Move an object in front of a thin converging or diverging lens and read off the image. The diagram uses the thin-lens equation 1/f = 1/u + 1/v and magnification m = −v/u. A positive image distance means a real image on the far side of the lens. A negative image distance means a virtual image on the same side as the object. A negative magnification means the image is inverted.
A converging lens brings rays parallel to the principal axis to a real principal focus. A diverging lens spreads them so they appear to come from a virtual focus. In the simulation a converging lens has a positive focal length and a diverging lens has a negative one.
Object beyond 2f: real, inverted, diminished image between f and 2f. Object at 2f: real, inverted image of the same size at 2f. Object between f and 2f: real, inverted, magnified image beyond 2f. Object inside f: virtual, upright, magnified image. A real image can be formed on a screen; a virtual image cannot.
For a real object, a diverging lens always produces a virtual, upright, diminished image on the same side as the object. The image sits between the lens and the virtual focus. That is why the image distance stays negative when you select Diverging.
The simulation reports m = −v/u. The size of m is image height divided by object height. The sign tells you the orientation: negative means inverted, positive means upright. GCSE often quotes the size only; A level and IB keep the sign.
GCSE and IGCSE ray diagrams cover the converging-lens cases and the diverging lens as a qualitative result. A level and IB use the lens equation and the sign convention to calculate v and m. The level switch on the page reveals the calculated values.
The lens is treated as thin, and only paraxial rays are drawn. Thickness, spherical aberration and the way different colours refract by different amounts are not included. A real lens only matches these numbers well for rays close to the axis.