Birefringence Questions | Help with Solutions

  • #1
chimneydials
2
0
Hi everyone,

I have some questions about birefringence. I have searched in vain on
the Internet and in a few books (it's tough to find books on
birefringence).

Usually determining how rays propagate after birefringence is simple
because the light is incident normally and the ne of the e-ray is
known.

But in my situation, I need to analyze light that is incident at an
angle. Moreover, the ne is also not known. As in only the maximum ne
is known but ne as you know varies with the angle between the o-ray
and the optic axis (if I am not wrong). So, ne should also change with
the incident angle. Does anyone have an equation that takes in the
incident angle, the max ne and the no and finds the walkoff angle +
refraction angle?

One more question: is birefringence expressed purely by differences in
ne and no? So, suppose I know the ne for a given situation, I should
be able to find the difference in angle between the o-ray and e-ray
using only Snell's law? Or is there a separate equation for Poynting
walkoff?

I am rather confused about these topics. Basically, I understand
birefringence conceptually but I have been unable to find appropriate
equations to apply for specific optical systems involving
birefringence. Hope some of you can help me with this. Even if you
don't know the answers to the questions, it would be helpful if you
can point me to some good sources either online or on paper.

Thanks.
 
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  • #2
chimneydials said:
Hi everyone,

I have some questions about birefringence. I have searched in vain on
the Internet and in a few books (it's tough to find books on
birefringence).
Birefringence is tackled in most books on nonlinear optics, and any book on anistropic media.
chimneydials said:
But in my situation, I need to analyze light that is incident at an
angle. Moreover, the ne is also not known. As in only the maximum ne
is known but ne as you know varies with the angle between the o-ray
and the optic axis (if I am not wrong).
Correct.
chimneydials said:
So, ne should also change with
the incident angle. Does anyone have an equation that takes in the
incident angle, the max ne and the no and finds the walkoff angle +
refraction angle?
n_e(theta) = (n_o)^2 .cos(theta) + (n_e max)^2 .sin(theta) from memory. Once you know n_e, you can use Snell's law and the equation in the link below to figure out the rest.
chimneydials said:
One more question: is birefringence expressed purely by differences in
ne and no?
In a uniaxial crystal, yes. In a given crystal, the cut of the crystal with respect to the crystal axis is also important.
chimneydials said:
So, suppose I know the ne for a given situation, I should
be able to find the difference in angle between the o-ray and e-ray
using only Snell's law? Or is there a separate equation for Poynting
walkoff?
Snell's law doesn't apply as it only gives you k.
http://www.rp-photonics.com/spatial_walk_off.html
chimneydials said:
I am rather confused about these topics. Basically, I understand
birefringence conceptually but I have been unable to find appropriate
equations to apply for specific optical systems involving
birefringence. Hope some of you can help me with this. Even if you
don't know the answers to the questions, it would be helpful if you
can point me to some good sources either online or on paper.

Thanks.
The RP photonics site that I linked to has a whole "photonics encyclopedia" that might be of some help.

Claude.
 
  • #3
Approach to calculating birefringent angle-HELP

Thanks for your help and the links. I think I have figured this out. Tell me if I am doing it right.[PLAIN]http://www.texify.com/img/\LARGE\!\frac{1}{(n_e(\theta))^2}%20=\frac{cos(\theta)}{(n_o)^2}%20+%20\frac{sin(\theta)}{(n_e)^2}.gif

I believe this is the equation that expresses the ne (not the max ne) in this case with respect to the theta. First, I am not sure what the theta is. Is it the angle of refraction of the e-ray with respect to the surface normal? Or is it the angle between the optic axis and the e-ray? I think it is the latter because the ne must depend on the angle between the polarisation state and the optic axis (that is what birefringence is all about). But I am not sure... so can someone tell which is right?

So, once I know what this angle is. I can solve the system of equations below to find the ne and the angle of refraction of the e-ray.

http://www.texify.com/img/\LARGE\!{n_{air}sin(\theta_{incident})}%20={n_esin(\theta_2)}.gif
[PLAIN]http://www.texify.com/img/\LARGE\!\frac{1}{(n_e(\theta))^2}%20=\frac{cos(\theta)}{(n_o)^2}%20+%20\frac{sin(\theta)}{(n_e)^2}.gif

After this, I can calculate the walkoff angle with the equation below. Here also, I am not sure which ne this is (I know the differential must be the first equation above but how about the other ne)? Is the max ne or is it the ne calculated earlier?

wo_angle.gif


Once the walkoff angle is calculated, I can sum up the two angles (angle of refraction and walkoff angle) to find the actual angle of the e-ray with respect to the normal, can't I?

Summary of steps:

1. Find refraction angle of e-ray using snell's law and ne equation
2. Find walkoff angle using ne equation, refracted angle and ne found earlier
3. Sum up two angles to obtain the actual angle of the e-ray with respect to the normal in any biref crystal

So, I should be able to find the angle of e-ray inside a birefringent prism depending on the angle it is incident at.

Is there anything wrong with any step in my approach?

Thanks for the help.
 
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  • #4
chimneydials said:
Thanks for your help and the links. I think I have figured this out. Tell me if I am doing it right.[PLAIN]http://www.texify.com/img/\LARGE\!\frac{1}{(n_e(\theta))^2}%20=\frac{cos(\theta)}{(n_o)^2}%20+%20\frac{sin(\theta)}{(n_e)^2}.gif

I believe this is the equation that expresses the ne (not the max ne) in this case with respect to the theta. First, I am not sure what the theta is. Is it the angle of refraction of the e-ray with respect to the surface normal? Or is it the angle between the optic axis and the e-ray? I think it is the latter because the ne must depend on the angle between the polarisation state and the optic axis (that is what birefringence is all about). But I am not sure... so can someone tell which is right?
Angle between the ray (or wave-vector if you like) and the crystal axis. A UNIAXIALLY birefringent crystal will have only one axis, the angle you are looking for is the angle the wavevector makes with that axis.
chimneydials said:
So, once I know what this angle is. I can solve the system of equations below to find the ne and the angle of refraction of the e-ray.

http://www.texify.com/img/\LARGE\!{n_{air}sin(\theta_{incident})}%20={n_esin(\theta_2)}.gif
[PLAIN]http://www.texify.com/img/\LARGE\!\frac{1}{(n_e(\theta))^2}%20=\frac{cos(\theta)}{(n_o)^2}%20+%20\frac{sin(\theta)}{(n_e)^2}.gif
Yes.
chimneydials said:
After this, I can calculate the walkoff angle with the equation below. Here also, I am not sure which ne this is (I know the differential must be the first equation above but how about the other ne)? Is the max ne or is it the ne calculated earlier?

[PLAIN]http://www.rp-photonics.com/eqn/wo_angle.gif[/QUOTE]
It's Ne not Ne max, as Ne is a function of angle, while Ne max is not.
chimneydials said:
Once the walkoff angle is calculated, I can sum up the two angles (angle of refraction and walkoff angle) to find the actual angle of the e-ray with respect to the normal, can't I?
Depending on whether the crystal is negative uniaxial (ne < no) or positive uniaxial (ne > no), the walk of angle may be a negative number, which could be construed as a subtraction in some formalisms - By keeping everything consistent by not omitting any negative signs, a simple summation should suffice.
chimneydials said:
Summary of steps:

1. Find refraction angle of e-ray using snell's law and ne equation
2. Find walkoff angle using ne equation, refracted angle and ne found earlier
3. Sum up two angles to obtain the actual angle of the e-ray with respect to the normal in any biref crystal

So, I should be able to find the angle of e-ray inside a birefringent prism depending on the angle it is incident at.

Is there anything wrong with any step in my approach?
Sounds good, just be careful by when you refer to the "ray". When you get walk off, the Poynting vector and the wavevector are not colinear (they are separated by the walk-off angle), so when you say "ray" you really should specify which vector you are referring to. It is only the Poynting vector that gets "deflected" if you like, the wavevector acts as normal, obeying Snell's law.
chimneydials said:
Thanks for the help.
You're welcome :smile:.

Claude.
 
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