# Double-slit diffraction: is $x$ measured from the centre to the fringe, or between the fringes on either side?

In using the formula $$\lambda = \frac{ax}{D}$$, is $$x$$ (the fringe separation) measured from the centre to the fringe, or between the fringes on either side of the centre? In other words, in the diagram below:

is $$x$$ the distance $$XY$$, or $$2 \times XY$$? I ask because most websites (e.g. Derivation of equation of path difference in double slit) seem to define $$x$$ as the distance from the centre to the fringe (i.e. $$XY$$ in my diagram), but the solution to a problem in the textbook I am working from uses $$2 \times XY$$.

The light and dark fringes are equally paced and so the fringe separation $$x$$ is the distance between adjacent intensity maxima and also the distance between adjacent intensity minima.

Update as a result of a comment from @Prasiortle

There are many ways to show that the path difference between rays $$BX$$ and $$AX$$ is approximately $$\dfrac{ax}{D}$$ if $$a\ll D$$.

For the n$$^{\rm th}$$ bright fringe $$n \lambda = \dfrac{ax_{\rm n}}{D}\Rightarrow x_{\rm n} = \dfrac {n\lambda D}{a}$$ where $$n$$ is an integer and $$XY = x_{\rm n}$$.

For the next, (n+1) $$^{\rm th}$$, bright fringe $$(n+1) \lambda = \dfrac{ax_{\rm n+1}}{D}\Rightarrow x_{\rm n+1} = \dfrac {(n+1)\lambda D}{a}$$.

So the fringe separation, which I shall call $$\Delta x$$ and you have called $$x$$, is found as follows.

$$\Delta x = x_{\rm n+1} - x_{\rm n} = \dfrac {(n+1)\lambda D}{a} - \dfrac {n\lambda D}{a}=\dfrac{\lambda D}{a}$$

and is independent of the order, $$n$$, of the fringe.

• Then why does physics.stackexchange.com/questions/210923/… define $x$ as "Perpendicular distance between interference of the rays to the medium point of the incident rays$? – Prasiortle May 21 at 5:04 • @Prasiortle I have updated my answer. – Farcher May 21 at 5:37 We get bright fringes when the path difference between the two rays is equal to $$n\lambda$$ where $$n$$ is an integer. There is a bright fringe at the central point, i.e. at $$Y$$, because the path difference is zero at that point. Then there is another bright fringe at $$X$$ (and also at $$-X$$) so the fringe spacing is the distance $$YX$$. If we draw the fringes on your diagram they look like this: • Your answer seems to contradict that of Farcher above, which says "$x$is the distance between adjacent intensity maxima and also the distance between adjacent intensity minima". It also gives the wrong answer according to my textbook. – Prasiortle May 21 at 5:41 • @Prasiortle no, Farcher and I are saying the same thing. Assuming$x$is the distance to a bright fringe (as I've drawn) then it is the distance$YX$. You don't show us the actual problem. Maybe it is defining$x$as the distance to the first minimum, i.e. dark fringe not to the first bright fringe. – John Rennie May 21 at 5:46 • The problem tells us that the wavelength is$8$mm, and says "Measure$a$and$D$on the diagram and use these values to find the percentage difference between the actual value of$x$and the value given by the approximate formula$\lambda = \frac{ax}{D}$." The measured value of the distance$XY$on the diagram is$25$mm, but the solution starts by saying "actual value of$x = 2 \times 25 \text{ mm} = 49 \text{ to } 51 \text { mm}\$". – Prasiortle May 21 at 5:55