# In Young's double slit experiment, why are the two theta values equivalent?

I've read several answers on here to similar questions, and I've also looked at several different picture interpretations to no avail. I can't wrap my head around it. I understand that under the assumption of L >> d, both rays from the slits are approximately parallel, but how are the two angles of $\theta$ equivalent? I'm assuming by alternate interior angles somehow, but I can't figure out how. Can someone draw it out for me or something? I'm at a loss.

• The angles of the two photon trajectories are obviously different. Oct 15, 2017 at 16:30

If you refer to the yellow-shaded triangle, yes, you're on the right path: it is purely geometric. Excuse me I'm not going to work much more than a cheap paint picture ^^

Steps:

1. (Green) If the angle $\theta$ is that one, you obviously have 90 degrees minus $\theta$ until the slit-wall.
2. (Blue) We are supposing that all rays are parallel.
3. (Pink) We draw a perpendicular to the rays.
4. (Red) Now focus on the upper triangle. If one angle is 90º and the other is 90º-$\theta$, the remainging one is neccesarily $\theta$ as well.

I hope you can see it now.

• Ah, I want to yell. I'm embarrassed that I didn't see this. Thanks. Great answer. Oct 16, 2017 at 0:57
• Don't worry. Things are always obvious AFTER the explanation, but not before. Even the simpler things can be difficult if nobody explains. Check that the yellow triangle can also be seen as a rotation of 90 degree of the blue one. Oct 16, 2017 at 13:08