How Does a Half Wave-Plate Affect Polarization at Various Angles?

[Niles]

Hi

Say I have a linearly polarized optical field traveling along z given by
$$E(z, t) = E(\cos(kz-\omega t)\hat x + \cos(kz-\omega t)\hat y)$$
In other words, it lies in the 1st and 3rd quadrant. Now the wave passes through a half wave-plate at an angle of 45 degrees (the wave-plate is along y pointing upwards), so the wave becomes
$$E(z, t) = E(\cos(kz-\omega t)\hat x - \cos(kz-\omega t)\hat y)$$
In other words, it now lies in the 2nd and 4th quadrant. Now I let is pass through a second half wave-plate as shown in the attachment (the red line is the fast axis of the retarder, and the black line is the optical field). Say the angle between the fast axis of the retarder and the polarization of the field is some Ω.

Now I know that after passing through the retarder, the polarization attains an angle of 2Ω besides the 45 degrees from the y-axis. There is two obvious ways for me to represent this:

1) I can do it in terms of a new coordinate system spanned by the axes parallel and perpendicular to the fast axis of the retarder.
2) Use the same "old" coordinate system as with the first retarder

Say I go with #2. What is the easiest way to do this? The only way I see how to do it is to do #1 first, but it seems quite tedious.Niles.

 

[eljose79]

Hi Niles,

Thank you for sharing your question and the interesting scenario you have presented. I would like to offer some insights on how to approach this problem.

Firstly, it is important to understand the concept of a half wave-plate and its effect on polarized light. A half wave-plate is an optical device that introduces a phase difference of half a wavelength between the two perpendicular components of a polarized light. This effectively rotates the polarization of the light by 90 degrees.

In your scenario, the initial polarized light is traveling along the z-direction, with components in the x and y directions. When it passes through the first half wave-plate at a 45-degree angle, the x-component is unchanged while the y-component is rotated by 90 degrees, resulting in a polarization along the x-axis. This is represented by your second equation, where the y-component has a negative sign.

Now, when this polarized light passes through the second half wave-plate at an angle Ω, it effectively introduces an additional phase difference of 2Ω between the x and y components. This results in a rotation of the polarization by an angle of 2Ω, as you have mentioned.

To represent this in terms of the "old" coordinate system, you can use the Jones matrix formalism. This involves representing the optical field as a 2x1 column vector, with the x and y components as the elements. The half wave-plate can be represented by a 2x2 matrix, with the elements representing the phase difference introduced by the device.

In your case, the first half wave-plate can be represented by a matrix [1 0; 0 -1], and the second half wave-plate can be represented by a matrix [cos(2Ω) sin(2Ω); -sin(2Ω) cos(2Ω)]. To obtain the final polarization in the "old" coordinate system, you can simply multiply the two matrices to get the resultant matrix, and then convert it back to the 2x1 column vector representing the x and y components of the polarization.

I hope this helps you in finding the easiest way to represent the final polarization in the "old" coordinate system. Let me know if you have any further questions.