# Fraunhofer diffraction at single slit derivation

My professor derived the conditions for maxima and minima on the screen by dividing the slits into equal parts.I ll try to summarize it:

1. The optical path distance between secondary waves emanating from the top and bottom of the slits is asinθ.

2.If asinθ=2 λ then the aperture AB can be imagined to be divided into 4 equal parts and the waves from the corresponding points separated by a distance a/4 in the two halves of each of the slit will differ in path by  λ/2 and this gives position of minimum intensity.

My question is why can't we divide the slit into two equal parts such that the optical path difference between waves emanating from corresponding points in each half differ by  λ and hence interfere constructively leading to maximum intensity?

Here is the situation you have described with the slit being split into two and the top half in advance by $\lambda$ compared with the bottom half.

Sources $A$ and $A'$ are in phase and the waves add together.

Sources $B$ and $B'$ are in phase and the waves add together.

However what do you notice about the phases of $A$ and $B$?
The path difference is $\frac \lambda 2$ and so they are exactly out of phase and the same is true of $A'$ and $B'$.

So what you are adding are the waves from sources $A$ and $A'$ which are in phase to the waves from sources $B$ and $B'$ which are also in phase but the waves from $AA'$ are exactly out of phase with those from the waves from $BB'$ which means that there is a minimum in that direction.

If you are going to cut up the waves from the slit the trick is that you must make sure that the sources in one segment differs from the sources in an adjacent adjacent segment by no more than a path difference of $\frac \lambda 2$.

First minimum - two segments and path difference between segments $\frac \lambda 2$
Second maximum - three segments and path difference between each adjacent segment $\frac \lambda 2$
Second minimum (the one you asked about) - four segments and path difference between each adjacent segment $\frac \lambda 2$
Third maximum - five segments and path difference between each adjacent segment $\frac \lambda 2$

etc

Sure, you can divide the slit so that you have pairs of points that have a path difference of one wavelength, and waves from the two points in this pair will interfere constructively but the waves from each of these pairs will, in general, not all be in phase. Waves from one pair can be cancelled by waves from some other pair.

Pairing up parts of the slit so that each pair cancels itself out eliminates this complication. A bunch of zeros always add up to zero.

• Thankyou for your reply,but doesn't that mean this explanation isn't satisfactory and I could counter by saying that waves from other pairs may interfere constructively and hence result in a maxima rather than a minima? Sep 15, 2016 at 15:41
• Add a bunch of zeros and you always get zero. Say you are adding eight vectors of the same magnitude and they are directed {up, right, down, left, up, right, down, left} if you add them (up + down) + (up + down) + (left + right) + (left + right) it's pretty easy to see the answer has to be zero. If you add them (up + up) + (right + right) + (down + down) + (left + left) the result is still zero but it's a little bit harder to spot right away. Sep 15, 2016 at 15:51