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So the general equation for the reflectivity at the interface between two materials is given by: $$R=\left(\frac{n_1-n_2}{n_1+n_2}\right)^2$$ in case of air/glass $n$ is real, but for, say, semiconductors or metals, where radiation is absorbed, $n$ is a complex number, with $\underline{n}=n_r-ik$. $k$ is described as the extinction coefficient and is related to the absorption coefficient with $\alpha=\frac{4\pi k}{\lambda}$, $\lambda$ being the wavelength.

I am looking to derive a formula for the reflectivity which only includes the real and imaginary parts of the complex refractive index. As far as I can tell, the equation above gives the reflectivity as long as the norm of the index is known, that is $$ n_1=\sqrt{n_{r_1}^2+k_1^2} \\ n_2=\sqrt{n_{r_2}^2+k_2^2} $$ in the above formula for the reflectivity, I replaced the norms of the complex numbers and not the numbers themselves,obviously. So doing that, I get a fraction where square root terms appear. On the other hand Wikipedia writes(https://en.wikipedia.org/wiki/Refractive_index) $$R=\left|\frac{n_1-n_2}{n_1+n_2}\right|^2$$which also makes sense and leads to $$R=\frac{(n_{r_1}-n_{r_2})^2+(k_1-k_2)^2}{(n_{r_1}+n_{r_2})^2+(k_1+k_2)^2}$$ Which formula is right?

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    $\begingroup$ As for me, the quantity $n = \sqrt{n_r^2 + k^2}$ does not make any sense. You don't want to mix the real and imaginary part of the refractive index as they describe different phenomena. $\endgroup$
    – Ilya
    Mar 24, 2017 at 16:45
  • $\begingroup$ $n$ is the norm of $\underline{n}$, $\left|\underline{n}\right|=n=\sqrt{n_r^2+k^2} $ , $n_r$ being the real and $k$ being the imaginary part. If $k=0$, there is no point writing an index $\small r$, since it's $n=n_r$ $\endgroup$
    – pitfermi
    Mar 24, 2017 at 16:48
  • $\begingroup$ Yes, I see that, but what is the physical meaning of this norm? The ratio of $n_r$ for two materials refers to the ratio of the wave speeds in them. $k$ is related to the decay rate of the amplitude. What about the norm? $\endgroup$
    – Ilya
    Mar 24, 2017 at 16:59
  • $\begingroup$ If I knew i wouldn't have asked in the first place, but mathematically, the complex refraction index is automatically assigned a norm. why do you assume that it's false to build a norm out of 2 constants which describe a different phenomena? both are constants. there is no problem with units etc. $\endgroup$
    – pitfermi
    Mar 24, 2017 at 17:10
  • $\begingroup$ @Travis gives a short answer that is correct for the special case of normal incidence. If that's what you want, fine. But if you want the reflectivity at some other angle of incidence the expression is quite a bit more complicated. See the Wikipedia page for Fresnel Equations $\endgroup$
    – garyp
    Feb 4, 2020 at 19:51

3 Answers 3

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The second formula is correct.

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  • $\begingroup$ It would be nice if you could add some additional details and insights for the community at large. Short answers like yours are best posted as comments. $\endgroup$ Mar 24, 2017 at 17:31
  • $\begingroup$ Why? This correctly answers the question. Why write a physics chapter here? $\endgroup$
    – my2cts
    Aug 7, 2018 at 12:06
  • $\begingroup$ The second formula is correct for normal incidence. The expression at angles other than zero is more complicated. $\endgroup$
    – garyp
    Feb 4, 2020 at 19:46
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The first equation (Fresnel reflectivity) is derived assuming you have a lossless system, meaning $n$ is always real. When you introduce absorption you get the second formula as all the $k$ go to zero.

This is more apparent when you examine the derivation of the first equation, where the EM field is always a plane wave, whereas the second equation also accounts for evanescent/exponentially decaying parts.

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a=a'+ia", b=b'+ib"

|(a-b)/(a+b)|^2=|a-b|^2/|a+b|^2=[(a'-b')^2+(a"-b")^2]/[(a'+b')^2+(a"+b")^2]

The two expressions are equivalent, they are both correct.

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