From the Fresnel equations and Snell’s Law, prove that, when θ = θB where tanθB = nt/ni, (θB is the Brewster angle); (a) Reflection coefficient = 0 , and (b) transmission coefficient = n/n’
reflection coefficient = (ntcosθi - nicosθt) / (ntcosθi + nicosθt)
transmission coefficient = (2nicosθi) / (ntcosθi+nicosθt)
Snell's Law : nisinθi = ntsinθt
Brewster's Angle: θi + θt = 90°
The Attempt at a Solution
I know that for Brewster's Angle θi + θt = 90°
Part a) is basically asking to derive the current version of the reflection coefficient, into it's more known tangent version:
tan(θi - θt) / tan(θi + θt)
For Brewster's Angle tan(θi + θt) = tan(90°) = 0
So reflection coefficient will equal 0.
Rearranging Snell's Law: ni= ntsinθt / sinθi
Subbing ni into the reflection coefficient and multiplying both numerator and denominator by sinθi / nt gives:
(sinθtcosθt - sinθicosθi) / (sinθtcosθt + sinθicosθi)
I don't know how to get from there to tan though. Any help would be much appreciated.