# Reflectivity off metals - polarization

1. Jul 1, 2011

### Nick89

Hi,

I am trying to understand the reflectivity from metal surfaces, but I'm stuck... I am finding contradicting results and I can't figure out where I'm going wrong.

I am reading a paper [1] about IRAS (Infrared reflection absorption spetroscopy) where an IR beam is reflected from a metal with a thin film of adsorbed material. The author first mentions the reflectivities $R_s$ and $R_p$ of the s- and p-polarized light respectively. Now, unless I am already really mistaken here, the s-polarized light is where the electric field is perpendicular to the plane of incidence, and the p-light is where the electric field is parallel to the plane of incidence.

He quotes the reflectivities from a clean metal surface (assuming $n^2+k^2 \gg 1$ which should be true for IR light) as:
$$R_s = \frac{\left( n - \sec \phi \right)^2 + k^2}{\left( n + \sec \phi \right)^2 + k^2}$$
$$R_p = \frac{\left( n - \cos \phi \right)^2 + k^2}{\left( n + \cos \phi \right)^2 + k^2}$$
where $\phi$ is the angle of incidence (measured from the surface normal as usual) and of course the refractive index is $n + ik$.

This seems fine to me.

The author then goes on to calculate the change in reflectivity resulting from the thin film of adsorbed material and finds a quite complex equation, that's not really relevant.
When he plots these changes in reflectivity however, he also plots the reflectivity of the clean metal surface. Example:

The R* (top graph) should be the reflectivity from a clean surface, ignore the others.

He does not state whether he is talking about $R_s$ or $R_p$, but I have assumed $R_p$ due to two reasons:
1. The paper is about IRAS on metals. In this case one uses a very large angle of incidence. The incoming and reflected light interfere in this case and the s-polarized light very nearly cancels out (due to a 180 degree phase shift). The idea of IRAS is to ignore s-light and use only p-light, so it wouldn't make sense that he would be talking about Rs...
2. A few lines later he mentions:
Equation 2.2 is the equation for $R_p$.

So it seems the reflectivity for p-light ($R_p$) has a 'dip' so it has a minimum at a certain high angle of incidence. So far so good... Until I tried to plot the function myself.

When I plot Rs and Rp using Maple, I get the dip for $R_s$ and certainly not for $R_p$!!! I don't get it...

Here's my work: image
The red graph is Rs and the blue graph is Rp.

So in my calculation $R_s$ has the dip and minimum in reflectivity, and not $R_p$...

I'm kinda lost at this point. Who's wrong? Is there something wrong with my calculations? Am I colorblind? :p

I decided to look for another source, didn't find much but I did find a similar graph in my old optics textbook. They don't talk about s- and p-light though, but rather about TE and TM modes. It's been too long, but as far as I understand TE mode is where the electric field is perpendicular to the plane of incidence, so this should be the s-light?
When they show the graph of reflectivity versus angle of incidence (for a metal) they show that the TM mode has the dip whereas the TE mode does not. Basically, their TM mode graph corresponds to my $R_s$ graph (the red one with the dip) and their TE mode corresponds to my $R_p$ graph (the blue one).

So, I am slightly more convinced that my calculations are correct and that it is indeed $R_s$ that has the dip in reflectivity, and not $R_p$ as the paper seems to state.

I am now absolutely confused though, because as far as I understand the whole point of using IRAS on metals is that the reflectivity $R_p$ is lowest at high angle of incidence (which it is not if my calculations are correct) while the CHANGE in reflectivity (due toe absorption by the thin adsorbed material) is highest. If that isn't true then I don't know what to believe anymore...

Any help on this matter?
Thanks!

[1] The paper might be hard to find, but:
F.M. Hoffmann. Infrared reflection-absorption spectroscopy of adsorbed molecules. Surface Science Reports, 3:107-192, 1983.

2. Jul 2, 2011

Nobody? :(

3. Jul 2, 2011

### sophiecentaur

Are you sure that this is not in standard text books on EM? I'm afraid that "it's been a long time . . . " but_
It looks to me that your curve is showing the Brewster angle (the value at the minimum), which applies to the VP (i.e. the polarisation for which the plane of polarisation is not parallel with the plane of the reflector.
Does that help? It gives you a key word to search on if you didn't already know it.

4. Jul 2, 2011

### Nick89

At the moment I am more and more convinced that the paper has switched around the definitions for Rs and Rp. All graphs I can find of reflectivities off metals show Rp to have the dip, but when I graph Rp as it is defined in the paper it doesn't have this dip (Rs does have it).

However I cannot find ANY other references that show or derive those equations... I'm a little confused, those must be quite common, but I can't find them anywhere. The paper does link to a reference but that's nearly 60 years old and I cannot find it anywhere...

I've been trying to derive them myself from the Fresnel equations (by simply replacing n with n+ik) but I've gotten nowhere; I'm not sure where and which assumptions I have to make (the paper mentions $n^2+k^2 \gg 1$ but that doesn't get my anywhere either...

All I can find is the reflectivity for normal incidence on metals, which is the same as both of my formulas for $\phi = 0$:
$$R = \frac{(n - 1)^2 + k^2}{(n+1)^2 + k^2} = \frac{(n-\cos(0))^2 + k^2}{(n+\cos(0))^2+k^2} = \frac{(n-\sec(0))^2 + k^2}{(n+\sec(0))^2+k^2}$$
but that doesn't help either... It just tells me that the formulas are correct, but I have a feeling that they switched around Rs and Rp. But I can't be sure without finding a reference or derivation of these formulas which I can't...

Last edited: Jul 2, 2011
5. Jul 2, 2011

### sophiecentaur

I can't be much help here but s-polarised should be the one with the dip at the Brewster angle. (what I would call Vertically polarised, with E fields having both V normal and parallel components to the surface). I am sure you a correct there.

HA HA you are amazed at a paper nearly 60 years old. It's not pre-Newtonian, you know. All this stuff was sussed out long before i went to Uni. You young things.

I could suggest you look at em theory books rather than optics books. Titles like "EM Waves and Propagation" would have what you need, I'm sure. There are bound to be lurking in some University Library, not far from you It was not long ago that I found useful stuff not far from where I live (Brighton UK)

6. Jul 2, 2011

### Nick89

Sorry I can't make out what you are saying... First you say it should be the s-polarized reflectivity that has the dip (meaning I would be wrong), then you say I am correct..? So which is it according to you?

And the fact that the paper is 60 years old merely means I cannot find it online where I can access it (using my universities access). I can find it via google but then I cannot access it unless I pay it out of my own pocket...

I'll take a look at some EM books.

7. Jul 2, 2011

### sophiecentaur

Sorry - muddled thinking.
I think p means what I would call VP (with the E vector in the plane of incidence - just cleared that up with good old wiki). That is the one with the dip I reckon. This would be because the resulting E field after refraction would have to be in the direction of propagation of a reflected ray i.e. no transverse component to form a ray.

There have been a few complaints, recently, about availability of info. But I can see why organisations want to charge. After all, they have to operate a site and they don't want advertisers. Personally, I would prefer not to have ads either. There is no such thing as a free lunch, is there. That's the beauty of textbooks. They are available to look at in most academic institutions and contain nearly all you want to know 'under one roof' (certainly of that vintage). No need to pay for copies of pages in a library if you have a half decent digital camera. The best ones are pricey to but - but they always were! That's why I can't bring myself to part with any of my books, purchased in the sixties. Some of them must have cost me at least £2 haha

8. Jul 2, 2011

### Bill_K

Nick89, Your formulas are given in the bible of optics, Principles of Optics by Born and Wolf. I couldn't find them at first because they aren't in the place I expected. They're in sect 13.4.1, which is pp628-629 in my copy.

EDIT: They're omitted from a lengthy section on a metallic surface, but then given two sections later in a discussion of metallic films. Also different notation: ρ2 instead of R2.

Last edited: Jul 2, 2011
9. Jul 2, 2011

### Nick89

Thanks for looking. I have a digital copy of that book, but I couldn't find the formulaes... I found a few that seem similar, such as
$$\rho_{12}^2 = \frac{(n_1 \cos \theta_1 - u_2)^2 + v_2^2}{(n_1 \cos \theta_1 + u_2)^2 + v_2^2}$$
(formula 6)
but when I tried to replace u2 and v2 by the formulaes given for them (formula 4) I ended up with a huge mess even Maple couldn't simplify...

10. Jul 2, 2011

### Bill_K

You don't need Maple. :uhh: In B&W's formula 6 put n1 = 1, u2 = n, v2 = k, and you've got Rp.

11. Jul 2, 2011

### Nick89

How can you justify that if the page before that they set
$$n^{^}_2 \cos \theta_2= n_2(1+i \kappa_2) \cos \theta_2 = (n_2+ik_2) \cos \theta_2 = u_2+iv_2$$
and then derive that
$$2u_2^2=n_2^2(1-\kappa_2^2)-n1_^2 \sin^2 \theta_1 + \sqrt{(n2_2(1-\kappa_2^2)-n1^2 \sin^2 \theta_1)^2 + 4n_2^4\kappa_2^2}$$
$$2v_2^2=-(n_2^2(1-\kappa_2^2)-n1_^2 \sin^2 \theta_1) + \sqrt{(n2_2(1-\kappa_2^2)-n1^2 \sin^2 \theta_1)^2 + 4n_2^4\kappa_2^2}$$

How can then u = n and v = k..? I must be missing something either really obvious or something fundamental that I've never seen before...

Edit: something wrong with the latex engine?

Last edited: Jul 3, 2011
12. Jul 2, 2011

### Bill_K

You stated your formula for Rp was an approximation, valid when n2+k2>>1. By Snell's Law this implies that θ2 ≈ 0. So B&W's formula (1) reduces to n2(1+iκ2) = u2 + iv2, and then just take the real and imaginary parts.

13. Jul 2, 2011

### Nick89

Ok, I didn't see that... Thanks, that makes sense.

So their equation for the TE mode comes out to what I wrote as Rp in my first post, and their equation for TM mode comes out to what I wrote as Rs (I checked it).

That doesn't help my confusion though. Another optics book I have shows the TM mode to have the dip, the minimum in reflectivity at some high angle of incidence. So apparently Rs should have the dip, (as it did in my calculations), so the formulas in my first post are correct...

Then I don't understand why the paper then goes on to show that Rp has the dip (well, that's how I read it anyway, it doesn't specify whether the graph I showed is Rp or Rs, but I think it should be Rp due to the two points I've already stated).

As far as I understand the paper on IRAS, the goal is to use p-polarized light at a high angle of incidence, where the reflectivity is lowest (not true, apparently it's s-polarized light that has a minimum there?!?!?) but the change in reflectivity (due to an adsorbed layer), which is what you measure, is highest.

14. Jul 2, 2011

### sophiecentaur

I don't think there is any doubt that the Brewster angle applies to the p polarisation - for a very simple argument I quoted before.
Am I wrong to be quoting the Brewster angle as relevant to this?

15. Jul 3, 2011

### Nick89

As far as I was aware the Brewster angle is that angle where the reflectivity is zero. That doesn't apply to metals.

This paper however seems to mention a 'pseudo-Brewster' angle where the reflectivity is minimum. It also quotes another formula for the reflectivity for p-polarized light and behold:
$$R_p(\vartheta) = \frac{(n^2+k^2) \cos^2 \vartheta - 2n \cos \vartheta + 1}{(n^2+k^2) \cos^2 \vartheta + 2n \cos \vartheta + 1}$$
Divide out $\cos^2 \vartheta$ and combine terms to get
$$R_p(\vartheta) = \frac{\left( n - \frac{1}{\cos \vartheta} \right)^2 + k^2}{\left( n + \frac{1}{\cos \vartheta} \right)^2 + k^2}$$
Now look at my first post... The other paper says this is $R_s$ instead..?!?!

I keep finding contradicting results so I'm completely lost...

16. Jul 3, 2011

### sophiecentaur

I have seen Brewster angle applied to non-zero reflectivity in the context of reflection of light from dielectrics (polarisers were, at one time, made using stacks of multiple glass plates because the polarisation by just one plate was far from complete) and radio waves being reflected from the ground. I guess it's just "usage".
I'm pretty sure that the basic argument I gave earlier must imply that the dip occurs for p polarisation. It's not a rigorous mathematical argument but is is a pretty convincing graphical one.

My textbook (Panofski and Philips) deals in more detail with dielectric reflection and the bit about metals is sketchy, I'm afraid. It says that calculating the reflection at metal surfaces "leads to rather complicated results" haha.

17. Feb 11, 2012

### iLIKEstuff

old post i know... but i think it's worth mentioning that there may have been a difference in convention especially given the age of the paper. complex refractive index is $\tilde{n} = n + ik$ in one convention and in the other, it is the conjugate, $\tilde{n} = n - ik$

last paragraph of the section http://en.wikipedia.org/wiki/Mathem...ex_refractive_index.2C_extinction_coefficient

also to note, units and conventions are important to keep in mind with older electromagnetics texts. A flurry of unit systems can be used: mks, cgs, gaussian, SI...