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?

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.

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.

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.

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.

18 minutes ago, Strange said:

Huh? Not sure where you get that mental model from. See my animation for how wavelength, etc is defined.

Again, reread the posts.

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.

I said that the way wavelength was calculated did not fit with the definition. That is where Klaynos came up with the objection that peaks/troughs belonged to the same wave.

You can now attempt to blame me for my confusing language, but I was not the one creating the confusion. And I still have no answer to my objection.

2 minutes ago, Strange said:

I don't know what you think the difference is. In John's drawing he has marked (with red dots) two consecutive peaks that represent the wavelength. (He has done this for four separate waveforms - not quite sure why, perhaps to show different points in time.)

Of course. Why wouldn't it be?

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.

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.

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.

4 minutes ago, studiot said:

 

Here is a true story about the Romans, who had conquered everybody and anybody in their path.

Whilst it is true that their superior military resources won through in the end, when they first came to Britain, they nearly lost everything because they rejected simple basic advice form others.

The tides in the Mediterranean are nearly non existant, so they failed to allow for the much higher tides in the Atlantic/North Sea and nearly lost all their boats.

 

It seems to me that you are doing the same.

 

 

I think I will keep ignoring you unless you come up with relevant arguments.

7 minutes ago, swansont said:

How is the second sentence in contradiction to the first?

I would certainly not say contradiction. It concerns two different models with different consequences. I do not think they are compatible with each other. Especially, if peaks of the same wave are considered, then the equation (frequency, wavelength, velocity) for the speed of light would not make any sense.

3 minutes ago, studiot said:

This was meant to help, not hinder.

That was my initial impression also. It has long vanished.

1 minute ago, studiot said:

 

So if you know the difference perhaps you can confirm what it is?

 

What is the difference between the frequency of oscillation and the frequency of a wave?

Why does a wave have a wavelength, but an oscillation doesn't

 

 

 

You are trying to prove your superiority by testing my knowledge. You would prove it better by answering my questions and showing how I am wrong.

10 minutes ago, studiot said:

Watching this thread it is immediately obvious that Dalo needs to study the basics of wave motion before he can appreciate the next steps.

What he is attempting to understand is far beyond his present knowledge.

Dalo, in your last thread I gave you a list of things you need to know before you could understand the answer to your question.

You chose not to ask anything about one of them.

Here is another shorter list.

The first thing you need to know is the difference between an oscillation, a wave and a sine curve.

Not understanding this trips many people up when they try to study waves.

 

You should not confuse disagreement with misunderstanding. This is a mistake often made in this forum because people are convinced that anybody who understands science perfectly can only agree with it. While it is always possible for people, for me also therefore, to misunderstand things, it should not be considered as automatic. Science thrives on disagreement, more than on consensus.

6 minutes ago, John Cuthber said:

You can.

Of course, but then you would have two frequencies, one in the direction the wave is going, the other along the wave. Is that what you mean?

1 hour ago, Strange said:

Yes. The frequency is the time between each peak of the wave

Here you make it sound that it is possible to calculate the frequency of peaks on the same wave which are all moving in the same direction at the same time.

19 minutes ago, Strange said:

Well the wave is moving from left to right (say) in that diagram. So if you look at a specific point in the diagram you will see the peaks and troughs moving past. The distance between them is the wavelength. The time between them is the frequency.

This is, once again, the water wave analogy, where the wavelength is the distance between two consecutive peaks or troughs, in the direction of the wave. And not two peaks or troughs belonging to the same wave (or wavy line in John's drawing)

16 minutes ago, Strange said:

Rays would be sent in all directions from each slit. So, some of those will be parallel but most won't be. (Rays are at right angles to the wavefront so where you have waves going out as concentric circles, the rays will be radial lines.)

Even if they start parallel?

16 minutes ago, Strange said:

Nt sure what you mean by transverse to one another. They are measuring different things. Wavelength is the distance between peaks. Frequency is the rate at which peaks pass some point. So they are obviously related to the speed of the wave.

The wavelength is calculated as the distance between two peaks of the same wave. Let us say on one and the same line as in John's drawing.

Frequency is calculated in the direction in which the wave is moving.

That make them transverse in my view.

13 minutes ago, Strange said:

Wavelength is the distance between the soldiers (distance between peaks).

Speed (of light or whatever) is the speed the soldiers are marching.

Frequency is the rate at which soldiers pass you.

It is what is known as an "analogy".

Then it is a very bad analogy for the model of wavelength as two peaks of the same wave. This would be a perfect analogy for the water wave model which you all have rejected.

11 minutes ago, John Cuthber said:

 

I don't understand how this is difficult.

Can you imagine a line of people standing one behind the other?

Can you imagine them all setting off walking?

Can you imagine standing next to that line.

They would walk past you.

If there was a gap of 1 metre between them, and they were walking at 2 metres per second then, in each second, 2 of them would walk past you.

see previous remark.

time out. Not everybody at the same time. I will answer soon.

3 minutes ago, John Cuthber said:

If a line of soldiers marches past you each moving at 2 meters per second and each one separated by 1 metre do you see why you are passed by 2 soldiers per second?

 

no idea what you mean by that.

9 minutes ago, Strange said:

Because they go through two different slits. So you end up with two waves from two different sources (the two slits). At different distances from the two slits, these two waves will have a different phase relationship. At some places the two peaks will coincide and the total amplitude will be double. In other locations, a trough and a peak will coincide and the total amplitude will be zero.

If you work out where the peaks coincide (either by using math or drawing diagrams) you will find that it forms radial lines (blue on my diagram). And the points where they cancel also form radial lines (in between these).

The diagram is perfectly clear, and interference is also perfectly understandable within the wave model. The ray model is what I have difficulty understanding. Since they all should have the same diffracting angle, shouldn't all rays, from whatever slit, be parallel to each other?

edit: this is assuming that they start as parallel rays from the source. Which is usually what is assumed in textbooks.

Also, I do not understand how we can use frequency and wavelength in the same equation to calculate the speed of light, when they are transverse to each other.

32 minutes ago, John Cuthber said:

Here's a picture of some waves (slightly lumpy because I can't draw). The distance between the red marks is the wavelength.

Since all light waves travel at the same speed, the time between each wave crest passing a given point is fixed for any given wavelength

and the number of peaks passing a given point in a second is also fixed.

 

I confess: I am really confused.

When I look at your drawing (mine are even worse!), each single wavy line could represent a single wave, followed by others.

At the same time, each wavy line could be considered as many waves following each other.

It just depends how you look at it. So, I will just opt for the first perspective, each line represents a wave with different peaks and troughs.

If such a line has to go through a diffraction grating, the different parts that got through will form one wave again. But since they are all diffracted at the same angle, why should they ever interfere? Maybe I am thinking too much in ray terms, as I imagine them going through the slits and being diffracted. I have no trouble understanding that because the different parts are not all at the same level, a trough may meet a peak, and vice versa. This is not difficult to understand.

It becomes more difficult when thinking in rays. Why should they ever meet?

Concerning the law of speed of light (c=frequency times wavelength), I also do not understand how it can be applicable to the model of wavelength as distance between two peaks/troughs of the same wave.

The wave is moving in one direction, the frequency, at least in the water wave model, also concerns the number of waves that pass a border within a certain period. But the wavelength is now measured in a transverse way relative to the frequency. Is it a legal mathematical operation to multiply wavelength by frequency in such a model?

1 minute ago, Strange said:

It is the distance between two peaks of the same wave. The peaks are represented by the red, green and purple dotted lines in my diagram. So the wavelength (shown as λ) is the distance between two red (or green or purple) dotted lines.

I must admit that I have never encountered this explanation of wavelength before. Every textbook or video I have consulted/watched, used the water wave model, and my questions were based on that model.

This opens a new line of questioning. Since they all have the same color (wavelength), and therefore the same diffraction property, shouldn't they be parallel to each other?

Also, I find the image of wavelength as the distance between peaks of the same wave very intriguing. In particular, I wonder whether the frequency of a wave also has to be calculated the same way?

2 minutes ago, John Cuthber said:

So we can calculate the wavelength.

But what is the wavelength? Is it the distance between two waves, or the distance between two peaks/troughs of the same wave?

I am just waiting for Klaynos' explanation.

4 minutes ago, swansont said:

different cycle, as the wave from one slit has to travel one extra wavelength compared to the other, to get constructive interference 

you sound like you are not following the discussion at all.

Let us start first with Klaynos' explanation, that wavelength concerns the distance between two peaks of the same wave. I would very much like to understand that before I look any further.

2 minutes ago, Klaynos said:

The wavelength is the distance between two peaks of the same wave. Not two different waves.

This is the first time I have ever heard of that.

4 minutes ago, John Cuthber said:

So the radii of the circles differ by exactly one wavelength.

Not by these calculations.

I see the drawing and it explains what you say. But it does not explain the link between the distance between the fringes and the distance between the waves prior to passing through the grating.

You have no way of measuring the distance between the peaks and the troughs on the right side. All you have is the distance between the grating and the screen.

The way the calculations are made may explain the effect of diffraction on the distance traveled by light to the screen, but it does not explain the distance between two waves.