The wonders of anisotropy

In order to get an EM wave with a polarization of $-\theta$ , we must invert the y-component of the E field of the original EM wave.

Thus, it also means that the EM wave going through the vertical refracting material should have a phase difference of of $\pi$ (this is also a path difference of $\frac{ \lambda }{2}$ ) with respect to the EM wave of the horizontal component.

The path difference when a EM wave enters a refracting material is $nt$

Therefore, the path difference of the two waves after they are split up by the birefringent material can be equated to

$(n_{1} - n_{2})l = \frac{\lambda}{2}$

l= length of the material

the two refractive indexes are $n_{1}$ and $n_{2}$ , (also known as "birefringence" if I'm not wrong)

$\lambda$ is the wavelength

Solving gives us $l = 255 \times 10^{-6} m$ which is 255 micrometres.

Before the electromagnetic wave enters the birefringent material, it is polarised, meaning that the $x$ - and $y$ -components of the electromagnetic field oscillate in phase. Both components are positive at the same time at every point along the wave, and both components are negative at the same time at every point.

When the wave enters the material, each component of the wave has a different wavelength. Thus, they travel different optical path lengths even though they travel the same distance through the material. If $z$ is the distance travelled by the wave, the optical path difference is given by: \begin{equation} \Delta = (n+0.001)z - nz = 0.001z \end{equation} The difference in phase between the two components is given by: \begin{equation} \phi = \phi_0 + \frac{2\pi}{\lambda}\Delta \end{equation} Where $\phi_0$ is the initial phase difference, which we know is zero. When the lights exits the material, the $x$ - and $y$ -components have different signs at every point, that is, the phase difference between them is $\phi = \pi$ . Inserting this into the above equation, as well as the earlier result for $\Delta$ , we get: \begin{equation} \pi = \frac{2\pi}{\lambda} 0.001z \Leftrightarrow \end{equation}

\begin{equation} z = \frac{\lambda}{2 \cdot 0.001} = 255 \mu {\textrm m} \end{equation} Of course, this is just the minimum distance the light must travel in the material to achieve this effect. Multiplying this distance with any positive odd integer will also work.

A light wave $\psi$ polarized at angle $\theta$ with respect to x-axis can be decomposed into two component waves: $\psi_x = \psi \cos \theta$ polarized in horizontal direction, and $\psi_y = \psi \sin \theta$ polarized in vertical direction.

The output light is polarized at angle $-\theta$ , which means its component waves are $\psi_x = \psi \cos (-\theta)=\psi \cos \theta$ , and $\psi_y = \psi \sin (-\theta)$ . This means that one of the component waves is lagging behind $\frac {1}{2}$ phase or half a wavelength.

We compare the wavenumber of $\psi_x$ and $\psi_y$ that pass through birefringent material of thickness $t$ . Let $k_x$ and $k_y$ , $\lambda_x$ and $\lambda_y$ , be the wavenumbers and wavelengths of $\psi_x$ and $\psi_y$ respectively; $n_x$ and $n_y$ be the two refractive indices along the respective axis.

Then we have, $k_x - k_y = \frac {1}{2}$ or $\frac {t}{\lambda_x} - \frac {t}{\lambda_y} = \frac {1}{2}$ . On the other hand, $\lambda_x=\frac {\lambda}{n_x}$ and $\lambda_y=\frac {\lambda}{n_y}$ where $\lambda$ is wavelength of input light.

Thus, $k_x - k_y = \frac {tn_x}{\lambda} - \frac {tn_y}{\lambda} = \frac {1}{2}$

$\therefore t= \frac {\lambda}{2\times (n_x-n_y)} = \frac {510 \times 10^{-9}} {2\times 0.001} = 255 \times 10^{-6} = 255 \mu m$ .

The polarization changes from $\theta$ to - $\theta$ . This can be considered as adding 180 degrees to it. (Relate it to sine of that angle)

Reading some stuff about this helps to get a crisper view of it.

The path difference between the waves along the two axes becomes,

((2*pi)/Wavelength) x path difference = Phase difference = (pi)

The path difference is optical path difference (difference in refractive index) * thickness.

Upon solving it thickness = Wavelength*500.

The phase difference of pi reverses the lets say upward part of light to make it downward, while the horizontal part remains same. This changes the angle.

Initially, the phase difference between the x and y components of the light is 0. In the end, the y component has turned negative with respect to the x component, so the phase difference is $\pi$ .

Consider two waves propagating with wavelengths $\lambda$ and $\lambda+\Delta\lambda$ . The phase of the first wave changes with distance as $\frac{2\pi L}{\lambda}$ while that of the second changes as $\frac{2\pi L}{\lambda+\Delta\lambda}$ . Hence the relative phase change is $\frac{2\pi L\Delta\lambda}{\lambda^2}$ . Equating this to $\pi$ , we obtain: $L=\frac{\lambda^2}{2\Delta\lambda}$

Since the wavelength of light in a medium is related to the wavelength in free space ( $\lambda_0$ ) by the refractive index ( $n$ ) as $\lambda=\frac{\lambda_0}{n}$ , we obtain that $\Delta\lambda=\frac{\lambda\Delta n}{n}$ . Since n is not given, we assume it is equal to 1. Thus we obtain: $L=\frac{\lambda}{2\Delta n}=255\mu m$

Assume the input light was of a form $E \sin\left(2\pi \left(\frac{z}{\lambda} - \frac{t}{T}\right)\right).$

The input light can be split into the fast and slow-axis components in the following way: $\begin{aligned} E_{fast} &= E \cos(\theta) \sin\left(2\pi \left(\frac{z}{\lambda_{fast}} - \frac{t}{T}\right)\right) \\ E_{slow} &= E \sin(\theta) \sin\left(2\pi \left(\frac{z}{\lambda_{slow}} - \frac{t}{T}\right)\right). \end{aligned}$

The wavelength changes with the refraction index because the wave propagation speed changes: $\lambda (n) = \frac{\lambda_0}{n}.$

Since the frequency of the light remains the same no matter what refraction index is, the part that creates a phase shift between the waves is wavelength-dependent. Therefore, the phase difference is $\alpha = \frac{2\pi L}{\lambda_0} (n_{slow} - n_{fast})$ where $L$ is the length of the material.

If the polarization was rotated for $2 \theta$ that means that the slow component flipped, or in other words, the phase shift was $\pi$ . Therefore $L = \frac{\pi \lambda_0}{2 \pi \Delta n} = 255~\mu\mbox{m}$ .