= Solution
For $u=8\,\mathrm{cm}$ and negative <focal length> $f=-8\,\mathrm{cm}$,
$$
\frac1v=\frac1f-\frac1u=-\frac18-\frac18=-\frac14\,\mathrm{cm^{-1}},\qquad\boxed{v=-4.0\,\mathrm{cm}.}
$$
The <transverse magnification> is $M=-v/u=+1/2$, so
$$
\boxed{h_i=+1.0\,\mathrm{cm}.}
$$
Thus the <virtual image> is upright, half the object's height, and \b[$4.0\,\mathrm{cm}$ from the lens on the object's side]. A parallel incident <light ray> diverges after the lens as though it originated at the near <optical focus>; the central <light ray> continues undeviated in the ideal <thin lens> model. Their backward extensions meet at the image tip, $(-4,1)\,\mathrm{cm}$.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021-q9-diverging-rays.png]
{title=<Ray tracing> for the diverging <thin lens>: backward extensions locate the upright <virtual image>}
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