Confusion about condition of dark fringes in YDSE I have read that the condition for dark fringes in YDSE is when the path difference is
$$y_n = \left(n + \frac{1}{2} \right) \frac{\lambda D}{d}$$
where $D$ is the distance between the double slit pane and $d$ is the distance between the two slits. I saw a question about the 5th dark fringe being formed opposite to one of the slits and I had to find out the wavelength of the light used. Pretty simple. I applied the above formula and reached an expression, but found that my answer was in fact incorrect (not even in the options - it was an MCQ) and this formula had been used in the solution:
$$y_n = \left(n - \frac{1}{2} \right) \frac{\lambda D}{d}$$
Now I'm confused. Which of the above formulae is correct?
 A: Mathematically, both these expressions are equally valid, it only differs in terms of how you label fringes, or rather, how you start counting $n$. One can always redefine a new $n' = n - 1$, start counting from $n' = 0, 1, 2 \ldots$ instead of $n = 1, 2, 3 \ldots$. The maths doesn't change. (Actually, the $n \lambda$ part is redundant, it is only the additional $\pm \lambda/2$ that matters, and one would have destructive interference in both cases, whether plus or minus.) Physically, of course, this variable change is absurd, since you would now have a "zeroth" dark fringe, an absurd title, since the central fringe is bright in this case.
Physically, it may be useful to have the following picture:
In YDSE, the central fringe is bright, let us leave this aside for the time being from the counting procedures. On either side, the path difference $\Delta$ grows to $\lambda/2$ first, and hence we have a dark fringe. Thereafter, $\Delta$ becomes $\lambda$, and we have our first bright fringe, aside from the central one. Still further, $\Delta = 3 \lambda/2$, and we have our second dark fringe.  

Thus, if you address this in terms of the physical basis, counting dark fringes as per the condition $\Delta = (m - 1/2) \lambda$, with $m = 1, 2, 3 \ldots$ seems more natural.
