diff --git a/config/en/mkdocs.yaml b/config/en/mkdocs.yaml
index 76728bf..2671f32 100755
--- a/config/en/mkdocs.yaml
+++ b/config/en/mkdocs.yaml
@@ -127,7 +127,7 @@ nav:
         - 'Geometric optics': physics/electromagnetism/optics/geometric-optics.md
         - 'Interference': physics/electromagnetism/optics/interference.md
         - 'Diffraction': physics/electromagnetism/optics/diffraction.md
-        - 'Polarization': physics/electromagnetism/optics/polarization.md
+        - 'Polarisation': physics/electromagnetism/optics/polarisation.md
     - 'Mathematical physics': 
       - 'Signal analysis': 
         - 'Signals': physics/mathematical-physics/signal-analysis/signals.md
diff --git a/docs/en/physics/electromagnetism/optics/polarisation.md b/docs/en/physics/electromagnetism/optics/polarisation.md
new file mode 100644
index 0000000..9ebb53e
--- /dev/null
+++ b/docs/en/physics/electromagnetism/optics/polarisation.md
@@ -0,0 +1,179 @@
+# Polarisation
+
+If we consider an electromagnetic wave $\mathbf{E}: \mathbb{R}^2 \to \mathbb{R}^3$ with wavenumber $k \in \mathbb{R}$ and angular frequency $\omega \in \mathbb{R}$ propagating in the positve $z$-direction given by
+
+$$
+    \mathbf{E}(z,t) = \exp i(kz - \omega t + \varphi_1) E_0^{(x)} \mathbf{e}_{(x)} + \exp i(kz - \omega t + \varphi_2) E_0^{(y)}\mathbf{e}_{(y)},
+$$
+
+for all $(z,t) \in \mathbb{R}^2$ with $E_0^{(x)}, E_0^{(y)} \in \mathbb{R}$ the magnitude of the wave in the $x$ and $y$ direction. We define $\Delta \varphi = \varphi_2 - \varphi_1$. 
+
+> *Definition*: the electromagnetic wave $\mathbf{E}$ is linear polarised if and only if
+>
+> $$
+>   \Delta \varphi = \pi m,
+> $$
+>
+> for all $m \in \mathbb{Z}$. 
+
+With polarisation angle $\theta \in [0, 2\pi)$ given by 
+
+$$
+    \theta = \arctan \Bigg( \frac{\max E_0^{(y)}}{\max E_0^{(x)}} \Bigg).
+$$
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+> *Definition*: the electromagnetic wave $\mathbf{E}$ is left circular polarised if and only if
+>
+> $$
+>   \Delta \varphi = \frac{\pi}{2} \;\land\; E_0^{(x)} = E_0^{(y)},
+> $$
+>
+> and right circular polarised if and only if 
+>
+> $$
+>   \Delta \varphi = -\frac{\pi}{2} \;\land\; E_0^{(x)} = E_0^{(y)}.
+> $$
+
+For every state in between we have elliptical polarisation with a polarisation angle $\theta \in [0, 2\pi)$ given by
+
+$$
+    \theta = \frac{1}{2} \arctan \Bigg(\frac{2 E_0^{(x)} E_0^{(y)} \cos \Delta\varphi}{ \big(E_0^{(x)} \big)^2- \big( E_0^{(y)} \big)^2} \Bigg).
+$$
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+> *Definition*: natural light is defined as light constisting of all linear polarisation states. 
+
+## Linear polarisation
+
+> *Definition*: a linear polariser selectively removes light that is linearly polarised along a direction perpendicular to its transmission axis. 
+
+We may concretisize this definition by the following statement, considered to be Malus law.
+
+> *Law*: for a light beam with amplitude $E_0$ incident on a linear polariser the transmitted beam has amplitude $E_0 \cos \theta$ with $\theta \in [0, 2\pi)$ the polarisation angle of the light with respect to the transmission axis. The transmitted irradiance $I: [0, 2\pi) \to \mathbb{R}$ is then given by
+>
+> $$
+>   I(\theta) = I_0 \cos^2 \theta,
+> $$
+>
+> for all $\theta \in [0, 2\pi)$ with $I_0 \in \mathbb{R}$ the irradiance of the incident light. 
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+For natural light the average of all angles must be taken, since $\lim_{\theta \to \infty} \cos^2 \theta = \frac{1}{2}$, we have the relation $I = \frac{1}{2} I_0$ for natural light. 
+
+## Birefringence
+
+Natural light can be polarised in several ways, some are listed below.
+
+1. Polarisation by absorption of the other component. This can be done with a wiregrid or dichroic materials for smaller wavelengths. 
+2. Polarisation by scattering. Dipole radiation has distinct polarisation depending on the position.
+3. Polarisation by Brewster angle, which boils down to scattering.
+4. Polarisation by birefringence, the double refraction of light obtaining two orthogonal components polarised.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+> *Definition*: birefringence is a double refraction in a material (often crystalline) and can be derived from the Fresnel equations without assuming isotropic dielectric properties. 
+
+If isotropic dielectric properties are not assumed it implies that the refractive index may also depend on the polarisation and propgation direction of light.
+
+Using the properties of birefringence, wave plates (retarders) can be created. They may introduce a phase difference via a speed difference in the polarisation direction.
+
+* A half-wave plate may introduce a $\Delta \varphi = \pi$ phase difference.
+* A quarter-wave plate may introduce a $\Delta \varphi = \frac{\pi}{2}$ phase difference. 
+
+## Jones formalism of polarisation
+
+Jones formalism of polarisation with vectors and matrices cna make it easier to calculate the effect of optical elements such as linear polarizers and wave plates.
+
+> *Definition*: for an electromagnetic wave $\mathbf{E}: \mathbb{R}^2 \to \mathbb{R}^3$ with wavenumber $k \in \mathbb{R}$ and angular frequency $\omega \in \mathbb{R}$ propagating in the positive $z$-direction given by
+>
+> $$
+>   \mathbf{E}(z,t) = \mathbf{E}_0 \exp i(kz - \omega t),
+> $$
+>
+> for all $(z,t) \in \mathbb{R}^2$. The Jones vector $\mathbf{\tilde E}$ is defined as
+>
+> $$
+>   \mathbf{\tilde E} = \mathbf{E}_0, 
+> $$
+>
+> possibly normalized with $\|\mathbf{\tilde E}\| = 1$.
+
+For linear polarised light under an angle $\theta \in [0, 2\pi)$ the Jones vector $\mathbf{\tilde E}$ is given by
+
+$$
+    \mathbf{\tilde E} = \begin{pmatrix}\cos \theta \\ \sin \theta\end{pmatrix}.
+$$
+
+For left circular polarised light the Jones vector $\mathbf{\tilde E}$ is given by
+
+$$
+    \mathbf{\tilde E} = \begin{pmatrix} 1 \\ i \end{pmatrix},
+$$
+
+and for right circular polarised light
+
+$$
+    \mathbf{\tilde E} = \begin{pmatrix} 1 \\ -i \end{pmatrix}.
+$$
+
+> *Definition*: Jones matrices $M_i$ with $i \in \{1, \dots, n\}$ with $n \in \mathbb{N}$ may be used to model several optical elements on an optical axis, obtaining the transmitted Jones vector $\mathbf{\tilde E}_t$ from the incident Jones vector $\mathbf{\tilde E}_i$ given by
+>
+> $$
+>   \mathbf{\tilde E}_t = M_n \cdots M_1 \mathbf{\tilde E}_i.
+> $$
+
+The Jones matrices for several optical elements are now given.
+
+> *Proposition*: the Jones matrix $M$ of a linear polariser is given by
+>
+> $$
+>   M = \begin{pmatrix} \cos^2 \theta & \frac{1}{2} \sin 2\theta \\ \frac{1}{2} \sin 2\theta & \sin^2 \theta \end{pmatrix},
+> $$
+>
+> with $\theta \in [0, 2\pi)$ the transmission axis of the linear polariser.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+<br>
+
+> *Proposition*: the Jones matrix $M$ of a half-wave plate is given by
+>
+> $$
+>   M = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix},
+> $$
+>
+> with  $\theta \in [0, 2\pi)$ the fast axis of the half-wave plate.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+<br>
+
+> *Proposition*: the Jones matrix $M$ of a quarter-wave plate is given by
+>
+> $$
+>   M = \begin{pmatrix} \cos^2 \theta + \sin^2 \theta & (1 - i) \sin \theta \cos \theta \\ (1 - i) \sin \theta \cos \theta & i(\cos^2 \theta + \sin^2 \theta) \end{pmatrix},
+> $$
+>
+> with $\theta \in [0, 2\pi)$ the fast axis of the quarter-wave plate.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+<br>
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diff --git a/docs/en/physics/electromagnetism/optics/polarization.md b/docs/en/physics/electromagnetism/optics/polarization.md
deleted file mode 100644
index 74a3b3c..0000000
--- a/docs/en/physics/electromagnetism/optics/polarization.md
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@@ -1 +0,0 @@
-# Polarization
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