# Image formed by photons

General question regarding how images are formed. As you move your eye "detector" around a illuminated room. The is image of lets say "a book" is in every position in the room at a given time correct--even before you "look" at it? The photons reflected off the book have formed an image of the book in every location in the room but it is not until a detector" the eye" is at a specific location to gather the information packed in the photon wave function that the image is generated. Is this a correct way of looking at this scenario?

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kuruman
Homework Helper
Gold Member
The is image of lets say "a book" is in every position in the room at a given time correct--even before you "look" at it?
It is the lens of the eye that forms an image on the retina which is interpreted by the brain as an object. The image does not otherwise exist. If you place just a piece of photographic film in a room, you will not get an image of the objects in the room on the film without using a camera that has a lens.

davenn
Understood with regards to the photographic film but nonetheless the the information of the image is everywhere in the room. Their are photons at every point in the room which have the information of the image, if that makes sense.

Place an “eye” anywhere in the room and information of every object in the room is “ stored in that point in space via a photons.

davenn
Gold Member
2019 Award
Understood with regards to the photographic film but nonetheless the information of the image is everywhere in the room. Their are photons at every point in the room which have the information of the image, if that makes sense.
no it isn't

no it doesn't

There

There are EM waves bouncing all around the room .... it's ONLY the EM waves, and I hate to use the word photons, that come directly from an object to your eye or directly to a camera that are detected and resolved into that object

Place an “eye” anywhere in the room and information of every object in the room is “ stored in that point in space via a photons.
nothing (EM or photons) is stored anywhere in the room

lekh2003
That is semantics with regards to “stored” and incorrect as well a photon or EM in a room obviously store energy. The fields have an associated energy. Don’t know what “ nothing is stored anywhere” even means ? Can you clarify

no it isn't

no it doesn't

There

There are EM waves bouncing all around the room .... it's ONLY the EM waves, and I hate to use the word photons, that come directly from an object to your eye or directly to a camera that are detected and resolved into that object

nothing (EM or photons) is stored anywhere in the room

That is semantics with regards to “stored” and incorrect as well a photon or EM in a room obviously store energy. The fields have an associated energy. Don’t know what “ nothing is stored anywhere” even means ? Can you clarify

davenn
Gold Member
2019 Award
That is semantics with regards to “stored”
no it isn't ... the word stored has a very specific meaning, with is totally irrelevent in the current discussion

Don’t know what “ nothing is stored anywhere” even means ? Can you clarify
read the above comment .... it is irrelevant .... no EM is stored in the room

Nugatory
Mentor
General question regarding how images are formed. As you move your eye "detector" around a illuminated room. The is image of lets say "a book" is in every position in the room at a given time correct--even before you "look" at it?
An image forms when you use a lens to focus the light onto a surface (retina of your eye, photographic film, ...) in such a way that the intensity at each point on the surface is correlated with intensity of the light reflected off the corresponding point on the book. So your description above isn't quite right; the one below (with the corrections I've made) where you say that the information is there but you need a device to gather it up and form the image is better.
The photons electromagnetic waves reflected off the book have formed an image of the book in every location in the room but it is not until a detector" the eye" is at a specific location to gather the information packed in the photon wave function time-varying electromagnetic fields at location that the image is generated.
Photons (which don't even have a wave function) don't come into this process at all - the whole thing is straightforward classical electrodynamics working with classical electromagnetic radiation and classical optics.

davenn
An image forms when you use a lens to focus the light onto a surface (retina of your eye, photographic film, ...) in such a way that the intensity at each point on the surface is correlated with intensity of the light reflected off the corresponding point on the book. So your description above isn't quite right; the one below (with the corrections I've made) where you say that the information is there but you need a device to gather it up and form the image is better.Photons (which don't even have a wave function) don't come into this process at all - the whole thing is straightforward classical electrodynamics working with classical electromagnetic radiation and classical optics.
Thank you for insight. And my mistake saying photons wave function which is incorrect. The main point I wanted clarification on is whether the “information” which has the potential to form an image exists in a given point in the room.

jbriggs444
Homework Helper
2019 Award
Thank you for insight. And my mistake saying photons wave function which is incorrect. The main point I wanted clarification on is whether the “information” which has the potential to form an image exists in a given point in the room.
At a "point" of zero size: no. There is zero energy passing through a single point and zero information available.

In an arbitrarily small region around any point you pick, yes. Stick a lens there. The smaller the region (i.e. the smaller the aperture) the longer you have to wait (i.e. the longer the exposure) before an image can become detectable. [And for smaller apertures, diffraction makes image decoding more challenging]

waves and change
lekh2003
Gold Member
Imagine you are blind. When a little ball hits you and you figure out that a ball is coming from somewhere and that it has just hit you, you became aware of an object or source where that ball came from. In some sense, a physical possible location and image of the source of that ball has just been made. When you are not in the room and the ball is bouncing around wildly not hitting into you, it has no stored information because there is nothing there to interpret something from its movement.

Photons are the same. They are actually just arbitrary EM waves which humans have evolved to detect. Their bouncing about is interpreted by the retina into an image. Otherwise, they are just balls bouncing about with no actual purpose.

Staff Emeritus
2019 Award
Photons are the same.
Photons are not little balls. See Nugatory's message.

sophiecentaur
Gold Member
Their are photons at every point in the room which have the information of the image, if that makes sense.
Your picture of Photons is very misguided. As has been pointed out many times on PF, they are nothing like the little bullets that people imagine. There is no particular "point" at which a photon exists until it actually interacts with an object. You have made the common mistake of assuming that introducing the photon idea is of any earthly use in discussing optics or that it enhances understanding. All of Optics is far better dealt with, using waves. Waves are not a 'second best' description. They are the way to do it. If you want to claim otherwise then you would, at least, have to give a credible reference and demonstrate how your little bullet model gives the right result.
This whole question is far better explained by introducing the phenomenon of Diffraction which describes how waves are modified as they pass through or by obstructions. An 'Image' is formed when waves, emanating in different directions from an object, come together in phase and arrive together in a pattern that maps the pattern of the object. The formation of a 'good' image requires an aperture of some finite width and also some mechanism (e.g. a suitable lens or converging mirror) to bring the waves together coherently.
Light Waves are emanating from every point in every direction. All those waves carry information about the brightnesses and positions of parts of the room (etc). If you took a lens, say, to any point in your room, it would accept waves from all directions, including your point of interest on the wall, and produce a detailed image in one particular place on a screen (when "focussed" for optimum effect). A lens placed elsewhere could also produce an image by gathering and 'assembling' the waves hitting it.
You could usefully re-state "Their are photons at every point in the room which have the information of the image," as "Their are waves travelling past every point in the room which have the information of the image."

lekh2003
Gold Member
Photons are not little balls. See Nugatory's message.
I meant that photons can be thought of like this in this situation. Sorry if it was misleading.

sophiecentaur
Gold Member
I meant that photons can be thought of like this in this situation. Sorry if it was misleading.
They can't, if they aren't in fact what you imply.
Why not do this 'properly' (i.e. in a way that actually works)?

lekh2003
Staff Emeritus
2019 Award
Reinforcing an incorrect picture is not helpful.

Photons are not little balls. See Nugatory's message.
Photons can be though of as “ little balls” in the sense that they act as particles when they interact with matter... ie the photoelectric effect. The formalism that best describes the way light travels is to regard it as a wave but to consider it as a particle when it interacts with matter Ie ....a photomultiplier

Your picture of Photons is very misguided. As has been pointed out many times on PF, they are nothing like the little bullets that people imagine. There is no particular "point" at which a photon exists until it actually interacts with an object. You have made the common mistake of assuming that introducing the photon idea is of any earthly use in discussing optics or that it enhances understanding. All of Optics is far better dealt with, using waves. Waves are not a 'second best' description. They are the way to do it. If you want to claim otherwise then you would, at least, have to give a credible reference and demonstrate how your little bullet model gives the right result.
This whole question is far better explained by introducing the phenomenon of Diffraction which describes how waves are modified as they pass through or by obstructions. An 'Image' is formed when waves, emanating in different directions from an object, come together in phase and arrive together in a pattern that maps the pattern of the object. The formation of a 'good' image requires an aperture of some finite width and also some mechanism (e.g. a suitable lens or converging mirror) to bring the waves together coherently.
Light Waves are emanating from every point in every direction. All those waves carry information about the brightnesses and positions of parts of the room (etc). If you took a lens, say, to any point in your room, it would accept waves from all directions, including your point of interest on the wall, and produce a detailed image in one particular place on a screen (when "focussed" for optimum effect). A lens placed elsewhere could also produce an image by gathering and 'assembling' the waves hitting it.
You could usefully re-state "Their are photons at every point in the room which have the information of the image," as "Their are waves travelling past every point in the room which have the information of the image."
Your claim that photons are nothing like the bullets people imagine is not correct. A photon interacts with matter like a particle. If your claim was true how would you explain the photoelectric effect which treats a photon as a “bullet”.

ZapperZ
Staff Emeritus
Photons can be though of as “ little balls” in the sense that they act as particles when they interact with matter... ie the photoelectric effect. The formalism that best describes the way light travels is to regard it as a wave but to consider it as a particle when it interacts with matter Ie ....a photomultiplier
Actually, it can't. "Little balls" have a lot of physical connotations that are not supported by physics when applied to photons. The ONLY similarities here is that each photon carries a specific amount of energy. But this doesn't imply that that they are like "little balls". You might as well say that an egg is like a cow, simply based on the characteristic that they are both edible.

The photoelectric effect, while it is a strong evidence for the photon picture, does NOT completely disprove the wave scenario (do a search on here, it has been discussed numerous times in numerous threads). The which-way experiments and the anti-bunching properties are more definitive.

BTW, a "photomultiplier" is nothing more than the photoelectric effect, with the initial photoelectrons being multiplied by an additional structure. It is not a property of an interaction with matter.

Zz.

sophiecentaur
Gold Member
Your claim that photons are nothing like the bullets people imagine is not correct. A photon interacts with matter like a particle. If your claim was true how would you explain the photoelectric effect which treats a photon as a “bullet”.
Quantisation and bullets are not related in any obvious way. If a photon is like a bullet, how can it be involved in diffraction? Of course there no question about the quantised behaviour of EM but it's risk and nonsense to go the whole way or we're back in the Corpuscular Theory. Bullets have mass; do photons?
If you want to explain a wave phenomenon the use waves.

Actually, it can't. "Little balls" have a lot of physical connotations that are not supported by physics when applied to photons. The ONLY similarities here is that each photon carries a specific amount of energy. But this doesn't imply that that they are like "little balls". You might as well say that an egg is like a cow, simply based on the characteristic that they are both edible.

The photoelectric effect, while it is a strong evidence for the photon picture, does NOT completely disprove the wave scenario (do a search on here, it has been discussed numerous times in numerous threads). The which-way experiments and the anti-bunching properties are more definitive.

BTW, a "photomultiplier" is nothing more than the photoelectric effect, with the initial photoelectrons being multiplied by an additional structure. It is not a property of an interaction with matter.

Zz.
Your answering your own questions. I never disregarded the duality with regards to photons behaving as waves when propagating. I simply said they act particle like when they interact with matter such as knocking electrons off a material- a photomultiplier was just an example.

Quantisation and bullets are not related in any obvious way. If a photon is like a bullet, how can it be involved in diffraction? Of course there no question about the quantised behaviour of EM but it's risk and nonsense to go the whole way or we're back in the Corpuscular Theory. Bullets have mass; do photons?
If you want to explain a wave phenomenon the use waves.
I was simply attempting to get the point across that photons interact with materials such as solar cells in a particle like way. Yes, a bullet is not a good example because as you said it propagates in a way best formalized using wave like behavior.

ZapperZ
Staff Emeritus
Your answering your own questions. I never disregarded the duality with regards to photons behaving as waves when propagating. I simply said they act particle like when they interact with matter such as knocking electrons off a material- a photomultiplier was just an example.
First of all, how was I "answering my own questions" when I didn't ask any?

You need to look at the non-photon description of the photoelectric effect. There is no "act particle like" in this case. (read 1st paragraph of this paper, and references therein: http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf)

The photomultiplier example you gave was wrong, because you said that ".. consider it as a particle when it interacts with matter Ie ....a photomultiplier... " This is incorrect, because the example of a photomultiplier is NOT how photons interact with matter, i.e. there is often no "multiplication"! If you simply have stuck with your photoelectric effect, I wouldn't have touched it.

Zz.

First of all, how was I "answering my own questions" when I didn't ask any?

You need to look at the non-photon description of the photoelectric effect. There is no "act particle like" in this case. (read 1st paragraph of this paper, and references therein: http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf)

The photomultiplier example you gave was wrong, because you said that ".. consider it as a particle when it interacts with matter Ie ....a photomultiplier... " This is incorrect, because the example of a photomultiplier is NOT how photons interact with matter, i.e. there is often no "multiplication"! If you simply have stuck with your photoelectric effect, I wouldn't have touched it.

Zz.
Thanks for the reference. I understand the quantum nature of the interaction makes a particle like description less coherent. That being said; would you say that single electrons being knocked off a material if the photons energy exceeds the work function to be particle like?