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Equations for total internal reflection

  1. Apr 7, 2015 #1
    If a beam of light undergoes total internal reflection, are there any equations to determine the angle of reflection based on the angle of incidence?

    Snell's Law: ## n_1sinA = n_2sinB ## is to my knowledge only valid under medium-to-medium transmission (for B: [0, pi/2]), correct? If total internal reflection occurs, is there an analogous equation to determine the angle of reflection in this case? I believe I saw previously that ## n_1sinA = sinB ## but I believe I took it out of context and am a little uncertain on how to proceed from here.
     
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  3. Apr 7, 2015 #2

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    Yes, there is such and equation! And it's very simple; it's much simpler than even Snell's law.

    I'd tell you what it is, but I hesitate just giving you the answer (per the forum rules).

    It's easy enough to find with an Internet search engine by googling "Total Internal Reflection" (or you can look in your textbook). Suffice it to say that the equation is very simple. Some might say trivial.

    [Edit: Hint: it's the same thing as the relationship with normal reflection, such as reflection off of a mirror. (Assuming of course that the angle of incidence is greater than the critical angle such that Total Internal Reflection occurs in the first place.)]
     
    Last edited: Apr 7, 2015
  4. Apr 7, 2015 #3
    Is it simply that the angle of incidence equals the angle of reflection? If so, I was under the impression there was an interaction at the boundary between the two media resulting in some sort of shift (in phase and/or trajectory) but this might be it.

    If not, I'll keep looking. :)
     
  5. Apr 7, 2015 #4
    My main question essentially stems from here: https://www.physicsforums.com/threa...nternal-reflection-of-a-beam-of-light.715719/

    It's just one step I'm having trouble understanding, and that's going from:

    90 - θ1 > arcsin(n2/n1)

    Substituting for θ1

    90 - arcsin(sin(θa)/n1) > arcsin(n2/n1)

    This substitution doesn't quite make sense from what I see. Although it's likely I'm just overlooking something right in front of me. :)
     
  6. Apr 7, 2015 #5

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    It's as simple as that, yes. :smile:

    No shifting involved. It's as simple as [itex] \theta_i = \theta_r [/itex]. Again though, this assumes that the angle of incidence is large enough for total internal reflection to happen in the first place.
     
  7. Apr 7, 2015 #6

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    Without going through all the steps in the thread you quoted, realize that [itex] \theta_A [/itex] and [itex] \theta_1 [/itex] (see attached imagine in that thread) do not involve total internal reflection. Snell's law applies normally for those angles. On the other hand, [itex] \theta_0 [/itex] involves total internal reflection only if [itex] \theta_0 [/itex] is greater than the critical angle.

    So the crux of the problem in that thread is finding [itex] \theta_0 [/itex] such that it equals the critical angle, and then translating that back to the corresponding [itex] \theta_A [/itex].

    My point is that the problem in that thread does not actually involve total internal reflection at all, in particular; rather it involves the critical point at which total internal reflection fails.
     
  8. Apr 7, 2015 #7

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    Getting to the question you posed in that thread,

    Do you see how that is just a rehashing of Snell's law using different variable names such that
    [tex] n_1 \sin \theta_1 = n_A \sin \theta_A [/tex]
    where [itex] n_A \approx 1 [/itex] since it's air.
     
  9. Apr 7, 2015 #8
    Thank you for all the help! So just to clarify, is it safe to assume that ## n_A = n_{air} ##? I was under the impression ## n_A ## is arbitrary and that is why I had trouble understanding the solution, but I guess it is somewhat justified when considering the very first beam IF it's entering from the air. (which it now appears he does show). Please correct me if I'm mistaken.
     
  10. Apr 7, 2015 #9

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    The problem statement in the that thread in question did say, "Suppose a beam of light enters the fiber from air at an angle [...]," then in the attached figure it labeled that angle as [itex] \theta_A [/itex] and index refraction of the air as [itex] n_A [/itex].

    It's safe to assume that [itex] n_A = n_{air} = 1[/itex] for this particular problem (in the quoted thread) only. Only because it was written that way, specifying that the beam entered the fiber from air.

    Here is the image that I'm referring to in that thread:

    83p011-jpg.62742.jpg
     
  11. Apr 7, 2015 #10
    Thank you for the clarification! I really need to spend more time actually reading the question. :P
     
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