3D refraction of ray / vector operations

risingedge
risingedge used Ask the Experts™
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Dear experts,

  I've been working at a problem for a while.  If anyone can help, I'd really appreciate it.  I'm working on refraction of an ultrasound beam as it crosses a planar interface between a Rexolite wedge and a steel sample block.  I've got it mostly worked out, but am struggling with a (the) key parameter:  How to define the (3D) vector after refraction across the plane.  I can get the angle with relation to the plane-normal by Snell's Law, and I'm assuming the vector will be in the plane defined by the normal and the incident wave, and the line will contain the point at which it intersects the plane.  This seems like it should be easy given the information I've already calculated, but I just can't seem to get the keystone in the top of the arch.  20140227-BeamPathWork-CloseUp.pdf  
Can anyone help?

Thank You,
-Matt
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Commented:
If your unit incident ray is i and your unit normal is n,
and the ratio of wavelength between the two media is r,
then your refracted ray will be
ri+n(rn·i-sqrt(1-r^2+(rn·i)^2)))
risingedge

Author

Commented:
You are good.20140227-BeamPathWork-CloseUp-2.pdf

Thank You, now I can go to sleep.
risingedge

Author

Commented:
Sorry, I have a follow-up:  Does this equation have a name?  I wanted to look more into it to see if there were any assumptions (small-angle, etc.).  I'm getting angles that differ slightly from direct Snell results.
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ozo
Most Valuable Expert 2014
Top Expert 2015

Commented:
If the number in the square root is negative, you'd have total internal reflection.
What are the numbers you are using when you get slightly different angles?
risingedge

Author

Commented:
If I try the simple case:
i=<0,0,1>
n=<1/Sqrt[2],0,1/Sqrt[2]>
r=0.73

I get <-0.28445, 0., 1.01465> which is 29.3 degrees from -n

If I use, Snell's Law:
Theta2 = 45 Degree;
Theta1 = ArcSin[Sin[Theta2]*0.73] = 31.1

I'm sure it's user error, but I'm searching for it.
Thanks again!
-Matt
ozo
Most Valuable Expert 2014
Top Expert 2015

Commented:
sorry, I used r and n the wrong way, try
ri-n(rn·i-sqrt(1-r^2+(rn·i)^2)))
risingedge

Author

Commented:
Hmm... That gives the right angle, but looks to be the mirror image across the refraction plane (i.e. reflected).  BeamPathWork-3-copy.pdf

Do you have a source, or are you deriving?

Thanks for the help!
ozo
Most Valuable Expert 2014
Top Expert 2015

Commented:
Looks like I reversed the sign of i too.
risingedge

Author

Commented:
I'm sorry, can you spell it out for me?  I tried changing all of the i's to (-i)'s, and a few other permutations, but I failed.
ozo
Most Valuable Expert 2014
Top Expert 2015

Commented:
i=<0,0,1>
n=<0.707106781186548,0,0.707106781186547>
r=0.73
ir-n(ir·n-sqrt(1-r^2+(ir·n)^2)))=<0.240619517519044,0,0.970619517519044>
-ir-n(-ir·n-sqrt(1-r^2+(-i·rn)^2)))=<0.970619517519044,0,0.240619517519044>
Is this what you got, and does it make sense with your conventions?
risingedge

Author

Commented:
Vectors
I really do appreciate your considerations.
ozo
Most Valuable Expert 2014
Top Expert 2015

Commented:
I'm not sure what convention to use, in http:#a39894769, the norm was pointing with the incident ray, but in http:#a39896563, the norm is pointing against the incident ray, so the sign convention changes.

With
i={0,0,1}
n={0.70710678118655,0,-0.70710678118655}
r=0.73
the ray that makes sense to me when I draw it is
ri - n(ri·n + sqrt(1 - r^2 + (ri·n)^2)) = {-0.240619517519044,0,0.970619517519044}

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