According to Snell's law:

$$\sin \theta_t = \displaystyle \frac{n_1}{n_2} \sin \theta_i$$

where $\theta_i$ is the incidence angle in medium 1 and (after the ray crossed the interface between medium 1 and 2) $\theta_t$ is the resulting angle in medium 2.

If $n_1 > n_2$, there will be a critical incidence angle $\theta_c$ (such that $\sin \theta_c = n_2 / n_1$), which causes $\sin \theta_t = 1$. When $\theta_i > \theta_c$, $\sin \theta_t$ is forced to be a value greater than $1$, which is not achievable through a real value of $\theta_t$.

Even though this angle $\theta_t$ has no physical meaning, which is its exact mathematical value? And is it unique?

My attempt:

Near the end of this answer, the value $\theta = \pi / 2 + n \pi + iI$ ($I \neq 0$) is suggested to generate a value $\sin \theta > 1$. But it seems not to be the case: in fact,

\begin{align} \sin \theta & = \frac{e^{i \theta} - e^{- i \theta}}{2i} = \frac{e^{i \left( \frac{\pi}{2} + n \pi + iI \right)} - e^{- i \left( \frac{\pi}{2} + n \pi + iI \right)}}{2i} \\ & = \frac{i \cos(n \pi) e^{-I} + i \cos(n \pi) e^I}{2i} = \cos (n \pi) \cosh I \\ & = (-1)^n \cosh I \end{align}

It seems not correct. The value $\theta = \pi / 2 - iI$ seems more appropriate. In fact:

\begin{align} \sin \theta & = \frac{e^{i \left( \frac{\pi}{2} - iI \right)} - e^{- i \left( \frac{\pi}{2} - iI \right)}}{2i} & = \frac{i e^I + i e^{-I}}{2i} \\& = \cosh I \end{align}


$$\cosh I \geq 1, \forall I \in \mathbb{R}$$


$$\cosh I = \frac{n_1}{n_2} \sin \theta_i, \ I = \mathrm{arccosh} \mathopen{} \left( \frac{n_1}{n_2} \sin \theta_i \right)$$

So, the initial question becomes: is this procedure correct?


1 Answer 1


Frankly, complex angles just aren't worth the hassle. As the answer you've linked to points out, they're just a complicated way to re-parametrize the evanescent wave $$ e^{i\mathbf k\cdot \mathbf r} = e^{i(k_x,0,i\kappa)\cdot (x,y,z)} =e^{ik_x x}e^{-\kappa z} \ \ \stackrel{\rm as}{=} \ \ e^{ik(\sin(\theta),0,\cos(\theta))\cdot\mathbf r} . $$

For the total-internal-reflection case that you ask about, take the form above for the evanescent wave in medium $2$ for $z>0$, so that you're required to have $$ k_x^2 -\kappa^2 = k_2^2 = \frac{n_2^2}{n_1^2} k_1^2. $$ With that notation in place, any angle that fulfills the equations \begin{align} k_2 \sin(\theta) & = k_{x,1} \\ k_2 \cos(\theta) & = i\kappa, \end{align} or equivalently \begin{align} \sin(\theta) & = \frac{n_1}{n_2}\sin(\theta_i) \\ \cos(\theta) & = i\sqrt{ \frac{n_1^2}{n_2^2}\sin^2(\theta_i)-1 } , \end{align} is an acceptable solution $-$ so long as those two conditions are satisfied, even if you get different $\theta$s, the consequences for the real-world waveforms are exactly identical, and the $\theta$s cannot be differentiated on the basis of any physical observable. On the other hand, it's important to note that both of these conditions need to be satisfied, or you run the risk of having an incorrect sign for $\kappa$ and, with that, an exponentially-increasing wave where you were meant to have a quickly-decaying evanescent wave.

To fully characterize the complex angle, let's work our conditions some more, though, by expanding out the angle into its real and imaginary parts, $\theta = \theta_\mathrm{re}+i \theta_\mathrm{im}$ (with $\theta_\mathrm{re},\theta_\mathrm{im} \in \mathbb R$), \begin{align} \sin(\theta_\mathrm{re})\cosh(\theta_\mathrm{im}) + i\cos(\theta_\mathrm{re})\sinh(\theta_\mathrm{im}) & = \frac{n_1}{n_2}\sin(\theta_i) \\ \cos(\theta_\mathrm{re})\cosh(\theta_\mathrm{im}) - i\sin(\theta_\mathrm{re})\sinh(\theta_\mathrm{im}) & = i\sqrt{ \frac{n_1^2}{n_2^2}\sin^2(\theta_i)-1 } , \end{align} and then requiring that the real and imaginary parts of these equations be satisfied simultaneously, \begin{align} \sin(\theta_\mathrm{re})\cosh(\theta_\mathrm{im}) & = \frac{n_1}{n_2}\sin(\theta_i) \\ \cos(\theta_\mathrm{re})\sinh(\theta_\mathrm{im}) & = 0 \\ \cos(\theta_\mathrm{re})\cosh(\theta_\mathrm{im}) & = 0 \\ \sin(\theta_\mathrm{re})\sinh(\theta_\mathrm{im}) & = - \sqrt{ \frac{n_1^2}{n_2^2}\sin^2(\theta_i)-1 } . \end{align} Since $\theta_\mathrm{im}$ is real, which guarantees that $\cosh(\theta_\mathrm{im})>0$, we conclude from $\cos(\theta_\mathrm{re})\cosh(\theta_\mathrm{im}) = 0$ that $$ \cos(\theta_\mathrm{re}) = 0 $$ and therefore that $$ \theta_\mathrm{re} = \frac{\pi}{2} + n \pi, \ \ n\in \mathbb Z, $$ as a necessary condition, which then implies that $\sin(\theta_\mathrm{re}) = (-1)^n$, so we still need to solve both of \begin{align} \cosh(\theta_\mathrm{im}) & = (-1)^n\frac{n_1}{n_2}\sin(\theta_i) \\ \sinh(\theta_\mathrm{im}) & = - (-1)^n\sqrt{ \frac{n_1^2}{n_2^2}\sin^2(\theta_i)-1 } . \end{align} Here the second equation is always solvable, as $$ \theta_\mathrm{im} = - (-1)^n\mathrm{arcsinh} \mathopen{}\left(\sqrt{ \frac{n_1^2}{n_2^2}\sin^2(\theta_i)-1 }\right), $$ and this will guarantee the correct sign for $\kappa$, but the first equation is only possible if $(-1)^n\sin(\theta_i)>0$, where you can conclude that $n$ is even if the incidence angle is counted as positive. This is a reasonable convention, but it is not guaranteed, and a more general solution is to adjust the offset $n$ according to the sign of the incidence angle as $$ (-1)^n = \mathrm{sgn}(\theta_i), $$ giving $\theta_\mathrm{re} = \frac{\pi}{2} \ (\mathrm{mod} \ 2\pi)$ for $\theta_i> 0$, and choosing $n=-1$ as the principal representative for the opposite case, $\theta_\mathrm{re} = -\frac{\pi}{2} \ (\mathrm{mod}\ 2\pi)$ for $\theta_i< 0$. This means, then, that the correct solution is \begin{align} \theta_\mathrm{re} & = \mathrm{sgn}(\theta_i) \frac{\pi}{2} \\ \theta_\mathrm{im} & = - \mathrm{sgn}(\theta_i) \mathrm{arcsinh} \mathopen{}\left(\sqrt{ \frac{n_1^2}{n_2^2}\sin^2(\theta_i)-1 }\right), \end{align} or in other words $$ \theta = \mathrm{sgn}(\theta_i) \left[ \frac{\pi}{2} - i\: \mathrm{arcsinh} \mathopen{}\left(\sqrt{ \frac{n_1^2}{n_2^2}\sin^2(\theta_i)-1 }\right)\right] \mod 2\pi $$ as the general solution.

I've phrased things like this because they're a surefire way to guarantee the correct signs, but it's clunkier than it needs to be, so it can be simplified a bit to swap that arcsinh-of-a-square-root for a simpler arccosh: to do that, start by writing \begin{align} \cosh(\theta_\mathrm{im}) & = (-1)^n\frac{n_1}{n_2}\sin(\theta_i) = \frac{n_1}{n_2}\left|\sin(\theta_i)\right| \end{align} (since the $(-1)^n$ factor is there to guarantee positivity), allowing you to write \begin{align} \theta_\mathrm{im} = \pm \mathrm{arccosh} \mathopen{}\left(\frac{n_1}{n_2}\left|\sin(\theta_i)\right|\right) \end{align} up to a sign ambiguity which we've resolved above. With the sign resolution from above, though, we get as our final result $$ \theta = \mathrm{sgn}(\theta_i) \left[ \frac{\pi}{2} - i\: \mathrm{arccosh} \mathopen{}\left(\frac{n_1}{n_2}\left|\sin(\theta_i)\right|\right)\right] \mod 2\pi . $$

  • 1
    $\begingroup$ Thank you for the very detailed answer, which I am trying to follow and understand. A doubt: "Since $\theta_i$ is given and real, which guarantees that $\cosh(\theta_i) > 0$, we conclude that $\cos(\theta_{\mathrm{re}})=0$". I agree with the first part: a real $\theta_i$ generates $\cosh(\theta_i) > 0$, and even $\cosh \theta_i \geq 1$. But what is the relation with the second part? In the four equations preceding this statement there is no $\cosh \theta_i$. There are only $\cosh \theta_{\mathrm{im}}$ and $\sinh \theta_{\mathrm{im}}$. $\endgroup$
    – BowPark
    Nov 23, 2018 at 9:42
  • $\begingroup$ Also, "as a necessary condition, which then implies that $\sin(\theta_r) = (−1)^n$": maybe you were meaning $\sin(\theta_{\mathrm{re}}) = (−1)^n$. $\endgroup$
    – BowPark
    Nov 23, 2018 at 9:57
  • $\begingroup$ Ah, those are remnants of a previous draft with inconsistent notation, which I obviously didn't get rid of fully. Still, the conclusions are unaffected. $\endgroup$ Nov 23, 2018 at 10:02
  • $\begingroup$ I imagined this. Thanks! But I still can't solve the doubt in the first comment... $\endgroup$
    – BowPark
    Nov 23, 2018 at 11:19
  • $\begingroup$ @BowPark See edited answer. $\endgroup$ Nov 23, 2018 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.