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EM Wave Interference

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posted on Jan, 5 2015 @ 02:06 AM
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Hi Ya'll,
A Happy New Year to all.
Let's take for example em waves in the optical spectrum. A mirror reflects all waves incident upon it and we can view the objects as they are.
Yet all these waves are crisscrossing on the way to the mirror and also away from it after reflection. So why do these waves not cause interference patterns?
Just wondering what you all think about this phenomenon?




posted on Jan, 5 2015 @ 02:14 AM
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a reply to: Nochzwei

Ah a good morning physics riddle. I suck at it, so just an idea: Maybe because electrons have the same negative charge and therefor don't collide, so it is a zip division juncture thingy...



posted on Jan, 5 2015 @ 04:09 AM
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"The two waves must have the same polarization to give rise to interference fringes since it is not possible for waves of different polarizations to cancel one another out or add together."
There's a decent wiki on interference and wave propagation here... en.m.wikipedia.org...(wave_propagation) scrolling to the optical interference tab should give you your answer I think.
edit on 5-1-2015 by TopCat1 because: correction



posted on Jan, 5 2015 @ 04:09 AM
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originally posted by: Peeple
a reply to: Nochzwei

Ah a good morning physics riddle. I suck at it, so just an idea: Maybe because electrons have the same negative charge and therefor don't collide, so it is a zip division juncture thingy...
Lol, nice avatar. There are no electrons there to collide, only photons or wave packets.



posted on Jan, 5 2015 @ 04:14 AM
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a reply to: Nochzwei

You are on exactly the right track. All these crisscrossing light waves do create interference patterns all the time!

Why do we not see the interference patterns, then?

First of all, the interference patterns you see in physics demonstrations are made so that you only need to see one interference pattern from one single source at a time. In the everyday wordl most of the time you will see a scene that is illuminated by an uncountable number of reflections, diffusions, and separate light sources. All of these interference patterns overlap randomly, and they average out to a medium brightness of the scene. Only in a situation where you can cancel some of the light you can see the interference pattern on the wall, the floor, on your screen.

Secondly, where do you expect to see these interference patterns? The way you wrote your OP feels like you expect them to float in the clear air. But how can a pattern of light and shadow float in the air if there is nothing there that will scatter the light to your eyes? The situation is just the same as with any shadow: you do not see the shadow projecting through the air unless there is a lot of dust floating in the air. The shadow only becomes visible when it hits a diffusely scattering surface.

This is how I see it. I am a radar engineer and former instructor in microwave systems.



posted on Jan, 5 2015 @ 04:22 AM
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I guess it may help you to understand that light acts bot as particles and waves both at he same time but only one event can be seen at a time/per observer.

www.youtube.com...



posted on Jan, 5 2015 @ 04:27 AM
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a reply to: Nochzwei

Yeah drum roll please. Well, I said I suck and I proved it...



posted on Jan, 5 2015 @ 04:38 AM
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a reply to: Nochzwei

I'm not sure of your single mirror setup but here is what I know of interferometer that might be of interest for this thread (I'm building a Michelson interferometer these days).

The shortest summary is: Light must be interfered upon itself within it's coherence length. If your using laser light, getting an interference pattern is not very difficult due to the often long length of coherence. However if using white light, the coherence length is minuscule. It mean each "path" of light used must be rigourously the same length. For a white light Michelson it can be difficult. Remember that used mirror must be very flat and correctly aligned.

If you still want to easily observe white light interference, you can put together thin glass piece like microscope coverslide, or use the reflection from one of those used like an optical flat. But the easiest method is to put a thin film over a liquid; try a small drop of oil on water.



edit on 2015-1-5 by PeterMcFly because: (no reason given)



posted on Jan, 5 2015 @ 04:45 AM
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originally posted by: [post=18838207]Pirvonen Secondly, where do you expect to see these interference patterns?

In the mirror of course. But your reply does answer that anyway. Thanks yr very pertinent reply



posted on Jan, 5 2015 @ 04:52 AM
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a reply to: Nochzwei

Your mirror would need to produce double reflection, like a flat glass piece. But then again, both surface flatness need to be extremely flat and the thickness of the glass be within coherence length of light used; white light = extremely thin.



posted on Jan, 5 2015 @ 07:23 AM
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Thanks all for your replies


originally posted by: PeterMcFly
a reply to: Nochzwei

Your mirror would need to produce double reflection, like a flat glass piece. But then again, both surface flatness need to be extremely flat and the thickness of the glass be within coherence length of light used; white light = extremely thin.
Ah ok. but, i would think there should be multiple images reflected



posted on Jan, 10 2015 @ 09:15 AM
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If you ever wear polarized film sunglasses, you'll start seeing interference patterns in a lot of places. One of the obvious (and sometimes annoying ones) is the glass on some car windows. Seems like other conditions have to be met first as well. (Polarization in the car window glass?)



posted on Jan, 10 2015 @ 11:50 AM
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The answer is because the EM field is connected to itself.



posted on Jan, 10 2015 @ 01:37 PM
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originally posted by: ImaFungi
The answer is because the EM field is connected to itself.
EM field is an entity on its own, so how can it be connected to itself.
Its like saying I am connected to myself, how can that be.



posted on Jan, 10 2015 @ 03:23 PM
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originally posted by: Nochzwei

originally posted by: ImaFungi
The answer is because the EM field is connected to itself.
EM field is an entity on its own, so how can it be connected to itself.
Its like saying I am connected to myself, how can that be.


Well, I dont really know the answer, but I thought I felt it out, but now I am left wondering in the same guise as the OP in this sense; Think about standing in front of a mirror, and perhaps there is a light overhead; Electrons are not being shot out of the bulb, light is created by vibrating the electrons, the EM field already exists everywhere, the electrons being shaken are just shaking the EM field, like when a bunch of people grab the ends of a big flag and wave it up and down. So light, EM radiation, is the electrons shaking the field, space, between the bulb and the mirror, which when the electrons in the mirror are disturbed by the Field they are coupled to being shaken, they start shaking too,...Well, I suppose first the light from overhead, bounces off the electrons in your face, and then they transmit the molecular 3d expression of your face through space, the details and information of which are translated now into the details in EM wave, which then hit the mirror, and due to the nature of this particle matter, molecular configuration, the material of the mirror, naturally, transmits the signal back away back through space to your eyes.

Where I understand the OPs concern, though it is simply said in physics that light, em radiation, cannot interact with itself, and it is self referentially explained, coming down to an answer of 'because', because its classified as a certain particle which infinite numbers of them can exist in the exact same space. But there must be a physical reason as to why and how this is possible.

Anyway. I guess the answer might have to do something about the tremendous speed of light, and the extremely relatively close proximity in which these situations are taking place, that of a person standing right in front of their mirror. If they hold an object in front of their face, then we can see the sort of disruption that takes place, the light can no longer bounce off their face and hit the mirror....in their frame of reference...but it is possible someone looking at a different angle can see their face. interesting and confusing topic, wish I knew why exactly and really light cannot collide with itself, what that physically means.

If light is reflecting off your face to the mirror, and off the mirror to your face, in a very near continuous stream at the same time, why do they not cancel out? there is only so much surface area of the mirror and your face, and a small bulb with an amount of shaking electrons 'lights up the whole room', which requires light to reflect off of every item that you are able to see, and go back into your eye, seemingly continuously.

Maybe it has something to do with the roundedness of eye balls...

Something I wondered about once, is how at night a relatively strong flashlight pointed into the sky, you can see the beam of the light. That means the electrons are shaking in the light, the EM field is being disturbed upwards and away, and at each point of that beam you are seeing, the EM field is disturbing the air molecules there, which happen to enough times, reflect the disturbance in the field, your way.. now is it really reflecting it in all ways, yes, I must suppose, as people standing all around the flashlight holder can see the beam.

But that means, that the true nature of the EM field shaking, is that it is not only shaking in that beam, it is shaking all around the beam, because it has to be, in order to get to your eye, your eye doesnt leave your face and touch the beam, the wave leaves that area, and comes to you, and is going all around. So that is the lessen that, the perhaps obvious lesson that, one can only see what enters their eye, though this doesnt mean that the means in which one uses to see, 'shaking em field', is not occurring elsewhere.. I suppose this has to do with the planes and polarities of EM radiation, which I have tried to question the ATS Physics knowers about, to no avail.



posted on Jan, 10 2015 @ 04:25 PM
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a reply to: Nochzwei because in wave form the electrical field component and magnetic field component oscillate over each other. it is a dual natured entity. the two fields tumble over each other and push pull each other along or push off each other; and as a particle (photon)well why can a baseball travel?



posted on Jan, 10 2015 @ 04:33 PM
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I'm a little tired but light interacting with itself after bouncing off a mirror while being observed(double slit) would be a particle and not have a problem of being in the same place at the same time. Knowing that and somehow taking away the observer then the em on the mirror might, IDK, interact but trillions of a second delayed. Knowing that if you had a mirror that sped up trillions of a second(they can slow down liht locally, can they speed it up locally?) as an em wave then yes they would interphere. I am not making sense so I'm off



posted on Jan, 10 2015 @ 11:42 PM
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a reply to: ImaFungiHi Fungi, the op was actually a ques posed in my high school exam way back and I couldn't do full justice to it.
But though em wave has both attributes B And H, they really don't interact or push pull on each other and as for putting the electrons of the mirror in motion, im not so sure.



posted on Jan, 11 2015 @ 01:13 AM
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originally posted by: Nochzwei
and as for putting the electrons of the mirror in motion, im not so sure.




The mirror is a large percentage composed of electrons, I assumed, I assumed the electrons dont absorb the light, they reflect it, thus, a mirror is a mirror.



posted on Jan, 11 2015 @ 08:25 AM
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originally posted by: ImaFungi

originally posted by: Nochzwei
and as for putting the electrons of the mirror in motion, im not so sure.




The mirror is a large percentage composed of electrons, I assumed, I assumed the electrons dont absorb the light, they reflect it, thus, a mirror is a mirror.
Oops Lol. Aber mach nichts.
But technically you are right, electrons are always in motion and exist as electron cloud about the nucleus. However this motion changing with an incident ray of light is possible depending on energy of the incident light ray.



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