Geometric optics quest?

  • Thread starter Poirot
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  • #1
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Homework Statement


Background from previous parts of the question: A simple isotropic dielectric occupies the region x>0, with vacuum in region x<0. I've found the wave equations for the electric field Incident, reflected and transmitted to prove Snell's law (Sinθ/Sinθ = c/c' = √εr) and the law of refraction, but the last part, below, has me stumped.
By imposing the relevant boundary conditions on the electric field across the boundary, or otherwise, show that if there is no reflected wave then the angle of incidence (θ) and the angle of refraction (θ) satisfy
θ + θ = π/2
I know it's something to do with Brewster's Angle, but all the proofs I've found online are doing it in a different way to how we've been asked.

Homework Equations


EIn= EIn0 exp[iωt - iω/c (x cosθ + y sinθ)]
ERef= ERef0 exp[iωt - iω/c (-x cosθ' + y sinθ')]
ETrans= ETrans0 exp[iωt - iω/c' (x cosθ + y sinθ)]

The Attempt at a Solution


I tried looking at the boundary conditions as if there is no Reflected wave, and with dielectrics the parallel component of E is continuous across the boundary and so:
EIncosθ|x=0=ETranscosθ|x=0
And looking at when y=0 and t=0 separately and together, but I keep going round in circles with no solution in sight.

Any help would be greatly appreciated, thank you!
 

Answers and Replies

  • #2
blue_leaf77
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Which coordinates between x and y that you specify to lie along the interface?
 
  • #3
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The interface lies at x=0
 
  • #4
blue_leaf77
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EIncosθ|x=0=ETranscosθ|x=0
You can start from that equation. Note that both fields appearing in that equation are the components of the field which lies in the incident plane (from now on I will call it the TM component, as it is usually referred to in textbooks). To find the connection between EIn and ETrans, you have to know the transmission coefficient for the TM components.
 
  • #5
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Do you mean √εr ?
 
  • #6
blue_leaf77
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No, I am talking about Fresnel equation. Do you know these equations?
 
  • #7
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Umm, No. The extent of geometric optics my module has covered is this derivation.
 
  • #8
blue_leaf77
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this derivation.
The derivation of Fresnel equation?
 
  • #9
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No sorry, I meant we'd done the derivation of Snell's law and the law of reflection using this set up, I've not heard of Fresnel's Equations until I started googling about how to solve the problem I originally posted.
 
  • #10
blue_leaf77
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Actually I feel it strange when one brings up Brewster angle but never introduce the concept of reflection and transmission coefficients beforehand. The case of Brewster angle happens only when the incident light lies in the so-called incident plane, which is called TM polarization. But since you have found the solution, I think there is nothing more to argue about.
 
  • #11
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Actually I feel it strange when one brings up Brewster angle but never introduce the concept of reflection and transmission coefficients beforehand. The case of Brewster angle happens only when the incident light lies in the so-called incident plane, which is called TM polarization. But since you have found the solution, I think there is nothing more to argue about.
I haven't actually found a solution for this, this question came up in a past paper my lecturer set and said it had something to do with Brewster's angle but didn't elaborate and so I don't have any knowledge of this. In theory the solution shouldn't be particularly physics heavy, but I assume using some maths, due to the fact it's not been taught explicitly.

Thanks for trying to help
 
  • #12
blue_leaf77
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Well, I don't know other ways to solve this if one doesn't assume the knowledge of Fresnel equation. This equation governs the portion of light being reflected and transmitted at an interface. Since at Brewster angle, the TM polarized light will not be reflected/fully transmitted, obviously the relevant Fresnel equation must be involved.
 

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