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Polarisation in non sequential mode

  • 22 December 2022
  • 3 replies
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Hi All,

I am using a source diode which is propagating the light. The light rays pass through a laser rod from the side (cross sectionally) and the laser rod is in Z-plane(Longitudinally).

 

 

So if I want to assign TE polarisation to my source diode rays, is the following setup is right?

Thanks in advance

Akhil

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Best answer by Ethan 27 December 2022, 22:26

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3 replies

Userlevel 3
Badge +2

Hi @Akhil,

The orientation of the electric or magnetic field also depends on your chosen Polarization Reference Method. The X Axis Reference is set by default since many systems use the Z-Axis as the direction of propagation. However, in your case, the rays propagate along the X-Axis so using a Y- or Z- Axis Reference would be a better choice. 

 

When you define Jx and Jy (Jones vector components), OpticStudio uses the following definitions to compute the appropriate E fields, where Jx is along S and Jy is along P:

  1. X Axis Reference: The P vector is determined from K cross X, and S = P cross K.
  2. Y Axis Reference: The S vector is determined from Y cross K, and P = K cross S.
  3. Z Axis Reference: The S vector is determined from K cross Z, and P = K cross S.

Therefore, if we choose Y Axis Reference and we set Jx = 1 / Jy = 0, then you will have a TE polarization for each ray hitting the structure.

For more detailed information about defining polarization, please see the Help File: The Setup Tab > System Group (the Setup Tab) > System Explorer > Polarization (System Explorer) > Defining the Initial Polarization. 

Best,
Ethan

Hi @Ethan

Thank you so much for your response. I want to clarify about one more thing.

If  my ray travells on X-Y plane with an angle (for eg: 20° from the X - Axis and 70° from the Y-Axis), then can you please explain to me how can I assign the TE polarisation to the didode rays?

Kind regards

Akhil

Userlevel 3
Badge +2

Hi @Akhil,

Please see this post for my answer: Polarisation defining | Zemax Community.

Best,
Ethan 

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