This is a bit hand-wavey in its' description of light reflecting. One of the most mind blowing bits of physics I've ever learned is that photons do not actually "bounce" off of a surface like little balls. They are always absorbed and re-emitted. The actual photons that hit an object are not the same ones that eventually enter your eyes. The atoms in an object are stimulated by photons hitting them to then emit a new photon via the photo-electric effect, which we perceive as light "bouncing".
Having thought about it, I'm not sure if I agree. Light can reflect off of transparent crystals, which do not have the right electron energy levels for absorption (that's why they're transparent.) It's possible that the Feynman diagram of a reflection might involve the photon "going away and another one replacing it" (I don't know what the diagram looks like), but Feynman diagrams cannot be interpreted as sequences of events. Instead, they describe an instantaneous flow of amplitude between different quantum states. I would only go along with the absorption and re-emission interpretation if there was always a time delay between the two steps (which there cannot be if the mirror does not have any energy level deltas suitable for storing the energy).
The other side of the debate would be that if the mirror is moving towards or away from the light, the reflection will be Doppler-shifted to a higher or lower frequency. Does this mean that the reflected photons are not the same photons as the incident photons, or does it mean that the same photons have had their energy changed? I think there is no meaningful distinction because every two particles with the same name in quantum mechanics are identical anyway. There's no telling which are which. If I showed you a photon, then took it back and showed you another, you would never be able to tell whether I had opened the same box twice or if I had taken the old one out and captured a new one from my desk lamp.
They are not the same photons. They are emitted by the electron changing energy states after excitation of the incoming photon. But your intuition that there is some quantum weirdness is correct. Quantum Electrodynamics is the theory you're looking for. It unifies quantum mechanics and special relatively in the context of light/matter interaction. Given you seem to be familiar with Feynman. You're in for a treat! His book on QED is meant to be an accessible explanation.[1]
Hydrogen atoms can't store photons that do not have the same wavelength as any of its spectral lines. Nonetheless, hydrogen gas has an index of refraction that is a continuous function of frequency, and as a result it can reflect light of any wavelength. Therefore, hydrogen does not store any energy for any length of time in the course of typical reflection. Even if the Feynman diagram for reflection looks like a combination of absorption and re-emission, if the absorption and emission process can't occur at separate times, then it is not right to say that one is followed by the other.
That's a good catch, but H2's spectral lines are still discrete. It's not that gasses can't absorb photons, it's that they can only do so at very specific frequencies.
You didn't address the concern. A given atom cannot absorb any photon. They can only absorb photons of certain energy levels/wavelengths. Electrons can occupy only certain energy levels. So they can only absorb photons that match the difference if two energy levels.
A photon doesn't need to carry a UUID for people to conceptualize a difference between a photon going through free space and being absorbed and re emitted. The parent comment asks the right question, what is being meant by "the same" is required to know before being able to say the discussion makes no sense.
Taking your first literal interpretation as the only way the discussion could have value is no more helpful than this exchange:
"I'm going to the store, do you want the same hot dogs we have in the fridge?"
"That doesn't make sense, the store can't have the same hot dogs we have in the fridge".
> "I'm going to the store, do you want the same hot dogs we have in the fridge?" "That doesn't make sense, the store can't have the same hot dogs we have in the fridge"
Yeah but hot dogs belong to classical mechanics, while photons to quantum mechanics, and in quantum mechanics you can actually have the same hot dogs in different places, until you observe them
That a photon can be in superposition is even a better example than my hot dog example. Nobody is impressed if you can find a corner case interpretation that makes the question sound dumb they are impressed if you can try to figure out what the person really wanted to know and answer it to the best of current understanding.
In the hotdog example the point is the person is actually trying to figure out if you want them to buy you more hotdogs not whether or not the exact same hotdogs can be found at the store and fridge at the same time. In this actual discussion about the photon it's not about whether or not the photon actually carries an ID card with it it's about what actual interactions are happening to photons when the light is reflected or refracted in different ways.
Sometimes what a person is trying to figure out may not line up with how things actually work, in fact take any science and add 100 years of knowledge and pretty much all discussions are this way. That does not mean it was nonsensical to talk about it just means the details are more exact and complicated than the asker knew how to phrase (and again, it's always like more exact and complicated than any of us know as we don't know everything there is to know about the universe).
The point is, if A) photons carry no individual identity, and B) the interaction happens instantaneously, then there is no meaningful difference between saying "the photon bounced off like a ball" and "the photon was absorbed and a new one was instantly re-emitted". It's literally a distinction without a difference.
As humans living in the classical world, we hear that and think "oh, that just means we don't know what's really going on, it either bounces or re-emits, it just looks the same to us." But no - reality is just an interaction of quantum fields and we're putting these human stories on top of it. It's like debating whether or not submarines swim.
I'll try again in another way and if this doesn't work I may need to come back to trying to communicate what I mean.
When you encounter an XY problem question/statement only stating "this doesn't make any sense. Computers are very complicated and don't work like that in the first place" is useless to everyone except maybe a comment or that wants to boost their ego. Taking the XY question, explaining why it's wrong, and trying to explain how things actually work at a deeper level is helpful. I think your comment about them being field interactions and needing to think about them that way is a good start on that kind of answer rather than the ones above.
I understand that my previous comment lacks of explanation, but I'm not a native English speaker so it's really hard for me to dig into details. I beg your pardon
>The parent comment asks the right question, what is being meant by "the same" is required to know before being able to say the discussion makes no sense.
This is true, but I think the comment you are replying to is correctly identifying a fundamental conceptual misunderstanding of the comments preceding it. In the sense that they likely mean, it's not really a valid question. Sort of like the "are you still beating your wife?" question. Asking it is making incorrect assumptions.
"Knowing" for me means, that there is no way, that we can be wrong about something. Otherwise it's only "believe", not "knowledge".
It would be good to say the following, instead of "it is so": "So far experiments did never show any sign of identity of photons." (Were there any such experiments?) Oh and what about quantum entanglement of photons? Could that not be interpreted as a sign of identity?
But that's just not true. The electron will only get excited if the incoming photon has a very specific wavelength, so most of the light that hit your house's walls, for example, is reflecting. The photoelectric effect is real, but it only applies in specific circumstances...
There will be a transfer of energy to the system if there is an interaction, but will only absorb the incident photon at specific energies. In the case of reflection in this classical view, it simply re-emits the energy back out as there are no valid energy transitions. The reality is far more complex and this thread is mixing multiple theories together, wave and particle views and simultaneously comparing single atom mechanics with those of lattices/solids. This is all causing a lot of confusion that's hard to address in comments.
I think this is called, "knowing enough to be dangerous". You clearly have some understanding of light, but you are missing some key facts that would lead you to the correct understanding.
>Light can reflect off of transparent crystals
No, the point of the above post is that a photon is not "reflected", but captured and re-transmitted by photo-electric effect.
Light travels at maximum speed C, only when in a vacuum. Otherwise it travels through a medium. We can see that the different speeds of light cause dispersion such as when it enters and leaves a clear prism (changing mediums, thus spreading out the different speeds/wavelengths of light). Since we know that light is "traveling in a medium" when it travels in air or in the prism, then what do we mean by this? It is traveling slower, so it must have some information about the medium...
So how does the light "know" it's in a material, without interacting with the atoms of the material?
Of course, it couldn't know. It is interacting with the material. The behavior of the particle is fully explained by the photo-electric effect.
Absorption and emission are not the only two processes by which light can interact with matter, reflection is another. The argument is that if the material can't store the energy then you can't really call the event an absorption. Feynman diagrams aren't sequences of events so just because you see the photon coming to, disappearing, and then departing an electron doesn't mean it happens in that order.
If light is reflecting off of something, it's precisely because that thing isn't transparent[0].
> if the mirror is moving towards or away from the light, the reflection will be Doppler-shifted to a higher or lower frequency.
This does not refute OP's assertion. If the photon is absorbed and a "different"[1] one emitted, the emitted photon is still emitted from a reference frame that is moving with some velocity, therefore the emitted photon will still be Doppler-shifted... The photon is still emitted at the correct energy level. The photon appears to have higher energy in the stationary reference frame.
[0] Disregarding human-centric definitions that have to do with visible spectrum.
[1] Thinking of the photons as "different" doesn't actually mean much when it comes to quantum mechanics; they're indistinguishable bosons.
It doesn’t really make sense to talk about “the same” photon because photons don’t really have identity. Even in a vacuum you can send a photon in one end and see one on the other and not really know that the one going in was “the same” as the one going out. It can turn into other particles on its way and then recombine into a photon and there is no way to know this based on the observation that a photon came out of the end of your vacuum.
That's one of the things, that annoys me with questions about physics, which me might not know well enough yet: When someone asks about whether something "is", people answer with what we can in general know at our current level of understanding and then make it seem as though that is the same, as answering the question, what really "is".
> Even in a vacuum you can send a photon in one end and see one on the other and not really know that the one going in was “the same” as the one going out.
I don't really care, if _I or current physicists_ cannot _distinguish_ them, when I ask, whether they _are_ the same. I would rather have an answer like as follows:
"We do not know this yet. We do not have the means of telling, whether it is really the same photon or not. However, even if they were not the same, it would make no difference (according to our current understanding of physics!), as their effect on the surroundings would still be the same, because ..."
no, what dan said is better, it might annoy you, but physics is such that wording is important. Even then, nearly all wording is losing information and in some cases changes it from the reality. You are asking a question which you want an answer but the question itself has a troubling fit with reality as you try to relate to things in terms that seem familiar.
No, physics is such that semantics are important. What’s the observable difference between absorption/emission and reflection? If the latter doesn’t happen at all the distinction is meaningless.
I don't think it's hand wavey. It's just a different (simpler) model. To describe this phenomena (wet things are darker), you don't need to go into details about photons, or quantum mechanics or whatever.
Geometrical optics is really sufficient here, it explains it perfectly, and is much more intuitive than other more complex models.
> The actual photons that hit an object are not the same ones that eventually enter your eyes.
This part is philosophy, not physics. Photons are “indistinguishable particles”. You cannot tell two photons apart. This has a rigorous meaning: if you swap two photons, the state of the universe is unchanged. (This is in contrast to fermions — if you swap two fermions, the state of the universe has its phase shifted 180 degrees.) If photons we’re distinguishable, then swapping two photons would swap them, and the resulting state would be different. The matters from a statistical perspective. Look up Bose-Einstein statistics if you’re interested.
As for whether photons are absorbed and re-emitted or merely reflected, this surely depends on the surface on question.
This is why two fermions can’t be in the same state: if you swapped them, you’d be back where you started, but that somehow also had to phase shift 180 degrees, and the only complex amplitude that is the same as its own negation is zero.
Metals and, generally, homogenous materials (dielectrics, conductors, etc). Light reflects off of a conductor or a dielectric due to a relatively straightforward solution to Maxwell’s equations.
In contrast, anything phosphorescent most definitely absorbs and re-emits, sometimes hours later. Imagine glow-in-the-dark pigment.
This probably depends on your perspective. If you think of a metal as a homogenous linear medium (which is a good approximation for long wavelengths but is not exact), then light is a wave, photos are an unnecessary detail, and the wave mostly reflects and is absorbed as heat. If you look the metal as a whole bunch of atoms and a sea of mobile, discrete electrons, then the incoming photons excite the sea of electrons and, when the interaction settles down, photons are leaving. If you insist on modeling the electrons with QED, congratulations, you’ve made your life unnecessarily complicated unless the photons have such high energies that electrons are created or end up with relativistic velocities. If you believe in string theory, maybe the photons were only an approximation in the first place.
Physics is all about choosing an appropriate model of the world that captures the detail you need without being more complicated than needed. Is a person on a playground swing a pendulum, or is it a bag of molecules sitting on a flexible piece of rubber suspended by a bunch of rusty metal links from a flexible steel bar that is, in turn, supported by many geological layers in the Earth’s crust?
How does absorption and reemission preserve phase information?
What about total internal reflection? Why does the light travel through the entire body of a glass prism without issue and then suddenly get absorbed and reemitted at the surface? Glass (and water) are transparent specifically because they don't absorb photons in the visible spectrum, so how is it that they absorb and reemit?
Because the electro-magnetic field polarization is preserved in many cases (specular). That doesn't mean there isn't a plasmon generated in between or that the sign of the electric field is maintained. What is interesting is that in the case of metalic (from low to high index of refraction) there is an inversion of the electric field. This does not happen at the reflection from a high index to low index (e.g. under water looking up to see total internal reflection the polarization remains the same). You can see this same effect bouncing water waves off of a concrete wall.
IANAP but from my limited understanding (mostly reading Richard Feynman)
light beam = stream of photons, they don't travel through entire body of glass prism without issue, they get absorbed and re emmitted along the way too (Which is what explains the different speed of light in different mediums)
I'm not convinced but even if we accept that explanation, why do all the atoms along the way emit the light in the direction it was travelling and then the ones at the surface suddenly emit it in a different direction?
I can recommend Hecht’s Optics textbook for this. I found it at my local library. I have had a number of electromagnetics courses over my life and none explained the unifying scattering principles of reflection, transmission, and refraction concept so clearly as that book
They don't! That's why the thickness of th glass of affects it's reflectivity, and not the simple way you might guess. Read Feynman's QED book or watch the YouTube vidoes for a lay intro.
Probably it's just how the math works out. The atoms emit in all directions, however the waves traveling in all the other non-correct (according to reflections and refraction laws) directions cancel each other out, and only the ones with the new correct direction are still there / visible.
Thanks for correcting me and thanks for the link. But what have I been talking about so far? I don't understand.
Also I just watched this video I'd recommend that explains the idea: https://www.youtube.com/watch?v=NumSE2LvSmQ
That doesn’t look right at all to me. Why the photon is then re-emitted in exactly the same direction that it would have if it was reflected rather than a random direction?
1. A gentle lead-in to the subject, Feynman starts by discussing photons and their properties. 2. What are reflection and transmission, and how do they work? 3. Feynman diagrams and the intricacies of particle interaction. 4. What does it mean, and where is it all leading?
Update: have now watched 3 of the 4 videos. Can see why you loved it so much! I've watched a lot of Feynman talks and lectures, and read most of his books. ..but this has particularly good explanations of the material, distinctly clearer than a couple of talks of his I'd seen before. Plus a few priceless references to NZ and the NZ national character! (filmed Auckland 1979)
The photon has some amplitude for reflecting in all possible directions, it happens that for certain directions these amplitudes are very close to the same value and so reinforce and become more probable, and at wider angles tend to cancel out and become less probable. When this is not the case and wider angles instead reinforce you can observe this directly in for instance a diffraction grating.
Yes, I agree. But what this has anything to do with the probability of emitting a photon in a specific direction? The parent was speaking about absorption and re-emission, not reflection.
If the emitted photon always had the same momentum as the absorbed photon, the light would go right through the mirror, passing on its way without changing direction.
Conservation of momentum doesn’t mean that the momentum of the emitted photon has to be equal to that of the absorbed photon. It means the sum of the momentum of the emitted photon and the mirror must equal the momentum of the absorbed photon, assuming the mirror had no momentum to begin with.
If I recall a high school physics class correctly the colour of the photons that are emitted from the object (the colour we see) are the ones the object does not absorb. If so, is the object the colour we see or is it the colour it absorbs.
The little balls model perfectly explains why light reflects off a mirror at the same angle at which it arrived. How does the absorb-and-reemit model handle this?
For example.. if I'm looking at a star, and thinking it's the 'same' photon that came off of that star billions of years ago... that might be true in the vacuum of space, but since it's passed through the medium of the atmosphere, does that mean it's constantly been absorbed/new photons being emitted?
> The atoms in an object are stimulated by photons hitting them to then emit a new photon via the photo-electric effect, which we perceive as light "bouncing".
That's how electrons and photons interact. It doesn't make sense to say that they are different photons or the same ones, since you can't observe the interaction.
> The actual photons that hit an object are not the same ones that eventually enter your eyes.
For anything opaque, you should expect this, because otherwise how could you see something's color if you only saw the same photons as the light source it reflected?
Light is not a wave. Photons are particles. Light as a wave is a convenient abstraction that explains a lot of light's behaviors, but not all of them. I recommend Feynman's QED lectures (or the book), where he explains this.
A "quant" is someone who studies quantitative finance, typically found in investment fields. The word for this wave-packet with particle properties is "quantum". It is a singular noun; its plural is "quanta".
> Once you spill water on the shirt, that part of the shirt is now covered with a thin film of water. So, any light which has to reflect off that part of the shirt has to go through water.
> Before water is spilt, 100% of the light travelling towards that part of the shirt will hit the surface. But now only a fraction of the light moving towards it will hit its surface. This is because the light now has a layer of water to go through. And due to the reflectance of water, not all light at the air-liquid-interface (border between air and water) goes through the water. Some of it is reflected.
This is not clear, or at least needs another step. It is saying that part of the reason the shirt looks darker is because some of the light never had a chance to reflect off the shirt, because it reflected off the water first. But the observer only cares whether light is reflected at all, not whether it reflected off the shirt or the water.
I assume the unwritten part is that this specular reflection is only reflecting light in one direction, instead of diffusely, so if the observer is not in that line of reflection, they won't see the light. But this explanation seems wrong to me, as it implies that a wet shirt will have some brighter highlights at some angles, and yet I have never seen this.
I guess the internal reflection bit explains why less total light energy might escape; if it hits the shirt and bounces off the water-air boundary and hits the shirt again, then it had two chances to get absorbed by the shirt instead of one.
But yes, you're right that shiny things tend to seem darker from most angles, because the observer is usually not in line with most of the reflected light.
In computer graphics, Blinn highlights are a pretty decent approximation for shiny surfaces. (Interestingly, the highlights generally don't take on the color of the object unless it's metallic.)
Car headlights on wet pavement are kind of a worst case. You don't get much benefit from your own lights because the vast majority reflects off at a useless angle, but it does reflect back up at oncoming traffic.
(Impractical startup idea: headlight drones, that fly ahead of your car in rainy weather and illuminate the road in front of you from a more useful angle. Powered by microwaves beamed from the car.)
Let's follow energy, after all darker = less energy. For simplicity let's consider the sum of energy over all observer angles, to get rid of the "it's shiny only from one angle" problem.
Reflection on the surface of the water is "free" both ways (from air to air bounce, and from water to water bounce), no energy loss. Travelling through water - similar, lite energy loss, though not exactly zero. The only way we can really lose energy is on the cloth reflection. Therefore (almost) all of the darkness effect should be attributable to increased number of cloth reflections, via the total internal water reflection mechanism.
This also explains why colors are more vivid when the cloth is wet - after all the same ray of white light bounces off the cloth several times, increasing the "color filtering properties" of the cloth.
Yes, wouldn't that only apply to soaking wet cloth? If it's just a little wet it still looks darker but not shiny, and there is no specific angle you can look at it and see the light source reflected. So it appears to still be diffuse in that state. Where does the light go?
It goes through, doesn't it? I can't find hard numbers, but basically Wet Clothes Don't Stop Sunburn. The UV rays don't bounce at all, unless it's off the skin. It mostly all turns to heat eventually (AFAIK).
Refractive index of a material is the ratio between speed of light in vacuum and speed of light in that material. Light tends to bounce back when encountered with a sharp change in refractive index. Being wet means that there's a water film covering the material, mediating the change in refractive index, resulting in reduced reflection.
Apart from index mediation, the water film does something else. For rough/fibrous surfaces, the reflection will be diffuse, i.e. visible from all directions. When a water film is present, the surface becomes smooth, and the reflection will be specular, and only visible in one direction. So in most directions, the material will appear darker.
Conductors are a completely different beast. The reflection off of metals are not solely dictated by the refractive index.
> When a water film is present, the surface becomes smooth, and the reflection will be specular, and only visible in one direction. So in most directions, the material will appear darker.
Yes, that's precisely the part I was addressing in my last paragraph. If it's specular reflection, then in "most directions" it will appear darker, as you say, but in one direction it should appear brighter, even shiny. But I've never seen a damp rag be shiny in any direction.
(...and it shouldn't be that hard to see, if that effect is really true. With any other shiny object (polished car, CD, balloon) you see the specular reflection frequently.)
> But I've never seen a damp rag be shiny in any direction.
The surface of a cloth is still much less smooth than a polished car, cd, etc, so is it possible that the specular reflection happens, but at a contrast that is too low for our eyes to detect, or in enough disjoint sections that we can't perceive it as a single effect?
There are plenty of phenomena that fall out of the range of our unaided perception.
We would, however, probably observe the specular reflection of instead of water we used a thicker transparent liquid, like clear glue.
If you find this kind of thing interesting, I strongly recommend the book [The Nature of Light and Colour in the Open Air](https://store.doverpublications.com/0486201961.html) by Marcel Minnaert. He discusses why wet things are darker, why shadows of leaves have little circular gaps, how to see the "green flash" that sometimes precedes a sunset, and dozens of others, some you might have noticed and wondered about, and some you won't realize you noticed until he points it out.
This is the theme of Bon Jovi's ill-fated concept album "Dark When Wet", the followup to their smash hit "Slippery When Wet". Such commercial success must have gone to their heads to attempt such an audacious second project as a sequel based on Quantum Electrodynamics. It was doomed to failure.
Re-interpretations of their classics, such as "Wanted, Dead or Alive - Schrödinger Redux" failed to capture the live animal energy of the original, and "Livin' on a Fermion", with lyrics such as "Pauli used to work on the docks/That's not even wrong/He'd have fallen right through/It's tough" left not only their core audience baffled, but indeed all sentient beings.
Hardly surprising that Mercury Record buried all mention of the album (and will deny it to this day). They returned to more familiar ground with "Jersey Shore", but lament the death of the Concept Album with this last, brave attempt.
Index of refraction matching (or, effectively impedence matching). Snow is white because you have a bunch of air/ice/air/ice/air/ice interfaces. The amount reflected at each interface is determined by the Fresnel Equations:
https://en.wikipedia.org/wiki/Fresnel_equations
Replace the air gap with liquid water, and the slush will be much darker because the difference in index of refraction at each intereface (now liquid/ice/liquid/ice etc) will be much less.
Basically, in electrical terms, less impedance mismatch so less reflection.
On a related note, I find it fascinating that modern CGI cannot quite simulate wet fabric. I watched Frozen 2 and the rendering is absolutely breathtaking on most materials. Water, fabric, hair, etc. All impeccable.
But the minute fabric gets wet, it looks so unrealistic. It looks plastic-y or greasy. I can only assume this is an active area of research, but we’re clearly not there yet.
For me, when I google "cgi wet fabric" the top result is your comment here referring to the top result that you found in Google when googling "cgi wet fabric"!
I am not an expert on this, but I suspect that the "diffuse reflection" picture is not telling the whole story here, as it shows reflections purely from the surface. From what I've seen, in most diffuse materials it's subsurface scattering that dominates. So I think the main contribution of wetness is that it decreases the internal scattering because the indices of refraction are more matched. The absorption remains similar, and I believe reflectance is mostly a function of the ratio of scattering to absorption (Kubelka-Munk theory).
This article and all the comments seem like fancy and longwinded and sometimes misleading theories of something that basically boils down to "a lot of light gets trapped at diffuse angles in the short distance between the rough surface of the fabric and the surface of the water on top of it."
I think this is a good reference for anyone who touches computer graphics. It's not really clear enough for Physics (see all the comments here), but as tips for understanding how computer tend to model things, it's fine.
For example, a raytracer can simulate specular (mirror) or diffuse (matte) reflections by simply playing with the incident angle when the ray hits an object. The more jitter, the more matte the object appears. Similarly, you can get very nice "milk" or "skin" coloration by following the incoming ray slightly below the surface before reflecting back out, the same way this discusses having a thin layer of water would prevent some of the light from reflecting back out.
The actual short version is: diffuse reflection from the surface means that some of the light leaving the surface is at angles which makes it get totally internally reflected at the boundary between the water film and air, giving it another chance to be absorbed by the surface
Wow, is this the same reason why you're more prone to sunburn if you have water on you or are in a swimming pool? Does the light refracted under the surface of the water bounce around and hit your skin multiple times?
I think it is more what mannykannot said, combined with the fact that chlorine/salt/sand will also irritate skin, as well as the fact that sun reflecting off of water/glass/metal is not just reflecting visible light (sometime in the past few months there was a great thread on here with a guy talking about how they use mirrors to reflect sun onto solar panels where he works in Canada I think).
Snow sunburn is also a thing, IIRC snow can reflect as much as 90% of the UV. You can actually go 'snowblind' from this uv reflection.
I am unconvinced by this argument. After all, the same light flux is hitting your body. Only thing a lens would do is make it focus in different spots than it otherwise would, but then you'd get spots with burns and spots without burns. This is not what I see happening at all.
It does happen, but due to inflammation and the fact that the “lens” wobbles since it is made out of water, you get enough consistency that it’s not noticeable without a detailed inspection.
It’s possibly still noticeable via the naked eye, but you’d have to compare and contrast to notice the slight mottling.
Nobody doubts that a drop of water can act as a kind of lens. The question is then if a large number of very small water-drop-lenses contribute to sunburn.
The argument is that the light flux (the total amount of energy) hitting your body is the same. A lens just focuses the energy of a wide area (the size of the lens) onto a smaller spot, it doesn't add any energy to the equation. If the lenses are kept totally still, you would get a higher flux (more of a sunburn) in the focal point, and less energy (less of a sunburn) around the focal point. If the lenses move around a lot, it all averages out into not having any effect at all.
There's a counterargument to the above in that thread: the droplet-lenses have a larger surface area than the area they're covering, which lends them the ability to gather more light flux than the area they're covering. This seems rather theoretical, though.
Start with a dry surface and consider the cone of light, scattered from a point on it, that will pass through your iris. If, as shown in the final diagram, the water, when added, forms a flat surface, and you are looking at it fairly perpendicularly, then none of the rays that are totally internally reflected would have been in that cone and ended up entering your eye anyway - they are too oblique. They do not contribute to the brightness either before or after the water is added.
On the other hand, when water is added, the above cone of light is broadened, on account of refraction, so it is no longer the case that all the light within it will enter your eye.
These considerations must be modified (I am not sure how) if the water forms a thin, capilliary-adhering film that conforms to the roughness of the substrate.
Finally, if the substrate is transparent or translucent (e.g. sand (quartz crystals)), the change of refractive index at its surface is lessened in the presence of water, causing more transmission / absorbtion and less reflection / scattering there.
This is not the full explanation, as can be checked with a simple experiment. Dip a piece of cloth in a glass of water and hold it there submerged beneath the surface. The submerged part will still look darker than the dry part, despite the fact that there is no film of water covering it.
Other comments here have put forward the explanation that the water matches the index of refraction of the material much more closely, meaning that light is more likely to pass straight through the material instead of bouncing off. This explanation seems much more likely to be correct to me.
Another experiment: Put some wet spots on a cloth and hold it up to a light source. More light will pass through the wet spots than the dry cloth. This certainly suggests that the reason wet cloth reflects less light is because more is passing through. (I have tried both these experiments with paper towel used as the cloth.)
I thought this was going to be kinda dumb but this was actually really neat. You never really take the time to understand the everyday things such as objects getting darker when wet.
>This is why you can’t see things in dark; no light means that there is no way that you can channel the information about the surroundings into the person’s eye.
This is not exactly accurate to my understanding. As carl Sagan once said "we are star stuff", and just like stars wherever there is a human to see - even in the "dark" - there is light.
Take the image of the sun/sun rays, moon and eye from the article, its important to remember all matter is emitting light (that includes the moon and the eye and human attached to the eye) not just the Sun.
Even in the "dark" everything including the person will be emitting light (electromagnetic radiation). Humans mostly emit electromagnetic radiation in the infrared wave length, but humans also emit some electromagnetic light in "visible" wave lengths. The issue is obviously humans have not evolved eyes that see "infrared" light like other animals, and similarly the visible light humans emit is below the intensity human eyes evolved to see. However, if human eyes were sensitive enough all humans would appear as shining stars emitting their own light.
Not sure its a thought experiment, but I always thought to understand human sight look at your hand. Then pretend you saw infrared and imagine what you hand looks like (or google a hand in infrared), then do the same as though you saw x-ray wave length light. Now try to imagine if you could see all 3 spectrum at once...what would that look like?
>> This is why you can’t see things in dark; no light means that there is no way that you can channel the information about the surroundings into the person’s eye.
> The issue is obviously humans have not evolved eyes that see "infrared" light like other animals, and similarly the visible light humans emit is below the intensity human eyes evolved to see. However, if human eyes were sensitive enough all humans would appear as shining stars emitting their own light.
In other words, the quote you pulled is completely accurate. It didn't say there was no way to channel the information about the surroundings into any photosensitive device. It says quite clearly that you can't channel the information into a human eye.
> Not sure its a thought experiment, but I always thought to understand human sight look at your hand. Then pretend you saw infrared and imagine what you hand looks like (or google a hand in infrared), then do the same as though you saw x-ray wave length light. Now try to imagine if you could see all 3 spectrum at once...what would that look like?
Probably much like looking at a solid object embedded in colored but not opaque glass. The ability to see things inside other visible things is not foreign to the visible light spectrum.
> It didn't say there was no way to channel the information about the surroundings into any photosensitive device.
No it specifically said
>"This is why you can’t see things in dark; no light means...".
Not sure how to make the distinction any simpler, maybe you can follow:
If there is no light, then it is dark (true) - thats what you claim is being said, but thats not what is said, what is said is
If it is dark, then there is no light (false) - any time you have been in the dark, there has always been light.
>Probably much like looking at a solid object embedded in colored but not opaque glass. The ability to see things inside other visible things is not foreign to the visible light spectrum.
Sure the non-imaginative approach is to say just superimpose 3 images on top of one another with each image having some transparency. And sure that may make sense for objects like bone inside the persons outer skin...but emitted heat is not a solid object inside another solid object, it neither embedded in the object (its emitted) nor solid.
When did "I pretend otherwise"? My comment(s) have nothing to do with the meaning of light. I'm not sure what you get out of pretending otherwise.
Either way in the dark when you can't see, the author is wrong, that doesn't mean there is no light...there is light, at minimum the person who can't see is emitting it. Many people don't know that and find it interesting.
Now I see the problem and confusion on your end...you don't think humans emit visible light.
Like I said, most people don't know and are fascinated, that is why I shared it...I don't find many who dispute it, but here are some articles about the studies that confirmed it:
Differently from diffusive reflection: specular reflection.
In this type of reflection, the angle at which light hits the surface is the same angle at which the light leaves the surface (reflected). This type of reflection tends to happen on smooth surfaces like a mirror or glass.
I've always just assumed the increased surface area from expansion results in more light getting absorbed, not unlike how more particulate in the air (dust, smoke, sand) reduces the amount of light hitting something.
Perhaps a related thought, but I find it fascinating that everything would reflect like a perfect mirror if there weren't these small imperfections in / the roughness of surfaces.
I'd also like an answer to this one. It seems like most of these explanations would also apply to plastic or metal getting wet, so why does cloth get so much darker than other materials?
It feels like the fact that water is absorbed has to matter here--wood, dirt, cloth and sand all darken when they absorb water but not when the water remains in a bubble on the surface (as it does on cloth/sand/wood treated with a hydrophobic coating).
If this were true, a glass of water sitting on top of concrete would look darker than a puddle of water on the same concrete. That’s why we know that this is not the explanation.
I think it's a funny coincidence that he says "Light acts as a medium of information transfer here.", because IMO this article is only a medium of info transfer from the source he links at the bottom (to be fair, at least he links it) without adding much.
I don't think there's anything wrong with making personal blog posts about learning well established concepts, but promoting them is the reason we have a million Medium/etc articles that are just regurgitating a paper without adding anything of substance.
