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: enter image description here

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.

  • $\begingroup$ 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$? $\endgroup$ – Prasiortle May 21 at 5:04
  • $\begingroup$ @Prasiortle I have updated my answer. $\endgroup$ – 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:


  • $\begingroup$ 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. $\endgroup$ – Prasiortle May 21 at 5:41
  • $\begingroup$ @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. $\endgroup$ – John Rennie May 21 at 5:46
  • $\begingroup$ 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 $x = 2 \times 25 \text{ mm} = 49 \text{ to } 51 \text { mm}$". $\endgroup$ – Prasiortle May 21 at 5:55

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