If a perfectly circularly polarized wave of light is incident on a dielectric medium (coming from air) at Brewster's Angle, what will the polarization state of the transmitted wave be? I am aware that the reflected wave will be linearly polarized, but will the transmitted wave be linearly polarized as well because the light was perfectly circularly polarized to begin with, and the perpendicular component was completely eliminated by the fact that the angle of incidence is Brewster's Angle?
[Physics] Circularly polarized light incident at Brewster’s Angle
electromagnetismpolarizationreflection
Related Solutions
With circularly polarised waves, the effect can only be observed interferometrically, because the Faraday rotation simply becomes the imposition of different phase delays on the two circularly polarised components. To understand this, witness that the Faraday rotation on a general polarisation state expressed with linear polarisation state basis is:
$$\left(\begin{array}{c}x\\y\end{array}\right)\mapsto \left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)\left(\begin{array}{cc}x\\y\end{array}\right)$$
and then we transform to circular polarisation basis states $f_+,\,f_-$ by:
$$\left(\begin{array}{c}f_+\\f_-\end{array}\right)= \frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)\left(\begin{array}{cc}x\\y\end{array}\right)$$
so that our Faraday rotation through angle $\theta$ above becomes:
$$\left(\begin{array}{c}f_+\\f_-\end{array}\right)\mapsto \left(\begin{array}{cc}1&i\\1&-i\end{array}\right)\left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)^{-1}\left(\begin{array}{cc}f_+\\f_-\end{array}\right) = \left(\begin{array}{cc}e^{i\,\theta}&0\\0&e^{-i\,\theta}\end{array}\right)\left(\begin{array}{cc}f_+\\f_-\end{array}\right)$$
Indeed, this is the essence of the spin-1 particle: its eigenstates pick up a phase of $e^{i\,\theta}$ on rotation through angle $\theta$ (unlike an electron, which would pick up a phase of $e^{i\,\frac{\theta}{2}}$).
Edit: Your understanding expressed in your question seems to be right. In the above I give the Jones matrix, i.e. the linear transformation wrought on the vector of superposition weights, for the two cases where (1) linear polarisation states and (2) circular polarisation states are used as the state space basis: in the former case, the effect is a rotation of the plane of polarisation through $\theta$ radians, in the latter the right handed circular polarised component's phase is increased by $\theta$ radians, the left hand component's phase is decreased by $\theta$ radians. In all cases, this $\theta$ is given by $\theta = V\, B_\parallel \,L$ where $L$ is the crystal's length, $B_\parallel$ the magnetic induction component along the optical axis and $L$ the crystal's length.
Edit 2: Well, the results for circularly polarised loght follow from: (1) experimental measurements for linearly polarised light and (2) experimental confirmation of the system's linearity; once we have (2) then circular / linear description is simply a change of basis. Moreover, you can do the experiments for circularly polarised light, you're simply looking for changes in phase. So you can do the experiment interferometrically. Or, you can do an experiment as below with waveplates and the rotator.
The rotator works exactly the same with linearly polarised input light whether or not the waveplates are in place, confirming the linearity of the system. So it is acting on the circularly polarised light states exactly as my basis trasnformation equations above foretell. At a deeper level, the waveplate's action as a transformer to and from circularly polarised light can be checked by the following experiment in 1936 by Richard Beth:
Simply, at Brewster's angle, the $\pi$ - component of incident electromagnetic wave always transmits 100%.
What is left is the $\sigma$ - component. The $\sigma$ - component of EM wave (which is already lying in xz - plane) striking on the second plate with respect to that plate is equal its amplitude times cosine of the angle the glass plate B will rotate.
The reflected wave is equal to that product times reflection coefficient, called Fresnel reflection coefficient.
The average reflected intensity is that thing to the square time $\epsilon_{0} \cdot 0.5$.
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Best Answer
To calculate the transmitted and scattered wave state at any angle from a plane interface, you need to resolve the incident field into s- and p-linear polarised wave amplitudes and multiply the two complex amplitudes by the amplitude Fresnel transmission and reflexion co-efficients. By "amplitude Fresnel co-efficient I mean the ratio of complex wave amplitudes rather than modulus squared of the Fresnel co-efficients, which gives the fraction of power reflected / transmitted and which are also sometimes called the transmission and reflexion co-efficients.
There is nothing "special" about the polarisation of the throughgoing wave calculated by this approach at the Brewster angle. It is simply a general elliptical polarised state which must be represented by two complex amplitudes $\vec{a}$ of the chosen basis polarisation states, or as the Stokes parameters $s_j = \vec{a}^\dagger \sigma_j \vec{a}$ where $\sigma_j$ are the Pauli spin matrices. The latter approach only represents the field to within a factor of $\pm 1$ (both $\vec{a}$ and $-\vec{a}$ have the same Stokes paramters).