Order of a dark fringe in Young's double slit experiment

[FranzDiCoccio]

TL;DR

Formulas for the angular position of a costructive/destructive interference fringe are often given in terms of an integer m. I can understand that this is the "order" of the fringe for constructive ones. I am not sure of the convention for destructive fringes.

So the angular position for constructive fringes is
<br /> d \sin \theta = m \lambda = (2m) \frac{\lambda}{2}, \qquad m=0, \pm 1, \pm 2, \ldots<br />
whereas for destructive fringes we have
<br /> d \sin \theta = m \lambda = (2m+1) \frac{\lambda}{2}, \qquad m=0, \pm 1, \pm 2, \ldots<br />

I can see that $m$ denotes the order of the fringe in the first formula, where $m=0$ is the central fringe, $m=\pm 1$ are first-order fringes, and so on.

I am not sure about this convention for dark fringes. Is there a zero order fringe? Is there one, or are there two?
Personally, I'd call "first order" the two dark fringes delimiting the central bright fringe, "second order" the two subsequent ones, and so on and so forth.
I mean, after all $2m+1$ is just a way of saying "an odd number". Also, $m=4$ would give $9/2 \lambda$, while $m=-4$ would give $-7/2 \lambda$, which I would not consider of the same order... The latter is closer to the central bright fringe than the former.
So it seems to me that $m$ is not as related to the order as for bright fringes. I'd list the order according to the corresponding odd number: (±1 are first order, ±3 are second order,...).
Or, maybe, I'd use $±(2m-1)$ but starting from $1$ and not from $0$.

This might seem like a pointless question, but when one has to find the angle corresponding to the fourth-order dark fringe (e.g. in a problem), it is a matter of definition. Is it the fourth from the central bright one, or is it the fifth?

Thanks a lot for your insight.

 

[ZapperZ]

Is there a reason why you don't think m=0 exists for the dark fringes?

If you look at the equation, for m=0, you have a non-zero angle on either side of the central bright fringe. So for the bright fringe, m=0 is at θ = 0, but for the dark fringe, m=0 is at a non-zero angle.

Besides, this is nothing more than a "counting index".

Zz.

 

[FranzDiCoccio]

Is there a reason why you don't think m=0 exists for the dark fringes?

There actually is, but it might very well be personal taste.

In some places, e.g. https://opentextbc.ca/physicstestbook2/chapter/youngs-double-slit-experiment/, i find a formula that is equivalent to the one above
<br /> d \sin \theta = (m+\frac{1}{2}) \lambda,\qquad m=0,\pm 1, \pm 2,\ldots<br />
This confuses me. As I explained above, the "first-order" fringes, $|m|=1$, would be at different distances from the center, which makes little sense, in my opinion.

I like the following option slightly better
<br /> d \sin \theta = \pm (m+\frac{1}{2}) \lambda,\qquad m=0, 1, 2,\ldots<br />
because at least here fringes with the same order are symmetric wrt the central bright fringe.

However I do not see a real reason for counting from $0$.
In that case, the 1st order fringes would be the 2nd from the center, the 2rd order fringes would be the 3rd from the center... Possible, and perhaps natural if you are a coder and work a lot with arrays, but probably confusing for most of the people.

Perhaps I'm wrong, but I link the "order" of the fringe with its "distance" from the center.
Then it's sort of natural that the central bright fringe is assigned $0$ distance from itself.
By the way, that fringe is qualitatively different from the others, in the sense that it is there for any $\lambda$. Also, for white light, that fringe is white whereas all the other bright fringes are "small rainbows".

So, all in all, I would introduce formulas for lateral fringes (the only fringes for which calculating an angle makes sense). That is
<br /> d \sin \theta = \pm (m-\frac{1}{2}) \lambda,\qquad m= 1, 2, 3, \ldots<br />
for the dark (lateral) fringes and
<br /> d \sin \theta = \pm m \lambda,\qquad m= 1, 2, 3, \ldots<br />
for the bright lateral fringes.

I do agree that $m$ is just a "counting index", but I am wondering whether counting from 0 is really convenient. I'd say that the 1st dark fringes immediately following the central fringes are 1st order.
In my opinion zeroth order for dark fringes is kind of artificial, and comes from using $2m+1$ instead of (the perfectly equivalent) $2m-1$ for "an odd number".

 

[sophiecentaur]

Besides, this is nothing more than a "counting index".

but it might very well be personal taste.

These are the appropriate (best) answers to the thread. When describing maxes or mins, there is no actual need to use the 'order', if you are not confident with your personal choice. That sort of demonstrates that it's not an important matter to worry about. If a teacher or supervisor wants to argue about it being important then just let 'em; you know better now.

 

[FranzDiCoccio]

Hi sophiecentaur,
thanks for your reply. I surely agree with you. If you're interacting with a human, you can actually explain yourself and agree on a notation.
I guess that, whenever I can, I'll refrain from using the term "order" and say something like "the nth (bright/dark) fringe to either side of the central one". This is unambiguous.

However, suppose that I find a question (e.g. in a test) asking for the angular position of the second-order dark fringe. Is this the second from the central one, or is it the third?
If you start counting from zero, it would be the third.
A common notation should be established for that question to be unambiguous.

By the way, the formula for a single slit is
<br /> d \sin \theta = m \lambda, \qquad m = \pm 1, \pm 2,\ldots<br />
Hence, in that case, there seems to be no problem in starting from 1.
It seems that the first-order dark fringe is the first one on the side of the central bright fringe for a single slit, but the second one for a double slit? Very confusing.

My feeling is: order primarily refers to bright fringes, which are the most visibile ones by definition.
I guess that the concept has been "extended" to dark fringes in (some) textbooks, mainly for having some assortment in exercises. I'm not sure dark fringes are ever used in actual experiments.
This generalization has been a bit too casual, and someone started using 2m+1 for odd numbers, which forces you to start counting from 0.

I have the feeling that "order" mainly applies to bright fringes, where there is no ambiguity, and one should be careful in extending this concept to dark fringes. I'm thinking of a diffraction grating, e.g.