##### Oblique Incidence
emdm.emwm.refl.oi

We now consider an interface between two media. Media \(1\) (for \(z < 0\)) has light velocity \(v_1\) while \(2\) (for \(z > 0\)) has \(v_2\).

Incident wave: \[ {\boldsymbol E}_I ({\boldsymbol r},t) = {\boldsymbol E}_{0_I} e^{i ({\boldsymbol k}_I \cdot {\boldsymbol r} - \omega t)}, \hspace{1cm} {\boldsymbol B}_I ({\boldsymbol r},t) = \frac{1}{v_1} \hat{\boldsymbol k}_I \times {\boldsymbol E}_{I} ({\boldsymbol r}, t). \] Reflected wave: \[ {\boldsymbol E}_R ({\boldsymbol r},t) = {\boldsymbol E}_{0_R} e^{i ({\boldsymbol k}_R \cdot {\boldsymbol r} - \omega t)}, \hspace{1cm} {\boldsymbol B}_R ({\boldsymbol r},t) = \frac{1}{v_1} \hat{\boldsymbol k}_R \times {\boldsymbol E}_{R} ({\boldsymbol r}, t). \] Transmitted wave: \[ {\boldsymbol E}_T ({\boldsymbol r},t) = {\boldsymbol E}_{0_T} e^{i ({\boldsymbol k}_T \cdot {\boldsymbol r} - \omega t)}, \hspace{1cm} {\boldsymbol B}_T ({\boldsymbol r},t) = \frac{1}{v_2} \hat{\boldsymbol k}_T \times {\boldsymbol E}_{T} ({\boldsymbol r}, t). \] All waves have the same frequency \(\omega\). Since \(\omega = k v\), the three wavevectors are related by

\[ k_I v_1 = k_R v_1 = k_T v_2 ~~\longrightarrow~~ k_I = k_R = \frac{v_2}{v_1} k_T = \frac{n_1}{n_2} k_T \tag{RTobliquek}\label{RTobliquek} \]

These forms for incident, reflected and transmitted wave can be substituted in the boundary conditions disc_nfc. Since these must be valid for any \(x\) and \(y\) (on the interface at \(z=0\)), we must have that the \(x\) and \(y\) components of the wavevectors coincide for all the waves: \[ k_{I_x} = k_{R_x} = k_{T_x}, \hspace{10mm} k_{I_y} = k_{R_y} = k_{T_y} \]

From now on we will orient the axes so that \({\boldsymbol k}_I\) lies in the \(xz\) plane. This means that \({\boldsymbol k}_R\) and \({\boldsymbol k}_T\) also lie in that plane. This is the

First law of reflection: the incident, reflected and transmitted wave vectors form a plane (called the plane of incidence) which also includes the normal to the surface.

Specializing RTobliquek to our notations, we have \[ k_I \sin \theta_I = k_R \sin \theta_R = k_T \sin \theta_T \] with the incidence (\(\theta_I\)) and reflection (\(\theta_R\)) angles and the angle of refraction (\(\theta_T\)) obey the following laws:

Law of reflection \[ \theta_I = \theta_R \] Law of refraction (Snell's law) \[ n_1 \sin \theta_I = n_2 \sin \theta_T \]

This takes care of the spatially-dependent exponential factors in the boundary conditions. The coefficients must further obey

\begin{align} \varepsilon_1 \left({\boldsymbol E}_{0_I} + {\boldsymbol E}_{0_R} \right)_z &= \varepsilon_2 \left({\boldsymbol E}_{0_T} \right)_z, & \hspace{10mm} \left({\boldsymbol B}_{0_I} +{\boldsymbol B}_{0_R} \right)_z &= \left({\boldsymbol B}_{0_T}\right)_z, \nonumber \\ \left( {\boldsymbol E}_{0_I} + {\boldsymbol E}_{0_R} \right)_{x,y} &= \left({\boldsymbol E}_{0_T}\right)_{x,y}, & \frac{1}{\mu_1} \left({\boldsymbol B}_{0_I} + {\boldsymbol B}_{0_R} \right)_{x,y} &= \frac{1}{\mu_2} \left({\boldsymbol B}_{0_T}\right)_{x,y}. \tag{EBRT}\label{EBRT} \end{align}

Further treatment depends on the polarization of the incoming wave. The two cases of polarization parallel and perpendicular to the plane of incidence must be treated separately. We thus divide our incident electric field as follows: \[ {\boldsymbol E}_{0_I} = {\boldsymbol E}_{0_I}^{\parallel} + {\boldsymbol E}_{0_I}^{\perp}. \]

Polarization in plane of incidence: in this case the first equation of EBRT gives \[ \varepsilon_1 \left(-E_{0_I} \sin \theta_I + E_{0_R} \sin \theta_R \right) = -\varepsilon_2 E_{0_T} \sin \theta_T. \] The second equation is a trivial \(0=0\). The third is \[ E_{0_I} \cos \theta_I + E_{0_R} \cos \theta_R = E_{0_T} \cos \theta_T \] while the fourth gives \[ \frac{1}{\mu_1 v_1} \left( E_{0_I} - E_{0_R} \right) = \frac{1}{\mu_2 v_2} E_{0_T} \] Given the laws of reflection and refraction, the first and fourth equations are the same and reduce to \[ E_{0_I} - E_{0_R} = \beta E_{0_T}, \] while the third equation becomes \[ E_{0_I} + E_{0_R} = \alpha E_{0_T}, \hspace{10mm} \alpha \equiv \frac{\cos \theta_T}{\cos \theta_I} \] Writing everything in terms of the incident amplitude, we get

Fresnel's equations for reflection and transmission amplitudes (parallel case)

\[ E_{0_R} = \frac{\alpha - \beta}{\alpha + \beta} E_{0_I}, \hspace{10mm} E_{0_T} = \frac{2}{\alpha + \beta} E_{0_I} \tag{Fresnel}\label{Fresnel} \]

Amplitudes for transmitted and reflected wave: depend on angle of incidence: \[ \alpha = \frac{\sqrt{1 - \sin^2 \theta_T}}{\cos \theta_I} = \frac{\left[1 - \left(\frac{n_1}{n_2}\right)^2 \sin^2 \theta_I\right]^{1/2}}{\cos \theta_I} \] Behaviour: for \(\theta_I = 0\) we recover ERT. For grazing waves \(\theta_I \rightarrow \pi/2\) we have that \(\alpha \rightarrow \infty\) and the wave is totally reflected. The most interesting angle is the one at which \(\alpha = \beta\) and the reflected wave has zero amplitude. This is known as

Brewster's angle (at which the reflected wave amplitude vanishes) \[ \theta_B = \arcsin \left[ \frac{1 - \beta^2}{(n_1/n_2)^2 - \beta^2} \right]^{1/2} \tag{Brewster}\label{Brewster} \]

Power per unit area striking the interface: \({\boldsymbol S} \cdot \hat {\boldsymbol n}\) and thus \[ I_I = \frac{\varepsilon_1 v_1}{2} E_{0_I}^2 \cos \theta_I, \hspace{5mm} I_R = \frac{\varepsilon_1 v_1}{2} E_{0_R}^2 \cos \theta_R, \hspace{5mm} I_T = \frac{\varepsilon_2 v_2}{2} E_{0_T}^2 \cos \theta_T. \]

Reflection and transmission coefficients: \[ R \equiv \frac{I_R}{I_I} = \frac{E_{0_R}^2}{E_{0_I}^2} = \left( \frac{\alpha - \beta}{\alpha + \beta} \right)^2, \hspace{5mm} T \equiv \frac{I_R}{I_I} = \frac{\varepsilon_2 v_2}{\varepsilon_1 v_1} \frac{E_{0_T}^2 \cos \theta_T}{E_{0_I}^2 \cos \theta_I} = \alpha \beta \left( \frac{2}{\alpha + \beta} \right)^2. \] Of course, we get \(R + T = 1\) as expected.

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42