Help on Standing Waves, related to the Silvertooth experiment

[lidal]

 

Does the value of the voltage stays the same?

The amplitude of the voltage will stay the same. But the phase of the voltage will change, which is not important because an RF detector placed at that point will only measure the amplitude.

[John Cuthber]

 

 

So explain the observation.

Actually, I think that's your responsibility.

Why does your la=er behave differently from all the lasers I have seen?

Of course, maybe it's my observation that's peculiar.

Does anyone else see the phenomenon that Studiot is describing?

The amplitude of the voltage will stay the same. But the phase of the voltage will change, which is not important because an RF detector placed at that point will only measure the amplitude.

Phase with respect to what?

If it's measured WRT the original source then that explains why the difference to the source makes a difference.

 

(having said that, you are wrong about this " an RF detector placed at that point will only measure the amplitude" .

An oscilloscope, for example, could show the phase relationship.

[studiot]

 

John Cuthber

Why does your la=er behave differently from all the lasers I have seen?

 

 

 

How many construction lasers have you seen, used and experimented with?

What do you know about the focusing optics of these devices?

 

The one I am referring to was bought for the UK Department of Transport from the Swedish Geodimeter company.

It was the first one of its type in the UK.

The variation was an oddity I noticed at the time and showed it to several colleague Charterd Engineers.

As it caused no difficulty with the job in hand we did not take it further at the time.

 

This actually was not the most dramatic problem with alignment lasers I have come across in my career.

The most dramatic was probably the occurrence in steps in the line due to double refraction in a pipe laying trench.

Edited September 6, 2015 by studiot

[John Cuthber]

OK, so what you did was describe some odd sort of laser as "standard".

That's not going to get us very far in terms of answering the original question.

[studiot]

OK, so what you did was describe some odd sort of laser as "standard".

That's not going to get us very far in terms of answering the original question.

 

Skeptical is one thing, but if all you want to do is sneer this discussion is a waste of time.

[John Cuthber]

Derailing it by adding stuff about the odd behaviour of some obscure bit of kit is more of a waste of time.

However, if we keep on about that it's just going to upset the mods.

Should we return to discussion of the OP's claims?

 

 

If you feel that your comment

"You can see for yourself what happens with a standard construction laser.

Point the laser in some safe but convenient direction.

Move a target along in the beam.

The laser line will impact the target as a small disk of light.

This diameter of this disk waxes and wanes as you move the target along in the beam.

When using such as laser for levelling or alignment purposes you need to take this phenomenon into account."

was helpful in term of moving the discussion along, please PM me to explain why.

[swansont]

The amplitude of the voltage will stay the same. But the phase of the voltage will change, which is not important because an RF detector placed at that point will only measure the amplitude.

 

How does the phase change, but the value stay the same?

[lidal]

Actually, I think that's your responsibility.

Why does your la=er behave differently from all the lasers I have seen?

Of course, maybe it's my observation that's peculiar.

Does anyone else see the phenomenon that Studiot is describing?

Phase with respect to what?

If it's measured WRT the original source then that explains why the difference to the source makes a difference.

 

(having said that, you are wrong about this " an RF detector placed at that point will only measure the amplitude" .

An oscilloscope, for example, could show the phase relationship.

 

What changes is phase with respect to the signal source. You can have two systems, with two signal sources with phases synchronized ( and same frequency, of course) and different cable lengths. Then you can see the two voltages on the same oscilloscope. Their phases will be different.

 

But the phase difference does not make any difference for the experiment I mentioned in my opening post.

[John Cuthber]

 

What changes is phase with respect to the signal source. You can have two systems, with two signal sources with phases synchronized ( and same frequency, of course) and different cable lengths. Then you can see the two voltages on the same oscilloscope. Their phases will be different.

 

But the phase difference does not make any difference for the experiment I mentioned in my opening post.

In the opening post you said

"But assume that the mirror and the standing wave sensor (SWD) are mounted on a common linear stage that can be moved relative to the laser source, so that the distance between the SWD and the mirror is always constant as the stage is moved relative to the laser. The question is: will the voltage at the SWD vary as the stage is moved relative to the laser, along the path of the light beam? The obvious answer is: NO. "

And now you accept that with the very similar system of a shorted cable the phase (measured at some set distance from the short circuited end) changes when you vary the length of the feed cable because that phase it is measured wrt the source and obviously, if the distance to the source changes, then the phase will change. And if two signals are in phase they will add up to something that's not zero, while if they are (exactly) out of phase they will add up to zero, so the sum depends on the phase so the voltage measured at some point not at the short the voltage will depend on the phase and so it will depend on the length of the cable to the generator.

 

Those two statements of yours can't both be right.

[lidal]

Consider again a laser source S, detector (SWD) and mirror. The laser beam is pointed towards the mirror at ninety degrees. The detector is placed between the mirror and the source, at distance L from the mirror and at distance D from the laser. We can remove the mirror and put a mirror image S' of the laser source. Now the detector is between S and S', at distance D from the source S and at distance D + 2 L from S'. We can consider S' as a physical source, 180 degrees out of phase with S.

 

For example, if L= N λ/2, where N is an integer, the amplitude of the wave will be maximum at the detector because the waves from S and S' will interfere completely constructively at the detector.

 

Now if the laser source S is moved back by distance ΔD away from the detector, S' will also move away from the detector by the same amount ΔD. The phase of the (incident) wave at the detector due to S will be delayed because S is moved away from the detector. But the phase of the (reflected) wave due to S' will also be delayed by the same amount. Hence the phase of the resultant (standing) wave will be delayed but the phase relationship of the incident and reflected waves at the detector will not change, because both are delayed by the same amount. The amplitude of the wave at the detector is determined by the phase relationship.

 

Hence the phase of the standing wave will change, but its amplitude will not change.

 

Changing the laser distance will result in change of absolute phase of the standing wave. But the amplitude of the standing wave will not change because the position of the detector with respect to the standing wave pattern hasn’t changed. The position of the detector changes relative to the standing wave pattern if the distance between the detector and the mirror is changed.

 

Regarding dithering the mirror:

 

Suppose that the detector is at a position between S and S’ such that the waves from S and S’ arrive in phase at the detector and hence maximum amplitude. (This occurs when L = N λ/2 . ) . Now if the mirror starts moving slowly towards the detector, which is equivalent to S’ moving towards the detector, the voltage at the detector starts decreasing because the waves will become out of phase, and the phase difference continues to increase as the mirror continues moving towards the detector ( we consider small movements less than half wavelength).

 

Next consider the case of the detector placed between S and S’ at a point of complete destructive interference. This occurs when L = Nλ . The voltage detected will be zero. Now if the mirror is moved slowly towards the detector, the voltage at the detector starts increasing because the two waves now start to be more and more ‘in phase’ as the mirror moves.

 

In the first case, the DC voltage from the detector starts decreasing and in the second case it starts increasing, for the same motion of the mirror. This means that if we have two systems, the detector placed at a point of complete constructive interference in one system and the detector placed at a point of complete destructive interference in the other system, and display the two voltages on the same oscilloscope, they will be out of phase.

 

My problem is that how/why change in phase of the SWD creates a change in phase of the AC voltage detected at the SWD, as the mirror is dithered. There is a helpful article by the same author, but I couldn't understand it too. A glass is placed infront of the laser, which will be equivalent to moving the laser away from the mirror.

http://www.conspiracyoflight.com/Onewayonelaser/Standingwave.html

In the opening post you said

"But assume that the mirror and the standing wave sensor (SWD) are mounted on a common linear stage that can be moved relative to the laser source, so that the distance between the SWD and the mirror is always constant as the stage is moved relative to the laser. The question is: will the voltage at the SWD vary as the stage is moved relative to the laser, along the path of the light beam? The obvious answer is: NO. "

And now you accept that with the very similar system of a shorted cable the phase (measured at some set distance from the short circuited end) changes when you vary the length of the feed cable because that phase it is measured wrt the source and obviously, if the distance to the source changes, then the phase will change. And if two signals are in phase they will add up to something that's not zero, while if they are (exactly) out of phase they will add up to zero, so the sum depends on the phase so the voltage measured at some point not at the short the voltage will depend on the phase and so it will depend on the length of the cable to the generator.

 

Those two statements of yours can't both be right.

 

I meant that the phase of the standing wave is not important. The standing wave is itself a result of incident and reflected waves. we can't think of the standing wave 'interfering' with any other wave, because it is the result of interference itself, if I understood what you mean. You have two sources S and S'. The two sources create two waves, the incident and reflected waves. The interference of the two waves creates a single wave, which is a standing wave. So we can't talk about this resultant wave interfereing with another wave because the standing wave is itself a result of interference.

[Strange]

For example, if L= N λ/2, where N is an integer, the amplitude of the wave will be maximum at the detector because the waves from S and S' will interfere completely constructively at the detector.

 

Now if the laser source S is moved back by distance ΔD away from the detector, S' will also move away from the detector by the same amount ΔD.

 

What am I missing? Surely, you will only get a standing wave if D is a multiple of λ. So, as soon as you move S, you destroy the standing wave. The position of the detector has nothing to do with it.

[lidal]

 

The obvious answer is yes, because if you move the source relative to the mirror, you lose the standing wave. It is not "locked" to the mirror. Move the source a half-wavelength away and the signal drops to zero.

 

 

You can see for yourself what happens with a standard construction laser.

 

Point the laser in some safe but convenient direction.

Move a target along in the beam.

The laser line will impact the target as a small disk of light.

This diameter of this disk waxes and wanes as you move the target along in the beam.

 

When using such as laser for levelling or alignment purposes you need to take this phenomenon into account.

 

 

 

The incoming wave's phase is dependent on the distance of the source from the mirror. If it's not an integral number of half-wavelengths, you will get some destructive interference, and that will reduce the signal.

 

 

 

Sure it will, if you are sending coherent signal in and detecting the amplitude. If you have the detector at a node, there is no signal from the incident wave. If the signal hits the mirror at an antinode, there is no reflected signal. You have destructive interference. That means a reduced signal.

 

This is how resonant cavities work. There are no supported signals unless the separation between the two ends are an integral number of half-wavelengths.

 

 

 

Does the value of the voltage stays the same?

 

 

 

What am I missing? Surely, you will only get a standing wave if D is a multiple of λ. So, as soon as you move S, you destroy the standing wave. The position of the detector has nothing to do with it.

 

 

 

I was stuck with this problem because I missed the fact that the optical system described acts as a resonant cavity, as you pointed out. Changing the distance between the laser source and the mirror affects the light intensity at the detector (SWD). It is true that the first maxima (anti-node) of the standing wave always occurs at quarter wavelength from the mirror, but its value changes as the distance between the laser and the stage is changed. Other anti-nodes are at integral multiple of half wavelength from the first anti-node.

 

Now I can explain why the phase of the SWD voltage, which is an AC voltage due to dithering of the mirror, changes as the stage is moved. i.e. why moving the mirror towards (away from) the laser source causes an increase of light intensity at the SWD for some laser-mirror distances and a decrease of light intensity for other laser-mirror distances.

 

Intuitively, if the movement of the mirror makes the laser-mirror distance closer to being an integral multiple of the wavelength, the light intensity at the detector increases because the optical system will be operating closer to resonance. If movement of the mirror makes the laser-mirror distance farther from integral multiple of the wavelength, the optical system will be detuned and the light intensity at the SWD will drop.

Thank you all for your invaluable ideas. If you have any other ideas or comments, you are welcome.