Background from previous parts of the question: A simple isotropic dielectric occupies the region x>0, with vacuum in region x<0. I've found the wave equations for the electric field Incident, reflected and transmitted to prove Snell's law (Sinθ/Sinθ = c/c' = √εr) and the law of refraction, but the last part, below, has me stumped.
By imposing the relevant boundary conditions on the electric field across the boundary, or otherwise, show that if there is no reflected wave then the angle of incidence (θ) and the angle of refraction (θ) satisfy
θ + θ = π/2
I know it's something to do with Brewster's Angle, but all the proofs I've found online are doing it in a different way to how we've been asked.
EIn= EIn0 exp[iωt - iω/c (x cosθ + y sinθ)]
ERef= ERef0 exp[iωt - iω/c (-x cosθ' + y sinθ')]
ETrans= ETrans0 exp[iωt - iω/c' (x cosθ + y sinθ)]
The Attempt at a Solution
I tried looking at the boundary conditions as if there is no Reflected wave, and with dielectrics the parallel component of E is continuous across the boundary and so:
And looking at when y=0 and t=0 separately and together, but I keep going round in circles with no solution in sight.
Any help would be greatly appreciated, thank you!