How to deal with a circular light boundary conditions

[Othman0111]

Homework Statement

Calculate the transmission and reflection coefficients as functions of the incident angle for right-handed circularly polarized light.

Are they the same or different for left-handed circularly polarized light?

Relevant Equations

included

Hi everyone,

in the attached file I tried to find the transmitted and the reflected coefficients.

I ran into trouble applying the boundary conditions to the linear components of the electric field.

Check the outlined boxes and see if they make sense.

Thanks

 

[hutchphd]

You realize they should sum to 1 ( R+T=1) ? Also what do you expect for part b?

 

[Othman0111]

You realize they should sum to 1 ( R+T=1) ? Also what do you expect for part b?

I'm sure my answer is not correct because of the boundary conditions, they don't make any sense. For part b I think they are the same.

 

[hutchphd]

I'm sure my answer is not correct because of the boundary conditions, they don't make any sense. For part b I think they are the same.

What about the boundary conditions isn't sensible? In particular there are conditions on the in- plane (parallel) and normal (perpendicular) components of E and B at the surface.. Have you just made an error you can't find or is it worse? I'll go through it if necessary (but not quite yet!)
Wait I just looked. Why do you say k⋅z is 0?

 

[hutchphd]

This is a nontrivial effort but quite important to understand. It is done (more or less completely) in every undergrad EM text. (I like Griffiths). Also I think you don't have the geometry quite correct in your head. Take another look and get back.

 

[Othman0111]

What about the boundary conditions isn't sensible? In particular there are conditions on the in- plane (parallel) and normal (perpendicular) components of E and B at the surface.. Have you just made an error you can't find or is it worse? I'll go through it if necessary (but not quite yet!)
Wait I just looked. Why do you say k⋅z is 0?

Because the plane of the medium is at z=0.
Something about the first boundary condition (D1 = D2) I don't have two equations (Ex and Ey) I only have one!

 

[Othman0111]

This is a nontrivial effort but quite important to understand. It is done (more or less completely) in every undergrad EM text. (I like Griffiths). Also I think you don't have the geometry quite correct in your head. Take another look and get back.

I looked through Griffiths and Jackson and (4+ hours of searching the internet) nobody has worked the boundary conditions for a right-handed circularly polarized light between two mediums! All that I found is working the linear E field as two cases (perpendicular and parallel)
If you find any please send me a link. Thanks

 

[hutchphd]

But if you have the solution for the two incident perpendicular linear polarizations, then each circular incident polarization is just linear combination of the two and all the resultant scattered amplitudes (with phase ) are the same linear combination. The coefficients R and T are mod squared and so they need to be worked from these new amplitudes.
How do we write down circular polarized light?

 

[Othman0111]

But if you have the solution for the two incident perpendicular linear polarizations, then each circular incident polarization is just linear combination of the two and all the resultant scattered amplitudes (with phase ) are the same linear combination. The coefficients R and T are mod squared and so they need to be worked from these new amplitudes.
How do we write down circular polarized light?

The right cricular one
$\displaystyle \overrightarrow{E_{R}} \ =E_{0} \ \left(\hat{x} -i\hat{y}\right) \ e^{i\left(\vec{k} .\vec{z} -\omega t\right)} \$

The left circular one
$\overrightarrow{E_{L}} \ =\ E_{0} \ \left(\hat{x} +i\hat{y}\right) \ e^{i\left(\vec{k} .\vec{z} -\omega t\right)}$

The y componenet is 90 degree out of phase with x component.

 

[hutchphd]

Exactly. So x can represent the ⊥ direction and y the || (or however you want to set it up to match your literature) . Shamelessly copy their results for transmitted and reflected waves for the ⊥ and || cases (be sure you understand what they did !). Form the same linear combination for the two scattered cases. Solve for the ratios you need.
Does Jackson do the whole thing? I think my ancient green one (1^st ed) has it all worked out for linear polarizations..

 

[Othman0111]

Exactly. So x can represent the ⊥ direction and y the || (or however you want to set it up to match your literature) . Shamelessly copy their results for transmitted and reflected waves for the ⊥ and || cases (be sure you understand what they did !). Form the same linear combination for the two scattered cases. Solve for the ratios you need.
Does Jackson do the whole thing? I think my ancient green one (1^st ed) has it all worked out for linear polarizations..

That what I did. But sadly the professor said it is wrong! Jackson Did the in-phase waves and in problem 7.29 I supposed to do it in out of phase situation (also didn't find a solution to this online).

I'm going to present this solution to the class tomorrow. What the worse can happen :)

 

[hutchphd]

That what I did. But sadly the professor said it is wrong! Jackson Did the in-phase waves and in problem 7.29 I supposed to do it in out of phase situation (also didn't find a solution to this online).

I'm going to present this solution to the class tomorrow. What the worse can happen :)

I am busy today but did work it out earlier. It is easy (almost trivial when you see it) to show that for either RHCP or LHCP the result for R (and T) is the root mean square average of the R_⊥ and R_ll (and T) for the linear case. This is then self-evidently unitary. The details of the angles I chose not to look at! Did you get to this result? Have fun...

 

[Othman0111]

I am busy today but did work it out earlier. It is easy (almost trivial when you see it) to show that for either RHCP or LHCP the result for R (and T) is the root mean square average of the R_⊥ and R_ll (and T) for the linear case. This is then self-evidently unitary. The details of the angles I chose not to look at! Did you get to this result? Have fun...

I made the assumption when a circular light hit, it reflects elliptical and transmits elliptical. I'm going to continue and see where that will lead me. Thanks

 

[hutchphd]

I made the assumption when a circular light hit, it reflects elliptical and transmits elliptical. I'm going to continue and see where that will lead me. Thanks

Yes that will get you there. Choose ll axis and ⊥ plane as basis directions for all waves. You can then use all the results in the books for linear polarization for your solutions directly. It should give the result I got. Also you need a normalization 1/√2 in your definitions of CP if you do the linear correspondences directly. I'm gone for the day...

 

[Othman0111]

I have completed all the work. Again, I'm not sure if it the right solution, but I'm going to present it tomorrow nonetheless.