Evanescent field of a waveguide

[Sciencestd]

If we look to the figure, in several articles they mentioned that the part of the mode field, the tail of the mode field diameter, travel in the cladding, this maybe I can understand that because of a little change between the refractive indices of the core and the cladding, then the transverse mode field will encounter less change. Then they mentioned that the other side they took off the cladding so now let say it is air, then the length of the evanescent field in that side obey to the equation as in the second picture, and the field is rapidly decaying... Why?! What is the physical reason? which equations include that? why should the transvers part changed to this situation?
Images source: https://www.researchgate.net/publication/44067714 and scirp.org

 

[Paul Colby]

If I take the equation at face value (it looks real sensible)

$n_{\text{Si}}^2 \sin^2 \theta - n_{\text{gas}}^2 \gt n_{\text{Si}}^2 \sin^2 \theta - n_{\text{Sapphire}}^2$

so that

$d_{\text{gas}} \lt d_{\text{Sapphire}}$ since $n_{\text{gas}}\approx 1 \lt d_{\text{Sapphire}}$

In a wave guide one can always separate (as a product of functions) the transverse wave from the longitudinal because of translational symmetry along the guide. The transverse mode function obeys a wave equation in each medium each with a different propagation factor.

 

[Sciencestd]

Thank so much for your help! :)
I have two questions: The first equation I can seek from Snell's law, but why to square them?
The second question is why then the mode field tail in sapphire can reach several micrometers but the evanescent field in low medium like gas just few nanometers (it's wavelength dependent in the equation "d=..")?

 

[Paul Colby]

The first equation I can seek from Snell's law, but why to square them?

Let's assume planar geometry with scalar field, $\psi(x,y,z)$. The wave equation is,

$\nabla^2 \psi + k_i^2 \psi = 0$

where $i$ is Gas, Si, or Sapphire depending on where the point $(x,y,z)$ is. The wave number, $k_i = \omega n_i$, so the square comes from the wave equation. $\omega$ is the angular frequency, $\omega = 2\pi f$.

[sorry for all the edits]

The second question is why...

Use separation of variables (I can do this because I chose plain geometry)

$\psi(x,y,z) = \chi(x)\eta(y)e^{-i\gamma z}$

and grind. I'd look at the case $\eta(y) = 1$ so it drops out. Note that one must solve for $\gamma$ the waveguide propagation constant. The wave equation for $\chi$ is one dimensional,

$\frac{d^2\chi}{dx^2} + (k^2_i - \gamma^2)\chi = 0$

$\gamma$ is a constant for any given solution of the full problem. What happens when $\gamma \gt k_i$ is $\chi$ becomes exponentially damped. The bigger the dampening the shorter the distance.

 

[Sciencestd]

I'm really very grateful to you... thank you so much! I have now the way to think about it... thank you :) for now I don't have questions..