# Griffiths Electrodynamics Problem 9.39: How can $\sin(\theta_T)$ be greater than one?

When an electromagnetic wave strikes an interface between two linear media, Snell's law states that $$\frac{\sin(\theta_T)}{\cos(\theta_I)} = \frac{n_1}{n_2}$$ where $$\theta_I$$ is the angle of incidence, $$\theta_T$$ is the angle of transmission, $$n_1$$ is the index of refraction of the first medium, and $$n_2$$ is the index of refraction of the second medium.

In the case where $$n_2 > n_1$$, we can then see that $$\sin(\theta_T) = \frac{n_1}{n_2} \sin(\theta_I)$$

From this we can derive a critical angle $$\theta_C$$ such that when $$\theta_I = \theta_C$$ we have $$\theta_T = 90 \deg$$. Consider, for instance, the case where light travels from water with an index of refraction of $$n_1=1.35$$ and air with an index of refraction of $$n=1$$. Then we find that $$\theta_C=47.8 \deg$$.

By taking $$\theta_I = \theta_C + \varepsilon$$ where $$\varepsilon$$ is some small positive value, we find $$\sin(\theta_T) = \frac{1.35}{\theta_C+\epsilon} > 1$$. That is, the sin function is taking a value outside of it's range!

Normally I would chalk this up to the problem being "out of the bounds" of the mathematical model of Snell's law, but Griffiths uses this fact to derive evanescent waves:

The only change is that $$\sin(\theta_T) = \frac{n_1}{n_2} \sin(\theta_I)$$ is now greater than $$1$$, and $$\cos(\theta_T) = \sqrt{1-\sin^2(\theta_T)} = i\sqrt{\sin^2(\theta_T)-1}$$ is imaginary. (Obviously, $$\theta_T$$ can no longer be interpreted as an angle!)

How is it possible that $$\cos(\theta_T)$$ is imaginary? What does it mean that $$\theta_T$$ cannot be interpreted as an angle?

• I should mention that $\sin(\theta_T)>1$ for all $\theta_C < \theta_I < 90\deg$. – Mason Hargrave May 7 at 9:17

Using Euler's formula $$e^{ix}=\cos{x}+i\sin{x}$$ one can write $$\cos{x} = \frac{e^{ix}+e^{-ix}}{2}$$ If $$x$$ is real, then $$\cos{x}$$ is real. However if you allow complex numbers $$x=u+iv$$ then $$\cos{(u+iv)} = \frac{e^{iu-v}+e^{-iu+v}}{2}$$ and you see that the cosine (and sine as well) can be complex valued and greater than one. Once you allow complex numbers you can't interpret the argument as an angle.
• So for imaginary inputs such as $x=iv$ where the real component $u$ is equal to zero and the imaginary component $v>.881$, I can see that $\sin(x) = \frac{e^{ix}-e^{-ix}}{2} >1$ and is real valued. In that exact situation, $\cos(x)$ is also real valued and greater than 1. In order for $\cos(x)$ to be imaginary, $x$ must be pure real. How then am I supposed to interpret that fact that $\sin(\theta_T) > 1$ implies that $\theta_T$ is pure imaginary with $v>.881$ while simultaneously $\cos(\theta_T)$ is imaginary which implies that $\theta_T$ is pure real valued? – Mason Hargrave May 7 at 9:59
• You need to divide by $i$ in the sine formula! – John Donne May 7 at 10:16
• In particular the condition you're looking for is $\cos{u}=0$. This will give you an imaginary cosine and a real sine – John Donne May 7 at 10:28