# Plotting curve of Fresnel coefficients

I am studying the Fresnel equations. I know that the Brewster's Angle can be found when the reflected coefficient:

$$\begin{equation} r_{\parallel}=\frac{\tan(\theta_{i}-\theta_{t})}{\tan(\theta_{i}+\theta_{t})} \end{equation}$$

is zero.

However, how should I plot this curve? Should $$\theta_{i}$$ or $$\theta_{t}$$ be the independent variable? Say it should be $$\theta_{i}$$, what happens with $$\theta_{t}$$?

• The transmitted angle follows from Snell's law.
– user137289
Nov 24 '18 at 12:30
• @Pieter I just thought about this. Using Snell's law yields (suppose $n_{i}=1$) $r_{\parallel}=\frac{tan(\theta_{i}-arcsin(sin(\theta_{r})/n_{t}))}{tan(\theta_{i}+arcsin(sin(\theta_{r})/n_{t}))}$. Is this correct? Nov 24 '18 at 12:44
• The coefficient is zero when the transmitted and the reflected beam are at right angles.
– user137289
Nov 24 '18 at 12:47
• I'm sorry, I've edited my comment to correct that. The equation does vanish, but I get something like an inverse sine function. I believe it should look like this Nov 24 '18 at 12:54
• You might find this useful geogebra.org/m/wKk62nUk Nov 24 '18 at 15:56

Should $$\theta_{i}$$ or $$\theta_{t}$$ be the independent variable? Say it sould be $$\theta_{i}$$, what happens with $$\theta_{t}$$?
The independent variable is the angle of incidence, $$\theta_i$$. The transmission angle $$\theta_t$$ is then obtained from Snell's law, $$n_i \sin(\theta_i) = n_t \sin(\theta_t),$$ as $$\theta_t = \arcsin \left( \frac{n_i}{n_t} \sin(\theta_i) \right).$$