Total Internal Reflection

Now we will consider the case $\sin\theta''_c = (n_1 / n_2 )\sin\theta_c = 1$. Here

$$\theta_c = \sin^{-1} \frac{n_2}{n_1}$$ (118)

and should $n_2 < n_1$. For water to air reflection $\theta_c = 48.66$ degree.

The transmitted light is along the surface direction. What happens if $\theta > \theta_c$ in that situation.

$$\cos\theta'' = \sqrt{1-\sin^2 \theta''} = i\sqrt{\sin^2 \theta'' -1}= i \sqrt{(n_1/n_2)^2 \sin^2\theta -1}$$ (119)

The transmitted wave can be written as

$${\bf E}'' = {\bf T}_0 e^{i({\bf k}'' \cdot {\bf r}-\omega'' t)}$$ (120)

$$= {\bf T}_0 e^{i(k'' \sin\theta'' x + k'' \cos\theta'' z -\omega'' t)}$$ (121)

$$= {\bf T}_0 e^{i(k'' \sin\theta'' x - \omega'' t)} e^{-k'' \sqrt{(n_1/n_2)^2 \sin^2\theta -1} z}$$ (122)

$$= {\bf T}_0 e^{i \frac{n_2 \omega \sin\theta'' }{c}x}e^{-i\omega'' t} e^{-\frac{n_2 \omega \sqrt{(n_1/n_2)^2 \sin^2\theta -1}}{c}z}$$ (123)

The Poynting vector ${\bf S}$ in the z direction in the medium 2,

$${\bf z}\cdot \langle {\bf S} \rangle = \frac{1}{2\mu_0 \mu}{\rm Re}({\bf E}\times{\bf B}''^* )\cdot {\bf z}$$ (124)

$$= \frac{1}{2\mu_0 \mu \omega} {\rm Re}[{\bf E}\times({\bf k}'' \times{\bf E}'' )^* ]\cdot {\bf z}$$ (125)

$$= \frac{1}{2\mu_0 \mu \omega} {\rm Re}[{\bf k}'' \vert{\bf E}'' \vert^2 - {\bf E}^{''*} (\underbrace{{\bf k}'' \cdot{\bf E''}}_{=0})\cdot {\bf z}$$ (126)

$$= \frac{1}{2\mu_0 \mu \omega} {\rm Re} (\underbrace{{\bf z}\cdot{\bf k}'' }_{k'' \cos\theta ={\rm pure \ imaginary}} \vert{\bf E}'' \vert^2 )$$ (127)

$$= 0$$ (128)

So the energy flux to ${\bf z}$ direction is zero. This wave is called evanescent wave.