Refraction at a spherical interface

A spherical interface intersects the optical axis at $(x=0,y=0)$. The interface has a radius $R$ and is centered at $C$. For $R>0$ the interface is convex and for $R<0$ the interface is concave. The index of refraction is $n_1 = 1$ to the left of the interface and $n_2$ to the right of the interface. A light ray (red) leaves an object $o$ on the left of the interface and strikes the interface at point $P$. Part of the ray is reflected and part is refracted. The angles between the red rays and the gray normal to the interface obey Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$. The incident ray and the normal to the interface define a plane. The reflected ray and the refracted ray lie in this plane. For the drawings below, the $z-$components of the incident ray and the normal vector are set to zero so that the plane is always the $x-y$ plane.

Question

To calcuate the direction of the refracted ray, decompose the incident ray into a component parallel to the normal and a component perpendicular to the normal $\vec{oP} = \vec{oP}_{\parallel} + \vec{oP}_{\perp},$ 
$\vec{oP}_{\parallel} = (\hat{n}\cdot\vec{oP})\hat{n}$  and  $\vec{oP}_{\perp} = \vec{oP} - \vec{oP}_{\parallel}$, 
where $\hat{n} = \vec{CP}/R$ is the unit normal to the interface at the point where the ray hits the interface. The angle $\theta_1$ between the incident ray and the normal can be calcuated from

$$ \vec{oP}\cdot\vec{CP} = |\vec{oP}||\vec{CP}|\cos\theta_1,$$

and the angle between the refracted ray and the normal can be calculated by Snell's law, $n_1\sin\theta_1 = n_2\sin\theta_2$. The refracted ray will lie in the plane defined by $\vec{oP}_{\parallel}$ and $\vec{oP}_{\perp}$. Any vector in this plane could be written as $a\hat{e}_{\parallel}+b\hat{e}_{\perp}$ where $a$ and $b$ are constants and $\hat{e}_{\parallel}$, $\hat{e}_{\perp}$ are the unit vectors in the direction of $\vec{oP}_{\parallel}$ and $\vec{oP}_{\perp}$. Snell's law will be satisfied if $\frac{b}{a} = \tan\theta_2$. For convenience, we take $a=1$ and then $\hat{e}_{\parallel}+\tan\theta_2\hat{e}_{\perp}$ is in the direction of the refracted beam.






	Physik für Geodäsie 511.018 / Physik M 513.805