# Reverse Vector Snell's Problem

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.

• Geometric Algebra and dual quarternions Sep 25 '21 at 1:38
• have a look here : youtube.com/watch?v=ichOiuBoBoQ Sep 25 '21 at 1:38

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

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

this implies that

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

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 $$$$\mathbf{n}\boldsymbol{=}\dfrac{\mathbf{t}\boldsymbol{-}\mu\mathbf{i}}{\Vert\mathbf{t}\boldsymbol{-}\mu\mathbf{i}\Vert} \tag{03}\label{03}$$$$

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

• Welcome to PSE. Well done, PeterBaumgart, it was so simple. Sep 25 '21 at 7:24