Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Section B / 9 / c / Solution

Use positive object distance $u=20\mathrm{cm}$ and focal length $f=4\mathrm{cm}$. The thin lens equation givesThe transverse magnification is $M=-v/u=-1/4$, so the signed height isThe real image lies $5.0\mathrm{cm}$ beyond the lens, is inverted, and has positive height magnitude $0.50\mathrm{cm}$.