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Electromagnetic wave incidence on interface

  1. Oct 20, 2015 #1
    Consider the problem shown in the "wave_incidence_1.png" attached image. An electro-magnetic wave is travelling towards an interface between its current medium and a new medium, which has a refractive index [itex]n_2 \neq n_1[/itex]. The interface is represented by the [itex](x,y)[/itex] plane.

    The electric field [itex]\mathbf{E}[/itex] is parallel to that plane and it is directed outwards with respect to the screen, like the [itex]x[/itex] axis. It is orthogonal to the plane of incidence, which is the [itex](y,z)[/itex] plane. The magnetic field instead lays in the plane of incidence, but for this problem only the [itex]H_y[/itex] component has to be considered. [itex]E_x, H_y[/itex] constitute a field which only has tangential components to the interface plane [itex](x,y)[/itex].

    It is also possible to represent this problem as a transmission line with characteristic impedance

    [itex]\eta^{(1)} = \displaystyle \frac{E_x}{H_y} = \displaystyle \frac{\displaystyle \sqrt{\frac{\mu_0}{\epsilon_1}}}{\cos \theta_i}[/itex]

    (so computed only with [itex]E_x, H_y[/itex]) terminated over a load which represents the second medium, as shown in the "wave_incidence_equivalent_line.png" attached image. The load impedance can be computed as

    [itex]\eta^{(2)} = \displaystyle \frac{\displaystyle \sqrt{\frac{\mu_0}{\epsilon_2}}}{\cos \theta_t}[/itex]

    So, one can obtain the reflection coefficient [itex]\Gamma[/itex] for the field, by computing it for the line:

    [itex]\Gamma = \displaystyle \frac{\eta^{(2)} - \eta^{(1)}}{\eta^{(2)} + \eta^{(1)}}[/itex]

    In [itex]z = 0[/itex] we have

    [itex]E_0 e^{-j k_y^{(1)} y} + \Gamma E_0 e^{-j k_y^{(1)} y} = T E_0 e^{-j k_y^{(2)} y}[/itex]

    and so the entire problem is solved with the transmission-line approach (and the equality [itex]k_y^{(1)} = k_y^{(2)}[/itex] will be implied).

    But why just the tangential components of the fields to the interface should be considered? [itex]H_z[/itex] in fact is not taken into account. I know that the tangential fields must be continuous over an interface between two dielectrics; but the orthogonal fields also have their boundary conditions!
     

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    Last edited: Oct 20, 2015
  2. jcsd
  3. Oct 20, 2015 #2

    blue_leaf77

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    This sentence is confusing, it contradicts your later question which questions why only the tangential components should be considered. From the picture, it's clear that ##H_z## is the tangential component of the magnetic field, not ##H_y##.
    If there is no free current on the interface, the orthogonal components of magnetic field are continuous. At the same time, the E field in this problem does not have tangential component.
     
  4. Oct 20, 2015 #3
    I think you are confusing the plane of incidence and the interface plane. The [itex](x,y)[/itex] plane is the interface plane, which separates the two media, the left one with [itex]\epsilon_1, \mu_0[/itex] and the right one with [itex]\epsilon_2, \mu_0[/itex]. [itex]H_y[/itex] is the magnetic field component which is tangential to the interface plane. The plane of incidence is the plane [itex](y,z)[/itex] instead. So the sentence is coherent with the question.

    For the same reason written above, the entire [itex]\mathbf{E}[/itex] field is tangential to the interface plane and orthogonal to the plane of incidence.

    The boundary condition for the magnetic field without a current flowing into the [itex](x,y)[/itex] plane is

    [itex]\mathbf{\hat{n}} \times \mathbf{H}_1 = \mathbf{\hat{n}} \times \mathbf{H}_2[/itex]

    and so [itex]H_y[/itex] is continuous. Moreover,

    [itex]\mathbf{\hat{n}} \cdot \mathbf{B}_1 = \mathbf{\hat{n}} \cdot \mathbf{B}_2[/itex]
    [itex]\mu_0 \mathbf{H}_1^{orth} = \mu_0 \mathbf{H}_2^{orth}[/itex]

    and so the orthogonal magnetic field ([itex]H_z[/itex]) is also continuous. But my question is: why is [itex]H_z[/itex] discarded?

    Representing this problem as a circuital (transmission-line) problem, the characteristic impedance for the left medium is just [itex]E_x / H_y[/itex] and not [itex]E_x / |\mathbf{H}|[/itex].
     
  5. Oct 21, 2015 #4

    ehild

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    No, it is wrong. The normal component of B is continuous, but BH and μ is different in the different media. It is true that [itex]\mu_1 \mathbf{H}_1^{orth} = \mu_2 \mathbf{H}_2^{orth}[/itex], the normal component of H is not continuous at the boundary between the two media.
    The definition of the characteristic impedance or "Optical Impedance" ensures that it is a continuous function in a stratified medium.
     
  6. Oct 21, 2015 #5
    Yes, but in this particular case [itex]\mu_1 = \mu_2 = \mu_0[/itex]: the two media differ only for the refractive index; [itex]\mu[/itex] is the same. Take a look at how I computed the wave impedances [itex]\eta_1, \eta_2[/itex] in the first post.

    Sorry, but I can't understand what you mean with this statement. This is a stratified medium, but [itex]\eta[/itex] changes along [itex]z[/itex]: it is [itex]\eta^{(1)}[/itex] in the first medium and [itex]\eta^{(2)}[/itex] in the second one, so it is not continuous.
     
  7. Oct 21, 2015 #6

    ehild

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    The magnetic permeabilities might be very close, but they differ in principle. And the definition of the characteristic impedance involves the components of E and H which are parallel with the interface. Both are continuous, so their ratio is also continuous. A continuous function is very useful to calculate the optical behaviour of thin films. My last sentence refers to this.
    If something is defined in a way, you can not ask why is it not defined by an other way.

    The refractive index is the ratio of the speed of light in vacuum over the speed of light in the medium n=c/v. And the speed of light involves both ε and μ : ##v=\sqrt{\frac{1}{εμ}}## .
     
  8. Oct 21, 2015 #7
    Several times I have seen computations involving two media with [itex]\epsilon_1 \neq \epsilon_2[/itex] but [itex]\mu_1 = \mu_2 = \mu_0[/itex], it is a (maybe purely mathematic) simplification and it is useful to focus only on the dielectric constants. You may know much more than me these materials, but I don't believe that you never put [itex]\mu_1 = \mu_2 = \mu_0[/itex]. Anyway, there is another point:

    You basically are stating that always

    [itex]\eta^{(1)} = \displaystyle \frac{\sqrt{\displaystyle \frac{\mu_1}{\epsilon_1}}}{\cos \theta_i} = \eta^{(2)} = \displaystyle \frac{\sqrt{\displaystyle \frac{\mu_2}{\epsilon_2}}}{\cos \theta_t}[/itex]

    This would mean that it is always

    [itex]\Gamma = \displaystyle \frac{\eta^{(2)} - \eta^{(1)}}{\eta^{(2)} + \eta^{(1)}} = 0[/itex]

    for every interface?
     
  9. Oct 21, 2015 #8

    ehild

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    Yes, it is usual to set μ=μ0, but it does not change the definition, that the characteristic impedance involves the transversal field components.


    No, I do not. The formula for η refers for a single travelling wave. That is, why its is called wave impedance. η1 is not equal to η2.
    Remember that there are two waves in the first medium, the incident wave and the reflected one, while there is only a single travelling wave, (the refracted wave) in the second medium. The continuity of the transversal field components is valid to the whole fields. That is, at the boundary, the sum of E(1)x+ in the incident wave and E(1)x- in reflected wave equals to E(2)+ in the refracted wave in the second medium. The same holds for the H components.
     
  10. Oct 22, 2015 #9
    Yes, I definitely agree. So, just for the sake of completeness, what is the quantity that you said is continuous? Maybe

    [itex]\displaystyle \frac{\left| E_x^+ + E_x^- \right|}{\left| H_y^+ + H_y^- \right|} = \displaystyle \frac{\left|E_2^+ \right|}{\left|H_2^+ \right|}[/itex]

    ?
     
  11. Oct 22, 2015 #10

    ehild

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    Yes, the fraction is continuous, just like the numerator (transversal component of E) and the denominator (transversal component of H.
     
  12. Oct 27, 2015 #11
    Thank you.
    My original question anyway was not about this. I was not arguing against the wave impedance definition, but about the physical behaviour of the wave. Why in this problem just the tangential components [itex]E_x, H_y[/itex] of the fields are considered? (Tangential components mean: components that are tangent to the interface plane [itex](x,y)[/itex].)
    What does happen to the orthogonal components of the fields ([itex]H_z[/itex]) during this reflection/transmission and why are they not significant in the computations above?
     
  13. Nov 4, 2015 #12

    ehild

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    The electric field incident upon the interface between two mediums, can be decomposed into transversal electic (TE) and transversal magnetic (TM) waves. In case of a TE wave, the electric field is normal to the plane of incidence and tangential to the interface. The same is true for the magnetic field in a TM wave.
    Maxwell equations give the relation between the electric and magnetic fields. If you know the electric field of the TE wave, you know the magnetic field, both its components. The same holds in case of the TM wave.
     
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