Wavelength Measurement

[swansont]

22 minutes ago, Dalo said:

Let me tell you, again, how I understand John's drawing.

If you want to calculate the frequency of a group of waves, you will need at least two waves following each other. The time it takes both waves to cross some line will give you the frequency or periods per second.

That's not two waves, or a group of waves. That's multiple peaks of one wave. 

22 minutes ago, Dalo said:

You can also consider one wave, and only one, but with more than one peak or trough (a wavy line). This wave, as a whole, moves in a certain direction,  not as it is usually depicted, with its peaks and troughs alternatively crossing a boundary, but as a horizontal line moving vertically.. That is how I understood the claim that the wave length concerns two peaks of the same wave.

There is no horizontal line when you draw sin(x)

22 minutes ago, Dalo said:

I repeatedly said that I had never heard of such a model, and that I did not understand it. But starting with Klaynos, everybody was intent in presenting it as the only reasonable model.

But now, you seem to suggest that there is in fact only one model, and apparently, it is also the model I have always known about.

So, maybe we have been discussing a non-issue? If that is the case, I do not think that I am to blame. Please reread the different posts.

Nobody said anything about a horizontal line moving vertically 

[Dalo]

2 minutes ago, swansont said:

That's not two waves, or a group of waves. That's multiple peaks of one wave. 

Don't blame me for a language that was perfectly clear.

The point was and still is: when you calculate frequency by measuring the distance between fringes, you are using another direction than the one defined, being two consecutive (in the direction of the beam) troughs or peaks. People were unable to respond to this simple objection and came up with confusing models to hide their own inadequacy. Instead of admitting that a simple change of perspective (looking at a wave as a wavy line moving left or right, instead of a group of lines moving in the forward direction) did not change a iota to the problem, they used it to disarm my objection. Confusion is their reward.

So don't point the finger at me.

[Strange]

14 minutes ago, Dalo said:

My objection against the concept of wavelength was that it was, to use intuitive terms, in the direction the light is going. But when measuring, it was measured on the screen, which is like a horizontal direction to the vertical direction of the light, or vice versa.

The distance between the interference fringes is not wavelength. Although it can be used to calculate wavelength as described earlier. 

14 minutes ago, Dalo said:

And I still have no answer to my objection.

What objection?

Edited November 27, 2017 by Strange

[swansont]

3 minutes ago, Dalo said:

Don't blame me for a language that was perfectly clear.

The point was and still is: when you calculate frequency by measuring the distance between fringes, you are using another direction than the one defined, being two consecutive (in the direction of the beam) troughs or peaks. People were unable to respond to this simple objection and came up with confusing models to hide their own inadequacy. Instead of admitting that a simple change of perspective (looking at a wave as a wavy line moving left or right, instead of a group of lines moving in the forward direction) did not change a iota to the problem, they used it to disarm my objection. Confusion is their reward.

So don't point the finger at me.

Giving you accurate information and correcting statements is not finger-pointing. Consider that trying to figure out a person's misconception is not an easy task. 

[Klaynos]

2 hours ago, swansont said:

That's not two waves, or a group of waves. That's multiple peaks of one wave. 

Two pages and a few hours of my sleep later and this (well the op statement that resulted in this response) is still the misconception that I think is fueling the confusion here. 

Dalo, a sin wave is a representation of a single continuous wave, not a series of waves. 

Take water, again, each crest (colloquially each wave) is not technically a single wave, bit just one peak on a series of peaks in a single wave. 

 

[studiot]

7 hours ago, Klaynos said:

Two pages and a few hours of my sleep later and this (well the op statement that resulted in this response) is still the misconception that I think is fueling the confusion here. 

Dalo, a sin wave is a representation of a single continuous wave, not a series of waves. 

Take water, again, each crest (colloquially each wave) is not technically a single wave, bit just one peak on a series of peaks in a single wave. 

 

 

Yes misunderstanding the sine curve was prophetic, +1

I try to avoid talking about a sine wave unless I am absolutely certain of my audience because it is loose terminology.

 

I would add a very basic description of simple harmonic motion (SHM) to my three prerequisites for waves since it helps explain the difference between oscillation and wave motion.

 

Dalo please note that both oscillation always has a frequency, but never has a wavelength.

Wave motion  has a frequency, except for solitary waves, but always has a wavelength.

[Strange]

1 hour ago, studiot said:

Wave motion  has a frequency, except for solitary waves, but always has a wavelength.

By "solitary waves" do you mean a single cycle of the waveform? 

But if it has a wavelength and a speed, then there must be an associated frequency. (Although, at the risk of causing more confusion, this will obviously a wide range of frequencies if the waveform is truncated to a single cycle.)

[studiot]

33 minutes ago, Strange said:

By "solitary waves" do you mean a single cycle of the waveform? 

But if it has a wavelength and a speed, then there must be an associated frequency. (Although, at the risk of causing more confusion, this will obviously a wide range of frequencies if the waveform is truncated to a single cycle.)

This is a good time to examine the formula

 

Velocity  =  frequency x wavelength.

 

Not all waves have a velocity, some are stationary or standing waves.

Not all waves have a frequency. Frequency has no meaning for something that only happens once.

Yes solitary waves only happen once and therefore have no frequency. Solitary waves is the original name given by the person (Russell) who identified them and they are not single cycles chopped out of a solution to the usual linear wave equation. Solitary waves are non linear and the mathematics constructs a single event. Often the Korteweg-de Vries equation.
This last bit is deeper than I really meant for Dalo but Strange also deserves an answer.

All waves occupy some space so we can define or identify the region occupied with a space measurement i.e. a wavelength.

[Strange]

Ah, I see. You mean solitons?

[studiot]

9 minutes ago, Strange said:

Ah, I see. You mean solitons?

Yes that is another name for them, though some reserve it for when they model particle like behaviour after the originators (Zebrusky and Kruskal) of that term about 100 years after Russell.

[Dalo]

23 hours ago, Dalo said:

Here is a genuine question:
To measure the wavelength, we are told to measure the distance between the points on a screen of (monochromatric) light that has come through two or more slits.
This is a horizontal distance, comparable to the width of the waves hitting the sand. But we want the distance between the crests or the troughs. Which is, seen from the perspective of someone standing on the beach, a vertical distance.
Imagine erecting a barrier somewhere in the water, with a diffraction grating, and measuring the points where the water comes through.
Apparently, you would be measuring points along the width of the wave, and not the distance between crests or troughs.
So, what is the real explanation when dealing with light?

 

19 hours ago, Dalo said:

Green light, when measured the way described above, gives a certain figure. And so does each other color, each color having its own figure. And, that is what we call wavelength.

If that is the case, then the mode of calculation does not fit the definition. You cannot define wavelength as the distance between two (monochromatic) waves, and then use another property of waves to calculate it.

Because of diffraction, the distances from the grating to the screen are particular to each color, but that has nothing to do with the original definition.

At least, I find it very difficult to link them together.

The calculations are legitimate and give us a very useful way of identifying colors, but they do not seem to follow from the definition.

 

18 hours ago, John Cuthber said:

OK, here's the picture from the site I cited earlier.

 

The picture is a plan view of light coming in from the left, hitting the grating and then spreading out from the holes in that grating as a series of circles (in green)

The green circles represent the peaks of the waves as they spread out. The distance between the circles is the wavelength of the light. The centres of the circles are the gaps in the grating. 

In reality the grating have thousands of lines and they are separated by something like a thousandth of a millimetre.

Does that make sense so far?

 

 

18 hours ago, Dalo said:

Yes, but it is entirely irrelevant to the question: how does the distance between the fringes relate to the distance between the waves, assuming that all the waves hitting the screen at the same time originate from one and the same wave front going through the slits?

(it would be even more complicated if they did not)

Your diagram would be valid even if we changed the distance between two crests or troughs. The results would be the same, or so it seems to me.

 

[Strange]

Not sure what the point of that post was but ... in that diagram, the wavelength is the distance between the blue lines, not the red arrows.

 

[Dalo]

 

11 minutes ago, Strange said:

Not sure what the point of that post was but ... in that diagram, the wavelength is the distance between the blue lines, not the red arrows.

 

It does not matter. The diagram makes it look like it would be possible to measure the distance between the (blue or red) lines, but that is a false suggestion. All we have is the measurement of the distance between the different points on the screen.

Edited November 27, 2017 by Dalo

[swansont]

5 minutes ago, Dalo said:

It does not matter. The diagram makes it look like it would be possible to measure the distance between the (blue or red) lines, but that is a false suggestion.

Since the label on the arrows is "direction" I'm not sure how one arrives at that interpretation.

5 minutes ago, Dalo said:

All we have is the measurement of the distance between the different points on the screen.

And we have an equation that relates that to wavelength. Why are we still going over this? Is this issue somehow not settled?

[Dalo]

2 minutes ago, swansont said:

Since the label on the arrows is "direction" I'm not sure how one arrives at that interpretation.

The diagram is justified by the calculation. Not the other way around. That is the whole issue.

[Strange]

40 minutes ago, Dalo said:

All we have is the measurement of the distance between the different points on the screen.

And from that measurement, one can calculate the wavelength.

(There are ways of directly measuring the wavelength as well.)

Edited November 27, 2017 by Strange

[Dalo]

43 minutes ago, Strange said:

(There are ways of directly measuring the wavelength as well.)

I'm afraid I don't know them. Which are they?

[Strange]

Here is one: http://www.planet-science.com/categories/over-11s/physics-is-fun!/2012/01/measure-the-speed-of-light-using-chocolate.aspx (I have never tried this; I don't have a microwave oven.)

Another (much more accurate) method is to use an interferometer: http://physical-optics.blogspot.co.uk/2011/06/michelsons-interferometer.html

[Dalo]

27 minutes ago, Strange said:

Here is one: http://www.planet-science.com/categories/over-11s/physics-is-fun!/2012/01/measure-the-speed-of-light-using-chocolate.aspx (I have never tried this; I don't have a microwave oven.)

Another (much more accurate) method is to use an interferometer: http://physical-optics.blogspot.co.uk/2011/06/michelsons-interferometer.html

I had seen the chocolate story but never really watched it. I will now.

As far as the interferometer is concerned, it is the same kind of set up as discussed in this thread, only the way to produce the image on the screen is different. 

 

[Strange]

26 minutes ago, Dalo said:

As far as the interferometer is concerned, it is the same kind of set up as discussed in this thread, only the way to produce the image on the screen is different. 

The only thing it has in common is the use of interference. It is a lot simpler and it is a direct measurement of the wavelength: the distance you move the mirror to go from one peak of constructive interference to the next is a wavelength (well, half).

[studiot]

1 hour ago, Dalo said:

I'm afraid I don't know them. Which are they?

 

I already gave you one.

 

Lecher Lines.

[Dalo]

3 minutes ago, Strange said:

it is a direct measurement

No, it is not. It presupposes wavelength and frequency, and a speed of light. Once you accept that at least two of the terms are correct, you get the third one automatically.

But that is the whole point of the discussion here. Is the way the wave length is calculated legitimate? I did no read any compelling argument besides strong a strong belief in the method.

I am afraid that I have nothing new more to add.

[Strange]

2 minutes ago, Dalo said:

It presupposes wavelength and frequency, and a speed of light.

What do you mean "presupposes"? Are you suggesting that light is not a wave phenomenon?

3 minutes ago, Dalo said:

But that is the whole point of the discussion here. Is the way the wave length is calculated legitimate?

Is that the point? I didn't realise. I thought it was just your inability to understand what a wave was and how a diffraction grating worked.

But, of course it is legitimate. It is just simple schoolboy geometry. 

5 minutes ago, Dalo said:

I am afraid that I have nothing new more to add.

Good.

[studiot]

16 minutes ago, Dalo said:

25 minutes ago, Strange said:

it is a direct measurement

No, it is not.

 

Complete rubbish.

I just looked at Strange's link (+1 for finding this) and I note that is aimed at 11 year olds.

 

 

They are asked to measure (with a ruler) the half wavelength at about 6cm.

You can't get much more direct than that.

 

The (wave)length is the only part of the experiment that is an actual measurement and not taken on trust.

Edited November 27, 2017 by studiot

[Dalo]

2 minutes ago, studiot said:

 

Complete rubbish.

I just looked at Strange's link (+1 for finding this) and I note that is aimed at 11 year olds.

They are asked to measure (with a ruler) the half wavelength at about 6cm.

You can't get much more direct than that.

 

The (wave)length is the only part of the experiment that is an actual measurement and not taken on trust.

they are definitely measuring something, and calling it "wavelength". In this sense, it is a direct measurement.

But is it the same "wavelength" that is supposed to represent the distance between two peaks/troughs in the direction of the beam?

My answer is no, feel free to disagree.

But then, you have already done so. 

Phi? Would you please close this thread? It is going nowhere.