# In single-slit interference, why don't we consider all combinations of points and their contribution to the diffraction pattern?

The explanation I was taught back in college was that we take points $$a/2$$ spaced apart where $$a$$ is the width of the slit, and at certain angles the difference in distance traveled will be half a wavelength apart causing destructive interference, causing dark fringes. Then repeat this for for the points $$a/4$$, $$a/8$$, etc. spaced apart (uhh...what? I thought if the points are half as close, then the difference in distance traveled will also decrease and it wouldn't be fully destructive). But the major issue with all this is that we aren't considering each pair of points along the single slit, and I feel like every pair of points would contribute to the diffraction pattern on the wall. Why is this not the case?

• "... where a is the length of the single slit ..." I think you mean the width. Commented Dec 26, 2023 at 18:33
• physics.stackexchange.com/questions/257768/…. This answer maybe helpful! See the last answer ... Commented Dec 28, 2023 at 17:49

"... we aren't considering each pair of points along the single slit ..." No, but we are considering each point, and we're choosing to add the contributions from each point in the most convenient way, selecting the points a pair at a time, such that the contributions from each pair cancel to zero. [Real number analogy: $$\text{If } a+d=0,\ b+e=0,\ c+f=0,\ \text{then }a+b+c+d+e+f=0$$.]
We know that waves leaving the one edge of the slit (width $$b$$) at an angle to the normal of $$\theta=\arcsin{(\lambda/b)}$$ will cancel to zero with waves leaving from the middle of the slit. So if we work our way along the first half of the slit width-wise from that edge, and along the second half, starting at the middle, corresponding points in the two halves will, at angle $$\theta$$, produce waves that also cancel to zero. This is because the path difference is $$\lambda/2$$ in each case. So the contributions from every point add to zero.