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Derivation of Fresnel equations / minus sign

  1. Aug 6, 2015 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    For an s-polarized wave (E and B fields are orthogonal to the plane of incidence) passing from medium to medium 2, I'm not understanding a minus sign.
    The matching conditions are ##\hat n \times (\vec E_2 - \vec E_1)=\vec 0## and ##\hat n \times (\vec H_2 - \vec H_1 )=\vec 0##, where ##\vec E_1 = \vec E_R + \vec E_I## and ##\vec E_2 = \vec E_T##. So far so good: the transmitted E field is equal to the incident one plus the reflected one, at the interface between the 2 media. The same applies for the H field.

    2. Relevant equations
    Matching conditions.

    3. The attempt at a solution
    The idea is that ##\vec E_I## is known and so the 2 matching conditions should allow us to solve for ##\vec E_R## and ##\vec E_T##. We know the relationship between H and E: ##Z_i\vec H_i = \hat k \times \vec E_i##, where Z is the impedance of the medium. So far so good.
    Soon comes the problem: using this last relation, I obtain that ##\hat k \times (\vec E_I + \vec E_R)Z_2=\hat k \times \vec E_T Z_1##. So that in terms of magnitude, ##(E_I -E_R)Z_2=E_TZ_1##. That's because ##\vec E_I## and ##\vec E_R## points in opposite directions.
    However if I use this same argument then the 1st matching condition becomes ##(E_I-E_R)=E_T## which is wrong. It should be ##(E_I+E_R)=E_T##. I don't understand why.
     
  2. jcsd
  3. Aug 6, 2015 #2

    RUber

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    Let's say you have an incident field in the x-y plane (transverse to z) defined by:
    ##\vec{E_I} = \hat z E_Ie^{-i(x\cos\theta + y \sin \theta) } ##
    And it reflects off an interface at y = 0.
    Then,
    ##\vec{E_R} = -\hat z E_Re^{-i(x\cos\theta - y \sin \theta) } ##
    Where ##E_R = \Gamma E_I## and ##\Gamma>0##.
    In this way, your exterior field has magnitude ##E_I - E_R## with fields defined by ##\vec{E_I}+\vec{E_R}##.

    When you say "in terms of magnitude" you are implying that the scalar terms ##E_I, E_R## are both positive.
    When you use the vector notation, you can add opposing vectors, as in ##\vec{E_I}+\vec{E_R}##.

    Does that address your question, or is there something I am missing?
     
  4. Aug 6, 2015 #3

    fluidistic

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    Hmm where the time dependence? Also let's keep it simple and let's consider normal incidence so that ##\theta=0##, cos theta becomes 1 and sin theta becomes 0.
    As you said, in region 1 the E field is worth ##E_I-E_R## in magnitude, while from a vector point of view we have that ##\vec E=\vec E_I + \vec E_R##.
    So then why the matching condion ##\hat n \times (\vec E_2 - \vec E_1) =\vec 0 \Rightarrow \vec E_2 = \vec E_1 \Rightarrow \vec E_T= \vec E_I + \vec E_R## simplifies as ##E_I+E_R=E_T## from a magnitude point of view instead of ##E_I-E_R=E_T##?
     
  5. Aug 6, 2015 #4

    RUber

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    I have seen some notation that keeps the negative sign inside term ##E_R##, so that ##|E_R| = -E_R##.
    The important thing is that the energy in is equal to the energy out, so ##|\vec E_I| = |\vec E_R| + |\vec E_T| ##. As long as that requirement is satisfied, you can have faith in your notation.
     
  6. Aug 6, 2015 #5

    fluidistic

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    I am still completly lost. The book (and everywhere else) get ##E_I+E_R=E_T## and ##Z_2(E_I-E_R)=Z_1E_T##.
    I don't understand how come there is a plus in one equation but a minus in the other one.
     
  7. Aug 6, 2015 #6

    RUber

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    I did a little more digging and found a good reference:
    http://ocw.mit.edu/courses/electric...ring-2011/lecture-notes/MIT6_007S11_lec29.pdf
    This shows that
    ##E_R = rE_I, \quad E_T = t E_I ##
    which allows for ##E_R## to be negative, so not "magnitude" per se.
    The second condition is derived from the H field, which in terms of the E field has a differential relationship, so the reflected field will have a negative sign in front of it.
    In normal incidence, time-harmonic, scenario:
    ##\vec E_I = \hat x E_I e^{ -i k_1z} ##
    ##\vec E_R = \hat x rE_I e^{ i k_1z} ##
    ##\vec E_T = \hat x tE_I e^{ -i k_2z} ##
    And ##\vec H = \frac{1}{\mu} \nabla \times \vec E = \text{sign}(i) \hat y \frac{E}{Z}e^{ -i kz} ##
    The continuity condition for the H field is:
    ##\hat n \times(\vec H_1 - \vec H_2 ) = 0## Which leads directly to the condition you refer to with ##Z_2 (E_I - E_R) = Z_1E_T.##
     
  8. Aug 6, 2015 #7

    fluidistic

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    Ok I get it. The thing is that for s-polarized waves, ##\vec E_I## and ##\vec E_R## (and ##\vec E_T## for that matter) have the same direction, while ##\vec H_I## and ##\vec H_R## don't. Hence the conditions ##E_I+E_R=E_T## and ##(E_I -E_R)Z_2=E_TZ_1## so I was wrong in my 1st post to say that ##\vec E_I## and ##\vec E_R## point in opposite directions.
     
  9. Aug 6, 2015 #8

    RUber

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    Sorry I was on the wrong track at first. I was thrown off by the discussion of magnitude. Glad to get it squared away.
     
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