Fresnel equation and Snell's law

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SUMMARY

The discussion focuses on demonstrating how the Fresnel equation (Eq 1) can be expressed in terms of Snell's law and the tangent function (Eq 2). The equations involved are Eq 1 = (ncosθ - n'cosθ') / (ncosθ + n'cosθ) and Eq 2 = (tanθ' - tanθ) / (tanθ' + tanθ). The participant successfully derives a relationship between the two equations by substituting n' with nsinθ / sinθ' and manipulating the expressions using trigonometric identities. The key insight is the division by cosθ' * cosθ to simplify the equation.

PREREQUISITES
  • Understanding of Snell's law (nsinθ = n'sinθ')
  • Familiarity with trigonometric identities (tanθ = sinθ/cosθ)
  • Basic algebraic manipulation skills
  • Knowledge of the Fresnel equations
NEXT STEPS
  • Study the derivation of the Fresnel equations in optics
  • Explore advanced applications of Snell's law in wave optics
  • Learn about the implications of the Fresnel equations on reflection and refraction
  • Investigate the relationship between trigonometric functions and wave behavior in different media
USEFUL FOR

Students of physics, particularly those studying optics, as well as educators looking for clear explanations of the Fresnel equations and Snell's law applications.

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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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