Answer to Question #184758 in Optics for Mellisa

Question

 Show that the Jones matrix representing a linear polarizer whose transmission axis is at an arbitrary angle 𝜃 with respect to the horizontal is given by

 𝑀(𝜃) =(  𝑐𝑜𝑠²𝜃        𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃    )          

      𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃  𝑠𝑖𝑛²𝜃 

Answer 1

The polarizer acts on a plane wave with arbitrary polarization. The electric field of our plane wave may be written as

$E(z,t)=(E_x \hat x+E_y\hat y)e^{i(kz-\omega t)}$

Let the transmission axis of the polarizer be specified by the unit vector $\hat e_1$ and the absorption axis of the polarizer be specified by $\hat e_2$ (orthogonal to the transmission axis). The vector $\hat e_1$ is oriented at an angle θ from the x-axis. We need to write the electric field components in terms of the new basis specified by $\hat e_1$ and $\hat e_2$. By inspection of the geometry, the x-y unit vectors are connected to the new coordinate system via:

$\hat x=cos\theta \hat e_1- sin\theta \hat e_2 \\\hat y=sin\theta \hat e_1+ cos\theta \hat e_2$

So, the electric field will be

$E(z,t)=(E_1 \hat e_1+E_2\hat e_2)e^{i(kz-\omega t)}$

After traversing the polarizer, the field becomes

$E_{after}(z,t)=(E_1 \hat e_1+\xi E_2\hat e_2)e^{i(kz-\omega t)}$

We now have the field after the polarizer, but it would be nice to rewrite it in terms of the original x–y basis.

$\hat e_1=cos\theta \hat x- sin\theta \hat y \\\hat e_2=sin\theta \hat x+ cos\theta \hat y$

Substitution of these relationships into together with the definitions for $E_1$ and $E_2$ yields

$E_{after} (z,t) = [(E_xcos\theta+E_ysin\theta)(cos\theta\hat x+sin\theta\hat y)+ \xi[(-E_xsin\theta+E_ycos\theta)(-sin\theta\hat x+cos\theta\hat y)e^{i(kz-\omega t)}$

If we represent the electric field as a two-dimensional column vector with its x component in the top and its y component in the bottom (like a Jones vector), then we can rewrite

$E_{after}(z,t)=\begin{bmatrix} cos^2\theta+\xi sin^2\theta & sin\theta cos\theta-\xi sin\theta cos\theta \\ sin\theta cos\theta-\xi sin\theta cos\theta & sin^2\theta+\xi cos^2\theta \end{bmatrix}\begin{bmatrix} E_x \\ E_y \end{bmatrix}e^{i(kz-\omega t)}$

The matrix here is a properly normalized Jones matrix, even though we did not bother factoring out $E_{eff}$ to make a properly normalized Jones vector. We can now write down the Jones matrix for a polarizer by inserting ξ = 0 into the matrix:

$M(\theta)=\begin{bmatrix} cos^2\theta & sin\theta cos\theta \\ sin\theta cos\theta & sin^2\theta \end{bmatrix}$ (polarizer with transmission axis at angle θ)