Fresnel equation and Snell's law

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Homework Help Overview

The discussion revolves around demonstrating the relationship between two equations using Snell's law in the context of optics, specifically focusing on the Fresnel equations. The original poster presents two equations that need to be shown as equivalent through manipulation and substitution.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the manipulation of the given equations, particularly focusing on substitutions and algebraic transformations. There is a discussion about dividing terms to simplify the expression and how to relate the trigonometric functions involved.

Discussion Status

The discussion is ongoing, with participants providing suggestions on how to approach the problem. Some have offered specific algebraic manipulations, while others are questioning the steps taken and suggesting alternative ways to view the problem. There is no explicit consensus yet, but productive lines of reasoning are being explored.

Contextual Notes

Participants are working under the constraints of using Snell's law and the definitions of trigonometric functions. There is an emphasis on ensuring that the manipulations adhere to the relationships defined by the equations presented.

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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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