Fresnel Equations and Snells Law

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    Fresnel Law
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SUMMARY

The discussion focuses on the relationship between the Fresnel Equations and Snell's Law in optics. The specific Fresnel Equation presented is \(\frac{n_t \cos(\theta_i) - n_i \cos(\theta_t)}{n_t \cos(\theta_i) + n_i \cos(\theta_t)}\), which is connected to the equation \(\frac{\tan(\theta_i - \theta_t)}{\tan(\theta_i + \theta_t)}\) through trigonometric manipulation. A participant successfully demonstrated the reverse derivation by expanding the tangent functions, confirming the mathematical connection between these two important optical principles.

PREREQUISITES
  • Understanding of Fresnel Equations
  • Knowledge of Snell's Law
  • Familiarity with trigonometric identities
  • Basic calculus for manipulating equations
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  • Study the derivation of the Fresnel Equations in detail
  • Learn about Snell's Law and its applications in optics
  • Explore trigonometric identities and their proofs
  • Investigate advanced optics concepts such as total internal reflection
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Students and professionals in physics, particularly those specializing in optics, as well as educators looking to clarify the relationship between Fresnel Equations and Snell's Law.

girlinphysics
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I have seen in many textbooks that the following Fresnel Equation [tex]\frac{n_t{cos(\theta_i)} - n_i{cos(\theta_t)}}{n_t{cos(\theta_i)} + n_i{cos(\theta_t)}}[/tex] and then they say that via Snells Law we get the following equation [tex]\frac{tan(\theta_i - \theta_t)}{tan(\theta_i + \theta_t)}[/tex]

I can see that some trig manipulation was done, and I know that [itex]{tan(\theta)} = \frac{cos(\theta)}{sin(\theta)}[/itex] but I cannot seen to get to that final result correct. If anyone is able to show me the maths behind it that would be great.
 
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When I first encountered this, it bothered me too because I couldn't connect them. But at last I did it with a trick. I started from the second one to reach the first one, which is pretty easy. And then the proof you want is the reverse of that. Just expand the tangents and the rest is almost obvious.
 

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