Total Internal Reflection

Total Internal Reflection
관련코스 현대광학
소분류 물리
선행 키워드
연관 키워드 Goos-Hanchen shift

When light penetrates from dense medium to rare medium, the reflection coefficient from Fresnel equations can have complex solution when $\theta \gt \sin^{-1}n$.

$r_s = \frac{\cos\theta-i\sqrt{\sin^2\theta-n^2}}{\cos\theta+i\sqrt{\sin^2\theta-n^2}}$
$r_p = \frac{-n^2\cos\theta+i\sqrt{\sin^2\theta-n^2}}{n^2\cos\theta+i\sqrt{\sin^2\theta-n^2}}$

Note that $|r_s|^2 = |r_p|^2 = 1$, and all the energy the EM wave carries is bounced back.

Evanescent wave

Now, let's talk about the wave on the side of rarer medium in the condition of total internal reflection. Although real solution for transmission angle $\phi$ doesn't exist, still we can try to write it down and see what happens.

The transmitted wave is represented as

$\mathbf{E}_t = E'' e^{i(\mathbf{k}''\cdot\mathbf{r}-\omega t)}$

where

\begin{align}\mathbf{k}''\cdot\mathbf{r} &= (k''\sin\phi, -k''\cos\phi)\cdot(x, y) \\ &= k''x\sin\phi - k''y\cos\phi \\ &= \left(k''\frac{\sin\theta}{n}\right)x - i \left(k'' \sqrt{\frac{\sin^2\theta}{n^2}-1}\right)y \end{align}.

Therefore, we have a wave that travels in x direction and decays in -y direction:

$\mathbf{E}_t = E'' e^{-\alpha|y|} e^{i(k_1 x-\omega t)}$

This wave, travelling along the boundary with speed $\omega/k_1$ and exponentially decaying with the distance from the boundary, is called evanescent wave. See for yourself what has to come in the place of $\alpha$ and $k_1$.

Phase change during TIR

The fact that reflection coefficient becomes complex value for TIR case means that we have phase changes between reflected and incident fields.

Let us write the reflection coefficient in the following form (remind its magnitude is 1):

$r_s = \frac{\cos\theta-i\sqrt{\sin^2\theta-n^2}}{\cos\theta+i\sqrt{\sin^2\theta-n^2}}\equiv \frac{a e^{-i\alpha}}{a e^{i\alpha}}\equiv e^{-i\delta_s}$
$r_p = \frac{-n^2\cos\theta+i\sqrt{\sin^2\theta-n^2}}{n^2\cos\theta+i\sqrt{\sin^2\theta-n^2}}\equiv -\frac{b e^{-i\beta}}{b e^{i\beta}}\equiv e^{-i\delta_p}$

Then we have the following:

$\tan \frac{\delta_s}{2} = \frac{\sqrt{\sin^2\theta-n^2}}{\cos\theta}$
$\tan \frac{\delta_p}{2} = \frac{\sqrt{\sin^2\theta-n^2}}{n^2\cos\theta}$

The fact that TE and TM waves take on different amount of phase shift implies that a linear polarized light will become an elliptically polarized light upon total internal reflection.

Fresnel's rhomb

The fact that s-wave and p-wave suffer different amount of phase shift upon total internal reflection means that one can make a device similar to waveplate using total internal reflection. This kind of device has advantages over conventional waveplate since its phase retardation does not depend on wavelength (broadband).