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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 $\theta_{t}$?

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    $\begingroup$ The transmitted angle follows from Snell's law. $\endgroup$
    – user137289
    Nov 24 '18 at 12:30
  • $\begingroup$ @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? $\endgroup$
    – IchVerlore
    Nov 24 '18 at 12:44
  • $\begingroup$ The coefficient is zero when the transmitted and the reflected beam are at right angles. $\endgroup$
    – user137289
    Nov 24 '18 at 12:47
  • $\begingroup$ 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 $\endgroup$
    – IchVerlore
    Nov 24 '18 at 12:54
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    $\begingroup$ You might find this useful geogebra.org/m/wKk62nUk $\endgroup$
    – ProfRob
    Nov 24 '18 at 15:56
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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 $$ \theta_t = \arcsin \left( \frac{n_i}{n_t} \sin(\theta_i) \right). $$

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  • $\begingroup$ I used Snell's law, but I obtain a curve that resembles an inverse sine (it vanishes in multiple points). Shouldn't it look like this? $\endgroup$
    – IchVerlore
    Nov 24 '18 at 12:56
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    $\begingroup$ @IchVerloren You seem to have a global-sign discrepancy, but other than that I get this, which is a good match to what it needs to be. Check your plotting carefully. (But note that we're not here to de-bug your code.) $\endgroup$ Nov 24 '18 at 13:01
  • $\begingroup$ Forget about that. I just realized the result is in radians. I post this in case someone makes the same mistake. I got what I wanted. $\endgroup$
    – IchVerlore
    Nov 24 '18 at 13:02

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