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Fresnel's equation with complex n 
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#1
Jul511, 12:06 PM

P: 550

Hi,
I posted a similar question a while back but never got a decent answer, most probably due to having too many details that obscured the real problem. This time around I'll try it more simply. Fresnel's equation for reflectivity for s and ppolarization are, according to Wikipedia, given by: [tex]R_s = \left( \frac{n_1 \cos \theta  n_2 \sqrt{1  \left( \frac{n_1}{n_2} \sin \theta \right)^2}}{n_1 \cos \theta + n_2 \sqrt{1  \left( \frac{n_1}{n_2} \sin \theta \right)^2}} \right)^2[/tex] [tex]R_p = \left( \frac{ n_1 \sqrt{1  \left( \frac{n_1}{n_2}\sin\theta \right)^2}  n_2 \cos\theta }{ n_1 \sqrt{1  \left( \frac{n_1}{n_2}\sin\theta \right)^2} + n_2 \cos\theta } \right)^2[/tex] Now suppose we have a vacuum to metal interface so that the light travelling through the vacuum strikes and reflects off a metal surface. I think in this case the vacuum should have an index of refraction equal to unity while the metal gets a complex index of refraction: [itex]n_1 = 1[/itex] and [itex]n_2 = n(1 + i \kappa) = n + ik[/itex] where [itex]k[/itex] is the extinction coefficient (or is it [itex]\kappa[/itex]?). Apparently, by substituting these values and making one assumption, that [itex]n^2+k^2 \gg 1[/itex], one should be able to arrive at the following much simpler expressions: [tex]R_s = \frac{\left( n  \cos \theta \right)^2 + k^2}{\left( n + \cos \theta \right)^2 + k^2}[/tex] [tex]R_p = \frac{\left( n  \sec \theta \right)^2 + k^2}{\left( n + \sec \theta \right)^2 + k^2}[/tex] Graphing these formulas tells me that [itex]R_p[/itex] has a minimum reflectivity at some high angle, similar to nonmetals (where this minimum actually goes to 0 for the Brewster angle) while [itex]R_s[/itex] is a smoothly increasing function. The problem is as follows... I have been looking everywhere for a derivation, or even a simple statement of these formulas but I haven't found a decent reference anywhere. The only times I can find solid references to the reflectivity of a metal is for normal incidence, for [itex]\theta = 0[/itex] so that both [itex]\cos \theta = \sec \theta = 1[/itex], so that doesn't help. I did find a few other references, but they either contradict themselves (describe in detail that Rp has the minimum in reflectivity, then show a graph and write Rs in it ...?!) or contradict other papers (some seem to mention Rp has the minimum, some mention it is Rs). Some papers / books also don't mention s or ppolarization at all but merely TE or TM modes. The same problem here: they contradict each other, some papers say Rs = TE, others seem to say Rs = TM... The questions are simple (but maybe not to answer): 1. What's the correspondance between TE/TM mode and s/ppolarization? Or is there no one definition and does it depend on something else..? I would think TM = p polarization and TE = s polarization, or vice versa. Which is it? Same goes for perpendicular and parallel polarization. Perpendicular = TE or TM, s or p? My view is this: s (senkrecht) = perpendicular = TE p (parallel) = parallel = TM. Correct, or not? 2. Are the final formulas I've given for Rs and Rp correct, or did I switch them around (formula for Rs labeled Rp and vice versa)? 3. Finally: how does one get to that result.....? I don't see how the assumption that n^2+k^2 is large is any help, I've been trying for days to derive it but I am getting nowhere. I did find one other 'derivation' (not in detail, which is what I want to see) which arrived at [tex]R_p = \frac{(n^2+k^2) \cos^2 \theta  2n\cos\theta + 1}{(n^2+k^2) \cos^2 \theta + 2n\cos\theta + 1}[/tex] after dividing out [itex]\cos^2 \theta[/itex] and completing the square that does yield [itex]R_p[/itex]. The problem is that another statement (without derivation) I've got shows this to be [itex]R_s[/itex] instead?! Any help is MUCH appreciated! Thanks!! 


#2
Jul711, 01:31 AM

Sci Advisor
PF Gold
P: 1,756

You could probably glean far more insight into this problem by just simply deriving these equations from the boundary conditions. One text that discusses the reflection coefficients is Chew's "Waves and Fields in Inhomogeneous Media." For the case of reflection off of a planar boundary lying in the xy plane, we can derive the following ODE:
[tex] \left[ \mu \frac{d}{dz} \mu^{1} \frac{d}{dz} + \omega^2\mu\epsilon  k_x^2 \right] e_y = 0 [/tex] [tex] \left[ \epsilon \frac{d}{dz} \epsilon^{1} \frac{d}{dz} + \omega^2\mu\epsilon  k_x^2 \right] h_y = 0 [/tex] where e_y is the amplitude of the TE wave and h_y is the amplitude for the TM wave that depend upon [itex]e^{\pm ik_xx}[/itex]. It is easy to see from this that for constant permittivity and permeability that we have plane wave solutions fo the form [itex]e^{\pm ik_zz}[/itex] where [itex]k_z = \sqrt{\omega^2\mu\epsilon  k_x^2}[/itex]. In addition, by inspection of the ODE we can deduce that the following boundary conditions must be made [tex] e_{1y} = e_{2y}, \qquad \mu_1^{1}\frac{d}{dz}e_{1y} = \mu_2^{1}\frac{d}{dz} e_{2y}[/tex] [tex] h_{1y} = e_{2y}, \qquad \epsilon _1^{1}\frac{d}{dz}e_{1y} = \epsilon _2^{1}\frac{d}{dz} h_{2y}[/tex] From this you can derive the Fresnel reflection coefficients for any form of the permittivity and permeability, [tex] R^{TE} = \frac{\mu_2k_{1z}  \mu_1k_{2z}}{\mu_2k_{1z} + \mu_1k_{2z}}[/tex] [tex] R^{TM} = \frac{\epsilon_2k_{1z}  \epsilon_1k_{2z}}{\epsilon_2k_{1z} + \epsilon_1k_{2z}}[/tex] What is s and p? I have no idea but you should be able to easily verify for yourself from the above TE and TM equations. [itex]n = \sqrt{\frac{\mu\epsilon}{\mu_0\epsilon_0}}= \frac{k}{\omega\sqrt{\mu_0\epsilon_0}}[/itex] relates the index of refraction with the permittivity and permeabilities above. Now \theta is the angle between the normal and the direction of incidence so [itex]k_x = k\sin\theta[/itex] and [itex]k_z = k\cos\theta[/itex] I believe. Barring the mathematical exercise, my guess is that p is TM since the Brewster angle only occurs for the TM wave if the material is nonmagnetic (\mu_1 = \mu_2). But to your more salient point, I think the reason why you cannot reduce your equations to those given in your paper is because the equation from Wikipedia is WRONG when using complex numbers. It is obvious that if you plug in a complex index of refraction into the first set of equations that the result is complex. But reflectivity is real, it is the magnitude squared of the reflective coefficients so of course we are going to see some problem in just blind simple substitution. Once you account for this it comes out trivially. Just take the first two terms of the Taylor's expansion of the square root. But instead of squaring the numerator and denominator, multiply them by the complex conjugate of themselves so that you get the magnitude squared. Then it will fall out in place (I've already checked this by hand myself). 


#3
Jul711, 03:09 AM

P: 550

Thanks for your reply.
If I understand you correctly you are saying that I should approximate the square root by [tex]\sqrt{1\left(\frac{\sin \theta}{n+ik}\right)^2} \approx 1\frac{1}{2} \left( \frac{\sin \theta}{n+ik} \right)^2[/tex] Then use [tex]r = \frac{1\frac{1}{2} \left( \frac{\sin \theta}{n+ik} \right)^2  (n+ik) \cos \theta}{1\frac{1}{2} \left( \frac{\sin \theta}{n+ik} \right)^2 + (n+ik) \cos \theta}[/tex] And finally [tex]R = r \bar{r}[/tex] I have been trying this but still getting nowhere... Problem is that I cannot take the complex conjugate unless I get r in the form a+ib which I cannot see how... Then I read your post again and saw that you said I should do this to the numerator and denominator separately, does that mean this? [tex]x = 1\frac{1}{2} \left( \frac{\sin \theta}{n+ik} \right)^2  (n+ik) \cos \theta[/tex] [tex]y = 1\frac{1}{2} \left( \frac{\sin \theta}{n+ik} \right)^2 + (n+ik) \cos \theta[/tex] [tex]r = \frac{x}{y}[/tex] [tex]R = \frac{x \bar{x}}{y \bar{y}}[/tex] That doesn't seem right to me mathematically... But even that couldn't get me anywhere. As for the rest of your post, I will look into deriving them manually. 


#4
Jul711, 03:26 AM

Sci Advisor
PF Gold
P: 1,756

Fresnel's equation with complex n
Whoops, sorry, I should have said that you take only the first term of the Taylor's expansion. The fact that n^2+k^2 >> 1 it looks obvious that we drop the second term (and working it out with the second term we find that it can be dropped in the end). So anyway, let's just look at the numerator for the reflectivity of s.
[tex] \left n_1 \cos\theta  n_2\sqrt{1  \left(\frac{n_1}{n_2}\sin\theta\right)^2} \right^2 = \left \cos\theta  \left(n+ik\right) \sqrt{1  \left(\frac{1}{n+ik}\sin\theta\right)^2} \right^2 \approx \left \cos\theta  (n+ik) \right^2[/tex] [tex] \left \cos\theta  (n+ik) \right^2 = \left(\cos\theta(n+ik)\right)\left(\cos\theta(nik)\right) = \cos^2\theta2n\cos\theta+n^2+k^2 = \left(n\cos\theta\right)^2+k^2[/tex] And you can work out the rest of them from that. Sorry about that mistake earlier. If you had taken the second term included you would get something like [tex] \left(n\cos\theta\right)^2+k^2 + \frac{\sin^2\theta}{n^2+k^2} \left[ k^2n^2+n\cos\theta+\frac{1}{2}\sin^2\theta\right][/tex] By virture of the fact that n^2+k^2 >> 1 we can drop these additional terms. EDIT: As for your talk about the complex conjugates and the like. Yes, you can split them up accordingly. Take for example that any complex number can be written as follows: [tex] a+ib = Ae^{i\theta_1}[/tex] Now let us just multiply this number to our heart's content: [tex] X = \frac{(a+ib)(c+id)}{e+if} = Ae^{i\theta_1}Be^{i\theta_2}C^{i\theta_3} = ABC e^{i(\theta_1+\theta_2+\theta_3)} [/tex] If we take the complex conjugate of the number X, [tex] X^* = ABCe^{i(\theta_1+\theta_2+\theta_3)} = Ae^{i\theta_1}Be^{i\theta_2}C^{i\theta_3} = \frac{(aib)(cid)}{eif} [/tex] So we note that if we have a product or quotient of complex numbers, then the complex conjugate of the final number is equal to the product and quotient of the conjugates themselves. 


#5
Jul711, 03:48 AM

P: 550

Thanks... I finally got it :) At least now I am sure that Rp is the reflectivity with the minimum. The paper I found these formulas in actually had them reversed, they stated Rs as the formula with sec theta and Rp with cos theta... I knew they were wrong, I just couldn't prove it to myself because I couldn't find any other decent references or derivations. But now I can finally see how you can get to those approximations and I get exactly the opposite (Rp is with sec theta while Rs is with cos theta). Thanks!!



#6
Jul2912, 08:18 AM

P: 1

Why do you not have a look in the relevant textbooks? E.g. Born/Wolf, Principles of Optics,7.Edition, Cambridge University Press 1999Capter 14.214.4.
Good luck H. Weber 


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