and the image stays on the same side as the object. Writing $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with $f<0$, as the object recedes from the pole to infinity the image moves out only as far as the focus, $v\to f$, never beyond it. The curve that keeps the image distance smaller in magnitude than the object distance and approaches the focal length asymptotically is the one that represents this relation.