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3D refraction of ray / vector operations

Posted on 2014-02-27
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Last Modified: 2014-03-01
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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Question by:risingedge
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12 Comments
 
LVL 84

Accepted Solution

by:
ozo earned 2000 total points
ID: 39893989
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)))
0
 

Author Comment

by:risingedge
ID: 39894079
You are good.20140227-BeamPathWork-CloseUp-2.pdf

Thank You, now I can go to sleep.
0
 

Author Comment

by:risingedge
ID: 39894168
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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LVL 84

Expert Comment

by:ozo
ID: 39894191
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?
0
 

Author Comment

by:risingedge
ID: 39894769
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
0
 
LVL 84

Expert Comment

by:ozo
ID: 39896325
sorry, I used r and n the wrong way, try
ri-n(rn·i-sqrt(1-r^2+(rn·i)^2)))
0
 

Author Comment

by:risingedge
ID: 39896363
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!
0
 
LVL 84

Expert Comment

by:ozo
ID: 39896369
Looks like I reversed the sign of i too.
0
 

Author Comment

by:risingedge
ID: 39896463
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.
0
 
LVL 84

Expert Comment

by:ozo
ID: 39896507
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?
0
 

Author Comment

by:risingedge
ID: 39896563
Vectors
I really do appreciate your considerations.
0
 
LVL 84

Expert Comment

by:ozo
ID: 39896851
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}
0

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