Reading about total internal reflection, and I understand that it occurs when the sine of the refracted angle is greater than 1. What I don't see is how you ever have a $\sin \theta > 1$. Jackson says

This means that $\theta$ is a complex angle with a purely imaginary cosine...

how does this help explain it? What, for that matter, is the physical explanation for a complex angle?

  • $\begingroup$ You can rewrite $sin(x+iy) = sin(x)cosh(y)-isin(x)sinh(x)$. If you have a refraction between $n_1 > n_2$ your refracted angle is greater then the incoming angle. There exists an angle where your refracted light beam is parallel to the surface for higher angles you get a total reflection. $\endgroup$
    – Alpha001
    Feb 17, 2017 at 23:55
  • $\begingroup$ All other things are easy to calculate. See for example: en.wikipedia.org/wiki/Total_internal_reflection $\endgroup$
    – Alpha001
    Feb 17, 2017 at 23:57
  • $\begingroup$ Doesn't Jackson explain what this means? If you are questioning something which Jackson says then you really ought to post an image of the relevant text so that we can see what argument he is making. Clearly you understand TIR. What you are asking about here is Jackson's method, not the physical phenomenon itself. $\endgroup$ Feb 18, 2017 at 3:49
  • $\begingroup$ If you naively apply Snell's law of refraction, you have $\sin \theta_2 = (n_1/n_2) \sin\theta_1$. If $n_1 > n_2$, then as $\sin\theta_1$ increases towards $1$, you will get $\sin \theta_2 > 1$. How you interpret that mathematical result is another question, of course. $\endgroup$
    – alephzero
    Feb 18, 2017 at 8:58

1 Answer 1


Look at it this way: We can express a plane wave as

$$\mathcal{E}_z(x,y,t) = \cos(k_x x + k_y y - \omega t)$$

where I've assumed a $z$-polarized wave propagating in the $x$ and/or $y$-directions. In phasor notation this can be written

$$E_z(x,y) = e^{i(k_x x + k_y y)}$$

where I've used the convention

$$\mathcal{E}_z(x,y,t) \equiv \text{Re}(E_z(x,y) e^{-i \omega t} )$$

In order for this wave to satisfy the wave equation, we must have

$$k_x^2 + k_y^2 = k^2$$

where $k$ is the propagation constant in the medium, and is real and positive. For a propagating wave, $k_x, k_y$ are real and $|k_x|, |k_y| \le k$, and they can always be written

$$k_x = k \cos(\theta)$$

$$k_y = k \sin(\theta)$$

for some real angle $\theta$.

What happens when, $|k_y| > k$? This can be the case for the solution in the refracted medium of the interface problem, and corresponds to total internal reflection. $k_x^2 + k_y^2 = k^2$ must still be satisfied, which means that $k_x$ must be purely imaginary. The full time-dependent signal then looks like

$$\mathcal{E}_z(x,y,t) = \cos(k_y y - \omega t) e^{ik_x x}$$

Here the electric field is decaying, or evanescent, in the $+x$-direction and there isn't any energy propagating. Physically, this means that the wave is oscillating in the $y$-direction faster than what is physically supported by the medium, which is why it must die out in the $x$-direction. There's no longer a direction of propagation, so there's no apparent reason to find a $\theta$.

In my opinion, that's pretty much the whole story. However, there is also Snell's law which states $n_1 \sin (\theta_1) = n_2 \sin (\theta_2)$, where $\theta_1$ is the angle of incidence, $\theta_2$ is the angle of refraction, and $n_1, n_2$ are the indices of refraction. In the case of an evanescent wave in the refracted medium, if we wanted to solve for $\theta_2$, we would find that $|\sin(\theta_2)| > 1$, which is equivalent to $|k_y| > k$. Clearly, that also implies that $k_x$ and therefore $\cos(\theta_2)$ are imaginary.

Obviously, for this to be true, we must use the following generalizations of sine and cosine which allow complex arguments

$$\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$$

$$\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$

Note that if we choose complex $\theta = (\pi/2 + n\pi) + i I$, where $I \ne 0$, we will get real $\sin(\theta)$, where $|\sin(\theta)|>1$ and imaginary $\cos(\theta)$.

Is there a physical reason to calculate such a complex $\theta$? Maybe not, but the math is clever. And it's nice to see Snell's law hold even for the case of total internal reflection.

  • $\begingroup$ As regards the last two sentences, why did you choose to include the term $n \pi$ in $\theta = (\pi / 2 + n \pi) + iI$? It seems not correct to obtain a unique result for $\sin \theta > 1$. If you can, please check out my question here. $\endgroup$
    – BowPark
    Nov 22, 2018 at 17:57
  • 1
    $\begingroup$ @BowPark You're correct, I was imprecise about the sign of $\sin\theta$. I fixed the answer to make correct. I was simply trying to be complete in showing examples of values of $\theta$ which yield $|\sin\theta| > 1$. Including the $n\pi$ term wasn't strictly necessary. $\endgroup$
    – LedHead
    Nov 22, 2018 at 22:10

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