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B Refraction formula confusion

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  1. Nov 23, 2017 #1
    Hi!

    I have a problem understanding what happens when a plane electromagnetic wave hits a surface consisting of two medias of different optical density.

    My old school litterature tells me two formulas which I really don't understand where they come from.

    The same teacher starts by telling me these facts:
    1) E1t=E2t
    2) H1t=H2t
    3) D1n=D2n
    4) B1n=B2n

    where n stands for normal to the surface and t stands for tangential to the surface.

    In my drawing I have tried to use 1 and 3 to wind up with my teacher's curious formulas like

    [tex]\frac{E_{rn}}{E_{1n}}=-\frac{sin(\theta_1-\theta_2)}{sin(\theta_1+\theta_2)}...1.1[/tex]

    and

    [tex]\frac{E_{rt}}{E_{1t}}=\frac{tan(\theta_1-\theta_2)}{tan(\theta_1+\theta_2)}...1.2[/tex]

    but I get a totally different answer like for instance for the tangential part

    [tex]E_1sin(\theta_1)=E_2sin(\theta_2)+E_rsin(\theta_r)...1.3[/tex]

    which may be rewritten as

    [tex]\frac{E_{rt}}{E_{1t}}=\frac{E_rsin(\theta_r)}{E_1sin(\theta_1)}=\frac{E_1sin(\theta_1)-E_2sin(\theta_2)}{E_1sin(\theta_1)}...1.4[/tex]

    or

    [tex]\frac{E_{rt}}{E_{1t}}=\frac{E_rsin(\theta_r)}{E_1sin(\theta_1)}=1-\frac{E_2sin(\theta_2)}{E_1sin(\theta_1)}...1.5[/tex]

    which isn't even close to my teacher's formula.

    So what am I doing wrong?

    Best regards, Roger
    PS
    I like complex numbers and by using them I kind of think that my failure of understanding might not be regarding the physics but regarding the math, my thoughts goes like this:

    [tex]K=\frac{E_{rt}}{E_1}=1-\frac{E_2sin(\theta_2)}{E_{1t}sin(\theta_1}...2.3[/tex]

    Here I have tried to comply with my teacher's formula by testing this

    [tex]K\approx 1-\frac{e^{j\theta_2}}{e^{j\theta_1}}=1-e^{j(\theta_2-\theta_1)}...2.4[/tex]

    Here I am guessing wildly and state that there can be no single real 1 in a complex number, which makes

    [tex]K\approx e^{j(\theta_2-\theta_1)/2}(e^{-j(\theta_2-\theta_1)/2}-e^{j(\theta_2-\theta_1)/2})...2.5[/tex]

    or

    [tex]K\approx -2je^{j(\theta_2-\theta_1)/2}\frac{(e^{j(\theta_2-\theta_1)/2}-e^{-j(\theta_2-\theta_1)/2}}{2j}...2.6[/tex]

    which equals

    [tex]K\approx -2je^{j(\theta_2-\theta_1)/2}sin((\theta_2-\theta_1)/2)...2.7[/tex]

    where we have a constant (complex) amplitude which only varies in phase and has been added 90 degrees in phase due to j, in any case the substraction of the angles is here correct and I am starting to wonder if the transponate also should be considered and in that case perhaps it "stays" in the nomenator of 2.4 which would yield

    [tex]K\approx -2je^{j(\theta_2-\theta_1)/2}\frac{sin((\theta_2-\theta_1)/2)}{sin((\theta_2+\theta_1)/2)}...2.8[/tex]

    still, I only get half the angles and it should also read tangens.

    This is the best I can do about this, I have given up so please help. Reflection.PNG
     
    Last edited: Nov 23, 2017
  2. jcsd
  3. Nov 23, 2017 #2
    Those initial equations relating the parallel and tangential components are the boundary conditions for the fields. The fields are discontinuous between different media and are related across the boundary by those conditions. The forms you have how are in the case that there is no free charge density or current density. They are derived form Maxwell's equations.

    I see a number of things wrong here. You absolutely cannot try to work backward from your teacher's answer. You need to go through the derivation logically and step by step starting with the boundary conditions. The form your teacher has arises much later in the derivation after an approximation and then an application of Snell's Law to the Fresnel equations, but you haven't even derived the basic set of equations yet and so you are far from getting that form. At this stage, you should still have several of the important constants around and cosines from the tangential and parallel projections, not sines.

    Also, your work with the complex exponentials is wildly off, so I suggest using the regular trig form for the time being and then reviewing complex numbers when you get a chance. You absolutely cannot get rid of the real numbers in your equations just because you're using the complex representation of trig functions.
     
  4. Nov 26, 2017 #3
    Perhaps I should not write any answer but I like to write in the english langage.

    I thank you for your answer, it was educational.

    I study physics on my own spare time and it was 21 years ago when I got my Master's degree in Electronic Engineering but I have been sick in my head for quite some time and has lost almost all of my knowledge, now I am just trying to get the most interesting part back and it feels like I am beginning from zero.

    I love the book Field and Wave Electromagnetics (Cheng) and I will reread it when I'm finished with my teacher's book.

    While Cheng explains and derives the bondary equations I feel I do not really have to know how they are derived right now, but I do need to know the next step that is how the tangential and normal bondary equations gives rise to the formulas we are discussing because if I don't understand how the equations are derived then there is really no point in continuing studying, at least this is my point of view.

    In other words, I can accept the bondary conditions but I can not accept the sine/tangens equations without knowing how they are derived.

    If I can not get to understand these equations and how they are derived, I will skip studying optics totally (and move on to Fluid Mechanics).

    It was interesting to hear that I just can't get rid of a real number in a complex expression, but if you consider frequency components like in a mixer

    [tex]Ae^{jw_1t}*Be^{jw_2t} \propto e^{j(w_1+w_2)t}...2.9[/tex]

    that is that the sum terms fall out directly BUT we know (and use) the difference terms also and where did they go? They come from the transponate i.e

    [tex]Ae^{jw_1t}*Be^{jw_2t^*} \propto e^{j(w_1-w_2)t}...2.10[/tex]

    so the proper expression is

    [tex]Ae^{jw_1t}*Be^{jw_2t} \propto e^{j(w_1+/-w_2)t}...2.11[/tex]

    and in the same time the fundamentals are also there that is that the number of frequencies are w1, w2, w1+w2 and w1-w2 ant the complex analize is complete (but not without the transponate).

    Remember that all this is just what i believe.

    Best regards, Roger
     
  5. Nov 27, 2017 #4
    I think I now understand a small part of the problem, the incident "beams" have ortogonal field vectors (EXB) which is what the bondary equations actually referes to so my sine really turns to cosine as RedDelicious says above.

    This gives me

    [tex]E_{1y}cos(\theta_1)=E_{2y}cos(\theta_2)-E_{ry}cos(\theta_1)[/tex]

    [tex]\frac{E_{ry}}{E_{1y}}=\frac{E_{2y}cos(\theta_2)}{E_{1y}cos(\theta_1)}-1[/tex]

    How far from the truth am I now?

    Best regards, Roger

    Reflection_E.PNG
     
  6. Nov 28, 2017 #5
    I'm not following what you're doing. What about the boundary conditions?

    For example, let's take a look at the case of perpendicular incidence.

    From the boundary condition for the magnetic field, we know that the tangential component must be continuous.

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

    We can relate this to the respective E components by,

    [tex]B_i = \frac{E_i}{v_i}
    \\
    B_r = \frac{E_r}{v_r}
    \\
    B_t=\frac{E_t}{v_t}
    [/tex]

    For the incident and reflected portions, we know that [itex]v_i = v_r[/itex] and [itex]\theta_i = \theta_r [/itex] because they're in the same medium. and so we then have

    [tex]

    \frac{1}{\mu_i v_i}(E_i-E_r)\cos{(\theta_i)}=\frac{1}{\mu_t v_i}E_t \cos{(\theta_t)}

    [/tex]

    However at the interface y = 0, the E field of the plane waves will have the same cosine arguments and so
    [tex]

    \frac{n_i}{\mu_i}(E_{0i}-E_{0r})\cos{(\theta_i)}=\frac{n_t}{\mu_t}E_{0t} \cos{(\theta_t)}

    [/tex]


    From here you can solve for the amplitude ratios and use the approximation that [itex]\mu_i \approx \mu_t \approx \mu_0[/itex].

    See if you can reduce it. If done correctly, applying Snell's law after will yield the first equation in your original post.

    The parallel case follows in a similar manner.
     
  7. Nov 28, 2017 #6
    I can't begin to describe how thankful I feel for you taking the time and effort into explaining this to me.

    I am very happy!

    However, I do think I see two faults in your nice explanation but they are minor faults but they made me, who does not understand so well, confused:

    The signs in this equation does not follow my Poynting vectors:

    [tex]-\frac{B_i}{\mu_i}\cos{(\theta_i)}+\frac{B_r}{\mu_i}\cos{(\theta_r)}=-\frac{B_t}{\mu_t}\cos{(\theta_t)}...3.1[/tex]

    They should be negated if I am to understand anything, actually you do just that later on in this formula (E=vB)

    [tex]\frac{1}{\mu_i v_i}(E_i-E_r)\cos{(\theta_i)}=\frac{1}{\mu_t v_i}E_t \cos{(\theta_t)}...3.2[/tex]

    but the "real" and confusing (obs) fault is in the right side of that formula which should read

    [tex]\frac{1}{\mu_t v_t}E_t \cos{(\theta_t)}...3.3[/tex]

    which is just some small misstake from your side, right?

    Otherwise I could actually follow you (after a while), the part where vB=E was interesting to relearn because my teacher spoke of it recently in the littarature, I had just not fully understood it, obviously, but I could follow you there but the use of refraction index (or what's it called) got me confused for a while but I soon realized that v=c/n where c conveniently disappears from the formula.

    Thank you for leaving some "homework" for me but I doubt that I will be able to solve it even from here.

    However, I will for certain try but if the reduction comes before Snell I doubt that I will be able to solve it.

    Best regards, Roger
     
  8. Nov 28, 2017 #7
    The signs are that way because for light incident from the right, the tangential component of the magnetic field points leftward for the incident and transmitted wave. The reflected wave is of course the opposite. The standard convention is to have light coming from the left and +x to the right, and so those components are negative, and the reflected is positive. The signs should work out even if you do it differently so long as you are consistent. I can't see your diagram.

    Yes, you are correct. It should be.

    [tex]\frac{1}{\mu_t v_t}E_t \cos{(\theta_t)}[/tex]

    You otherwise wouldn't get the correct index of refraction for nt on the right hand side.

    I actually meant to include a note about the v=c/n part because I realized it wasn't obvious where the index of refraction came from, but I apparently forgot, so I'm glad you were still able to figure that out.
     
    Last edited: Nov 28, 2017
  9. Nov 29, 2017 #8
    Okey, let's try to use what I've learned

    [tex]\frac{n_i}{\mu_i}(E_i-E_r)\cos{(\theta_i)}=\frac{n_t}{\mu_t}E_t \cos{(\theta_t)}...4.1[/tex]

    This equation comes from 3.2 above when v=c/n is considered, then we have

    [tex]\mu_i\approx \mu_t...4.2[/tex]

    which yields

    [tex]n_i(E_i-E_r)\cos{(\theta_i)}=n_tE_t \cos{(\theta_t)}...4.3[/tex]

    rearranging gives

    [tex]n_iE_r\cos{(\theta_i)}=n_iE_i\cos{(\theta_i)}-n_tE_t\cos{(\theta_t)}...4.4[/tex]

    that is

    [tex]\frac{E_r\cos(\theta_i)}{E_i\cos(\theta_i)}=\frac{E_r}{E_i}=1-\frac{n_tE_tcos\theta_t}{n_iE_i\cos{\theta_i}}...4.5[/tex]

    using Snell's Law

    [tex]n_i\sin{\theta_i}=n_t\sin{\theta_t}...4.6[/tex]

    we get

    [tex]\frac{E_r}{E_i}=1-\frac{n_i\frac{sin(\theta_i)}{sin(\theta_t)}E_tcos(\theta_t)}{n_iE_icos(\theta_i)}...4.7[/tex]

    or

    [tex]\frac{E_r}{E_i}=1-\frac{E_t\cot(\theta_t)}{E_i\cot(\theta_i)}...4.8[/tex]

    This is the first time in my life I have used cotangens, also it does not comply with my teacher's formula.

    Best regards, Roger
     
  10. Dec 2, 2017 #9
    You're going to want to use the fact that [itex]E_{0i}+E_{0r}=E_{0t} [/itex] at the boundary. From there, it will be a simple exercise in algebra to find the ratios.

    Also, for the Snell's law simplification, you're going to want to have your equation in a form where you can see to use the sum and difference identities for sine, meaning everything in terms of sines and cosines.
     
    Last edited: Dec 2, 2017
  11. Dec 6, 2017 at 5:26 PM #10
    It is late and I am soon going to bed but I will try a reply (thanks for your reply, by the way):

    Using 4.8 and your statement that

    [tex]E_i+E_r=E_t...5.1[/tex]

    where I have skipped the zero in the indexing because it is only hard work codfing for it while we are discussing the bondary anyway.

    Also, I would like to rewrite this equation of yours as

    [tex]E_i-|E_r|=E_t...5.2[/tex]

    because isn't this what's it all about?

    The differense in incoming and reflected energy is the transmitted energy like the energy density

    [tex]w_E=\frac{1}{2}\epsilon E^2...[J/m^3]...5.3[/tex]

    that is, we have E squared "everywere" so that energy is lost in reflection but the rest is transmitted.

    Getting back to 4.8 and your statement, we have

    [tex]\frac{E_r}{E_i}=1-\frac{E_t\cot(\theta_t)}{E_i\cot(\theta_i)}=1-\frac{(E_i+E_r)\cot(\theta_t)}{E_i\cot(\theta_i)}...5.4[/tex]

    or

    [tex]\frac{E_r}{E_i}=1-\frac{\cot(\theta_t)}{\cot(\theta_i)}-\frac{E_r\cot(\theta_t)}{E_i\cot(\theta_i)}...5.5[/tex]

    Still far from my teacher's formula.

    Best regards, Roger
     
  12. Dec 6, 2017 at 7:54 PM #11
    No. The need for continuity in the tangential components at the boundary rules that form out. Your equations need to obey the boundary conditions. Physically, if you start from the beginning with your plane wave construction, it might make more sense if you think in terms of the combined waves on the left (reflected and incident) will join at the boundary and should equal the combined wave on right right (transmitted) indicating to sum them. You shouldn't worry about energy until you have the correct equations for the fields because you still need to find them to find the fields anyway. Your algebra is wrong there as well. Even if you took that approach, which you shouldn't, that's not the equation you'd get.

    You should start over from your equation 4.3 making the substitution in my last post and then solve for the ratio. I'm also going to repeat my very strong suggestion of keeping everything in terms of sines and cosines. You're not going to get anywhere you're using cotangents. You want to keep it in terms of sines and cosines so that it's clear to use the sum and difference identities for sine.

    The form you want to end up with is.

    [tex]\left(\frac{E_{0r}}{E_{0i}}\right)_\perp = \frac{n_i \cos(\theta_i)-n_t \cos(\theta_t)}{n_i \cos(\theta_i)+n_t \cos(\theta_t)}[/tex]

    From here you can easily apply Snell's law, do a little algebra, and then use the sum and difference identity for sine to get your teacher's form.
     
  13. Dec 7, 2017 at 9:50 AM #12
    Thank you very much for your answer!

    I will have think some about this before I get back to you with a serious answer but I can tell you that one kind of strange thing that my teacher told me about a rope reflecting against a wall (denser media, obviously) is that the calculation of the waves goes like this

    [tex]s_i(x,t)=Asin(wt-kx)...6.1[/tex]

    and

    [tex]s_r(x,t)=Asin(wt+kx+\phi)...6.2[/tex]

    and at x=0 there can be no disturbance (s) so that

    [tex]s_i+s_r=0...6.3[/tex]

    which both gives that

    [tex]\phi=\pi...6.4[/tex]

    AND that the incident wave and reflected wave are added which I feel is hard to grasp, I see an incoming wave and a reflected wave totally separated, I don't see them both, how could I?

    I can however accept that it is this way not to torture you anymore, and get down to the hints you have given me.

    Give me a day or two :)

    Best regards, Roger
     
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