Past exam of the physics 2 paper of the University of Oxford / 2022 / 6

The distance from the origin to the incident straight light ray is $d_a$ (taking $d_a\ge 0$). At its intersection with the sphere, let $i$ and $t$ be the angle of incidence and angle of refraction, measured from the radial normal vector. Elementary right triangle geometry gives $d_a=r_b\sin i$. By Snell's law,The refracted straight light ray inside the sphere has distance $d_b=r_b\sin t$ from the origin. ThereforeThe angle ${\theta }_b$ describes the normal vector of the new line; it need not equal the propagation angle. The existence of a crossing requires $|d_a|<r_b$. A signed convention for $d_a$ gives the same line equation after choosing the orientation of its normal vector consistently.

Approximate the atmosphere by concentric spherical layers of refractive index. At every boundary, Snell's law conserves $nr\sin \psi$, where $\psi$ is the angle between the light ray and the radius. Straight propagation between boundaries conserves the same quantity because $r\sin \psi$ is the distance of that line from the origin. In the smooth limit this is the spherical ray invariantAt the surface, a light ray apparently on the horizon is tangent, so $p=n_{s}R$. In vacuum far from the Earth, $n=1$, and the straight incident line has distance $p$ from the centre. Removing the atmosphere while retaining that incident line gives minimum altitudeThe detailed variation of refractive index with altitude changes the local curvature of the light ray, but not the conserved $p$. This conclusion assumes a spherically stratified atmosphere and a light ray which actually connects the surface to space; nonspherical structure or a trapping layer would require separate analysis.