Thickness of dark wet spots on drying dishes

[Orthoceras]

TL;DR Summary

What is the minimal thickness of dark wet spots on drying dishes?

A common observation when doing the dishes is that the wet spots on a drying dish are darker than the dry spots, especially if the dish is held in such a way that it reflects the light of the sky. Evidently the total reflection coefficient is reduced by the water film. The border between dark and bright is acute. My question is, at what thickess does the wet film stop being dark, would a thickness of a single molecule still be dark?

 

[.Scott]

I believe that, depending on the pattern of background lighting, the angle of observation, and the color and texture of the dish, that "wet" can appear as either light or dark.

But the key is that there is a difference in how the light will be reflecting off the surface of the dish.

The diameter of a water molecule is about 2.75 angstroms. The wavelength of visible light is about 4000 to 7000 angstroms.

Once the thickness of the water reaches a few wavelengths, the reflectivity should be constant.
During the transition from 0 to 20,000 angstroms, you might expect changes in the reflectivity. But I am noticing in your photo that as the water edge dries and become thin, it seems to ramp up quickly to the level of the main mass of water - the thicker water. You can judge the slope of the water surface by it reflectivity - and I am definitely seeing lightning and darkening along the very edge of the wet spots in your photo. This implies a beading of the water at the edge.

Even a very slight beading will cause the transition from a water depth of zero to 20,000 Angstroms to occur over a very short distance. For example, if that angle between the water surface and the plate is only 1%, then the transition from dry to 20,000 Angstroms will occur in about 1,200,000 Angstroms, 0.12 mm.

That would be about 0.1 pixels in your photo.
But in fact, that angle looks to be well more than 1 degree - so the transition period would be much less.
That angle would be dependent on the amounts of soap and other things in the water and the chemistry of the surface of the plate. It might also be affected by temperature.

 

[Orthoceras]

Ok, a glass surface against a black background is a better standard situation. In my experience the result is the same, the reflection of the sky is always darker on the wet surface, compared to the dry surface.

The wet surface is also darker on a convex glass surface such as the outside of a wineglass, also darker when using pure water instead of soap, and also darker when using acetone instead of water.

Wouldn't it be possible that a wet film is one molecule thick on a convex glass surface?

 

[sophiecentaur]

Isn't it largely the difference between diffuse and specular reflection? Unless there is an actual source of light, reflected into your eye, you see the generally darker background from a wet surface which is more level and conditions are 'more' specular. With a diffuse reflection, you get a diffuse level that's a mean between sources and non sources of light over the whole hemisphere.

 

[Orthoceras]

I do not see any diffuse reflection on either wet glass or dry glass

 

[sophiecentaur]

Do you mean to say that you see sharp, contrasty images in dry crockery? It does depend on the specific sort of crockery we're talking about. Terracotta is an extreme example of what I am referring to but, any ceramic wares that are not high gloss will have a 'visible' surface, due to diffuse reflection (even a mirror surface can be detected if there is dust or condensation on it.)

The OP refers to "dishes" and not glass, so we may be talking at cross purposes.

In hard water areas, there can be traces of calcium carbonate after any dish dries and it can boiled up over time. Not long ago, we bought a water softener and we live in a very hard water area. Even the soluble sodium salts can be seen on dried dishes - it's just that they wash off next time and don't build up.

 

[.Scott]

Wouldn't it be possible that a wet film is one molecule thick on a convex glass surface?

My guess would be that a 1-molecule (2.75Å) layer of water would have very little effect on reflectivity - because it is such a tiny portion of a wavelength.

A 2.75Å layer might be possible - especially under cryogenic conditions. Normally evaporation and condensation would keep it from ever being that uniform. Perhaps if you used a centrifuge. Or perhaps it you had a super hydrophilic surface that immediately pulled the water down.

 

[Orthoceras]

Do you mean to say that you see sharp, contrasty images in dry crockery? It does depend on the specific sort of crockery we're talking about. Terracotta is an extreme example of what I am referring to but, any ceramic wares that are not high gloss will have a 'visible' surface, due to diffuse reflection (even a mirror surface can be detected if there is dust or condensation on it.)

The OP refers to "dishes" and not glass, so we may be talking at cross purposes.

My starting post was inadequate, wineglasses are a better example of the phenomenon than dishes, because diffuse reflection can be ignored. So let me try again. Take a wineglas or flat window glass, against a black background. Make the surface partially wet and notice that the reflection of the sky is darker on the wet parts. The wet parts are darker because of the reflectivity R. If the wet film is thick, R = ((n₁-n₂)/(n₁+n₂))² + ((n₂-n₃)/(n₂+n₃))², independent of the thickness of the film.

I suppose my question is, what is the formula for the reflectivity if the thickness of the wet film is much smaller than the wavelength. Is the thin film reflectivity noticably different from the thick film reflectivity. What is the reflectivity if the layer is one molecule thick.

 

[sophiecentaur]

I do not see any diffuse reflection on either wet glass or dry glass

My starting post was inadequate, wineglasses are a better example of the phenomenon than dishes, because diffuse reflection can be ignored. So let me try again. Take a wineglas or flat window glass, against a black background.
etc. . . .
I suppose my question is, what is the formula for the reflectivity if the thickness of the wet film is much smaller than the wavelength. Is the thin film reflectivity noticably different from the thick film reflectivity. What is the reflectivity if the layer is one molecule thick.

View attachment 262839

That's better - thanks - we can talk about the same thing. now. My first comment is that we are dealing with quite small numbers here and you have to use a dark background to get the effect. I would say that it's only the same sort of phenomenon that you get with terracotta but at a much lower level. Any solid surface will have roughness and a liquid will always tend to be smoother. I would say that the spatial low pass filtering of the surface contours when there's water present will change the optics of the reflections. There will always be some flare on a surface and that's due to diffuse reflections. Your eye is pretty sensitive to this when it's presented in the right way (also, high quality photography with top end lenses).

If you want a great example of how surface finish can affect the 'depth of shine' then try to get a piece of silver to shine as well as a silver plated spoon, straight out of the shop and still in its plastic bag. You start off with a sawn surface, you file it to 'remove' the scratches, you use a series of finer and finer abrasive papers then you use a polishing mop with a buffing soap - it looks really good - until you start getting fussy. You then apply a rouge compound, which laps the surface and pulls all the tiny scratches that the buffing soap left and fills the valleys. That makes a stunning amount of difference to the appearance and you can get somewhere near the professional, machine finish on the shop spoon. Very good specular reflection.

Water on the glass surface will temporarily achieve what the buffing soap / rouge did to the silver surface. Compare the appearance of an expensive crystal wine glass that never went near dishwasher chemicals with the very best surface you get with a wet glass plate. That glass plate has had surface damage ever since you started using it - diffusion rules.

 

[Orthoceras]

Thanks. Your suggestion seems to be that I should restrict my question about the reflectivity of very thin wet films to polished glass surfaces. Ok. If the wet film on a polished glass surface is very thin (<<λ), the amplitudes of the reflections from both interfaces (air/water and water/glass) should be added. So R_wet&thin = ((n₁-n₂)/(n₁+n₂) + (n₂-n₃)/(n₂+n₃))² = 4% (assuming n1=1.00, n2=1.33, and n3=1.51). So the reflection of the sky on the thin wet surface looks like the reflection on a dry surface. This is unlike the reflection on a thick wet surface, which is darker.

 

[sophiecentaur]

I have a feeling that you are conflating two effects here. My interpretation of the phenomenon is based on a macroscopic / mechanical approach to the surface flatness. You mention the reflection coefficients of the two interfaces being relevant but that involves vector addition of the two reflected waves. You cannot just add the two magnitudes.

I assume you have already looked at the interference in oil films on water etc. because when the films are thin then there can be colouration at different angles. Your quoted formulae (Fresnel formula) is also affected by the actual film thickness, which will affect the phases of waves reflected at the air/water interface and the water/ glass interface. See the Wiki article
It is standard practice to use thin coatings on glass lenses which reduce reflections (called "blooming", in the old days). That can reduce reflection and improve transmission through the surface by an impressive amount. To get the best improvement, though, you have to start with a well ground and polished lens which doesn't cause significant diffuse reflections.

There is a problem with using water for this experiment and that is the difficulty of wetting a surface evenly. A much easier experiment to carry out is to put a small quantity of oil on a water surface in a beaker. The Fresnel equations work for that too. You can vary the thickness of film by using the appropriate volume of oil and both surfaces are inherently flat and well behaved. You then have no need to worry about diffuse reflections. You can buy optical glass flats but they cost you and they are not much use for anything else unless you are starting up an optics lab.

 

[Orthoceras]

Yes, the phase difference would matter if I hadn't assumed that the film thickness is <<λ. However, if the thickness is <<λ, the phase difference can be ignored.

Yes, to get the best results a polished lens would be best. However, cheap picture framing glass seems to be flat enough, as Newton's rings appear when pushing a glass microscope slide on the picture framing glass.

 

[sophiecentaur]

Now you're discussing a regular school lab experiment and you should get results that agree with the basic theory.
I still doubt that the water film will be easy to control, though.
You are right about the <<λ thing.
What are you actually wanting to measure / show with the experiment? You may find that oil is easier to control as it won't evaporate just when things get interesting.

 

[Orthoceras]

What are you actually wanting to measure / show with the experiment? You may find that oil is easier to control as it won't evaporate just when things get interesting.

My question while doing the dishes, as explained more or less in my starting post, was what the Fresnel equations predict for the reflectivity of the wet spots on dishes and glasses if the wet film gets thinner and thinner, up to a thickness of one molecule, on an ideally flat surface. I thought, as the reflectivity does not depend on the thickness d if d>>λ, it is probably the same for d<<λ, up to a thickness of one molecule. A few posts later in this topic I understood this is incorrect, because the amplitudes instead of the intensities should be added if d<<λ. So the reflection of the sky on a wet film on an ideal convex surface would be uniformly dark except for a central zone where d<<λ. This is probably what Scott said, although without explanation.

I think your suggestion to consider oil on water is maybe another topic, it seems to resemble petrol on water with beautiful rainbow colors. The difference is that oil has a higher index of refraction than water, and it floats on water.

 

[sophiecentaur]

Firstly, the rough glass of the dishes is a different issue and, as you don't see colours, there is no interference effect. The total effect of a wet upper and lower layer of water on rough glass will be additive with no interference so both sides wet will have half as much diffuse reflection and, consequently more contrast in the reflected image.
Secondly, there is no inherent difference between water and oil except that you will have problems controlling the water film. You can do the sums to find the max and min levels of the interference pattern for a 'theoretical' layer of water - just put in the values for water and glass in place of oil and water.
The refractive index of water is not very different from that of glass so you will not get the amount of reflection that oil on water produces.
So I ask again, what you are actually trying to measure / see/demonstrate (that the textbook can't predict for you)? See what the textbook tells you before you start trying to measure something that may be hard to see. (You see, I am convinced that what you are seeing is a roughness effect and not Fresnel Reflection.)

 

[Orthoceras]

Thanks, your insightful posts about the roughness effect have been very helpful!

 

[crashcat]

This has been touched on by a lot of people, but I think the main effect from wetting the plate is the reduction of scattering (because the water surface is much smoother).

Dry plate = high scattering and incident light from any direction can enter your eye. The result is that the plate looks bright and white-ish (color would come from either absorption or interference, which are weak effects in this case).

Wet plate = low scattering. Incident light is reflected like a mirror and only enters your eye if the angles are just right. The result is that it looks darker unless it's actually reflecting the ceiling light or something directly into your eye.

Think of wet tree trunks or the world in general after it rains. Scattering is reduced on all surfaces. The result is that they look darker, but ALSO that all the colors look richer. This is because you no longer see all the added white light from scattering, and only see the colored light after absorption of certain wavelengths in the tree trunk.

 

[Orthoceras]

Somewhere in this thread I became more interested in simple glass surfaces than in white glazed ceramic. On glass surfaces, such as microscope slides and float glass, wet areas reflect the sky darker than dry areas. Does anyone know the typical surface roughness (in nm) of such glass surfaces? It would be interesting to compare it to the average wavelength of the light.

 

[sophiecentaur]

Does anyone know the typical surface roughness (in nm) of such glass surfaces?

I think you are expecting other people to do your research for you in this case. I googled "surface quality of optical glass" and a few other terms. There were dozens and dozens of hits. I suggest you read some of them.

 

[crashcat]

Does anyone know the typical surface roughness (in nm) of such glass surfaces?

I think it's several nm RMS for standard slides, and up to tens of nm RMS for really cheap slides.

 

[Orthoceras]

Yes, or even less. In this study untreated glass slides had a surface roughness of about 1 nm. So if it is assumed that the reflection of the sky on the wet surface is darker due to a roughness effect, all wet film thicknesses > 10 nm would be equally dark. As a test I tried alcohol as wet film on a glass slide, which is convenient because the film gets automatically thinner and thinner due to evaporation, and at a certain thickness interference colors appear. So film thickness > 10 nm does matter. The reduced total reflectivity is present only in the middle where the thickness is a few wavelengths or more. The visible border of the alcohol film has a thickness of about λ/2. Surface roughness does not seem to be relevant here, maybe not surprising because the sky is a uniform extended source. Similar observations for alcohol on glazed ceramic.

It is possible to see the same color rings with water. First wet the surface with soap water, let it evaporate, breathe on the surface, and see the interference colors appear. Although this foggy condensation layer is probably composed of droplets instead of a film.

Alcohol film on glass slide

 

[crashcat]

I don't know why I didn't do this simple back of the envelope check earlier, but using the Fresnel equations you would expect about 3.4% reflection from a dry slide and about 2.2% reflection from a wet slide.

Assumptions: low angle of incidence so polarization doesn't matter (<10 deg), glass index 1.45, water index 1.33, no interference effects (which is valid for any film where you don't see the colors, because sun/room light has a super short coherence length)