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B Refraction formula confusion

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  1. Jan 1, 2018 #21
    When I uploaded the drawing the system somehow failed to upload my text also (it wasn't much though).

    Anyway, I wish to use my drawing and emphasize certain points:

    An incident plain wave may be written

    [tex]E_{iy}=E_{iy0}sin(wt-kx)...9.1[/tex]

    In the drawing, this is the part parallel to the normal of the interface and it is called "parallel"

    The perpendicular part does only exist if the wave is circularly polarized like

    [tex]E_{iy}=E_{iy0}sin(wt-kx)...9.2[/tex]

    [tex]E_{iz}=E_{iz0}cos(wt-kx)...9.3[/tex]

    which only is a special case of elliptically polarized waves.

    In this case we have a plane wave (i.e its electrical field is in one plane only, in this case the xy-plane) so that the discussion regarding what happens in the parallel plane (parallel to the plane of incidence) is also valid for the perpendicular plane but only if we consider a perpendicular polarized wave, right?

    In other words, we do not consider circular or elliptical polarized waves here, only plane waves, right?

    Because if this isn't right I really do not understand a thing.

    Happy New Year!

    Best regards, Roger
     
  2. Jan 4, 2018 #22
    Just trying to understand...

    For the parallel part we have

    [tex]E_{i\parallel}=E_{in}=E_{iy0}sin(\theta_i)=E_isin(\theta_i)[/tex]

    where the last equality is just to be conform with earlier discussions.

    For the perpendicular part we have

    [tex]E_{i\perp}=E_{it}=E_{iz0}cos(\theta_i)=E_icos(\theta_i)[/tex]

    where one might see the perpendicular part as strictly being mirrored into the tangential part of the interface, I don't really know what I am talking about here but it is simple to see the parallell part but not the perpendicular part yet draw some kind of conclusion that the perpendicular part then should lay in the tz-plane.

    Except for my amateur speculation I think I now got this part correctly, please tell me if I'm wrong.

    From Maxwell we then have
    1) E1t=E2t
    2) H1t=H2t
    3) B1n=B2n
    4) D1n=D2n

    These I will look up why they are as they are but right now I will accept them as is.

    Looking at the interface and thinking energy conservation I get

    [tex]E_i-Er=Et[/tex]

    The E-fields should be squared but the principle is the same, I think (obs).

    Using

    [tex]\theta_r=\theta_i[/tex]

    we then get for the perpendicular part

    [tex]E_icos(\theta_i)-Ercos(\theta_i)=E_tcos(\theta_t)[/tex]

    and for the parallell part we have

    [tex]E_isin(\theta_i)-Esin(\theta_i)=E_tsin(\theta_t)[/tex]

    But this is wrong, right?

    Time for late supper, good night!

    Best regards, Roger
    PS
    If it wasn't for you I would not know how to code for perp an parallel :)
     
  3. Jan 5, 2018 #23
    Let's begin from the beginning and use proper formulas

    For the parallel part:

    [tex]E_{i\parallel}=E_{in}=E_{iy0}sin(\theta_i)=E_isin(\theta_i)...10.1[/tex]

    which we rewrite as

    [tex]E_{i\parallel}=E_isin(\theta_i)...10.2[/tex]

    and for the perpendicular part we do the same and write

    [tex]E_{i\perp}=E_icos(\theta_i)...10.3[/tex]

    Now we know that the perpendicular part is in the interface plane, this means that we have to use

    [tex]H_{1t}=H_{2t}...10.4[/tex]

    but why can't we use

    [tex]E_{1t}=E_{2t}...10.5[/tex]

    ?

    Anyway, using 10.3 and 10.4 gives

    [tex]H_icos(\theta_i)-H_rcos(\theta_i)=H_tcos(\theta_t)...10.6[/tex]

    or

    [tex]H_rcos(\theta_i)=H_icos(\theta_i)-H_tcos(\theta_t)...10.7[/tex]

    while

    [tex]B=\mu H...10.8[/tex]

    we have

    [tex]\frac{B_r}{\mu_i}cos(\theta_i)=\frac{B_i}{\mu_i}cos(\theta_i)-\frac{B_t}{\mu_t}cos(\theta_t)...10.9[/tex]

    now

    [tex]E=vB...10.10[/tex]

    such that

    [tex]\frac{E_r}{v\mu_i}cos(\theta_i)=\frac{E_i}{v\mu_i}cos(\theta_i)-\frac{E_t}{v\mu_t}cos(\theta_t)...10.11[/tex]

    now

    [tex]v=\frac{c}{n}...10.12[/tex]

    which gives

    [tex]\frac{n_rE_r}{c\mu_i}cos(\theta_i)=\frac{n_iE_i}{c\mu_i}cos(\theta_i)-\frac{n_tE_t}{c\mu_t}cos(\theta_t)...10.13[/tex]

    or

    [tex]\frac{n_rE_r}{\mu_i}cos(\theta_i)=\frac{n_iE_i}{\mu_i}cos(\theta_i)-\frac{n_tE_t}{\mu_t}cos(\theta_t)...10.14[/tex]

    using

    [tex]\mu_t\approx \mu_i...10.15[/tex]

    gives

    [tex]n_rE_rcos(\theta_i)=n_iE_icos(\theta_i)-n_tE_tcos(\theta_t)...10.16[/tex]

    using

    [tex]n_r=n_i...10.17[/tex]

    gives

    [tex]n_iE_rcos(\theta_i)=n_iE_icos(\theta_i)-n_tE_tcos(\theta_t)...10.18[/tex]

    dividing this with

    [tex]n_iE_icos(\theta_i)...10.19[/tex]

    gives

    [tex](\frac{E_r}{E_i})_\perp =1-\frac{n_tE_tcos(\theta_t)}{n_iE_icos(\theta_i)}...10.20[/tex]

    using Snell's Law like

    [tex]n_i=\frac{n_tsin(\theta_t)}{sin(\theta_i)}...10.21[/tex]

    we get

    [tex]?[/tex]

    Am I close or far from it?

    Best regards, Roger
    PS
    I think I should inject your amazing formula

    [tex]E_{0i}+E_{0r}=E_{0t}[/tex]

    here but that equation is really hard to understand because how can there be a sum of electric field intensities when the wave is going away from the interface? Maybe one can see it as the total electric field intesity being the same on both sides of the bondary? Like there is a potential that needs to be the same regardless if one travels inside or outside of the bondary?
     
    Last edited by a moderator: Jan 5, 2018
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