Calculate intial angle of a refracted wave

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BOYLANATOR
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Please see attached image for problem and brief description of attempted solution.

Diagram of a seismic-style ray path from a source at top left to receivers along the top, with two layers of velocities V1 above and V2 below, depths z1 and z2, and angles theta1 and theta2; Snell's Law and the starting equation D/2 = z1 tan theta1 + z2 tan theta2 are written alongside.
 
BOYLANATOR said:
Please see attached image for problem and brief description of attempted solution.

[ ATTACH=full]84469[/ATTACH]
Yes this is solvable.
 
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.
 
BOYLANATOR said:
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.
Do you know how the velocities are related to index of refraction?
 
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.
 
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Lets say that D/2 is the sum of the horizontal distances traveled 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 traveled in the first layer.
 
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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?