Answer to Question #185710 in Optics for melvin

Question

 Write the normalized Jones vectors for each of the following waves, and describe completely the state of polarization of each. (a) 𝐸 ⃗ = 𝐸𝑜 cos(𝑘𝑧 − 𝜔𝑡)𝑥 ̂ − 𝐸𝑜 cos(𝑘𝑧 − 𝜔𝑡)𝑦 ̂    
(b) 𝐸 ⃗ = 𝐸𝑜 sin2π(𝑧/lamda − 𝑣𝑡)𝑥 ̂ + 𝐸𝑜 sin2π( 𝑧/lamda − 𝑣𝑡)𝑦 ̂ 

Answer 1

a)

$\vec{E}=E_x\vec{x}+E_y\vec{y},$

$E_x=-E_y,$ here phase difference between x- and y-components is $\pi,$

$E=\begin{pmatrix} E_x\\ E_y \end{pmatrix}=E_0\begin{pmatrix} 1 \\ -1 \end{pmatrix},$

$E^*\cdot E=1$

$E_0^2(1^2+(-1)^2)=1,$

$E_0=\frac{1}{\sqrt 2},$

normalized Jones vector will be

$\frac{1}{\sqrt2}\begin{pmatrix} 1 \\ -1 \end{pmatrix},$ hence it represents linearly –45° polarized light.

b)

$\vec{E}=E_x\vec{x}+E_y\vec{y},$

$E_x=E_y,$ here phase difference between x- and y-components is $0,$

$E=\begin{pmatrix} E_x\\ E_y \end{pmatrix}=E_0\begin{pmatrix} 1 \\ 1 \end{pmatrix},$

$E^*\cdot E=1$

$E_0^2(1^2+1^2)=1,$

$E_0=\frac{1}{\sqrt 2},$

normalized Jones vector will be

$\frac{1}{\sqrt2}\begin{pmatrix} 1 \\ 1 \end{pmatrix},$ hence it represents linearly 45° polarized light.