# How to calculate birefringence in lithium niobate LiNbO3

## Main Question or Discussion Point

I am trying to calculate what thickness of LiNbO3 would displace a beam of light 9 microns. I seek to make something like a microscope slide that would displace light in the vertical. I am confused about something else if someone can clarify; for the crystal, I think the "z" axis is "optical axis." I think from what I have read that light that travels this axis in any polarization is unaffected, Is this correct? If so it confuses specifying the dimensions of the slide, as I think of the z dimension as being the thickness. A "z-cut" crystal in this case would not displace at all would it?

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blue_leaf77
Homework Helper
Again, you need to know the crystal's optic axis orientation with respect to the incoming ray. For an arrangement such that the input plane is parallel to the so-called principal plane of the (uniaxial) crystal (like the one in the picture below) and is normal to the interface, one can prove that the deviation angle ##\theta_s## between e- and o-rays is given by
$$\cos \theta_s = \left( \frac{\cos^2\alpha}{n_o^2} + \frac{\sin^2\alpha}{n_e^2} \right) \left( \frac{\cos^2\alpha}{n_o^4} + \frac{\sin^2\alpha}{n_e^4} \right)^{-1/2}$$
where ##\alpha## is the angle subtended by the incoming ray and optic axis. The e- and o-rays will propagate at the same direction when ##\alpha = 0## or ##90## degree. This means if you want your crystal to exhibit double refraction, you cannot align it such that the incoming ray is parallel with the optic axis.

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I am thankful for anyone's time, let me try again.
Goal: a rectangular wafer of that when placed in front of a camera image sensor will displace light(visible) 9 microns vertically, and how to specify it to order it properly!
1) LiNBO3 crystals available for purchase off the shelf are specified as x-cut, y-cut, or z-cut. -which cut is appropriate?
2) With this cut/orientation, what is the calculation to achieve a 9 micron displacement? Could a helpful chap walk me through solving for thickness. I would like to understand and be able to solve for different displacements.
If you are interested in why I am trying to do this, it is to solve a problem my camera has. The camera sensor is a large megapixel one with many rows of pixels for high still resolution. In order to record video the camera must reduce the resolution to the smaller size, the camera throws out the information on 2 out of every 3 horizontal sensor lines. This causes Moire, and Aliasing. I seek to use a LiNBO3 wafer in front of the sensor and behind the lens to displace information that would normally resolve in the middle of the unrecorded lines to the recorded line. So both the information that would normally resolve on the recorded line and the information that would normally have resolved in the unrecorded area are combined (blurred) but only in the vertical. To achieve the a 9 micron shift- Too thick a crystal like calcite or quartz will affect the "back focus" resolving the image in front of the sensor or behind, so focus problems. Too thin, like yttrium and fragility becomes an issue.
I just want to try it as an experiment, and for the fun of solving a problem and learning something new. I am having a hard time pushing through from here and humbly ask for help, or even a referral to someone who might be able to, thanks

DrDu
Here are the X, Y and Z cuts .
X and Y cuts should be equivalent, optically.
However you need a cut which is between an X and a Z cut.
To be more precise, double refraction will only occur if the incoming light ray is neither parallel nor perpendicular to the optical axis. In any case the splitting of the rays will depend on the angle of the incident ray.

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blue_leaf77
Homework Helper
How thin do you want the crystal to be? The thinnest birefringent crystal wafer provided by newlight http://www.newlightphotonics.com/Birefringent-Crystals/Calcite-Crystals is 0.2 mm. Also, to obtain a specific beam displacement may require the crystal to be rotated, i.e. adjust the incident angle, and this wil cause the ordinary ray to be displaced too. Will such situation be acceptable in your application?

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DrDu
The calculation of the angle can be found in Landau Lifshitz, Electrodynamics of continuous media in the excercises for chapter 98. For normal incidence on a material whose optical axis is at an angle ##\alpha## with the incident ray, the angle ##\theta## of the extraordinary ray relative to the normal of the surface is determined as
## \tan \theta=\frac{(\epsilon_\parallel-\epsilon_\perp)\sin 2\alpha }{\epsilon_\parallel+\epsilon_\perp+(\epsilon_\parallel-\epsilon_\perp)\cos 2\alpha}##
The angle dependence of the denominator can be neglected and the numerator is maximal for ##\alpha=45^\circ##.
If ##d## is the thickness of the wafer, the displacement on exit relative to the ordinary ray is
##\Delta x=d \tan \theta##.
For Li NbO3, the birefringence is 0.09 and ##\epsilon=2.25##. For a wafer with ##\alpha=45^\circ## and d=500 ##\mu##m we find
##\Delta x=10 \mu m##
If you want to be able to adjust the displacement, you could use two plates. If they are oriented alike, the displacements add, while for anti-parallel orientation, they cancel. You can get any value inbetween for the displacement by just changing the angle.

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Andy Resnick