In single slit diffraction, why (according to the equation for $\theta$) do successive dark spots exist at whole number intervals ($n=1,2,3, \ldots$) and not half? The correct path difference for destructive interference is $1/2$ a wavelength, so why in $\theta = n(\lambda)/(\mbox{slit width})$ do we use $n = 1,2,3,\ldots$?
[Physics] SIngle slit diffraction and location of dark spots
diffractionopticswaves
Related Solutions
The method you are using is fairly standard if you cannot or do not wish to do an integration.
The slit is split into a number of sections and it is assumed that within each section there are secondary sources.
In each section there is a secondary source with a corresponding secondary source in the next section at the same relative position within the section.
The secondary sources emit secondary wavelets in phase and then these wavelets overlap (superpose) having differing phases because they travel different distance.
If the path difference is $\frac \lambda 2$ the superposing wavelets are exactly out of phase and so the resultant amplitude is zero.
This will be true of all corresponding wavelets from the two sections and so the resultant amplitude is zero.
With logic 2 you are missing out minima and inserting maxima in their place.
With logic 2 you are adding the amplitude of secondary sources between $AB$ to those between $CE$ and producing a maximum because the corresponding sources are in phase with one another.
Unfortunately you are adding zero to zero as the combined effect of the secondary sources in $AC$ is zero as is shown by using logic 1.
When you use this method you must ensure that the phase change across one of the segments does not exceed $\pi$ ($ \frac \lambda 2$).
The problem in single slit diffraction is that there is an exception when considering the maxima. The central maximum is actually twice as wide as the other maxima. If you take this exception into account however the same formula that is valid for the minima is also valid for the maxima.
All minima have a width
$\Delta y = \lambda L /d$
The same applies to the maxima except for the centre which has a width of
$\Delta y = 2 \lambda L /d$
Best Answer
You need to split the slit by small zones and sum up field from the zones. If the first and the last zones has destructive interference with each other than we have big total intensity. If the phase make a full turn we have a dark line. Therefore, angles of the dark lines are $\theta_n = n\lambda/b,\ \,n = 1,2,3,\dots$