- #1

says

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## Homework Statement

From the Fresnel equations and Snell’s Law, prove that, when θ = θ

_{B}where tanθ

_{B}=

*n*/

_{t}*n*, (θ

_{i}_{B}is the Brewster angle); (a) Reflection coefficient = 0 , and (b) transmission coefficient =

*n*/

*n’*

## Homework Equations

reflection coefficient = (n

_{t}cosθ

_{i}- n

_{i}cosθ

_{t}) / (n

_{t}cosθ

_{i}+ n

_{i}cosθ

_{t})

transmission coefficient = (2n

_{i}cosθ

_{i}) / (n

_{t}cosθ

_{i}+n

_{i}cosθ

_{t})

Snell's Law : n

_{i}sinθ

_{i}= n

_{t}sinθ

_{t}

Brewster's Angle: θ

_{i}+ θ

_{t}= 90°

## The Attempt at a Solution

I know that for Brewster's Angle θ

_{i}+ θ

_{t}= 90°

Part a) is basically asking to derive the current version of the reflection coefficient, into it's more known tangent version:

tan(θ

_{i}- θ

_{t}) / tan(θ

_{i}+ θ

_{t})

For Brewster's Angle tan(θ

_{i}+ θ

_{t}) = tan(90°) = 0

So reflection coefficient will equal 0.

Rearranging Snell's Law: n

_{i}= n

_{t}sinθ

_{t}/ sinθ

_{i}

Subbing n

_{i}into the reflection coefficient and multiplying both numerator and denominator by sinθ

_{i}/ n

_{t}gives:

(sinθ

_{t}cosθ

_{t}- sinθ

_{i}cosθ

_{i}) / (sinθ

_{t}cosθ

_{t}+ sinθ

_{i}cosθ

_{i})

I don't know how to get from there to tan though. Any help would be much appreciated.