Where is my image now (3)
Instead of a monochromatic ray box, try using, as I did, two red and one green laser pens. You will notice that the green beam does not coincide with the intersection point of the red mean. That is no surprise, Newton called it different refrangibility, and it was a central aspect of his analysis.

But, let us again attempt not to be blinded by the beams, and try and understand how an image of different colors is formed.

We have now three beams, two red one green, coming from laser beams which have been set along each other, facing a convex lens.

Optics has concentrated on the paths these light beams took and, I think, somehow ignored what happened to the image of the three little pigs.
Suppose now that we want the lens to show us, inverted or not, the image of those three pens. What should we do?

Well, if you asked a photographer he would tell you that you have to measure the light reflected off the pens themselves, and not their reflected beams. Again, just like taking a picture of the sun. You will have somehow to neutralize the strength of the solar rays before you can take a snapshot of the fiery planet. Accordingly, your exposure time will be much longer than it would be on the basis of the beams reaching the camera.

What is an essential question is whether it will be possible to take a picture in which all three laser pens, and the color coming out of them, would be as sharp. In other words, will we be able to take a sharp picture of the green pen, and, at the same time, in the same picture, show the red beams as sharply as their Martian brother?

That sounds like a stupid question, after all we constantly take pictures of multicolored objects, and we never have problems with that. All colors are depicted as sharply.

This brings us to a very familiar issue in Optics:

Chromatic aberration
One can find many examples of so-called CA on Youtube, but you will have to be very critical since not everything that is presented under this title truly deserves it. Still, it gives a realistic indication of the problems people get confronted with in their everyday use of cameras.

Let us analyze a picture of our three little pens and ask ourselves if we can make sense of the concept of CA.

The green beam is obviously a game breaker. It would be much easier if it just did what the two do, and met with them at the same intersection point.

The problem of course is whether that behavior could honestly be called an aberration. Mister Green is doing what it always has done, and what its nature compels it to do. And that is, when in the company of other colors, to refract according to its own nature.

Still, pictures which show such an effect are considered somehow as faulty and in need of amelioration. It is the most obvious at the edges and that is, when you come to think of it, not really surprising.

Remember that we have no problem taking pictures of multicolored objects or scene, and the accepted explanation is that CA is not perceptible when not at the edges.

My own humble opinion is that there is, in general, no CA at the center of a picture because having a sharp image of multicolored objects is never a problem.

The problem is when we are taking a picture where the beams themselves and their behavior is registered by the camera.

Let us go back to our simple example and I will try to explain it as clearly as I can.

There is a difference between a picture of the pens, and the color they propagate from their mouth, and a picture of the beams that come out of those mouths. The deviating behavior of the green beam is only present in the second case. We have no trouble depicting green and red objects with the same degree of sharpness if they are at the same distance from the lens.

In other words, when light is projected on a surface, there is no CA present, but when beams are registered we can be witness of the different degrees at which different colors are refracted.

One again, we have to make the distinction between images, and the path of light beams.

edit: I put a gray filter, a so-called ND-filter, between the beams refracted through the lens, and a white sheet of paper. I had not trouble getting a sharp image of two red and one green spots. Better lighting on the pens, and a more powerful filter, would have given a clear image of all three pens and the color coming out of them. Every pen and colored mouth as sharp as the other.

This clearly shows that the different refraction angle of each color does not play any role in the formation of the image. Only when it is registered as such, as a beam.

This is a sliver taken out of the first picture. Personally, it reminds of the images made of light going through gratings with hundreds of slits or more, and where the different bright and dark spots are explained by constructive and destructive interference. Here, we are dealing with collimated light that goes unimpeded through a hole and shows the same characteristics.
I think that people forget that laser lights are made of diodes, therefore of discrete objects that, however close to each other, keep a certain distance from each other. I find this a much simpler explanation than what the wave theory has to offer.

This post belongs in my other discussion Rømer and the speed of light 1676

Why is everybody so afraid?

I find it really curious that some issues are systematically ignored when it comes to the theory of light, while at the same time people try to prove me wrong at all costs, even to the point of making me sound ridiculous.

My challenge is very simple:
How come we can see illuminated objects while we are in a completely dark corner where no light reaches us?

If you are unable to answer this question, I cannot take your defense of the established theory of light seriously.

Please prove me wrong! It would make things much easier for me too!

How come we can see illuminated objects while we are in a completely dark corner where no light reaches us? — Hachem

We don't. If no light is reaching something from any direction, and producing no visible light of it's own, then that thing sits in darkness.

Light reaches us though, so we see things... I'm not sure what your challenge is supposed to mean...

I am glad you are back, let me understand it straight. If I am standing in darkness while you are standing somewhere illuminated, I cannot see you because you cannot see me?

I am glad you are back, let me understand it straight. If I am standing in darkness while you are standing somewhere illuminated, I cannot see you because you cannot see me? — Hachem

Consider the following:

Light comes from the sun and bounces off of me (and other places) and then some of it travels to your eyes, rendering me visible to you. Some of the light bouncing off of me and other sources travels to various parts of your exposed surfaces, and then reflects in many directions (diffuses), and then some of that light might make it back to my own eyeballs, making you visible to me also. It could be possible that the distance between us was is so great and there are so few other light sources that by the time the light from me bounces off of you and makes it back to me, it is then of too low intensity to detect. You could then say "you're in darkness". Observe below:



Darkness is just the absence of light, but "light" is everywhere. Electromagnetic waves get emitted from every object above zero kelvin in the form of thermal radiation. This is why thermal imaging cameras can see in the "dark"; they detect a portion of the light spectrum which our eyes cannot.

In order to put something into complete and total darkness, you would have to get it to reach absolute zero temperature, otherwise it would be bathed in it's own thermal radiation. For this you would need an outer barrier that reflects all light radiation and all conductive heat transfer. Then you would need a layer of coolant capable of reaching internal temperatures of absolute kelvin, and then you would need to exist inside of that, at zero kelvin, in order to achieve perfect darkness. Observe below:

very nice drawings. That is why it took you so long to answer.
The distinction between visible and invisible light is a nice touch. But tell me, when I am looking at you from my dark place, what is reaching my eyes, visible, or invisible light? After all, I am seeing you with my naked eyes, right?

But tell me, when I am looking at you from my dark place, what is reaching my eyes, visible, or invisible light? — Hachem

"Visible" depends on the quality of the apparatus, in this case, your eye. If the light has become too diffuse you might not be able to see it (maybe a bigger eye with a bigger lens could though. If you can detect it with your eyes, then it's visible with the naked eye. Remember, "visible" in this case means "visible with the human eye"...

P.S: the distinction between light the human eye can see and the full spectrum of electromagnetic radiation isn't so much a "nice touch" as it is the demonstrable and accepted scientific truth.

You are a real live textbook. So, let me take advantage of it. You reflect enough light back to me that I can see you, but not enough that you can see me. But my eyes could be anywhere. Light that enters my eyes could just as easily fall on any place of my face, or my whole body. And still, you cannot see me, while I would still be able to see you if the light illuminating you would become so weak as to only allow your silhouette to be visible.
Is that what you are saying?

You are a real live textbook. So, let me take advantage of it. You reflect enough light back to me that I can see you, but not enough that you can see me. But my eyes could be anywhere. Light that enters my eyes could just as easily fall on any place of my face, or my whole body. And still, you cannot see me, while I would still be able to see you if the light illuminating you would become so weak as to only allow your silhouette to be visible.
Is that what you are saying? — Hachem

The strength of light is drastically reduced when it diffuses against a surface and spreads out in many directions. So yes, it's possible that you could see me and that I would not see you, if the conditions were right.

The conditions required to create a silhouette can vary, contrast is required, which is different than seeing the light bounce off something directly such as the case we're discussing.

The point is that however much light falls on you, it just never seems to be enough to make me visible to you, while you remain visible to me however weak the light becomes. To the point where maybe I will not be able to see you anymore, but still be able to see the light (figuratively speaking).

The point is that however much light falls on you, it just never seems to be enough to make me visible to you, while you remain visible to me however weak the light becomes. To the point where maybe I will not be able to see you anymore, but still be able to see the light (figuratively speaking). — Hachem

This is of course not entirely correct. Within the same distance I will become visible to you at some point if the intensity of the light is raised high enough. I am assuming a limit to the maximum intensity of light reaching you so that I can still remain invisible. Then we must explain how I can remain invisible in such conditions.

Then we must explain how I can remain invisible in such conditions. — Hachem

You're not explaining the setup clearly at all, which leads me to believe you have not understood what I laid out earlier, but assuming you do understand, here is your answer:

The amount of photons that are statistically likely to strike my eyeballs, having bounced off you, from me, from the sun, can become so negligible that our eyes are not able to actually recognize it because the photons are too few and far between.

If you shine your laser pointer at a painted wall, you will notice that the beam reflects and shows a second red dot on another wall (usually), but it is much weaker than the main red dot. This is because most of the photons get reflected in random directions. The reason why your lazer pointer doesn't create reflection after reflection is because it loses strength too fast at each reflection, so only 2-3 dots are ever visible at a time...

The problem with such a picture is that even those weak photons are visible from all directions.
Allow me this link to something I wrote almost a year ago on the same subject.
https://philpapers.org/post/22794

The problem with such a picture is that even those weak photons are visible from all directions.
Allow me this link to something I wrote almost a year ago on the same subject.
https://philpapers.org/post/22794 — Hachem

Unfortunately "a handful of photons" is so inaccurate and unscientific a description that the entire piece "isn't even wrong".

Furthermore "weak photons" aren't visible from all directions. If you shine your laser pointer across the room, why aren;t you able to see the beam itself unless dust or smoke passes through it?

You are now being dishonest. A handful of photons is not a scientific term because the post is not a scientific formula but expressing a concept. Take the minimum number of photons that would create a visual impression of any kind and ask yourself how many people would be able to see it at the same time. That is what the miraculous multiplication of photons is all about. It expresses the fact that, despite all your valiant efforts, you cannot explain vision by photons entering our eyes. At least, all your waving with pretty pictures did not convince me one bit.

"A handfull of photons" is such a crude term that it demonstrates you don't understand what a photon is (it's an electromagnetic particle-wave).

How many photons per second do you think are emitted by a light-bulb? (hint: it's somewhere in the (10^20) range, which is 10 followed by twenty zeroes).

There is no such thing as photon multiplication. A single candle emits easily over a trillion photons per second. A single photon cannot magically duplicate itself. This would break the laws of thermodynamics.

It is funny that suddenly the number of photons has become insurmountable, when all scientific experiments concerning photons, among those the photoelectric effect, are aimed at controlling the number of photons emitted and received. I admit not being a scientist and I am sure I would get all kind of details wrong, but when I hear somebody as Richard Feynman speaking of a minimum of 5 photons to get a visual impression, and I hear your objections, then I am ready to take whatever figure will suit you. Just tell me which, approximate, minimum number of photons is necessary for a visual impression by one individual, and then explain to me then how so many people can get the same visual impression at the same time.

There is no such thing as photon multiplication. A single candle emits easily over a trillion photons per second. A single photon cannot magically duplicate itself. This would break the laws of thermodynamics — VagabondSpectre

That seems to me to be a hard truth, and I couldn't have said it better myself. So, if photons do not multiply, what are all the people seeing? You want to stick to the principle that photons make vision possible, then you must be ready to believe in miraculous photon multiplication.

The fact that we see a light beam that is directed away from us , should be a hint that we need something else but photons to explain vision. Unless you take the Maxwellian model and believe that when we are looking at a beam, say shining horizontally in front of us, we somehow perceive the magnetic and electric fields it is producing. After all, a collimated beam, like a laser beam, is not supposed to propagate any photons sideways. The fact that the beam hits dust particles which make it visible would of course explain the creation of new photons. But we do not see lines of light reaching out to us from the beam. We only see the beam itself.

It is funny that suddenly the number of photons has become insurmountable, when all scientific experiments concerning photons, among those the photoelectric effect, are aimed at controlling the number of photons emitted and received. I admit not being a scientist and I am sure I would get all kind of details wrong, but when I hear somebody as Richard Feynman speaking of a minimum of 5 photons to get a visual impression, and I hear your objections, then I am ready to take whatever figure will suit you. Just tell me which, approximate, minimum number of photons is necessary for a visual impression by one individual, and then explain to me then how so many people can get the same visual impression at the same time. — Hachem

Do you understand how eyeballs work? There is a membrane of photosensitive cells that the lens in your eye projects incoming photons onto. When enough photons strike a particular cell, then a "signal" can then be sent to the image processing parts of your brain which it then interprets into RGB light and a particular coordinate in your field of view.

So, if 5 consecutive photons strike a particular cell in your photo-sensitive membrane, let's assume that this is sufficient to cause it to fire off a signal with some color data. What you might actually see is a momentary blip of light. If there aren't enough photons striking particular photo-sensitive cells or if the photons in question are too weak to be detected then no "image" might be formed.

I'll ignore the massive host of issues with the following, but just to humor your desire for numbers, here you go:

Google says that the resolution of human eyesight is around 500 megapixels (or, 500 million individual pixels). In order to completely fill a human field of view, let's just guess that 500 million photons are required (per frame) in order cover our entire photo-sensitive membranes...

Let's also assume that the human eye sees at around 1000 frames per second (a generous estimate)...

So, if a single light bulb emits 10^20 photons per second, then a single viewer is going to consume 500 thousand million photons per second. All we need to do is divide the total number of photons emitted per second by the number of photons consumed by an eyeball per second:

1 000 000 000 000 000 000 000 / 500 000 000 000 = 2 000 000 000

This means that there are enough photons being thrown from a single light-bulb to fill the view of 2 billion viewers.

The real issue you're trying to get at is the sensitivity of various light detectors, the mechanics of which we could get into, but it has nothing to do with Romers experiments or contradicting the finite speed of light.

So, if photons do not multiply, what are all the people seeing? — Hachem

Photons are created at a given light source, and they bounce off of objects and are altered (thus reflecting aspects of color) or are absorbed by them (thus transferring some energy). For the intents and purposes of this thread: photons are created and by vibrating/excited electrons, and are also absorbed by them (which excites them)...

it has nothing to do with Romers experiments or contradicting the finite speed of light. — VagabondSpectre

I've had it with you. I am not responding to anymore of your posts. You are just too dishonest. Read again what I said about the speed of light. End of discussion

The fact that we see a light beam that is directed away from us , should be a hint that we need something else but photons to explain vision. — Hachem

We cannot see beams of light that are directed away from us unless something reflects it toward us (like dust particles or a solid surface).

Read again everything I've written in this thread. You've not addressed a singly solitary point of contention and instead have seen fit to constantly say things like "you obviously havn't read what I've wrote", or, "Read this other thing I wrote a year ago", or "you're clearly being dishonest", or "I'm not going to continue to put in effort because you clearly bla bla bla"...

Almost nobody responds to this thread, NOT because your posts are poorly communicated and poorly defended, but mainly because the ideas you communicate are themselves utterly malformed. Saying things like "you have to deal with miraculous photon duplication" is simply asanine. If you have no sweet clue how photons are generated (again, electromagnetic waves: look it up), fine, but maybe you should rethink your confidence in debunking a scientifically accepted theory that you have little to no introduction to and no demonstrable comprehension of. This was never a discussion. I tried to explain why your assumptions are demonstrably wrong and you tried to hold on to your intellectual dignity by shoving your head in the sand.

There's no visible light down there...

After all, a collimated beam, like a laser beam, is not supposed to propagate any photons sideways. The fact that the beam hits dust particles which make it visible would of course explain the creation of new photons. But we do not see lines of light reaching out to us from the beam. We only see the beam itself. — Hachem

You still don't understand: the particles reflect the light, they aren't generating visible light themselves. Particles will diffuse the light in random directions, which will provide a kind of ambient light effect. Individual photons going in many directions (reflected by particles) do not constitute a beam.

I will make another effort, but this time I will refrain from any controversial claim and simply express what I understood from the objections.
1) photons are in themselves invisible, they need to hit matter to create light.
2) When we are looking at a beam that is directed away from us, even if it is a collimated beam, photons are still propagated in all directions, and when they hit our eyes, we can see the beam, even if there is no visible line from the beam to us. That also explains why we can see someone in a bright place even if we are standing in complete darkness.

Here is what I understand of light in very general terms.
3) The Huygens principle states that everything goes in the same, forward direction, because each point is in itself a wave that drives the beam forward. On the sides, wavelets do the same thing, with a minimum loss of energy to lateral movements, contributing to the forward movement.
4) Maxwell states more or less the same thing, but the lateral movements have become magnetic and electric fields that stand perpendicular to the direction of the beam, one creating the other, producing an infinite movement.

Did I get it right?

1) photons are in themselves invisible, they need to hit matter to create light. — Hachem

1) Photons ARE visible light; they are the thing we use to to see. They are what our mechanical eyes gather and detect to produce and abstract visual data like color, shape, and position. Photons need to enter our eyes in order for us to see anything. If you want to see matter with your eyes, then it's required for photons bouncing of it (or being emitted from it) to enter your eyes.

2) When looking at a beam that is directed away from us, we cannot see the beam. It's only if somehow the photons are redirected and actually make it into our eyeballs that we can see it. The reason we can see someone who is in a bright place when we are in darkness is because light from the bright place bounces off the person (in all directions) and some of it is gathered by the eyeballs of the viewer in the darker place; light from the bright place travels to the dark place...

3) I'm sure there are many convoluted mechanics of nearly perfect collimated beams, but you've got to remember that the laws of energy conservation cannot be broken. As the nearly collimated beam that is your laser slowly expands/diffracts/diverges, it becomes less focused. Laser beams may be enhanced by some self-correcting phenomenon, but the energy involved is definitely not self-creating.

4) In a beam, each photon has it's own electromagnetic field and it's own direction. The energy required to create and send them on their way must be expended when they are originally created (from a bulb, a star, etc...). The only "infinite movement" I can think of is newtons Idea that an object in motion stays in motion unless acted upon by an outside force. If photons could duplicate and diverge without losing energy then energy conservation laws would be broken (energy will have been created from nothing), which is impossible.

4) In a beam, each photon has it's own electromagnetic field and it's own direction. The energy required to create and send them on their way must be expended when they are originally created (from a bulb, a star, etc...). The only "infinite movement" I can think of is newtons Idea that an object in motion stays in motion unless acted upon by an outside force. If photons could duplicate and diverge without losing energy then energy conservation laws would be broken (energy will have been created from nothing), which is impossible. — VagabondSpectre

I should have used the term wave instead of beam. After all, a beam is something we perceive, but the wave is what is supposed to explain the mechanics involved.
If I understand you correctly, each time we shine a light, we are in fact, as it were, using a bulb, and not a spotlight, as far as photons are concerned, since every photon has its own direction?

If I understand you correctly, each time we shine a light, we are in fact, as it were, using a bulb, and not a spotlight, as far as photons are concerned, since every photon has its own direction? — Hachem

Spotlight's have bulbs, but that's not the point... My point is that a beam consists of many photons which have been prepared to head in the same direction (that's what makes the "beam"). Each photon in the beam has it's own mass and it's own direction, although in such a beam all their individual directions happens to be the same or almost exactly the same.

Each photon in the beam has it's own mass and it's own direction, although in such a beam all their individual directions happens to be the same or almost exactly the same. — VagabondSpectre

You are walking with a friend at night through a wood, she is holding a torch and you follow her a little from behind. She is directing the torch forward, in front of both of you. How come you see the torch?
edit: I mean the beam coming out of the torch.