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Calculate intial angle of a refracted wave

  1. Jun 4, 2015 #1
    Please see attached image for problem and brief description of attempted solution.

    upload_2015-6-4_18-26-28.png
     
  2. jcsd
  3. Jun 4, 2015 #2
    Can you be more clear on what exactly is known?
     
  4. Jun 4, 2015 #3

    SammyS

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    Yes this is solvable.
     
  5. Jun 4, 2015 #4
    I'm looking for a general solution when you know the velocity contrast of the layers (to give the ratio for snells law) and you know the thickness of the two layers as well as the separation of the source and receiver.
     
  6. Jun 4, 2015 #5

    SammyS

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    Do you know how the velocities are related to index of refraction?
     
  7. Jun 4, 2015 #6
    Yes v2/v1 =n1/n2. But whether I use the ratio of refractive indices or the ratio of velocities doesn't really matter. Either way they are just two known variables that carry through.

    Without writing out my attempt at a derivation in full, what I did was re-write Θ2 in terms of Θ1. Then you get a nasty term of the form tan(arcsin(x)) which can be replaced by [itex]x/(sqrt(1-x^2))[/itex]. I get rid of the other tan(theta) in terms of opp/adj in an attempt to first work them out (getting theta back will be easy at the end) but then the equations are hard to solve.

    Sorry I don't have my workings here. I can write out some maths tomorrow.
     
    Last edited: Jun 4, 2015
  8. Jun 8, 2015 #7
    Lets say that D/2 is the sum of the horizontal distances travelled in the two layers (only considering the downgoing wave):

    [itex]D/2 = (d_1+d_2=) z_1tan\theta_1+z_2tan\theta_2[/itex]

    From Snell's Law:

    [itex]\theta_2=sin^{-1}(\frac{v_2}{v_1}sin\theta_1[/itex])

    [itex]D/2 = z_1tan\theta_1+z_2tan(sin^{-1}(\frac{v_2}{v_1}sin\theta_1))[/itex]

    There is a substitution of the form:

    [itex]tan(sin^{-1}(x))=\frac{x}{\sqrt{1-x^2}}[/itex]

    Therefore,

    [itex]D/2 = z_1tan\theta_1+z_2\frac{\frac{v_2}{v_1}sin\theta_1}{\sqrt{1-(\frac{v_2}{v_1}sin\theta_1)^2}}[/itex]

    From here, I wasn't sure what direction to go in. I tried swapping the angles for known distances using basic trig but the algebra became very long. The goal is really just to rearrange for either [itex]\theta_1[/itex] or if it's easier to swap out the angles then I would want to solve for either [itex]d_1[/itex] where [itex]d_1[/itex] is the horizontal distance travelled in the first layer.
     
    Last edited: Jun 9, 2015
  9. Jun 11, 2015 #8
    I needed the solution for this to run a computer model. I have since just solved the problem iteratively in MATLAB but I am interested to know if the maths leads to a reasonably neat solution. Any ideas?
     
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