Ordinary and extraordinary refractive index

In summary, the difference between ordinary and extraordinary refractive index is that extraordinary refractive index has a larger ratio of amplitudes of elliptical polarization.
  • #1
skrat
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Homework Statement


From an uniaxial crystal, a 0.7 mm thick plate is cut out in a way that the optical axis is parallel to the surface. A light falls perpendicular to the surface (circular polarization) with 500 nm. After the crossing of the plate, the light has elliptical polarization and the ratio of the amplitudes is 0.5 . What is the difference between ordinary and extraordinary refractive index?

Homework Equations

The Attempt at a Solution


The idea is:
Initial Jones vector is $$J_i=\binom{1}{i}$$ while the output is $$J_f=\binom{e^{ik_od}}{ie^{ik_ed}},$$ if "o" stands for "ordinary", "e" for "extraordinary" and ##d## is the thickness of the plate. Final Jones vector can be rewritten as $$J_f=e^{ik_ed}\binom{e^{i(k_o-k_e)d}}{i}\approx \binom{e^{i\delta }}{i}$$ for ##\delta=(k_o-k_e)d=k_0(n_o-n_e)d##.

Now the idea is that from the fact that after the crossing of the plate, the light has elliptical polarization with ratio of amplitudes ##1/2##, I am supposed to somehow find out how much ##\delta ## is. But have no idea how to do that. I tried to use the form of final Jones vector: $$E_x=e^{i\delta}e^{-i\omega t}$$ and $$E_y=ie^{-i\omega t}.$$ If I consider only the real parts and forget about the imaginary parts, this simplifies to $$E_x=\cos\delta\cos\omega t+\sin\delta\sin\omega t$$ and $$E_y=\sin\omega t.$$ Inserting the last identity into the previous one leaves me with general equation for ellipse $$\frac{E_x^2}{\cos^2 \delta}+\frac{E_y^2}{\sin^2 \delta}-\frac{2E_xE_x\sin\delta}{\cos^2\delta}=1$$ and I am not really sure how this helps me. I wanted to somehow express ##\cos\delta## or ##\sin\delta## with the amplitudes ##E_x## and ##E_y##...

Any help will make my day!
 
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  • #2
skrat said:
the ratio of the amplitudes is 0.5
If the fast and slow axes of the uniaxial crystal are along the directions in which the Jones matrix of the incoming field is defined (e.g. x and y directions), then the amplitude of the outgoing field should not change as can be check from the matrix form of the corresponding uniaxial crystal configuration. I think you actually wanted to refer to the ratio between the major and minor axes of the (possibly) rotated ellipse. If that's really the case, you will want to find first the expression of the major and minor axes of a rotated ellipse.
 
  • #3
blue_leaf77 said:
I think you actually wanted to refer to the ratio between the major and minor axes of the (possibly) rotated ellipse. If that's really the case, you will want to find first the expression of the major and minor axes of a rotated ellipse.

That is what I was referring to, yes, the major and minor axes of a rotated ellipse.
And again yes, my idea was the same - to find the expression of the major and minor axes of rotated ellipse.

BUT I finally managed to fined some equations about ellipse here http://glwaves.sourceforge.net/derivations/ . So I hope that below I have written a correct solution:
Following the link above, the phase difference is $$ \delta =arccos(\frac{Y\sin\alpha \cos\tau + X\cos\alpha\sin\tau }{Y\sin\beta \cos\tau + X\cos\beta \sin\tau}),$$ where ##X## and ##Y## are the major and minor axis, therefore ##X/Y=1/2##, ##\alpha =\arctan(-\frac{Y\sin \tau}{X\cos\tau})## and ##\beta= \arctan(\frac{Y\cos \tau}{X\sin\tau})##. Here ##\tau## is inclination and is the only parameter I need in order to finish with this problem.

Is there a way how one can calculate this with the data given?
 
  • #4
skrat said:
Inserting the last identity into the previous one leaves me with general equation for ellipse $$\frac{E_x^2}{\cos^2 \delta}+\frac{E_y^2}{\sin^2 \delta}-\frac{2E_xE_x\sin\delta}{\cos^2\delta}=1$$
That doesn't look like an equation for an ellipse formed from two superposing perpendicular E field with equal amplitude. I suggest that you look at equation (2.11) in here : http://www.diss.fu-berlin.de/diss/s...ionid=19537D2F32F0D3B282B1D2D885A7732A?hosts=
In accordance to that link, I suggest that you put all phase differences in one component such that in the expression of ##J_f## you take the ##e^{ik_od}## out of the column matrix, instead of ##e^{ik_ed}##. The expression of the total phase difference then also contains the 900 angle inscirbed in the y-component.

Regarding the link you gave there, it involves many symbols and it is not designed to extract the phase difference given the ellipticity. While you can pursue the way suggested in that link, I suggest that you try finding the expression for the major and minor axes by viewing the problem as an optimization problem. So you simply want to find the largest and smallest values of ##f(x,y) = x^2+y^2##, which denotes the distance from the origin to a point (x,y), with constraint ##x^2+y^2-2xy\cos{\delta} = \sin^2{\delta}## (the correct equation for an ellipse). Then use the fact that ##\sqrt{f_{max}/f_{min} }= \frac{1}{2}## (the known quantity). Hint: Using Lagrange multiplier method in executing the minimization/maximation problem can be extremely helpful. I believe that if you go along this way, you will have to do much less effort than you would have to when using your link as a reference.
 
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FAQ: Ordinary and extraordinary refractive index

1. What is the difference between ordinary and extraordinary refractive index?

The ordinary refractive index is the measure of how much light is bent when passing through a material, while the extraordinary refractive index is the measure of how much light is bent when passing through a material in a specific direction.

2. How is the ordinary and extraordinary refractive index measured?

The ordinary and extraordinary refractive index can be measured using a refractometer, which measures the angle of refraction when light passes through a material.

3. What factors can affect the ordinary and extraordinary refractive index?

The ordinary and extraordinary refractive index can be affected by the chemical composition, density, and temperature of the material. Additionally, the direction and polarization of the incident light can also impact the refractive index.

4. What is the relationship between the ordinary and extraordinary refractive index?

The ordinary and extraordinary refractive index are related to each other through the birefringence, which is the difference between the two values. Birefringence is a measure of how strongly a material refracts light and is dependent on the properties of the material.

5. How are the ordinary and extraordinary refractive index used in practical applications?

The ordinary and extraordinary refractive index are used in various applications, including optics, telecommunications, and material characterization. They are also important in the production of polarizing filters and liquid crystal displays, as well as in the study of crystal structures and properties.

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