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Wave reflection and refraction, relations between angles

  1. Oct 28, 2015 #1
    This post is strictly related to my previous one. Let's consider the same context and the same image. Regarding the oblique incidence of a wave upon an interface between two dielectric, all the texts and all the lectures write an equation like the following:

    [itex]e^{-j k_1 y \sin \theta_i} + \Gamma e^{-j k_1 y \sin \theta_r} = Te^{-j k_2 y \sin \theta_t}[/itex]

    where [itex]k_1[/itex] is the wavenumber in the medium 1 (left) and [itex]k_2[/itex] is the wavenumber in the medium 2 (right).
    The texts also say: if this relation has to be valid for all [itex]y[/itex], then the [itex]y[/itex] variation must be the same on all the terms and so

    [itex]k_1 \sin \theta_i = k_1 \sin \theta_r = k_2 \sin \theta_t[/itex]

    I know that this relation is true (it's Snell's law!) and it can be experimentally proved, but I have a doubt: is equation (2) really the only way to verify the equation (1) for all [itex]y[/itex]? I can't understand - neither physically nor mathematically - why.

    Why is this solution the only acceptable one and not just a trivial one?
  2. jcsd
  3. Oct 28, 2015 #2


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    It is based on the orthogonality of the sine function.
    ## x r sin(\theta) = x t \sin(\phi) ## for all x, implies that t = r, since that is your magnitude, and ##\phi = \theta + k 2\pi ## where k is an integer. Other solutions might have t = -r, and phi an odd number of pis offset from theta, but since k_1, k_2 are restricted to positive numbers and theta_i, theta_r, theta_t restricted to [0,2pi], there is only one solution.
    For more on orthogonality of sine ... check out this page on Fourier Series.
    ref: http://mathworld.wolfram.com/FourierSeries.html
  4. Oct 29, 2015 #3
    First of all thank you for your answer.
    In the link you suggested, the othogonality seems to be just for sines of the form [itex]\sin (mx)[/itex] and [itex]\sin (nx)[/itex] with [itex]n \neq m[/itex] and so this situation is quite different; anyway, the sine functions [itex]k_1 y \sin \theta_i[/itex] and [itex]k_1 y \sin \theta_r[/itex] are actually equal only in the cases you listed.
    What is still weird for me is that here, from a single equation (equation (1) of my post), multiple equations arise. Is it due to the features of the sine function, that imposes equality for both its amplitude and its phase, or what else?
    Last edited: Oct 29, 2015
  5. Nov 2, 2015 #4


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    Sorry I did not reply to this sooner. Let's look at this question in more detail.
    You have:
    ##\forall y, \quad Ae^{-j k_1 y \sin \theta_i } + Be^{-j k_1 y \sin \theta_r}=Ce^{-j k_2 y \sin \theta_t} ##
    Then, it must be true for y = 0.
    So A + B = C. Which gives:
    ##\forall y, \quad Ae^{-j k_1 y \sin \theta_i } + Be^{-j k_1 y \sin \theta_r}=(A+B)e^{-j k_2 y \sin \theta_t} ##
    Now, by enforcing equality of the imaginary parts when y is not equal to zero, you will see that the imaginary term of the right side is zero when
    ## y = \frac{ n \pi }{ k_2 \sin \theta_t}##
    Then you must also have the imaginary parts on the left be zero, so
    ##k_1 \frac{ n \pi }{ k_2 \sin \theta_t} \sin \theta_i = m \pi ##
    ##k_1 \frac{ n \pi }{ k_2 \sin \theta_t} \sin \theta_r = m \pi ##
    In order for this to be true, you will need
    ##\frac{ k_1 \sin \theta_i}{ k_2 \sin \theta_t} \in \mathbb{Z}## and ##\frac{ k_1 \sin \theta_r}{ k_2 \sin \theta_t} \in \mathbb{Z}##
    Noting that the two sides have to be zero at the same times, there can be no zeros on the left side for y between
    ##(\frac{ n \pi }{ k_2 \sin \theta_t}, \frac{ (n+1) \pi }{ k_2 \sin \theta_t})##
    This imposes the condition that:
    ##\frac{ k_1 \sin \theta_i}{ k_2 \sin \theta_t} \in \{1, -1\}## and ##\frac{ k_1 \sin \theta_r}{ k_2 \sin \theta_t} \in \{1, -1\}##
    Generally, there is an assumption that your theta values are measured relative to the interface, so ##k\sin\theta## is always positive. This reduces our solution set to simply:
    ##\frac{ k_1 \sin \theta_i}{ k_2 \sin \theta_t} =1## and ##\frac{ k_1 \sin \theta_r}{ k_2 \sin \theta_t} =1##
    which can be written as:
    ##k_1 \sin \theta_i = k_1 \sin \theta_r = k_2 \sin \theta_t ##
    If you didn't have the assumption of positive signs, it would look more like:
    ##k_1 \sin \theta_i = \pm k_1 \sin \theta_r = \pm k_2 \sin \theta_t ##
  6. Nov 13, 2015 #5
    Don't worry, I am late this time more than you. Thank you instead for all your detailed observations.
    I followed all your steps and I agree with them. But maybe there is another solution and I would like to ask you for your opinion.
    Consider your equation:

    [itex]A e^{-j k_1 y \sin \theta_i} + B e^{-j k_1 y \sin \theta_r} = A e^{-j k_2 y \sin \theta_t} + B e^{-j k_2 y \sin \theta_t}[/itex]

    and the following system

    [itex]\begin{cases} e^{-j k_1 y \sin \theta_i} = e^{-j k_2 y \sin \theta_t} \\ e^{-j k_1 y \sin \theta_r} = e^{-j k_2 y \sin \theta_t} \end{cases}[/itex]

    (It is obviously obtained by separately equating the [itex]A[/itex] and [itex]B[/itex] coefficients in the first equation).
    Is this system a necessary and sufficient condition in order for the first equation to be satisfied?
    The conditions

    [itex]\begin{cases} k_1 \sin \theta_i = k_2 \sin \theta_t \\ k_1 \sin \theta_i = k_1 \sin \theta_r \end{cases}[/itex]

    would immediately follow.
    Can it be a valid solution too or does it lack some details?
  7. Nov 15, 2015 #6


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    Yes that is a straightforward and valid way to approach the explanation. If you were unsure of whether any of those assumptions had logical holes in them, I feel that my explanation above should remove those doubts.
  8. Nov 18, 2015 #7
    Yes, it does, because it is very detailed. Thank you!
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