Past exam of the physics 2 paper of the University of Oxford / 2016 / 10 / c / Solution

A real image is a point where the actual outgoing rays converge; a screen placed there can intercept the focused light. A virtual image is a point from which the outgoing rays appear to diverge: only their backward extensions intersect, so a screen at that apparent point does not receive a focused real image from the same rays.

For a converging thin lens with $f>0$ and a real object a distance $u>0$ away, the lens relation gives $v=fu/(u-f)$. If $u>f$, then $v>0$: the image is real, on the opposite side, and inverted. If $0<u<f$, then $v<0$: the image is virtual, on the same side as the object, upright and enlarged, with $m=-v/u=f/(f-u)>1$.

At $u=f$, the emergent rays are parallel and the image is at infinity; there is no finite focused image. For an object very far away, $v\to f$ from above. These are conditions for a real object in the stated lens convention, rather than claims about all systems with virtual incident objects.