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?
Best Answer
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.