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There are two formulae for destructive interference. In which situation do we use them? $$(n+1/2)\lambda $$ and $$(n-1/2)\lambda $$

I'm confused as my book has mentioned the first one but In question both of them are used.

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    $\begingroup$ Are there any limitations on n? If so, then think about how that affects the results. And what physical quantity do these formulae represent? A position, a difference in position, a Doppler shift? You should be more clear in defining quantities. $\endgroup$ – Bill N Jun 10 at 18:46
  • $\begingroup$ “Option D is not correct because the first dark fringe (n = 1) occurs when the path difference is λ / 2.” There were 2 options of (n+1/2)or (n-1/2) I selected the plus one. I have quoted the examiner can you please explain this statement? The correct answer was (n-1/2) $\endgroup$ – Asad Jun 10 at 18:49
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Why it doesn't matter which you use

In both those formulas $n$ nominally represents any integer value so the $n \pm \frac{1}{2}$ (both of them) represent all the places half-way between the integers. They both describe the same set of values.

They are the same.

Why it could matter in class

If a question were too ask you for the path difference of "the second dark fringe" or something similar then you might expect to substitute 2 in for $n$, but then there is an ambiguity: you have to know which fringe is "first" to know how to proceed.

Here you are concerned about how you label the members of the set.

For this reason I prefer to use things like "the dark fringe between the second and third bright fringes" when I am writing questions. Because we usually use just one formula ($n\lambda$) for numbering the bright fringes.

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  • $\begingroup$ “Option D is not correct because the first dark fringe (n = 1) occurs when the path difference is λ / 2.” There were 2 options of (n+1/2)or (n-1/2) I selected the plus one. I have quoted the examiner can you please explain this statement? The correct answer was (n-1/2) $\endgroup$ – Asad Jun 10 at 18:34
  • $\begingroup$ This is exactly the issue I'm talking about in the second part of the answer. Assigning ordinals (first, second,...) to the fringes selects a convention for how you assign $n$ that differs between the two forms. Generally the "first" dark fringe is one located next to the central maximum. You can get that using the minus form and $n=1$ or the plus form using $n=0$. many people consider the former more natural and a few assert that it is the One True Way. As I said, I prefer to be more specific about what the question is. $\endgroup$ – dmckee Jun 10 at 21:03
  • $\begingroup$ It is worth mentioning that engineering disciplines often have standards that practitioners are effectively required to follow and they often specify these kinds of things to help with uniformity of communications and reduction of losses due to ambiguities. Academics and sufficiently insulated researchers are less likely to feel bound by such standards and sometimes follow their own notions of best practice. $\endgroup$ – dmckee Jun 10 at 21:07
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Destructive interference happens when the crests of one wave align with the troughs of another wave, the amplitude of the resulting wave will then be the absolute difference of the amplitudes of the source waves.

Over the span of every wavelength there is one crest and one trough, separated by half a wavelength. So to align crests with troughs, you have to shift one of the waves by half a wavelength. It doesn't matter in which direction you do this, because waves are periodic, nothing changes when you shift a wave by one wavelength. I don't exactly know what your formulas represent, but if they are the difference in path length necessary for two initially coherent waves to destructively interfere, you can see that the difference is exactly one wavelength

$$ \left(n + \frac{1}{2} \right) \lambda - \left(n -\frac{1}{2}\right) \lambda = 1 \lambda $$

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