# Does Snell's law still apply after total internal reflection has occurred?

Snell's law states that:

\begin{align} n_1\sin\theta_1 &= n_2\sin\theta_2\, ,\\ \sin\theta_1 &= \frac{n_2}{n_1}\sin\theta_2\, ,\\ \sin\theta_1&\propto \sin\theta_2\, . \end{align}

The critical angle $\theta_1 = \theta_c$ is reached when $\sin\theta_2 = 1$, assuming that $n_2 > n_1$.

Total internal reflection occurs when $\theta_1 > \theta_c$, and when $\theta_1$ increases between 0 and 90 so does $\sin\theta_1$, and since $\sin\theta_1 \propto \sin\theta_2$, $\sin\theta_2$ should increase also, which should be impossible since $\sin\theta_2 = 1$.

Does the proportionality (and therefore Snell's law) no longer hold once the critical angle $\theta_c$ has been reached, or have I got my maths wrong?

• Can you draw a diagram to illustrate your question? Snell's law gives the relationship between an incident ray and a refracted ray. In the case of total internal reflection, there's no refracted ray, so what are you expecting Snell's law to tell you ? – The Photon Jun 18 '17 at 18:01
• Or, to put it another way, suppose Snell's law tells you that $\sin \theta_1$ has a value greater than 1. What do you expect that means? – The Photon Jun 18 '17 at 18:04
• For $\theta>\theta_c$ Snell's law still holds; you just need to expand the domain of your trigonometric functions to the complex numbers so that you can have $\sin \theta>1$. – Mo_ Jun 19 '17 at 0:10
• – Emilio Pisanty Jun 19 '17 at 21:01

First, let's calculate the critical angle. As you noted, at $\theta_1=\theta_c$, $\sin\theta_2=1$, and thus $\theta_c=\arcsin(n_2/n_1)$. Obviously, for a critical angle to exist, we must have $n_1>n_2$, so that $n_2/n_1<1$.

Now, from Snell's law, we know that \begin{align} \sin\theta_1=\frac{n_2}{n_1}\sin\theta_2\end{align} As $\theta_1$ increases to $\theta_c$, $\sin\theta_2$ goes to $90$ degrees, as you noted:

\begin{align} \sin \theta_c =\frac{n_2}{n_1}=\frac{n_2}{n_1}\sin\theta_2\rightarrow \sin\theta_2=1 \end{align}

For $\theta_1>\theta_c$, we find that

\begin{align} \sin \theta_1 \equiv C = \frac{n_2}{n_1}\sin\theta_2\rightarrow \sin\theta_2=\frac{C}{n_2/n_1} \end{align}

where $C$ is some number larger than $(n_2/n_1)$ but less than 1, implying that $\frac{C}{n_2/n_1}>1$. Thus, you are correct; Snell's law will not be able to solve for the refracted angle for an incident angle larger than $\theta_c$. Instead, the system in this limit obeys the law of reflection. For more about Snell's law and total internal reflection, see here for a good reference.

• Thank you. I've read up on the law of reflection and I guess it makes sense now. I was still thinking of the reflected ray as being refracted, when in fact that's not true, hence my confusion. – Pancake_Senpai Jun 18 '17 at 23:55

Snell's law is obtained by applying the electromagnetic boundary conditions to the problem; therefore it holds under all circumstances where Maxwell's equations hold.

In order to see what happens for angles of incidence greater than the critical angle, we derive Snell's law. The electromagnetic boundary conditions take the following form for both TE and TM polarizations of the incident wave:

$$A_ie^{j\mathbf{k_+ \cdot r}} + A_re^{j\mathbf{k_- \cdot r}} = A_te^{j\mathbf{k'\cdot r}}$$

Snell's law is derived by taking care of the exponential parts for $$z=0$$:

$$k_x=k_x' \rightarrow k\sin \theta = k'\sin \theta' \rightarrow \boxed{n\sin \theta =n'\sin \theta'}$$

Now we obtain the wavevectors in the $$z$$ direction, $$k_z$$ and $$k_z'$$:

$$k_z^2+k_x^2=k^2=n^2 k_0^2$$

$$k_z'^2+k_x'^2=k'^2=n'^2 k_0^2$$

Using the Snell's law ($$k_x' = k_x$$, so $$k_x' = k \sin \theta = n k_0 \sin \theta$$) we find, from the second equation:

$$k_z'^2 = k_0^2 \left( n'^2 - n^2 \sin^2 \theta \right) \Rightarrow k_z' = k_0\sqrt{n'^2-n^2\sin^2 \theta}$$

Using this equation we can see what happens for angles of incidence greater than the critical angle. If $$n' the wavevector in the $$z$$ direction would become imaginary. $$k_z' = j \alpha$$, with

$$\alpha = k_0 \sqrt{n^2\sin^2 \theta - n'^2}$$

and the wave function in the right-hand medium would be

$$()e^{- \alpha z + j k_x' x}$$

Therefore, there exists a wave in the second medium even for $$\theta >\theta_0$$, but it decays exponentially with $$z$$ and doesn't carry any energy in the $$z$$ direction (an evanescent wave).