# Single-slit diffraction

1. Jul 29, 2009

### KFC

Hi guys,
I have a doublt about deducing the formular of dark fringe for single-slit diffraction. In the text, it reads "Consider two narrowstrips, one just below the top edge of the drawing of the slit and one at its center. The difference in path length to point P is $$(a/2)\sin\theta$$, where a is the slit width. Suppose this path difference happens to be equal to $$\lambda/2$$; then light from the two strips arrives at point P with a half-cycle phase difference, and cancellation occurs."

Why it said "happens to be ..." ? Why it must be the top point and the center of the slit? It seems not that convincible! What about I pick the top point of the slit and a point apart the top edge of the slit by a/2.25 and said "Suppose the path difference (from these two points) happens to be $$\lambda/2$$ ..." then I will also get a similar formula but with different $$\theta$$! So the text choose that specific points make sense?

2. Jul 29, 2009

### Born2bwire

A rough approximation with a wide slit is that light passing through the center of the slit does so unperturbed (think of it as in wavelets). However, the light that strikes the edge of the slit will be diffracted. This diffracted light at the edges will travel outward back towards the center of the slit and interfere with the light that originally passed through unperturbed.

3. Jul 30, 2009