Fresnel equation and Snell's law

AI Thread Summary
The discussion focuses on deriving Eq 2 from Eq 1 using Snell's law. The initial equations are presented, with the relationship n'sinθ' = nsinθ guiding the transformation. The user attempts to manipulate Eq 1 by substituting for n' and simplifying the expression. They explore dividing the numerator and denominator by cosθ'cosθ to facilitate the conversion to Eq 2. The conversation emphasizes the importance of recognizing trigonometric identities to achieve the desired result.
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Homework Statement


Use Snell's law to show fresnel Eq 1 can be expressed as Eq 2

Eq 1 = (ncosθ-n'cosθ') / (ncosθ+n'cosθ)
Eq 2 = (tanθ' - tanθ) / (tanθ' + tanθ)

Homework Equations


nsinθ=n'sinθ' (Snell's law)

The Attempt at a Solution


n' = nsinθ / sinθ'

Substitute n' into equation and then multiply numerator and denominator by sinθ'/n gives:

(cosθsinθ'-cosθ'sinθ) / (cosθsinθ'+cosθ'sinθ)

I'm not sure how to get from here to Eq 2 though. I know sinθ/cosθ = tanθ and cosθ/sinθ=1/tanθ.
 
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says said:
(cosθsinθ'-cosθ'sinθ) / (cosθsinθ'+cosθ'sinθ)
Divide numerator and denominator by an appropriate expression.
 
cosθ?
 
Consider the first term in your numerator: cosθ⋅sinθ'

What would you have to divide this by to get the first term in the numerator of Eq 2?
 
cosθ'/cosθ
 
says said:
cosθ'/cosθ
Not quite.
 
cosθ⋅sinθ'*1/cosθ'*cosθ=tanθ'
 
says said:
cosθ⋅sinθ'*1/cosθ'*cosθ=tanθ'
OK. So you divided by cosθ'*cosθ (not cosθ'/cosθ).

Good. This suggests seeing what happens if you divide the top and bottom of (cosθsinθ'-cosθ'sinθ) / (cosθsinθ'+cosθ'sinθ) by cosθ'*cosθ.
 
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