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I had gone through a few posts on this topic in this community,however the doubt i have is different from them a bit. There it was said that we do not actually see a laser beam unless they are relfected by dust particles. But i have one question.

When we see anything,it means light is reflected on them and then reaches our eye. Thus,what we are seeing is actually the object and not the light. So when light collides with the dust in air,shouldn't we then see the dust particles only? How are we then seeing the trajectory of light then since the light that is reflected to us isn't even travelling further anymore. Here as shown in the picture,after colliding with the dust,the light ceases when it reaches our eye,so we should have only seen the dust,isn't it? This is related to another question. If light hits an electron(like in compton effect),the electron moves away and if our eye was in such a position that we could see the reflected light,what would we see of the electron? Will we see the electron in the initial position and nothing else since the electron moved away and is not hit by light further?

Kindly enlighten me with the proper concepts.

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  • $\begingroup$ You can look at the emitter directly, can't you? $\endgroup$ Commented Oct 31, 2022 at 6:00
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    $\begingroup$ (don't do that, though) $\endgroup$
    – Polynomial
    Commented Oct 31, 2022 at 6:04

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So when light collides with the dust in air,shouldn't we then see the dust particles only? How are we then seeing the trajectory of light then since the light that is reflected to us isn't even travelling further anymore?

What you're describing would be true if the laser just emitted a single photon.

The laser emits many of many billions of photons per second. Only a small percentage of them hit dust particles, and only a few percent of those happen to reflect photons back in the direction of your eye.

Here as shown in the picture,after colliding with the dust,the light ceases when it reaches our eye,so we should have only seen the dust, isn't it?

Technically, yes, that's exactly what you're seeing. The "beam" you see in the air is not actually the laser beam itself, but all of the tiny particles floating in the air that happen to be illuminated by the beam and happen to reflect some light toward your eye.

The colour of the reflected light remains the same because the laser light is monochromatic and coherent, i.e. it consists of a single wavelength and all of the light emitted is in-phase. The dust is (very probably) not fluorescent, so the reflected wavelength will be the same as the incident wavelength (within a very small bound, due to dispersion).

If light hits an electron (like in compton effect), the electron moves away and if our eye was in such a position that we could see the reflected light, what would we see of the electron?

The photon doesn't bounce off the electron. The incident photon hits the electron, causes an interaction (possibly resulting in an energy state change), and a new photon is emitted. We cannot see the electron, we just see light with a specific wavelength, which is defined by the incident wavelength and the specifics of the interaction.

Will we see the electron in the initial position and nothing else since the electron moved away and is not hit by light further?

Insofar as one could "see" an electron, yes, in a certain sense. You wouldn't see the electron, but rather the single reflected photon from the interaction. This behaviour is subject to the observer effect and uncertainty principle.

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  • $\begingroup$ Thank you very much for replying. In your second last paragraph,you said that a new photon is emitted from the interaction. In our day to day life,light is hits an object and reaches our eye,isn't this "hit" a collision? In that case can't we say that collision of photon takes place with the molecules in the surface? Why don't we then day that a new photon emerges but instead we say the photon is reflected? Kindly clear out this confusion. $\endgroup$
    – madness
    Commented Oct 31, 2022 at 6:31
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    $\begingroup$ That's exactly what happens. A reflection is comprised of photons that resulted from an interaction between an incident photon and a surface of some kind. We just abstract that away when talking about reflective surfaces in day to day life because most people don't need to consider (or even know about) the underlying optics theory or quantum mechanics. $\endgroup$
    – Polynomial
    Commented Oct 31, 2022 at 6:34
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You are mixing a few ideas here that need to be disentangled. Firstly, when you see any object, what you see is light reflected from it or emitted by it. Specifically, the light enters your eye, where it causes signals to be propagated to your brain, and your brain interprets those signals to produce a mental model of the object, a model that allows you to infer properties of the object, such as how far away it is. The amount of detail you can make out varies according to the lighting conditions, the position of the object, etc. For example, if you look at a nearby tree, you can see the texture of the bark and the individual leaves- but if you look at a forest on a distant mountain you might not even be able to make-out individual trees, let alone their leaves.

When you see the path of a laser, what you are really seeing is the light reflected from microscopic particles in that path. Your brain interprets the light, and the mental image that generates is what you interpret as being the beam. You do not see the individual particles from which the light was scattered, in the same way that you do not see the individual leaves from which light was scattered to general the image of a distant forest.

You would not see the beam at all if it was passing through a pure vacuum.

The colour of the laser is an entirely mental construct. Light does not have any intrinsic colour. It is an accident of evolution that makes you interpret a particular wavelength of light as red, say.

Finally, an electron will recoil from a scattering event, so if you detect the location of the event it will no longer be the location of the electron.

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  • $\begingroup$ Now i am confused,you are saying one thing and @polynomial says the other regarding what we actually see as a laser beam. $\endgroup$
    – madness
    Commented Oct 31, 2022 at 11:52
  • $\begingroup$ Why exactly are you confused? Both I and @polynomial agree that what you see is light coming from dust etc in the path of the laser. $\endgroup$ Commented Oct 31, 2022 at 12:14
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When we see anything, it means light ... reaches our eye. Thus, what we are seeing is actually the object and not the light.

The word see is being used here with two different meanings. Meaning 1 - you see photons because these are what directly cause the chemical changes in the cells of your retina that lead to the perceptual sensation of sight. Meaning 2 - you see an object because that is the perception that the sensation of sight creates in your visual cortex.

Am I hearing pressure waves that hit my eardrum or am I hearing Bohemian Rhapsody ? Both are true.

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As always, I look at light as a stream of photons. Between their emission and their absorption, photons follow their geodesic, which is determined only by the gravitational potential along their motion. In our case, practically a straight line.

Your question about dust and laser beam

In a dust-free room, I see the emitted light of the laser only at the point of impact and only for a non-reflecting surface. Then the electrons reemit the light randomly in all directions out of the screen and some of the reemitted photons reach my eye.

A grain of dust that is in the path of the laser beam has no reflecting surface (not a mirror) and the part of the photons of the beam that hits it is diffusely reflected just as above and a small part of the photons reaches my eye. (Even for this, protective goggles may be required!)

So when light collides with the dust in air,shouldn't we then see the dust particles only? This is correct. A lot of dust let us see the beam, one particle only a bright dot in the air.

How are we then seeing the trajectory of light then since the light that is reflected to us isn't even travelling further anymore. Of the zillions of photons, some hit the dust and are reflected and others make it to the screen. Depending on the amount of dust in the air, perhaps at some point no light at all reaches the screen.

Here as shown in the picture, after colliding with the dust, the light ceases when it reaches our eye, so we should have only seen the dust, isn't it? Usually not. See the answers above.

Your question about electron and laser beam

If light hits an electron (like in compton effect), the electron moves away and if our eye was in such a position that we could see the reflected light, what would we see of the electron? When a single photon hits an electron of a gas molecule, most of the time nothing is seen because the re-emitted photon is almost never emitted in our direction. If this absorption/reemission occurs frequently, the same thing happens as observed with dust. The gas begins to glow for us.

Will we see the electron in the initial position and nothing else since the electron moved away and is not hit by light further? This question has too many ifs and buts to be answered here. Why don't you take it out of here and ask the question separately?

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