Reverse Vector Snell's Problem

Question

Is there a simple way to reverse the Snell's-Equation-in-Vector-Form problem? I.e., given the incident and transmitted vectors $\,\mathbf{i}\,$ and $\,\mathbf{t}\,$ (both with norm = 1), find the normal vector of the boundary surface (between media with refr. indices n1 and n2, resp.) required to achieve the desired change of direction through refraction? I haven't found a simple solution, yet, after trying for an hour and a half. (I get a messy system of coupled non-linear equations.) The "forward" problem is discussed and solved here: Snell's law in vector form

This could be useful for finding and orienting a prism to bend a given ray into a desired direction.

Answer 1

I think I found the answer: Since Snell's law can be written as

\begin{equation} \left(\mathbf{n}\boldsymbol{\times}\mathbf{t}\right)=\mu\left(\mathbf{n}\boldsymbol{\times}\mathbf{i}\right) \tag{01}\label{01} \end{equation}

this implies that

\begin{equation} \mathbf{n}\boldsymbol{\times}\left(\mathbf{t}\boldsymbol{-}\mu\mathbf{i}\right)\boldsymbol{=0} \tag{02}\label{02} \end{equation}

which means $\,\mathbf{n}\,$ and $\,\left(\mathbf{t}\boldsymbol{-}\mu\mathbf{i}\right)\,$ must be parallel to each other (as neither will have length zero).

Therefore the desired normal vector $\,\mathbf{n}\,$ is given by \begin{equation} \mathbf{n}\boldsymbol{=}\dfrac{\mathbf{t}\boldsymbol{-}\mu\mathbf{i}}{\Vert\mathbf{t}\boldsymbol{-}\mu\mathbf{i}\Vert} \tag{03}\label{03} \end{equation}

Make sure that the dot product of $\,\mathbf{i}\,$ and $\,\mathbf{n}\,$ is positive, or you are violating the geometrical assumptions, see Figure below.