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Why does gradually increasing refractive index coating reduce reflection?

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EDIT:

As @Michael Seifert nicely describes in his answer, reflection only occurs if there is an abrupt change in refractive index. Since for a gradually increasing refractive index, there is no abrupt change, it makes sense that there is very little reflection. However, what I don't understand is how thick this layer needs to be ? ...because I guess that if the gradient of the refractive index is too steep then it will almost look like a abrubt change, right ?

Any literatur reference is also appreciated.

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The question itself isn't entirely clear, but I believe this can be understood in terms of the Fresnel equations. Whenever a light wave hits an interface, some of the wave is reflected and some is transmitted. If the media are very similar to each other, then very little of the wave is reflected; in the limit that the indices of refraction are identical, there is no reflected wave at all. A continuously varying index of refraction can be understood as a succession of interfaces between thin layers, each with a slightly larger index of refraction; if the change in the index of refraction between each "layer" is sufficiently small, then there will be very little reflection at each layer, and (hopefully) very little reflection overall.

An analogy would be water waves coming in towards a beach. The phase velocity of surface waves in shallow water is proportional to the square root of the depth of the water ($c = \sqrt{gh}$.) This means that as the waves travel towards shore, their phase velocity is decreasing, just as the phase velocity of the light decreases as the index of refraction increases. But because the increase in the phase velocity is gradual, the waves never "reflect" off of the shallower water and head back out to sea.

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  • $\begingroup$ Somewhat unsatisfying, as it is not clear that as the number of layers approaches infinity (and the thickness approaches zero) that the net effect of the infinite number of non-zero reflections approaches zero. $\endgroup$ – garyp Sep 17 at 14:28
  • $\begingroup$ @garyp: Agreed; a full answer would involve using a geometric-optics approximation to Maxwell's equations. But I didn't have the time to go into all of that detail when I wrote the answer, and I wasn't sure whether it was really necessary (the OP's level of technical skill is unclear.) Feel free to write up a competing answer with more technical detail and I'll happily upvote it. $\endgroup$ – Michael Seifert Sep 17 at 17:04
  • $\begingroup$ A secondary effect, if the transition layer thickness is similar to or larger than the wavelength, is that the differetial reflections are going to have non-trival phase differences and should partially interfere. $\endgroup$ – dmckee Sep 17 at 19:51
  • $\begingroup$ Thank you very much for your answer, which already helps quite a bit. Please have a look at my updated question for my follow-up question. $\endgroup$ – james Sep 18 at 8:51
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I believe that reflection from a layer with variable refractive index can be treated as a direct analog with tapered impedance transmission lines, in the same way that single and multilayer anti-reflection coatings are analogous with impedance matching transmission line sections.

Much of the literature on tapered transmission lines was published in microwave engineering journals in the 1950s, for use in designing waveguide transitions, horn antennas, etc. The only accessible reference that I have on hand is the book by R. E. Collin, Foundations for Microwave Engineering, McGraw-Hill, 1966 (Chapter 5, Sections 5.12-5.15).

Qualitativly, the idea is that a tapered impedance matching section works somewhat like a high pass filter, allowing tranmission with reduced reflections at frequencies where the taper is gradual compared to the wavelength, that is when the length, $L \gg \lambda$. A quantitative discussion from Collin is sketched below.

Reflection from a differential length of transmission line, with characteristic impedance a function of length, $Z(x)$ is given by

$$d\Gamma = \frac{Z+dZ-Z}{Z+dZ+Z} \approx \frac{dZ}{2 Z} = \frac{1}{2} d(\ln Z)$$

To add up reflections at the input of the taper, first multiply by the phase factor appropriate for the distance of each differential reflection, and then integrate.

$$\Gamma_{in} = \frac{1}{2} \int_{0}^{L} e^{-j2\beta x} \frac{d (ln Z)}{dx}dx$$

Here $\beta = 2 \pi/\lambda$, and $j$ is the imaginary unit, since the derivation comes from EE and not physics or optics literature!. This expression for the reflection coefficient is not exact, since multiple reflections are neglected, but is a good approximation for a gradual taper.

To connect this theory with a varying optical index of refraction it is only necessary to note that wave impedance (using normal incidence for simplicity) for a media with refractive index, $n$, is given by $\frac{1}{n}Z_0$, where $Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}$. Using normalized impedances, the taper starts with $Z(0)=1$ and ends at $Z(L)=1/n$, where $n$ is the index of the material to be matched.

A simple example is an exponential taper where the log of the impedance varies linearly with length.

$$Z(x)= e^{(x/L) \ln Z_L}$$ $$\ln Z = \frac{x}{L} \ln Z_L$$ $$ Z_L = 1/n$$

The exponential taper can be integrated giving the result

$$ \Gamma_{in} = \frac{1}{2} e^{-j\beta L} \ln Z_L \frac{\sin \beta L}{\beta L}$$

Other taper profiles, such as a linear taper, are easily evaluated numerically. The following plot shows comparison of magnitudes for the reflection coefficient versus $L/\lambda$ of a linear and exponential taper to match a material with index of refraction, $n = 1.5$, and thus impedance, $Z_L=2/3$.

Reflection Coefficient Magnitude

Note the reduction in maxima of reflections as $L/\lambda$ increases. Also, in the long wavelength limit where the reflection coefficient magnitude is larger, there is a slight error due to neglect of multiple reflections. The result should approach $\Gamma = 0.2$ as $L \rightarrow 0$, but the computed value is slightly higher, approximately 0.203. Also, note the similarity between linear and exponential taper results. Minima of the linear taper reflection do not reproduce the zeros of the sinc function, and its maxima are very slightly larger. Practically, these differences are likely negligible.

Collin gives a more rigorous analysis of reflections from tapered impedance sections, including multiple reflections, resulting in the following differential equation for the reflection coefficient.

$$\frac{d\Gamma}{dx} = 2 j \beta \Gamma - \frac{1}{2} (1 - \Gamma^2) \frac{d (ln Z)}{dx}$$

This non-linear Ricatti type equation differs from the approximate analysis given above mainly by the $(1 - \Gamma^2)$ factor, that can usually be neglected if reflecions are small, as in most practical use. The exact equation can be solved in some special cases, and in general numerically. However, at least for the waveguide applications, there is more difficulty in specifying a valid characteristic impedance for cross sections that vary, unless the taper is gradual. So the exact equation is rarely necessary in practice.

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