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

Write $T$ for transit time and $\ell (y)=t_0-\alpha y^2$ for lens thickness to avoid using $t$ for both. In the thin lens approximation, the two geometrical distances to the lens plane are $\sqrt{a^2+d^2}$ and $\sqrt{b^2+d^2}$. Their sum counts the lens segment as vacuum. Replacing that segment by material of refractive index $n$ adds $(n-1)\ell (d)$ to the optical path length. Thus, to the retained thin-element accuracy,Using the Taylor expansion $\sqrt{1+u}=1+u/2+O(u^2)$ for $d\ll a,b$ givesThe omitted geometrical terms include $O(d^4/a^3+d^4/b^3)$ in optical path length, together with finite-thickness corrections discarded by the thin lens and paraxial approximations. The addition is $n-1$, because the underlying distance already includes propagation across the thickness once in vacuum.

Fermat's principle makes the optical path length, or transit time in a stationary medium, stationary under small variations of the light ray with fixed endpoints. It need not be a global minimum. In the result of part (a), the coefficient of $d^2$ isFor a family of paraxial light rays from the object to its image, stationarity with respect to $d$ requires this coefficient to vanish. Therefore the thin lens equation isFor a distant object, $a\to \infty$, and the image position $b=f$ defines the focal length. HenceThe axial member $d=0$ is stationary by symmetry for arbitrary $b$; requiring the neighbouring non-axial light rays to meet the same image is what determines $b$.

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.

For the convex vacuum void, use internal refractive index $1$, external refractive index $n$, and the original $\alpha >0$ in the immersed thin lens formula:A concave glass thin lens of the same curvature magnitude in vacuum has ${\alpha }_{\mathrm{concave}}=-\alpha$ and thereforeBoth elements diverge, but their physical focal lengths differ by a factor $n$ for $n>1$. For example, $n=1.5$ makes the void's focal length magnitude $1.5$ times that of the concave glass element. Their reduced optical powers $n_{\mathrm{outside}}/f$ are equal; physical focal length also depends on the surrounding refractive index.