Light-beam propagation across impassable barrier?

In summary, although the intensity of the wave falls to zero at that point the wave is still there. Introducing the barrier will block or reflect the wave.
  • #1
Pelion
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Hi all,

(See attached image file)

Two mutually coherent and collimated light beams intersect as shown, creating the stationary 'bright' and 'dark' fringes of fig.A. Suppose that, after the fringe pattern has formed, we insert a very thin (compared to the fringe-width) and (ideally) perfectly reflecting foil, as shown in fig.B. It appears as if we can "cut the light beams in two, across an impassable barrier" yet the light beams persist and freely propagate!
Is this possible??
LightBeamsFringesBarrier.jpg
 
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  • #2
Pelion said:
Is this possible??
No. Although the intensity of the wave falls to zero at that point the wave is still there. Introducing the barrier will block or reflect the wave.

Edit: What you have drawn in the second picture is just reflection.
 
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  • #3
NFuller said:
No. Although the intensity of the wave falls to zero at that point the wave is still there. Introducing the barrier will block or reflect the wave.

Edit: What you have drawn in the second picture is just reflection.
what you say makes no physical or mathematical sense, as far as what we mean by 'waves'...sorry, no good. try something else
 
  • #4
Pelion said:
what you say makes no physical or mathematical sense, as far as what we mean by 'waves'...sorry, no good. try something else
It sounds like you don't understand what is meant by waves. If you don't think this makes mathematical sense, then why don't you solve the wave equation for this set up and show us what part of the math you don't understand.

This is labeled as an "I" level thread, meaning you have taken undergraduate coursework related to this topic. Griffiths EM book gives a good introduction to this and you should start there.
 
  • #5
NFuller said:
It sounds like you don't understand what is meant by waves. If you don't think this makes mathematical sense, then why don't you solve the wave equation for this set up and show us what part of the math you don't understand.

This is labeled as an "I" level thread, meaning you have taken undergraduate coursework related to this topic. Griffiths EM book gives a good introduction to this and you should start there.
The math is well known, and simple, because this is a common and well-known configuration...what is there to solve? Look it up anywhere, in minutes you will find the relations giving the fringe pattern depicted, that's not the big deal...the deal, so far, is that you haven't provided a physical or mathematical justification for why the beams won't continue to "persist and freely propagate"...that is the question posed here, not the writing down of the very well-known and simple math that describes the simple configuration.
 
  • #6
I also just realized the interference pattern in your drawing is not quite right. The fringes will appear both vertically and horizontally.
Interferences_plane_waves.jpg
 
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  • #7
NFuller said:
I also just realized the interference pattern in your drawing is not quite right. The fringes will appear both vertically and horizontally.
View attachment 209854
now your talking...if that is the case, then case closed; now you have given a physical reason why what I claimed can't happen.
 
  • #8
NFuller said:
I also just realized the interference pattern in your drawing is not quite right. The fringes will appear both vertically and horizontally.
View attachment 209854
...hmmm, you may have things in reverse, but I am not sure yet...you may be showing us a 'negative'...in other words, the dark areas are the bright fringes...in that case, the case continues
 
  • #9
NFuller said:
I also just realized the interference pattern in your drawing is not quite right. The fringes will appear both vertically and horizontally.
View attachment 209854
ContinuousDarkFringes.PNG
 
  • #10
Pelion said:
looks much more promising for my case...although pics above are for slightly non-collimated (diverging) beams, the dark fringes are 'continuous' across the plane of the proposed reflective foil
 
  • #11
Pelion said:
Two mutually coherent and collimated light beams intersect as shown, creating the stationary 'bright' and 'dark' fringes of fig.A. Suppose that, after the fringe pattern has formed, we insert a very thin (compared to the fringe-width) and (ideally) perfectly reflecting foil, as shown in fig.B. It appears as if we can "cut the light beams in two, across an impassable barrier" yet the light beams persist and freely propagate!
Is this possible??

Is what possible? You can certainly insert a thin reflective surface in the space where the two beams intersect, but beyond that I'm not sure what you're asking.
 
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  • #12
Drakkith said:
Is what possible? You can certainly insert a thin reflective surface in the space where the two beams intersect, but beyond that I'm not sure what you're asking.
Well, it certainly is counter-intuitive and seems paradoxical that two freely propagating waves can be completely 'cut' by a 'wall' and persist as traveling waves! If this is the case, there may be interesting applications, and the same effect can be created for photons in the quantum-optical realm. The amplitude doesn't matter here, so: Don't you find it strange that we can have trillions of photons streaming along, on one side of the barrier, then they "go somewhere else and in reside some other state" (in the region of stationary fringes), so that an impassable barrier becomes 'invisible' to them, and they emerge and continue to stream along on the other side! Talk about a 'tunneling effect,' if this is true!
 
  • #13
Pelion said:
Don't you find it strange that we can have trillions of photons streaming along...
A beam of light is not "trillions of photons streaming along". Photons are not at all like what people imagine when they hear the English-language word "particle", and light simply cannot be understood as a stream of particles.

Problems like this one are properly analyzed using only classical electrodynamics. Photons only come into the picture when quantum mechanical effects are involved (and then you have to use the full and daunting mathematical formalism of quantum electrodynamics) and there are no quantum mechanical effects in this problem.
 
  • #14
Nugatory said:
A beam of light is not "trillions of photons streaming along". Photons are not at all like what people imagine when they hear the English-language word "particle", and light simply cannot be understood as a stream of particles.

Problems like this one are properly analyzed using only classical electrodynamics. Photons only come into the picture when quantum mechanical effects are involved (and then you have to use the full and daunting mathematical formalism of quantum electrodynamics) and there are no quantum mechanical effects in this problem.
...So, in the end, you do then agree that we can 'cut' the 'classical' light beams in such a bizarre fashion? The 'effect', if true, will still be there for a single photon in a superposition of being in one collimated beam or the other...and also for a pair of quantum mechanical 'coherent states' where trillions of photons can be present and the photon-number description of those states is well-known and widely used.
 
  • #15
No photons will bass through the barrier because it is a BARRIER. Let's assume it is a good barrier with a high potential such that tunneling is out of the question. A poor barrier could be a partially reflective surface, which let's some light through anyway so no mystery there.

If you stick a reflective surface in the middle of the crossing beams, then the two light beams will just reflect off of the surface. If you put a nonreflective surface in the middle, then no light will be reflected. In either case, the light does not pass "through" the barrier.
 
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  • #16
NFuller said:
No photons will bass through the barrier because it is a BARRIER. If you stick a reflective surface in the middle, then the two light beams will just reflect off of the surface. If you put a nonreflective surface in the middle, then no light will be reflected. In either case, the light does not pass "through" the barrier.
OK, at least you have a definite opinion...but, again, you don't provide justification...and, by the way, there was an experiment called the "Afshar Experiment" where exactly this type of effect was seen (placing an array of wires in the 'dark' fringes of a double-slit pattern); all the light made it to the screen, avoiding the wires, as mathematically predicted, as if the wires weren't there. The Afshar setup had implications for wave-particle duality, but was not paradoxical in the sense that light had to disappear/re-appear across a wall, as is the case in my setup.
 
  • #17
Pelion said:
light had to disappear/re-appear across a wall, as is the case in my setup.
Again, why do you think this has to happen?
 
  • #18
NFuller said:
Again, why do you think this has to happen?
because the mathematical description predicts that there will be (approximately speaking) "no light" to be reflected or absorbed in the region of the stationary dark fringes...but the region of these fringes is finite and it is possible, geometrically, to cut off the two 'propagating' portions of the beams with little effect on their phase and intensity..."no light" means nothing for a 'barrier' to reflect or absorb, and the barrier is situated only in this "no light" region of the freely propagating beams.
 
  • #19
Pelion said:
because the mathematical description predicts that there will be (approximately speaking) "no light" to be reflected or absorbed in the region of the stationary dark fringes...but the region of these fringes is finite and it is possible, geometrically, to cut off the two 'propagating' portions of the beams with little effect on their phase and intensity..."no light" means nothing for a 'barrier' to reflect or absorb, and the barrier is situated only in this "no light" region of the freely propagating beams.
...and remember, the fringe pattern first has to be created, then the barrier is inserted, and the slightest phase change in either of the beams will rapidly destroy the effect and both beams will then be reflected by the barrier, as one would have if barrier was put there before fringe pattern was established.
 
  • #20
Pelion said:
...and remember, the fringe pattern first has to be created, then the barrier is inserted, and the slightest phase change in either of the beams will rapidly destroy the effect and both beams will then be reflected by the barrier, as one would have if barrier was put there before fringe pattern was established.
...also, the effect of a phase change, such that a 'bright' fringe is moved into the plane of the reflective foil, is irreversible: the beams will reflect on both sides of the foil as normally expected, and then we have to remove the foil, re-create the fringe pattern, and then re-insert the foil.
 
  • #21
Pelion said:
because the mathematical description
What mathematical description? You have shown us no math anywhere to support your claims.
Pelion said:
but the region of these fringes is finite and it is possible, geometrically, to cut off the two 'propagating' portions of the beams with little effect on their phase and intensity..."no light" means nothing for a 'barrier' to reflect or absorb, and the barrier is situated only in this "no light" region of the freely propagating beams
I doubt you realize that the perpendicular component of the magnetic field ##\mathbf{H}_{\perp}## and the parallel component of the electric field ##\mathbf{E}_{||}## go to zero anyways at the surface of a perfectly reflective surface. The boundary conditions that must be satisfied at the surface with unit normal ##\mathbf{n}## are
$$\mathbf{n\cdot D}=\sigma$$
where ##\sigma## is the induced surface charge desnity
$$\mathbf{n\times H}=\mathbf{K}$$
where ##\mathbf{K}## is the induced surface current and
$$(\mathbf{B-B^{\prime}})\cdot\mathbf{n}=0$$
$$(\mathbf{E-E^{\prime}})\times\mathbf{n}=0$$
at the surface.
Now that you have the needed boundary conditions, do the math, solve the wave equation, and see what actually happens. If you are unable or unwilling to solve the problem, then you will have to just take the our word on what happens here.
 
  • #22
Pelion said:
Don't you find it strange that we can have trillions of photons streaming along, on one side of the barrier, then they "go somewhere else and in reside some other state" (in the region of stationary fringes), so that an impassable barrier becomes 'invisible' to them, and they emerge and continue to stream along on the other side! Talk about a 'tunneling effect,' if this is true!

As others have explained, this is not what is happening. Classically, you'll just be inserting a reflective barrier in between two beams of light and they will no longer interfere with each other.

Pelion said:
because the mathematical description predicts that there will be (approximately speaking) "no light" to be reflected or absorbed in the region of the stationary dark fringes...

The model predicts that the interference of the two beams will produce a dark fringe. If they are no longer interfering with each other then this is no longer an accurate description.

Pelion said:
by the way, there was an experiment called the "Afshar Experiment" where exactly this type of effect was seen (placing an array of wires in the 'dark' fringes of a double-slit pattern); all the light made it to the screen, avoiding the wires, as mathematically predicted, as if the wires weren't there. The Afshar setup had implications for wave-particle duality, but was not paradoxical in the sense that light had to disappear/re-appear across a wall, as is the case in my setup.

This Afshar setup is not the same as your proposed setup here in the thread. In the Afshar experiment a grid of small wires is inserted into the nodes of an EM wave propagating perpendicular to the wires. But these wires do not block the rest of the possible routes for the wave to take. Your setup completely blocks the path of both light beams.

Perhaps more importantly, your drawing of the interference pattern of the light isn't even correct. NFuller already posted the correct picture, showing that there is a number of nodes, not a single plane-shaped node. This is true regardless of whether the nodes are the light or the dark areas as can clearly be seen.
 
  • #23
Pelion said:
you just wasted your time writing down this stuff, which has no relevance to the situation described:
I think that @NFuller has a good point. You mention a mathematical description, but didn't post any math.

It is not clear to me that the fields you describe do in fact satisfy the conductive boundary condition mentioned above. It also is not clear to me that the Poynting vector implies energy flux across the destructive interference region. You didn't explicitly calculate either of those so it is certainly not clear.
 
  • #24
Pelion said:
...hmmm, you may have things in reverse, but I am not sure yet...you may be showing us a 'negative'...in other words, the dark areas are the bright fringes...in that case, the case continues

Note that none of the nodes pass through the middle of the volume of intersection between the two beams. You'd have to offset the barrier vertically, destroying the symmetry of your original drawing.

Pelion said:
looks much more promising for my case...although pics above are for slightly non-collimated (diverging) beams, the dark fringes are 'continuous' across the plane of the proposed reflective foil

I'm fairly certain the dark regions in your linked picture are not nodes. You don't usually color the areas of constructive interference the same color as the background.
 

1. What is light-beam propagation across impassable barrier?

Light-beam propagation across impassable barrier refers to the phenomenon where a light beam is able to pass through a physical barrier, such as a wall or solid object, without any apparent opening or gap.

2. How is light able to pass through an impassable barrier?

Light is able to pass through an impassable barrier because of its wave-like nature. When a light beam encounters an obstacle, it diffracts and bends around it, allowing the beam to continue propagating on the other side of the barrier.

3. Is light-beam propagation across impassable barrier a common occurrence?

No, light-beam propagation across impassable barrier is not a common occurrence in our everyday lives. It is typically observed in controlled laboratory settings or under specific conditions, such as using laser beams.

4. What factors affect the ability of light to pass through an impassable barrier?

The ability of light to pass through an impassable barrier depends on various factors, including the wavelength and intensity of the light, the size and material of the barrier, and the angle at which the light beam approaches the barrier.

5. What are the potential applications of light-beam propagation across impassable barrier?

The ability of light to pass through an impassable barrier has potential applications in fields such as telecommunications, remote sensing, and medical imaging. It can also be used in the development of new technologies, such as cloaking devices.

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