Linearly polarized light on a quarter wave plate

[macaholic]

Homework Statement

A linearly polarized beam propagates in the z-direction with its E-field
oscillating in the y-direction. It is incident on a quarter wave plate (QWP) located in the
x-y plane at the origin.

a. How should the fast and slow axes of the QWP be oriented if the beam emerging
from it is to be right hand circularly (RCP) polarized? State the directions.

b. Assume the light wave incident on the QWP can be written as

$E(z, t)= E_{0}\widehat{y} e^{i(kz -\omega t)}$. What is the complex unit vector that describes the
polarization of the wave leaving the QWP? Write this down explicitly in terms of $\widehat{x}$ and $\widehat{y}$ . Write down an explicit form of the wave.

c. Now another identical QWP is placed in the beam following the first one, and its
fast and slow axes are oriented in exactly the same manner as the first one. Write
down a form for the light wave that emerges from the second QWP.

Homework Equations

Not sure what should be here. There's always the form of the incident E-field: $E(z, t)= (E_{f}\widehat{f} + E_{s}\widehat{s})e^{i(kz -\omega t)}$ where the f and s axes are the quarter-wave plate's fast and slow axes.

The Attempt at a Solution

a. For this one I just sort of reasoned it out geometrically. I said the fast and slow axes had to be at 135 degrees and 225 degrees (second and third quadrant). I'm not sure if this is right though. I think this makes it right handed polarized from the point of view of the receiver.

b. I'm not entirely sure what to do about the vector describing the polarization. I know it has to be rotating, so my best guess would be: $(\widehat{x} + \widehat{y})e^{-i(\omega t)}$. I'm not sure what exactly defines the "polarization vector". Is it just the vector that points in the plane of the E-field at any given time?

I think since the fast and slow axes make a 45 degree angle with polarization of the light (y axis) then the form of the emitted wave should just be: $E(z, t)= \frac{E_{0}}{\sqrt{2}} (\widehat{x} + \widehat{y})e^{i(kz -\omega t)}$
Although I could be oversimplifying things.

c. This one is really confounding me. I think that if the polarizer was of the opposite handidness, then there would be zero emitted light (or I could be wrong on this too). However, since it's an identical polarizer, I'm not really sure what would happen. My instinct is that nothing would happen to the emitted light, but I don't have much to back this up with.

Thanks for the help.

 

[mark726]

a. You are correct in your reasoning. The fast and slow axes of the QWP should be at 135 degrees and 225 degrees (second and third quadrant) in order for the beam to emerge as right hand circularly polarized.

b. The polarization vector is defined as the vector that points in the direction of the electric field at any given time. In this case, the polarization vector can be written as \widehat{y}e^{i(\omega t)}. To find the wave emitted by the QWP, you can simply multiply the incident wave by the Jones matrix of the QWP, which is given by:

\begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix}

So the emitted wave would be:

E(z, t) = \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix} \begin{bmatrix} E_{0} \\ 0 \end{bmatrix} e^{i(kz -\omega t)} = E_{0}\widehat{x}e^{i(kz -\omega t)} + iE_{0}\widehat{y}e^{i(kz -\omega t)} = E_{0}\widehat{x}e^{i(kz -\omega t)} - E_{0}\widehat{y}e^{i(kz -\omega t)}

c. You are correct that if the second QWP is of the opposite handidness, then there would be no emitted light. However, since it is an identical QWP, the light would just pass through unchanged. So the form of the wave emerging from the second QWP would be the same as the incident wave: E(z, t) = E_{0}\widehat{y}e^{i(kz -\omega t)}