Hi, I'm a newbie in this forum, this is my first time. I'm hoping you'll be kindPlease Register or Log in to view the hidden image! This thread regards to intersecting two identical laser beam. As I was looking thru the net, researching for results of two beams (with same frequency) intersect & its outcome, I decided to post it instead. I have two identical laser beam sensor (both through-beam) that needs to be installed in a cross-manner. My definition for cross-manner is that both beam intersects @ a single point. As I was reading thru the Young's double slit experiment, I think the cross installation will create wave interference. I was just concerned that if both identical laser beam sensor collides at a certain point, one or both transmitted beam may become weak that it can no longer be detected by its respective receivers. I was trying to test it using two colliding beams of photoelectric, the result was one beam becomes weaker when collided with the other beam. I'm not sure with the laser, i think the latter has stronger beam than the former. What do you think? Will this kind of installation cause major trouble along the way?
Laser beams will cross without affecting each other at all. But don't try it with proton packs! Please Register or Log in to view the hidden image!
Why is that so? Laser beam is a wave & the Young double slit experiment is applicable to all waves (sound or light).
Yes, if a beam goes through a slit, it will diffract. No, if a beam goes through another beam, it will not diffract.
Two laser beams crossing each other will interfere in the region where they cross. However, outside that region they will continue propagating unaffected. Light beams cannot "collide" with one another.
True, the beams will not collide with each other. However, the total energy focused on the point of intersection will be double what either individual beam is. If you were to recombine the beams, and they were out of sync, then you would have regions where the amplitude of the beams would interfere or add with each other; but not more than double, or less than zero.
First, welcome. James would be referring spscifically to laser beams, or any photonic beams I think. Sound and water waves interact differently and you could expect some interference.
No, it's true of any type of wave*, including sound and water waves. They will interfere in regions where they cross but emerge unaffected after crossing. --- * Technical note: This applies to linear waves, which obey the principle of superposition. It is not true for non-linear waves such as solitons. Most familiar waves are more or less linear under most circumstances.
Thanx James, that was a bit technical for my poor braincell. Especially this time of night.. I've only got the one, so I keep it carefully pickled in cider..
Intersecting lasers I came here looking for, but did not end up finding (or maybe understanding) if an intersection between 2 lasers of the same frequency would be visible. For example, if I took two identical laser pointers, and crossed their beams, would I be able to see at least a faint increase in light where they cross?
Visible? No. You have to remember that, in clean air, laser beams are not visible until they hit a surface. Like people have been saying, the interference doesn't extend beyond the region where the beams overlap, so you wouldn't notice any effect of the intersection if you place a screen beyond the intersection point. If you place a screen at the intersection (or fill the room with light smoke, so the whole beam is visible), you'll see an interference pattern with alternating light and dark patches in the intersection area. But even though some patches will be lighter or darker than before, it's important to remember that the average brightness over the whole beam width has to be the same inside and outside the intersection. This is because of conservation of energy.
This (bold by me) above has confused me when reading explanations like this from different sources. But in the two slit experiment (sometimes illustrated using a water tank with a barrier with two slits), the waves from both slits spread out and where the troughs and crests interfere constructively the energy content only of those parts of the waves "adds together" to produce double-energy crests and waves on the "screen" detector. But in parts of the waves where crests and troughs do NOT "add" because they interfere destructively, the energy content of both waves at those parts "cancel out" to produce no-energy crests or waves on the screen detector. That is what confuses me. Where did the energy in those parts of the destructively interfering wave crests/troughs go? It didn't go to increase the constructive interference signals because those just added the two local parts of the waves (crest with crest, trough with trough) to give double, not more, than what was in those parts of the waves. We are pretty certain (aren't we?) that the destructively interfering parts both had "just energy", only now meeting directed in opposite directions at those "destructive interference" spots (no 'negative' versus 'positive' kinds of energy as such?)? But since we had energy and not positive versus negative energy, the energy content of both those parts should be somewhere unless that energy was not conserved? The energy that disappeared was not some positive and some negative energy, it was just energy which was just directed opposite. So where is the energy of both those destructively interfering parts "conserved" is what I am confused about? Can you help my naive understandings there please, Fednis48, or anybody else?
If memory serves, the intensity of light (which determines its brightness, as well as its energy content) goes as the square of the electric field amplitude. So when two fields interfere constructively to give double the original amplitude, you actually wind up with quadruple the original power. The incoming light can only supply two original beams' worth of energy, so the remaining two get pulled from the destructively interfering areas.
Actually, Fednis, with the screen at the intersection, the beams would look brighter because there are two of them combined. I think you are perhaps thinking of the "average intensity remaining the same" in the dual-slit experiment which only has a single photon source.
Good call. The total brightness (in units of luminosity \(\times\) area) would remain the same due to conservation of energy. But since two beams are squeezed into a single beam waist at the intersection, the average brightness would double.
This is the perfect opportunity to reference Mandel. You guys missed his original question, Which as we all know, independent sources of photons do not show coherence. Unless of course they are identical and are prepared specifically to show coherence. This requires splitting both beams and adding them at a second location where one reads the temporal coherence with a time resolved detector. The correlation function will have an interference term with two contributing factors, one temporal and one spatial. Like this exp{(k-k')*x + (v-v')*t} By post-selecting the observations that are "in phase" [where v=v' so that exp{(v-v')*t}=1] we eliminate the temporal term and these observations show interference. This post-selection amounts to selecting the temporal interference (an optical beat) where the beat frequency is at its lowest value. Voila, interference. A la Pfeegor-Mandel (1967,1968) or Magyar-Mandel (1963). These were the papers that prooved Dirac's famous quote "Each photon only interferes with itself."
