In deriving the single slit diffraction pattern,it is stated that the intensity is zero at angles of $\theta = n\lambda / a$,where $a$ is the slit width. Justify this by suitably dividing the slit to bring out the cancellation.

(N/A) Consider a single slit of width $a$. To find the condition for minima,we divide the slit into $2n$ equal parts.
The path difference between the secondary wavelets from the corresponding points of the upper and lower halves of the slit is $\Delta x = (a/2) \sin \theta$.
For the first minimum,we set the path difference between the top edge and the center of the slit to be $\lambda / 2$. Thus,$(a/2) \sin \theta = \lambda / 2$,which gives $a \sin \theta = \lambda$.
In general,for the $n^{th}$ minimum,we divide the slit into $2n$ equal parts. The path difference between the wavelets from the corresponding points of the two adjacent parts is $\lambda / 2$.
Since every point in the upper half of the slit has a corresponding point in the lower half such that their path difference is $\lambda / 2$,the waves from these pairs interfere destructively.
Consequently,the resultant intensity at these angles $\theta = n\lambda / a$ becomes zero.