I have read that photon "size" is proportion to wavelength, but I've also read that photon "size" isn't exactly a thing. So does this mean that different colors, which have different wavelengths and thus different "size" photons, will leave different size marks on a light sensitive screen when fired?

By "firing" photons, I mean the technology that results in a build up pattern of circular points, like here (although this one is with electrons): https://www.youtube.com/watch?v=hv12oB_uyFs

And by different color filters, I mean placing different colored filters in front of a light source, like in this video: https://www.youtube.com/watch?v=kcSYV8bJox8

So when you swap filter colors in front of whatever light source is being used to emit photons in a build up pattern experiment, do the circular dots that build up, have different sizes (or densities/intensities) dependent on the color of the filter used?

Or would it just be the build up pattern itself (like the interference pattern in a double slit experiment) that is different, and the individual build up marks themselves are all identical regardless of any colored filter that might be in front of the light?


3 Answers 3


The size of a photon means something entirely different than it would in classical physics. A photon is described by a wave function. From the wave function, you can calculate the probability of finding the photon at a particular position if you measure it. You measure it by interacting with it, which changes the photon.

One way a photon interacts when it hits a screen is to promote an electron in a single atom to a higher energy orbital. The photon is absorbed and ceases to exist. It does not matter how big the photon's wave function is. It will interact with just one atom this way.

Other ways of interacting can be more spread out. A mirror is a smooth metal surface. Conduction electrons in a metal do not stick to a single atom. They spread out. A photon interacts with the sea of spread out electrons. The result is a photon just like the incoming photon, but reflected away. This interaction is easiest to understand from a classical description. Classical light is an oscillating electromagnetic field. It causes surface electrons to vibrate, and the light is absorbed. A plane of vibrating electrons emit light. This creates the reflected photon.

For more about the nature of a photon, see How can a red light photon be different from a blue light photon?.

Also see Does the collapse of the wave function happen immediately everywhere?. This describe an electron, but it works the same for a photon.


In classical physics yes the size of the photon would vary, and hence the size of the mark it left on the light sensitive film. However, in quantum electrodynamics the photons are only interacting with electrons. So the size of the photon is more like the size of the area over which it could interact with an electron.

There are two different things at play here, I think imagining the case when only a single photon is fired at the light sensitive screen would be helpful.

So the wavelength of the photons would definitely affect the diffraction pattern (size of circular pattern), but the circular dots which are caused by a single electron being knocked from it's place by a photon would not be any different size. At some point if you increased the wavelength of the photons enough they would not have enough energy to kick an electron out of its place, because the energy of a single photon is inversely proportional to its wavelength.

  • $\begingroup$ "So the circular pattern itself would be a different size, but the circular dots which are caused by a single electron being knocked from it's place by a photon would not be any different size." Thanks. I have seen it said that when people say blue light has more energy than red light, what they really mean is that the photons of blue light have more energy than the photons of red light. But if the actual photons leave the same sized marks, how exactly do they vary in terms of energy? Is it the energy of the photon itself that varies, or just the frequency at which the photons hit something? $\endgroup$
    – Tristan
    Nov 1, 2021 at 14:51
  • 2
    $\begingroup$ Yeah, the blue light photons themselves have more energy than the red. You couldn't really tell from the size of the mark (because it is just one electron getting knocked out of it's place) but the blue photon could give the electron a much bigger kick and effectively send it flying out of place faster than the red photon. Some electrons might be hard to knock out of place, and the red photons wouldn't have enough energy to do this while the blue ones would. This discovery called the photoelectric effect is what lead to the invention of quantum mechanics and Einstein winning a Nobel prize! $\endgroup$
    – Hunter
    Nov 1, 2021 at 14:55
  • $\begingroup$ Thanks. Is this a probabilistic thing? Like, a blue photon is more likely to eject an electron than a red photon? I am not sure whether this is a statistical thing, or applies to all photons, e.g. all blue photons have more energy than all red photon (if from the same light source, but with a different filter). $\endgroup$
    – Tristan
    Nov 1, 2021 at 15:02
  • 1
    $\begingroup$ No it isn't probabilistic. All blue photons have more energy than red photons. In fact there is a really simple relationship between wavelength and energy. Energy=h/wavelength where h is Planck's constant. If you know the wavelength of a photon, you can calculate its energy exactly. $\endgroup$
    – Hunter
    Nov 1, 2021 at 15:05
  • $\begingroup$ Thank you! Do these blue and red photons have different geometries? Presumably they travel at the same speed, the speed of light, so if they have the same geometry, how can they vary in terms of energy? Or is that where the "wavelength" of electric field oscillations and the EM field associated with a photon come into play? $\endgroup$
    – Tristan
    Nov 1, 2021 at 15:13

Well, first of all, filters do not create photons of a particular wavelength (there are substances that can absorb light at one frequency and emit another frequency, but that's another phenomenon). A filter simply is selective about what light it lets through, and will let through light of a particular wavelength if that wavelength was already present in the incident beam.

Second, the wavelengths of visible light are microscopic. You're not going to be able to see with your naked eye anything on that scale. Generally, patterns from shining a light beam are from vast numbers of photons. If a single photon is creating a visible dot, it's because there is a cascading reaction where the photon interacting with one atom sets off a chain reaction that affects a macroscopic area, not because the photon interacted directly with that entire area.

Third, a photon interacts with only one atom. Its wavelength will affect how many atoms it can interact with; a larger wavelength means more atoms that it could have interacted with. But it doesn't affect how many it does interact with. You can kind of think of it as a skier who's weaving back and forth as they go down a slope. A larger wavelength is like the skier weaving back and forth a larger distance. There's a larger area over which they can crash into a tree, but once they crash into a tree, they're going to stop and not crash into any more.

A larger wavelength will generally mean more dispersion, so if you try to focus a beam of photons to a single point, a larger wavelength will result in a larger spread, but less average intensity per unit area over the area it spreads out over. If you're just taking a light bulb and passing it through a filter, this doesn't apply, as light bulbs don't aim photons in a narrow beam.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.