Reflections on a thin metal layer?

Main Question or Discussion Point

A one-way mirror is a thin aluminium coating on glass, and the thickness of the aluminium layer is one of the factors determining the reflection coefficient.

Does the reflection occur at the two interfaces (air-aluminium, and aluminium-glass)?

Does aluminium have an index of refraction, which can be used to calculate the reflection coefficient at each interface (Fresnel equations)?

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UltrafastPED
Gold Member
Yes, if it is thin enough to transmit light it will have an index of refraction, and the Fresnel reflection/transmission equations apply. Then both interfaces contribute - but also pay attention to the absorption for light loss. You will find that n changes rapidly with wavelength for thin metal films.

Here is a calculator - enter the material, the thickness, the wavelength, and click the "return arrow" key.

At 600 nm n=1.2 for Al.

1 person
Thanks. Where is the calculator?

UltrafastPED
Gold Member

Some cars have an 'athermic' windshield (infrared reflecting) with a blue or purple reflection color. Sometimes it is said that the infrared reflecting layer is composed of silver. According to the calculator, the thickness of a silver layer should be less than 10 nm for the layer to be transparent. That is too small for interference of visible light, so I suppose that means that those who say the reflection color is due to interference are incorrect?

The calculator also shows that the refractive index of silver is fairly constant in the visible spectrum (wavelength 400-700 nm), much smaller than the refractive index of glass, so that doesn't explain the blue or purple reflection color either. Is there still another explanation?

sophiecentaur
Gold Member
As far as I know, the everyday one way mirrors are nothing more than a thin layer of metal which is so thin that it lets about half the light flux through and reflects the other half. The 'One-wayness' is really just due to the difference in illumination levels on the side of the observer and the observed (i.e. you observe from a darkened room). The layer just represents a mis-match in impedance, with a consequential reflection at the surface. I don't think we're looking for an interference filter at work here. Also, remember, any simple linear medium will not have a preferred direction of propagation, so it cannot be truly 'one way', in the sense that a microwave circulator or isolator functions (using ferrites).

UltrafastPED
Gold Member
The reflections from an athermic windshield are just that - reflections - there is no interference involved. Your source was correct: the thin metal film is usually silver.

See the description here, in the athermic section: http://www.eclat-digital.com/visualization-of-car-windshields/

They also provide the reflection spectrum for a typical athermic windshield compared to a normal windshield. You can see from the spectrum that the blues are more strongly reflected: 400-500 nm, on into the green.

DrDu
It is possible to describe a metal in terms of a dielectric constant, but it is essential that this constant is complex.

DrDu
As far as I know, the everyday one way mirrors are nothing more than a thin layer of metal which is so thin that it lets about half the light flux through and reflects the other half. The 'One-wayness' is really just due to the difference in illumination levels on the side of the observer and the observed (i.e. you observe from a darkened room). The layer just represents a mis-match in impedance, with a consequential reflection at the surface. I don't think we're looking for an interference filter at work here. Also, remember, any simple linear medium will not have a preferred direction of propagation, so it cannot be truly 'one way', in the sense that a microwave circulator or isolator functions (using ferrites).
That's not the whole story. While it is true that the transmittance won't depend on direction, the reflectivity does. This is easiest to understand for a metallic mirror on a thick but weakly absorbing material with dielectric constant ##\epsilon_r\approx 1##. If you look at the metallic side, you will have ideally up to 100% reflection. If you look at the absorbing side, there won't be reflection at all but the material will look black. If neither of the two sheets is completely absorbing, then, even when illumination at both sides is equal, you will have higher contrast looking at the dark side than at the reflecting side.

Bobbywhy
Gold Member
There are other devices which behave like "one way mirrors":

"A beam splitter is an optical device that splits a beam of light in two. It is the crucial part of most interferometers."

http://en.wikipedia.org/wiki/Beam_splitter

sophiecentaur
Gold Member
That's not the whole story. While it is true that the transmittance won't depend on direction, the reflectivity does. This is easiest to understand for a metallic mirror on a thick but weakly absorbing material with dielectric constant ##\epsilon_r\approx 1##. If you look at the metallic side, you will have ideally up to 100% reflection. If you look at the absorbing side, there won't be reflection at all but the material will look black. If neither of the two sheets is completely absorbing, then, even when illumination at both sides is equal, you will have higher contrast looking at the dark side than at the reflecting side.
That's an interesting point. So, the energy is reflected one way and absorbed, the other. I'm trying to think of a transmission line equivalent. Reciprocity needs to apply - but, of course, that only refers to transmission through the line.

[Edit: thinking about this again; is this not just the 'sunglasses effect'? The darker reflected image at the back is surely just because of the absorption in the double path, compared with the single path for the transmitted light from the front(?). This improves the effect of flare and is useful when there is enough light from the scene for it to be visible.]

DrDu

[Edit: thinking about this again; is this not just the 'sunglasses effect'? The darker reflected image at the back is surely just because of the absorption in the double path, compared with the single path for the transmitted light from the front(?). This improves the effect of flare and is useful when there is enough light from the scene for it to be visible.]

Yes, mirrored sun glasses are certainly a good example.

sophiecentaur
Gold Member
OK, then. Just a predictable effect. I was looking for something more hi tech, that's all. For anything more, I believe you need polarisers and more bells and whistles - as in microwave circuits.

Some cars have an 'athermic' windshield (infrared reflecting) with a blue or purple reflection color. Sometimes it is said that the infrared reflecting layer is composed of silver. According to the calculator, the thickness of a silver layer should be less than 10 nm for the layer to be transparent. That is too small for interference of visible light, so I suppose that means that those who say the reflection color is due to interference are incorrect?
Hi there, I spent a few years designing silver coatings, so I know this very well!

I'm sure of the following:
- 10nm of silver does cause interference if you think it the Fresnel way. The first interface does reflect nearly 100% of light, and the second one reflects an opposed wave, reducing reflection, and allowing the evanescent wave to propagate on the other side. This causes the semi-transparent, semi-reflecting behavior of the silver thin film. This causes no coloring as you would have with dielectric film interference
- However, all these windshield layers have at least two silver films, sometimes three. They are separated by about 80nm of dielectric material of IOR ~ 2. This is highly interferential on purpose and creates a Fabry-Perrot interferometer, to destruct the reflected wave in the visible range. On the other hand, for infrared, the reflections are in phase and additive. This ensures the main properties of the windshield : lowest possible reflection in the visible range, highest possible in the infrared range. As a side effect, this causes the coloring.

Cheers!

Galinette

all these windshield layers have at least two silver films, sometimes three. They are separated by about 80nm of dielectric material of IOR ~ 2. This is highly interferential on purpose and creates a Fabry-Perrot interferometer, to destruct the reflected wave in the visible range. On the other hand, for infrared, the reflections are in phase and additive. This ensures the main properties of the windshield : lowest possible reflection in the visible range, highest possible in the infrared range. As a side effect, this causes the coloring.
Thanks for the information. It seems to disagree slightly with the etalon formulas, as the formulas predict the lowest possible reflection occurs for ultraviolet light, instead of visible light. The first order reflection minimum occurs at λ=2nL=320 nm (ultraviolet), if the distance between the silver layers is 80 nm and the index of refracton (IOR) is 2. At the red end of the visible range the reflectance is closer to maximal than minimal. Wouldn't an interlayer distance larger than 80 nm have been more effective?

(The transmittance of the etalon, T(λ), is between 1 and 1/(1+F), and the reflectance of the etalon, R(λ), is between 0 and 1-1/(1+F). F is the finesse of the etalon, F=4R'/(1-R')2, where R' is the reflectance of a single silver layer.)

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Thanks for the information. It seems to disagree slightly with the etalon formulas, as the formulas predict the lowest possible reflection occurs for ultraviolet light, instead of visible light. The first order reflection minimum occurs at λ=2nL=320 nm (ultraviolet), if the distance between the silver layers is 80 nm and the index of refracton (IOR) is 2. At the red end of the visible range the reflectance is closer to maximal than minimal. Wouldn't an interlayer distance larger than 80 nm have been more effective?
Actually, it's a little more complicated than this. The interlayer is maybe closer to 85nm, and IOR closer to 2.1-2.2, I gave rough figures. But most importantly, silver layer thicknesses are not negligible (>10nm) and must be taken into account. Also, the silver/dielectric/silver sandwich is itself encapsulated in dielectrics which have a minor but non-zero role in interference. So at the end, the minimum reflection is more around 500nm. The calculation requires a computer, but you got the base phenomenon right and your remark is true.

Also, triple Ag stacks have a double antireflection : 80nm and 80+80nm (that's again slightly more complicated as both interlayer are slightly different). This gives a broader and sharper antireflection band, which is more effective.

In this etalon model of an athermic windshield silver is just a metal with a reflection coefficient. Would aluminium instead of silver have resulted in about the same intereference colors? Why is silver better than aluminium, in an athermic windshield?

In this etalon model of an athermic windshield silver is just a metal with a reflection coefficient. Would aluminium instead of silver have resulted in about the same intereference colors? Why is silver better than aluminium, in an athermic windshield?
Likely not. In these metal/dielectric/metal structures, interference colors is extremely sensitive to any small change in refractive index (or more generally dielectric function). Even different silver qualities change colors.

Silver is better than aluminum, because it reflects long waves more. Infrared reflectivity is directly linked to metal sheet conductivity, and for this application silver is the best possible metal.

Silver thin films are also better, because they transmit more light and absorb less. But this is also closely linked to the previous property (conductivity/reflectivity)

Etienne

Some time ago I noticed similar interference colors in the windshield of trains (images here) which contain a conductive layer, composed of indium tin oxide (ITO), for defogging the window. I was wondering why silver is preferred in the windshield of cars while ITO is used in the windshield of trains. Is it the same answer: is silver, compared to ITO, better in reflecting long waves?

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