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Electromagnetic Wave Calculations

  1. Jun 23, 2015 #1
    1. The problem statement, all variables and given/known data

    A monochromatic, electromagnetic plane wave is travelling in a non-conducting, transparent medium. Its real electric field is described by:

    ##E=100 \ cos (7.62 \times 10^6 (x+y+z) \ - \ 2.98 \times 10^{15} t) \ \frac{1}{\sqrt{5}} (\hat{y} + 2 \hat{z}) \ V/m##

    (a) What would be the vacuum wavelength of the wave?

    (b) What is the refractive index of the medium?

    (c) In what direction is the wave travelling?

    2. Relevant equations

    ##\lambda = 2\pi/k, \ \ v=\omega/k, \ \ n=ck/\omega##

    Real electric field general expression:

    ##E(r,t)=E_0 \ cos(k.r-\omega t) \hat{n}##

    Where ##\hat{n}## represents the polarization.

    3. The attempt at a solution

    (a) I have tried to do this part using the two different equations: ##\lambda = 2\pi/k## and ##\lambda=c/\nu##. But each time I get a different answer:

    ##\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{7.62 \times 10^6}= 824.5 \ nm##​

    And since ##\nu = \omega / 2\pi## we have:

    ##\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{(2.98\times 10^{15})/2\pi} = 632.5 \ nm##​

    So which method is correct? :confused:

    (b) For refractive index I got:

    ##n= \frac{ck}{\omega} = \frac{(3 \times 10^8)(7.62 \times 10^6)}{2.98 \times 10^{15}} = 0.76##​

    But this doesn't look right. Shouldn't n be greater than or equal to 1?

    (c) Since the expression for electric field has the general form ##cos(kz-\omega t)##, I believe it is travelling to the right. But how do I explain the "x+y+z" part? What plane or line is the direction of travel parallel/perpendicular to?

    Any explanation would be greatly appreciated.
     
  2. jcsd
  3. Jun 23, 2015 #2
    As a potential bump in the right direction, keep in mind that the EM wave in this problem isn't necessarily traveling in vacuum, i.e., the speed of the wave isn't necessarily ##c##.
     
  4. Jun 23, 2015 #3
    (a) Note that the problem asks, "What would be the vacuum wavelength of the wave?" The wave is not propagating in a vacuum. So, the answer to this question is not the wavelength of the wave in the medium. So, the problem gives you a phase constant (sort of - you have to figure it out since the wave is propagating in an oblique direction). The problem also gives you the frequency of the wave. You know what the speed of light in a vacuum is. (And you should remember that the speed of light in a dielectric is not equal to the speed of light in a vacuum). You also know the frequency of the wave. So, what would be the vacuum wavelength of the wave?

    (b) You need to figure out the phase constant of the wave (be careful - the wave is traveling in an oblique direction). Then think about how the phase constant, the frequency, and the index of refraction are related to each other. If you don't know, think about the velocity of the wave (remembering that the velocity in the medium is less than the speed of light).

    (c) Clearly, the (x+y+z) indicates that the wave is traveling in an oblique direction. You need to figure out the unit vector specifying the direction of travel. Keep in mind that cos(-A) = cos(A). Set up a phasor representation of the E-field.
     
  5. Jun 23, 2015 #4
    So, my second method was correct? (i.e. ##\lambda_0 = c/\nu = 632.5 \ nm##)

    Vacuum wavelength is given by ##\lambda_0 = \lambda n##. As n is increased, ##\lambda## decreases because frequency is unchanged when speed decreases. Since the given frequency is same as vacuum frequency AND I know the vacuum speed, I just used the relationship ##\lambda = c/\nu##.

    So where is the phase constant? And how does a phase constant factor into the calculations for ##\lambda_0##?
    A sinusoidal wave has the form ##f(z,t)=A \ cos (kz-\omega t + \delta)##, where ##\delta## is the phase constant. So in the given equation for ##E##, I don't see a phase constant...

    I know we can use ##n=v_{medium}/c## (If we know the wave velocity in the medium). But why is it not possible to use ##n=ck/\omega## as I did?

    That was my question. How do we find the unit vector representing that direction? :confused:

    My textbook (Griffiths) does not explain this, and I've posted similar problems in other threads before but unfortunately I couldn't get any help with those. Any explanation or links on how to do this part would really be appreciated.
     
  6. Jun 24, 2015 #5
    Yes.

    I don't know what n is.

    No. k is your phase constant (rad/m). You need to figure out how the phase constant, the speed of the wave, and the frequency of the wave relate to each other.

    Consider this function: f(z,t) = cos(wt - kz). This has a value of 1 as long as wt-kz = 0. Now, as t increases (as it always does), z must also increase in order for f(z,t) = 1. That is, this wave is propagating in the positive z direction. But how fast? Well, v = dz/dt. So, if we make wt - kz = 0, then we are saying wt = kz, or wt/k = z.

    v = dz/dt = w/k.

    Excuse my notation. I don't know how to do equations on PF. If you could let me know how you do this, I would appreciate it.

    Let me know how to write equations on PF, and I'll be happy to explain this to you. It is not easy. I don't think that most students would be able to get it just by thinking really hard. But once it is explained, it is readily understandable - to most students. It is just hard to explain with words.
     
  7. Jun 24, 2015 #6
    To write your equations in LaTeX, simply include your code within double hash tags (##), or between [tex] [/tex].

    Here are the common LaTex commands.

    ##n## is the refractive index, of course.

    I used the equation relating k to ##\omega## to find it:

    ##n = \frac{c k}{\omega} = 0.7##​

    But "0.7" cannot be a valid refractive index. What's wrong? :confused:
     
  8. Jun 25, 2015 #7
    It seems like you have a couple of problems here.

    It appears that your value for k is 6.9533 x 10^6. Is that what you have? How did you get that?
     
  9. Jun 25, 2015 #8
    I couldn't figure out how to do this in latex (with all the superscripts and hats, etc.). But check out the attached. You need to recognize that the ##7.62(10^6) (x + y + z)## term in the argument is a dot product of your beta vector and your position vector. Given a position vector, you should be able to find a unit vector in the same direction (just divide the position vector by the magnitude of the position vector). Find your beta.

    Oh, and what I'm calling beta is often called k.
     

    Attached Files:

  10. Jun 25, 2015 #9

    blue_leaf77

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    Can you possibly mention the link or book from which you found this problem, because the propagation direction and polarization vector in
    are not perpendicular, as it should if the wave is propagating in an isotropic medium.
     
  11. Jun 25, 2015 #10
    I think we are talking about a guided wave here.
     
  12. Jun 25, 2015 #11

    blue_leaf77

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    In any isotropic medium, be it guiding or not, we have the relation
    $$ \epsilon_R \frac{\omega^2}{c^2} (\mathbf{k} \cdot \mathbf{\hat{u}})= 0$$
    where ##\mathbf{\hat{u}}## is the unit vector along the polarization, we see that in order to satisfy that equation the wavevector must be perpendicular to the polarization.
     
  13. Jun 25, 2015 #12
    blue_leaf,

    Are you saying that in any isotropic medium - even in waveguides - all electromagnetic waves are TEM?
     
  14. Jun 25, 2015 #13

    blue_leaf77

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    Maybe I should have been careful in saying that general statement, that equation can be derived simply by using Maxwell equations by requiring that the solution is a plane wave, that i,s the surface of constant phase is a flat surface, constant amplitude, and uniform wavevector, and in addition the permittivity is scalar not a tensor as it is in anisotropic media . These requirements will in turn give that equation.
    The difference between guiding and infinitely large extend medium is the presence of boundary conditions, and due to this the amplitude of the propagating E field should be a function of space, which is not the case in the original problem posted by the OP.
     
  15. Jun 25, 2015 #14
    I think the polarization of this wave can be ignored in this problem. It is not necessary to answer the questions asked.
     
  16. Jun 25, 2015 #15

    blue_leaf77

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    For part (a) probably it is not needed. But the OP's answer to part (b) looks fishy, therefore we need to know the exact problem and in which context he/she got it, only the OP can tell us.
     
  17. Jun 25, 2015 #16
    The OP's answer to part (b) is wrong. I worked the problem - ignoring polarization - and got the answer to part (b).
     
  18. Jun 25, 2015 #17

    blue_leaf77

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    Ah I failed to notice that he used the wrong value for ##k##.
     
  19. Jun 25, 2015 #18
    So, the unit vector is ##\frac{\hat{x}+\hat{y}+\hat{z}}{\sqrt{3}}##?

    So I need to multiply ##k=7.62 \times 10^6## with √3 to get the right k?

    This is what I got:

    ##n=\frac{c(k=7.62 \times 10^6)}{\omega=2.98 \times 10^{15}} (\sqrt{3}) = 1.2##

    Is this right? The answer looks more realistic (it's close to the refractive index of water, for instance)

    This is a question from an old exam paper.
     
  20. Jun 26, 2015 #19
    Your unit vector is correct. But you are not multiplying k by ##\sqrt{3}##. Rather you are multiplying ##7.62 \times 10^6## by ## \sqrt{3} ## to get k. That's your k. ##7.62 \times 10^6## is not your k.

    Somehow your answer is off though. I get ##n = 1.329##. Check your arithmetic.
     
  21. Jun 26, 2015 #20
    Alright, thank you so much for the help!
     
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