= Solution
Take the <thin lens> at $x=0$, with the object on the left and <light rays> travelling to the right. The positive object distance is $u=12\,\mathrm{cm}$ and the <focal length> is $f=8\,\mathrm{cm}$. The <thin lens> equation gives
$$
\frac1v=\frac1f-\frac1u=\left(\frac18-\frac1{12}\right)\mathrm{cm^{-1}}=\frac1{24}\,\mathrm{cm^{-1}},\qquad\boxed{v=24\,\mathrm{cm}.}
$$
The <transverse magnification> and signed image height are
$$
M=-\frac vu=-2,\qquad h_i=Mh_o=-2(5\,\mathrm{cm})=\boxed{-10\,\mathrm{cm}}.
$$
Thus the <real image> is \b[$24\,\mathrm{cm}$ beyond the lens, inverted and $10\,\mathrm{cm}$ high]. For the <ray tracing>, a <light ray> parallel to the <optical axis> emerges through the far <optical focus>; a <light ray> through the optical centre continues straight in the ideal <thin lens> model. They intersect at $(24,-10)\,\mathrm{cm}$, independently reading off both answers from the diagram.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022-q4-lens-rays.png]
{title=<Ray tracing> for the converging <thin lens>: object and image heights and both focal points}
The calculation and diagram use the ideal <paraxial approximation> implicit in the <thin lens> equation. The comparatively large object height does not supply enough information to calculate <optical aberrations> of a real lens.
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