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I have three layers of different complex refraction index. I am shining with a laser of wavelength lambda onto the layers.

enter image description here

Through etching, each layer is etched one by one, and hence one by one, the thickness is reduced of each layer until zero.

Experimentally, I am getting this intensity plot (the first and last layer have the same refraction index (blue line):

enter image description here

How can I obtain this with an analytic solution ? (or iterative process)

I am mostly interested in normal incidence !

The plot is just there for reference. It only shows the shape of the final solution. The y-Axis represents the light intensity. The x-Axis the time.

Blue: Line -> Light intensity

Yellow Line: It's second derivative

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  • $\begingroup$ How do we read this plot? What is x-axis, what is y-axis? What is yellow line? What is blue line? $\endgroup$ – 299792458 May 10 '17 at 13:27
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    $\begingroup$ @TheDarkSide I added some more information $\endgroup$ – henry May 10 '17 at 13:54
  • $\begingroup$ Isn't this device multiple Fabry-Pérot interferometres in serie? You can easily find the reflected and transmitted amplitudes (see en.wikipedia.org/wiki/Fabry%E2%80%93P%C3%A9rot_interferometer otherwise). $\endgroup$ – Spirine May 10 '17 at 20:04
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This isn't directly an answer since the method of doing the calculation is rather long and tedious, if basically straightforward, but I can tell you where to find the answer because I did precisely this as part of my PhD. The method is described in Optical Properties of Thin Solid Films by O. S. Heavens, Butterworths Scientific Publications, London 1955. It is on Google Books, though sadly not a scan, and I see there is a 1991 edition so at least you won't be trying to find the 1955 edition that I had to use.

Basically at each interface you calculate a relationship between the forward going and reflected waves either side of the layer, then start at the bottom layer where you know the forward going wave is zero and work your way back up.

It's not too bad for normally incident light but becomes very messy indeed for light incident at any significant angle to the normal because then you need to account separately for the two polarisations of the light.

I used it for films that were reacting so their thickness was changing because a layer of reaction product was growing, but the method will work just as well for films that are being etched.

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  • $\begingroup$ Thank you very much for your reference ! I will definitely have a look ! For now, I am mostly interested in normal incidence. So, if the calculation is straightforward, would you mind to post it here, or is there a publication of your PHD work also available ? Thank you very much. $\endgroup$ – henry May 2 '17 at 17:39
  • $\begingroup$ @DoHe: I'm afraid my thesis was published before PDFs or the World Wide Web existed. The calculation is straightforward but long and tedious, and it would take me all day to type up the details, even if I had them to hand (which I don't). But if you can lay your hands on the book it's easy enough to program. $\endgroup$ – John Rennie May 2 '17 at 17:43
  • $\begingroup$ Okay, yes, no problem. I will search for the book. The Google Book that I found, only goes up to page 54. Thank you ! $\endgroup$ – henry May 2 '17 at 18:10
  • $\begingroup$ Yes ! I just found a version of the book from 1955 in our University library. :) $\endgroup$ – henry May 2 '17 at 18:17
  • $\begingroup$ @DoHe I'm guessing that (at)JohnRennie is talking about a T-matrix approach. You just set up a series of matrices, one for each film and one for each interface, then multiply them all together. It's exactly as he says, "rather long and tedious, if basically straightforward" which means (at least in my case a long time ago) a lot of debugging unless you are pretty good at typing math into a program and getting it to run. Then you need to check your calculations against an independent program to make sure it's right. $\endgroup$ – uhoh May 7 '17 at 8:25

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