Dalo

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

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

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.

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.

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.

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.

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

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

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.

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. 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?

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.

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)

Even if they start parallel? 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. 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. see previous remark.

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

no idea what you mean by that.

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.

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?

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?

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.

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.

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