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I Field frequency and angle of incidence in optical waveguides

  1. Jan 27, 2016 #1
    Hello!
    In this previous post, most replies point out that it is not possible to predict the angle of refraction (and so the frequency) at a certain interface, given the wavelength of the original signal.
    In particular,
    But when dealing with optical waveguides, it seems to be different. I am referring for example to this link, slide 23: the lower the mode in the fiber, the lower the angle of incidence (and the angle of refraction will vary as a consequence, knowing that [itex]n_2/n_1[/itex] is a given value).

    It seems to be the same for the dielectric slabs and for the graded index optical fibers like this (Graded Index Fibers paragraph).

    I know that in a certain waveguide, if [itex]\beta[/itex] is the propagation constant along [itex]z[/itex] (the direction of propagation), the following relations exist:

    [itex]\omega^2 \mu \epsilon = k_c^2 + \beta^2[/itex]
    [itex]\beta = \sqrt{\omega^2 \mu \epsilon - k_c^2}[/itex]

    where [itex]k_c[/itex] is the transverse wavenumber. So, depending on the frequency [itex]\omega[/itex], there will be waves with a higher [itex]\beta[/itex] and a lower [itex]k_c[/itex] and vice-versa: this will vary the angle of incidence according to the mode. But what exactly is the relation between the wave frequency and this angle? And why the lowest modes always have an almost straight path and a high [itex]\beta[/itex], while the highest have a low [itex]\beta[/itex], so bouncing more frequently between the core-cladding interface?
     
  2. jcsd
  3. Jan 31, 2016 #2
    The relations that you posted are called dispersion relations (or equations of separation). They do not give the relationship between the wave frequency and the angle directly. Instead you need to impose a strict phase relation between two successive bounces on the same surface. The dervation is shown in detail in Section 8.2 Planar Dielectric Waveguides of Fundamentals of Photonics by Saleh 2nd edition. The relevant equation you are seeking is 8.2-4 which I will not produce here as I do not know LATEX. Note that the equation is trancendental and as such must be solved graphically or numerically.
     
  4. Feb 22, 2016 #3
    Thank you for your post. I found that relation, but still can't understand how it is justified.
    Let's look at this image: in Figure (a), a wave coming from left propagates with a certain [itex]\mathbf{k_u}[/itex]. It then bounces a first time (in the upper plate), as in Figure (b), and a second time (in the lower plate), as in Figure (c). Textbooks basically say that, given a point A in Figure (a) where the original wave has phase [itex]\phi_A[/itex], at the same point in Figure (c), after 2 bounces, the wave should have the same phase it had in Figure (a), [itex]\phi = \phi_A + 2 n \pi[/itex]. Then the condition you mentioned is derived.
    If it is satisfied, the wave is able to regenerate itself along the guide, because after every couple of bounces it constructively adds with itself.
    I know that when, for example, [itex]\phi = -\phi_A[/itex], the wave destructs itself; but a lot of other intermediate situations can occur, where [itex]\phi[/itex] is different both from [itex]\phi_A[/itex] (perfect constructive interference) and from [itex]-\phi_A[/itex] (perfect destructive interference). And at a first glance I can't state that these waves will cancel. Why in all these situations the wave always and anyway destroys itself, so that only waves with [itex]\phi = \phi_A + 2 n \pi[/itex] can be accepted?
     
  5. Mar 2, 2016 #4
    I don't think this is a difficult question, just I can't figure out the problem. The question can be asked in a simpler way.
    A plane-wave is propagating with a certain angle of incidence into a parallel-plate waveguide: it will bounce off the upper plate, then the lower plate, and so on. Why can this wave survive into the waveguide only if, after every couple of bounces, it regenerates itself?
    2gy6ecl.png
    Best case is [itex]\phi_B = \phi_A + 2 n \pi[/itex]: this is (according to textbooks) the only acceptable case.
    Worst case is [itex]\phi_B = -\phi_A + 2 n \pi[/itex] and the wave cancels itself.
    But between the above two limit-cases there is an infinity of intermediate situations, where in B the wave has a phase which is nor able to cancel itself, neither able to regenerate itself. If the wave is not completely cancelled, why cannot it propagate anyway along the guide?
     
  6. Mar 2, 2016 #5
    I think the key is not that it regenerates itself but rather that it is itself. The phase relation only ensures that you are describing the same wave and not a different one shifted in phase.
     
  7. Mar 8, 2016 #6
    Ok, it could make sense. So we are stating that a wave which can't be always itself along the guide, can't propagate there. Why are we so sure about it? I mean: a wave can change along the guide, without destroying itself.
     
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