Past exam of the physics 2 paper of the University of Oxford / 2022 / 11 / c / Solution

In surrounding refractive index $n^{\prime }$, the geometrical segments contribute $n^{\prime }[\sqrt{a^2+d^2}+\sqrt{b^2+d^2}]$ to the optical path length, while replacement by the lens adds $(n-n^{\prime })\ell (d)$. Fermat's principle therefore givesThe physical focal length measured in the surrounding medium isFor $\alpha >0$ and $n>1$, increasing $n^{\prime }$ from $1$ towards $n$ makes the positive focal length grow monotonically from its vacuum value to infinity: $df/d{n}^{\prime }=n/[2\alpha (n-n^{\prime }{)}^2]>0$. At $n^{\prime }=n$, there is no refraction at the boundary, so the element has zero optical power and no finite optical focus. If $n^{\prime }>n$, its optical power changes sign and the same shape becomes diverging.