Snell's law can be derived from Fermat's principle, which states that the light travels the path which takes the least time. It finds the path by setting dT/dx=0 to arrive at n_1 sin θ_1= n_2 sin θ_2. Since there is this one equation with two variables: θ_1 and θ_2, solutions are infinite where θ_2=〖sin〗^(-1) (n1/n2)sinθ_1. Assuming there are three light waves travel from air n_1 = 1 to glass n_2 = 1.5 between two fixed endpoints (a_1, b_2) and (a_1, b_2) with the following (θ_1, θ_2) solution pairs: (30°, 19.5°), (45°, 28.1°), (60°, 35.3°) as shown below.
Only W2 can end at point (a_1,b_2)? I know that this would work if we don't fix the endpoints and make W1 and W3 pass through W2's incident point. But just wonder how to interpret the least time path derivation.
[Edited] Thanks for showing me the link. From there, I was able to figure out my mistake. The Snell's law states the refraction angle relationship to medium indices at the incident point x. Fermat's Principle proves it by finding the minimum of total travel time T(x) by setting dT/dx = 0 where x can be expressed in terms of the begin and end points (a1, b1) and (a2, b2) as: (x - a1)/sqrt[(x - a1)^2 + b1^2] = (a2 - x)/sqrt[(a2 - x)^2 + b2^2] (n2/n1) which reduces to Snell's law. So we can solve x in terms of the positional values and it is UNIQUE. That's why in my graph, only W2 hits the target (a2, b2), not any other waves. Solving x from the above equation is numerically harder than Snell's law. The former is asking, given the begin and end points and without knowing θ1 or θ2, what is x, and the latter is asking, given an incident angle θ1, what is θ2.
