Posted on 2023-12-11 :: physics python example

Introduction: Simple Equation Derivation

We start with Fresnel Equations

$\tilde{E}_{0_R} = \left(\frac{\alpha-\beta}{\alpha+\beta}\right)\tilde{E}_{0_I}, ~~~~~~ \tilde{E}_{0_T} = \left(\frac{2}{\alpha + \beta}\right) \tilde{E}_{0_I}. \tag{1}$

where

$\alpha\equiv \frac{\cos\theta_T}{\cos\theta_I}, \tag{2}$

and

$\beta \equiv \frac{\mu_1 v_1}{\mu_2 v_2} = \frac{\mu_1 n_2}{\mu_2 n_1}. \tag{3}$

If we assume the special case where $\mu_1 \cong \mu_2 \cong \mu_0$ then

$\beta \cong \frac{n_2}{n_1}. \tag{4}$

Based on Snell's law we know that

$\sin\theta_T = \frac{n_1}{n_2} \sin\theta_I. \tag{5}$

Using identity $\cos^2\theta_T+\sin^2\theta_T = 1$ then the previous $ \alpha $ equation (2) becomes

$\alpha = \frac{\sqrt{1-\sin^2\theta_T}}{\cos\theta_I} = \frac{\sqrt{1-\left[(n_1/n_2)\sin\theta_I\right]^2}}{\cos\theta_I}.$

For reflectance and transmittance

$\newcommand{\rbrak}[1]{\left(#1 \right)} T\equiv\frac{I_T}{I_I}=\underbrace{\frac{\epsilon_2v_2}{\epsilon_1v_1}}_\beta\underbrace{\rbrak{\frac{E_{0_T}}{E_{0_I}}}}_\text{Eq. 1}\underbrace{\frac{\cos\theta_T}{\cos\theta_I}}_\alpha=\alpha\beta\rbrak{\frac{2}{\alpha+\beta}}^2.$

Introduction: Simple Equation Derivation

Source Code

Final Result

Video on Plotting Fresnel Equations