Light and electromangetic interference

[jacc123456]

Hello everybody.

I´m new in the forum, and i would like to ask a question.
If light is an electromagnetic wave, why is it not affected by electromagnetic interference.
For example, in electric engineering is used fiber optics cable when one wants to avoid EMI.
Why? I'm looking for a plain physics explanation.
I thank you all in advance for your kind attention.

Best regards.

 

[AJ Bentley]

Electromagnetic waves are not 'affected by electromagnetic interference' (not in the sense you mean).

Electromagnetic interference is a term used to describe a problem with equipment that is intended to listen to one source of electromagnetic waves being subject to a different source.

Much as you might have difficulty hearing music if someone is using a hammer-drill in the next room.

Because there are so many radio stations all crowded together, each trying to use a narrow part of the spectrum, radio interference is common. Same with TV and eventually cell-phones will begin to suffer as they are used more and more.

A fibre-optic cable on the other hand is pretty well a direct connection, nothing else can get in.

 

[jacc123456]

I appreciate your help, but u didnt quite anwser my question.
Im looking for a physics explanation for light not being affected by electromagnetic waves.

 

[AJ Bentley]

What makes you think that any other sort of electromagnetic wave is 'affected' by electromagnetic waves?

 

[Naty1]

Light is affected by EMI...check out the double slit experiment,,,,

and here:
http://en.wikipedia.org/wiki/Optical#Superposition_and_interference


Fiber optic cable light is an electromagnetic wave; As Bently says, most light can't get in from the outside, so its pretty immune to interference, but I don't think COMPLETELY.

In the simplist of terms, fiber optic cables are typically "immune" from EMI because much of it can't get into the cable. I think because fiber optic cables are typically clad which reflects and retains the light signal, external EMI is suppressed and blocked by opaque protective coatings. EMI generates heat so in theory could be used to burn a cable...that's not "immunity" to me...


Wikipedia does say fiber optic signals are "immune" to electromagnetic interference"...here...
I think that's a practical rather than a theoretical statement...

http://en.wikipedia.org/wiki/Optical_fiber

Would the EMI pulse from a nuclear explosion affect a fiber optic signal?? I sure think so.

You can read a little about that here:
http://en.wikipedia.org/wiki/Photonic_integrated_circuit#Advantages_of_photonic_circuits


In addition, noise is a form of EMI, an internal rather than external source, but still a consideration in fiber optic systems.

 

[AJ Bentley]

Wikipedia is correct to say 'immune'.
EMP cannot harm fibre-optic cable. EMP relies on inducing massive currents in electronic circuits, thereby overloading and damaging the wiring and devices. That cannot happen with fibre-optic (plastic or glass) cable because it's an insulator. (It can fry the components at the ends of the cable though)

Interference in the case of double slits is NOT the same thing as 'interference' in common usage (which is what the OP is referring to). It refers to the special case of a photon interfering with itself in the quantum mechanical sense.

In general, electromagnetic waves do not interfere with each other. A radio beam will pass through another unaffected, in the same way as two beams of light will do.

 

[Dale]

What AJ Bently is describing is called the principle of superposition. It is a direct result of the linearity of Maxwells equations.

 

[jacc123456]

Good afternoon.

I thank you both for your help, but i rather read an anwser from someone who really knows what is talking about. I'm not trying to set a discussion, i just want a plain explanation, which I'm pretty sure it exists.
Maybe i wasn't too clear, so i will try to explain for further opinions.
Lets consider this example. Two cables. One feeds a motor, the other is a bus for communication. If both were unshielded, the first would cause electromagnetic interference on the second. Why? Simply, because it would create an electromagnetic field too powerful to be "ignored" by the small amount of energy flowing on the bus cable, causing disruption of the signal.
Now, another example. The same power cable, and an optic fibre cable. This doesn't case interference. Why? As light is considered an electromagnetic wave, why is it not affected by electromagnetic interference?

 

[jtbell]

Lets consider this example. Two cables. One feeds a motor, the other is a bus for communication. If both were unshielded, the first would cause electromagnetic interference on the second. Why? Simply, because it would create an electromagnetic field too powerful to be "ignored" by the small amount of energy flowing on the bus cable, causing disruption of the signal.

In this case, the signal carried by the second cable depends on the motion of the conduction electrons in the conductor. The electromagnetic fields produced by the first cable affect the motion of the conduction electrons in the second cable, and thereby the signal carried by it.

Now, another example. The same power cable, and an optic fibre cable. This doesn't case interference. Why?

A fiber-optic cable is not a conductor. There are no conduction electrons to be affected by external electromagnetic fields. The mechanism of propagation of light through a transparent medium is different from electrical signals through a conductor.

 

[jacc123456]

In this case, the signal carried by the second cable depends on the motion of the conduction electrons in the conductor. The electromagnetic fields produced by the first cable affect the motion of the conduction electrons in the second cable, and thereby the signal carried by it.



A fiber-optic cable is not a conductor. There are no conduction electrons to be affected by external electromagnetic fields. The mechanism of propagation of light through a transparent medium is different from electrical signals through a conductor.

Thanks. This kind of makes sense, but then again, isn´t light considered an electromanetic wave? And how does light propagate in a transparent medium?

 

[NanakiXIII]

I think this may be the simple explanation you're looking for. It's not as complete as some of the other answers given, but I think those might be a little beyond what you're asking.

A signal carried electrically depends on electrons moving through a wire. An electromagnetic field can exert forces on electrons, so that an external electromagnetic signal can scramble your electrons and mess with your signals.

Light is not subject to those forces. Light does not interact with light. If you take two flash lights and shine them such that the beams of light cross each other, you see the same two spots of light on the wall as you would if you did not cross the beams.

 

[jacc123456]

Thank you nanaki, but it's not quite there yet. I need a little more complexity.
I'm an electrical engineer, so i can handle a little complexity. Not too much, because my area is not physics.
Tell me if I'm right (anyone). Again, i ask for anwsers of those who really understand about this issue.
In the example i gave about the two copper cables (power and communications), there is interference because electromagnetic fields interfere with electrons (i can easily understand this part cause it's my area).
In the second example, using a power cable and an optic fibre cable, there is no interference cause light is an electromagnetic wave, and two electromagnetic waves don't interfere with each other. Is this right? And why is this (why don't they interfere with each other?) ?
Thank you all for helping me in this.

 

[Naty1]

jacc: Apparently you did not read the Wikipedia references I posted?. Or maybe we have a different view of "interference" as Bently suggests.

Here is another insight: Wikiepdia:
"Fiber lasers have a fundamental limit in that the intensity of the light in the fiber cannot be so high that optical nonlinearities induced by the local electric field strength can become dominant and prevent laser operation and/or lead to the material destruction of the fiber. This effect is called photodarkening."

http://en.wikipedia.org/wiki/Laser_light

But this mechanism may be different than you had in mind...so...


As an EE, you likely understand modulation,right?? contrast your example with Bently's comment in post #2:

Because there are so many radio stations all crowded together, each trying to use a narrow part of the spectrum, radio interference is common.

and you will have your answer. (hint: consider power AND frequency)

 

[jacc123456]

It's already hard to understand physics, but trying to understand it in english, and with "mere opinions" is even harder. I posted a clear question, supported by a more than clear example, and none of you has given a clear anwser (although i appreciate everyone's help).
Once again...

"" In the example i gave about the two copper cables (power and communications), there is interference because electromagnetic fields interfere with electrons (i can easily understand this part cause it's my area).
In the second example, using a power cable and an optic fibre cable, there is no interference cause light is an electromagnetic wave, and two electromagnetic waves don't interfere with each other. Is this right? And why is this (why don't they interfere with each other?) ? ""

 

[Naty1]

You don't understand physics yet asked for an explanation in physics??

I virtually gave you an answer, in terms of modulation, that most any EE should be able to understand.

What don't you understand about my last post?

 

[NanakiXIII]

What you write is correct. If you want to know why light doesn't interfere with light, you'd probably have to do Quantum Electrodynamics. I'm not familiar enough with QED to explain it to you, but maybe someone else can give you a simple explanation.

A less complete simple explanation might be to simply consider the fact that you are dealing with waves. Make the analogy with any other familiar type of waves, like waves in water. If you have two sources of surface ripples in a body of water, and the waves cross each other, they will combine to make complex patterns as they cross paths, but when they move away again, the waves look exactly the same as they would have if they had not crossed the waves from the other source.

Note that if someone was measuring these waves at the point where they cross each other, he would have a problem, because the waves are interfering (they add up) in the sense that Naty means. But if you're measuring the waves after they cross each other, you're fine. When dealing with fiber optics, you'll generally not have this problem. The electric field generated by a nearby wire will probably (don't quote me on this) be a lot weaker than the laser light and of a different frequency which the photodetector at the end of the fibre may not even be sensitive to. Moreover, optic fibres are designed to keep the signal in and keep other signals out.

 

[jacc123456]

So they don't interfere (cause loss of signal) with each other because they have different frequencies?

 

[NanakiXIII]

No, let me try again. You need to distinguish two types of interference. One is what electromagnetic fields do to electrons: they disrupt your signal because they mess with the electrons in your wire. This type of interference does not happen between light and light: if you have an optic fiber and you place it in an electromagnetic field, the laser light inside the fiber will simply travel through the field and come out undisturbed.

The other type is what two light signals do if they cross each other: they are waves, so they add up if they enter the same region of space. If you happen to be measuring light waves at that point in space, then you will be measuring both signals.

Fundamentally, this has nothing to do with frequencies. Electromagnetic waves of any frequency do what I just described: they do not interact (they pass each other without disturbing), but they do interfere (they add up if they are in the same region of space). However, if you have a detector at the end of an optic fiber, it will not be sensitive to all frequencies, most likely. It will be sensitive to the frequency of the laser light, because that is what it is designed for. If the electromagnetic field from a nearby wire has a very different frequency, your detector may not be able to detect it. In that case, there can be no interference (in your measured signal, that is, of course physically, there can be). For example, red light has a frequency of like 10^14 Hz. If your wire is running standard AC of 50/60Hz, that's a big difference and chances are your photodetector cannot see that. But, again, there is nothing fundamental about this, just a practical remark.

 

[jtbell]

The other type is what two light signals do if they cross each other: they are waves, so they add up if they enter the same region of space. If you happen to be measuring light waves at that point in space, then you will be measuring both signals.

And if you measure the light waves after they have intersected and then separated from each other again, you observe no effect on either wave, compared to the waves before the intersection.

As an example, aim two laser beams at right angles to each other, intersecting at a single location. Turn one beam on and off. There is no change in the other beam, past the intersection point.

(At least to a very very very good approximation. It's theoretically possible for photons to interact with each other by producing intermediate virtual electron-positron pairs via http://en.wikipedia.org/wiki/Delbruck_scattering" , but this is such an extremely tiny effect that it has never been observed experimentally.)

 

[jacc123456]

Fundamentally, this has nothing to do with frequencies. Electromagnetic waves of any frequency do what I just described: they do not interact (they pass each other without disturbing), but they do interfere (they add up if they are in the same region of space).

This doesn't make sense. You say any frequency electromagnetic waves pass over each other without disturbing, but they add up.
Ok, maybe it even makes sense, it's just me who's not being able to get the message through.
I know that if there's two radio frequencies (for example) and they cross each other, they don't change their frequency, but they can "mess up" the signal, as they may add or subtract their signal. This can be "fixed" at reception point by using a simple filter (removing the unwanted frequency). I know all this.
But I'm just considering the signal point of view. And if light is an electromagnetic wave, composed by an electric and a magnetic field, could (and should) it not disrupt another signal (carried by electrons) ?

 

[Naty1]

Two completely different questions:

Post # 1

If light is an electromagnetic wave, why is it not affected by electromagnetic interference.

[In general as I think Bently posted, electromagnetic waves in free space do not interfere with each other...nobody knows "why" not. Yet radio transmissions do "interfere" and that's why the FCC assigns broadcast frequencies to specific users...radio and TV and microwave for example.]

Post 20:

And if light is an electromagnetic wave, composed by an electric and a magnetic field, could (and should) it not disrupt another signal (carried by electrons) ?

yes, of course it does...think sunspots as an example which affect microwave radio transmissions for example. Or a powerful electronic magnetic pulse as from a nuclear or "E" bomb.


As Bently suggested, electromagnetic waves ( or photons) if you prefer, do not interfere with each other in free space..that's why Maxwell's equations do not require a tensor forumlation; On the other hand gravitons DO interact, so general relativity requires a tensor formulation. In one of my posts I gave a reference indicating the potential for electromagnetic interference with an optical signal due to the introduction of non linearities. I also believe electromagnetic waves of sufficient strength, such as a nuclear EMP can dislodoge ions in optical fiber material and that can also disrupt optical signals, probably via induced noise.

 

[NanakiXIII]

This doesn't make sense. You say any frequency electromagnetic waves pass over each other without disturbing, but they add up.
Ok, maybe it even makes sense, it's just me who's not being able to get the message through.
I know that if there's two radio frequencies (for example) and they cross each other, they don't change their frequency, but they can "mess up" the signal, as they may add or subtract their signal. This can be "fixed" at reception point by using a simple filter (removing the unwanted frequency). I know all this.
But I'm just considering the signal point of view. And if light is an electromagnetic wave, composed by an electric and a magnetic field, could (and should) it not disrupt another signal (carried by electrons) ?

I'm not really sure what your remaining question is at this point. As for your new question, yes, light can interact with electrons. You don't really need to worry about light from a lamp messing with a signal going through a wire, though. It will probably just be reflected or absorbed at the surface of the wire and not penetrate it. I'm not very familiar with the details of these interactions, though, so if you want to know more (I'm not sure this is what you were asking) maybe someone else can tell you something about it.

 

[jacc123456]

Hi.
I still have questions because one says one thing, and then comes another saying another thing. I will try again to explain my doubt.
Imagine a light signal traveling through a fibre optic cable. It goes as happy as he can be. Then comes a nasty power cable next to him. This cable is strong and mean.
And as it is controlling a motor, it has lots of electromagnetic waves (radiation) coming out from him (he's unshielded). Now that we have a cenario, the question is: could the electromagnetic waves (radiation, interference, or whatever you want to call it) disrupt the optical signal? Forget the light interfering with the electrons. It's the opposite here. Not hte electrons, but the electromagnetic "something" causing problems on the light signa. AND...if electromagnetic waves can "add up" to each other, and if the electromagnetic waves (radiation, interference, or whatever you want to call it) from the power cable were to be too strong, shouldn't they "add up" with the light signal, causing disruption (problems) of that same signal?
The problem here is that you come up with all the theory (and I'm not saying it's not right) but don't anwser the questions i pose.
I apologize for my english. I know sometimes i have trouble expressing myself correctly (i'm portuguese). Thank you all for your pantience.

 

[NanakiXIII]

Yes, the waves from the power cable will add up with the light. However, if the fiber is just passing through, the light will come out undisturbed once it is out of range of the power cable. If you are trying to measure the light signal near the power cable, you may be picking up both signals. However, as I pointed out, the detector likely won't be sensitive to the waves coming from the power cable.

By the way, people are not disagreeing with each other, as you seem to think. They were mostly just pointing to different phenomena, which is where some confusion arose, I suspect. I tried to clarify and summarize this in post #18.

Also, I'm not sure what it is you want to know if not "all the theory". What kind of answer were you looking for?

 

[jacc123456]

Ok, i think i got it now.

""For example, red light has a frequency of like 10^14 Hz. If your wire is running standard AC of 50/60Hz, that's a big difference and chances are your photodetector cannot see that. But, again, there is nothing fundamental about this, just a practical remark.""

What did you meant by this (nothing fundamental)? This is correct theory right?

Naty also seems to have a different idea:

Post 20:
""And if light is an electromagnetic wave, composed by an electric and a magnetic field, could (and should) it not disrupt another signal (carried by electrons) ?

yes, of course it does...think sunspots as an example which affect microwave radio transmissions for example. Or a powerful electronic magnetic pulse as from a nuclear or "E" bomb.""

 

[NanakiXIII]

I meant it is not fundamental because it has nothing to do with the electromagnetic waves themselves, but with the properties of the detector. I wanted to emphasize that the waves themselves do add up, their not being of the same frequency does not matter. It is only because the detector is not designed to detect the low-frequency signal that you don't have a problem seeing your light signal.

Naty was talking about electric signals being disrupted, I assume, because that is what you asked about: "(carried by electrons)".

 

[jacc123456]

Naty was talking about electric signals being disrupted, I assume, because that is what you asked about: "(carried by electrons)".

I don´t think so, because he mentioned sunspots (whatever that is) affecting microwave radio transmissions. Maybe, he meant that the sunspots disturb the receiver, and not the electromagnetic waves themselves. Do you think that was it?

 

[NanakiXIII]

That would be my guess, yes. Of course, he might clarify better himself.

 

[jacc123456]

Ok, thanks nanaki, and all of you who helped me.