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This maybe a bit of a silly question of my part, but I am trying to rewrite the Reflectivity, $R$ as a function of the complex dielectric function $\epsilon = \epsilon_r + i\epsilon_i$. Starting from the generic reflectivity equation, assuming a vacuum interface ($n_2$ =1) \begin{equation} R = \frac{(1-n)^2 + k^2}{(1+n)^2 + k^2} \end{equation} And then using that $n + ik = \sqrt{\epsilon}$ and the algebraic form for the square root of a complex number, we obtain: \begin{equation} n = \frac{1}{\sqrt{2}}\sqrt{\sqrt{\epsilon_r^2 + \epsilon_i^2} + \epsilon_r} \end{equation} \begin{equation} k = \frac{1}{\sqrt{2}}\sqrt{\sqrt{\epsilon_r^2 + \epsilon_i^2} - \epsilon_r} \end{equation} However, I now have an issue in that I obtaining negative R values when one plots $R$ as a function of $\epsilon_r$ and $\epsilon_i$. I have plotted the resulting equation below. Is there something wrong with the approach I have done here? Is a negative reflectivity physically meaningful or is it indicating that equation can only be used for certain values of $\epsilon_r$ and $\epsilon_i$?

Plot

UPDATE: This was a calculation error on my side, by rewriting my code. I believe that accidentally put a minus sign in the numerator, leading to negative values. This new graph is obtained: enter image description here

tjsmert44
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  • Are you familiar with the Fresnel equations? – my2cts Jun 13 '21 at 20:08
  • Yes, I am. This is where I obtained the original equation from, in the case where you have a complex refractive index. I am also assuming that you are striking the surface at normal incidence. – tjsmert44 Jun 13 '21 at 20:59
  • In the first equation $R=1$. – my2cts Jun 13 '21 at 21:22
  • It is a good approach. Section 13: Optical properties of solids. I don't know why you are getting negative reflectivities. My first thought is complex dielectric function handles a phase shift. Perhaps reflectivity is negative when the phase shift > $\pi$? – mmesser314 Jun 13 '21 at 21:32
  • @my2cts Thanks for pointing that out, I have changed it accordingly. – tjsmert44 Jun 14 '21 at 07:23
  • @mmesser314 I think you may be correct, considering that $R$ technically refers to the ratio of the reflected and incident electric fields. So having a negative value would just correspond to a phase shift. However, I am unsure in this case, considering that we have absorption in our system, see: https://physics.stackexchange.com/questions/353917/why-does-a-negative-reflection-coefficient-correspond-to-a-phase-difference-of – tjsmert44 Jun 14 '21 at 07:45
  • Negative $\varepsilon_i$ need not to be considered, otherwise your approach is good. Please re-check your formulae for $n$ and $k$ as well as how the calculation of R is performed in your software wrt complex numbers. – sleepy Jun 14 '21 at 08:49
  • By definition R is positive. – my2cts Jun 14 '21 at 09:23
  • @my2cts I believe that the values of R obtained result from the resulting phase shift that would occur under certain values of n (ie n<1). Similar results have been noted previously here: https://physics.stackexchange.com/questions/353917/why-does-a-negative-reflection-coefficient-correspond-to-a-phase-difference-of – tjsmert44 Jun 14 '21 at 09:32
  • @sleepy All my code does is calculates the definition of $R$ given the equations above for $R$, $n$ and $k$ as a function of the dielectric components. – tjsmert44 Jun 14 '21 at 09:33
  • @tjsmert44 The problem is obviously in how everything is calculated. You have two options. 1. find out where exactly things go wrong, i.e. do it step by step and plot every intermediate result, 2. use $R=|(n-1)/(n+1)|^2$ instead (here $n$ is complex) – sleepy Jun 14 '21 at 10:04
  • OK this is definitely a calculation error, I have rewrote the code and I have now obtained the newly added plot. – tjsmert44 Jun 14 '21 at 10:20

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