Question

2 point sources, S1 and S2, oscillating in phase send waves into the air at the same wavelength, 1.98 m. Given that there is a nodal point where the 2 waves overlap, find the smallest corresponding path length difference.
Jemurray helped me here already, but I wanna know if a certain formula works too. |PS2−PS1|=(n−1/2)λ/d

Answer 1

MY answer was 1m. The correct answer is 0.99m. This is not a rounding problem.

Answer 2

um isn't the smallest path length difference 0. S1=S2?

Answer 3

Sounds agreeable. But then the answer would be less then the correct answer, 0.99.

Answer 4

like you said, the only cases where two sine waves share common nodal points (zeros) is when they are in perfect phase or perfectly out of phase. so | S1 - S2 | = n λ or | S1 - S2 | = (n-1/2) λ, where the wavelength λ is 1.98.

Discarding the trivial case where S1=S2 gives you 1.98/2 = 0.99

Answer 5

But I want 0.99....

Answer 6

And the formula |PS2−PS1|=(n−1/2)λ/d works in this case right?

Answer 7

what is d?

Answer 8

Distance, like in every other physics equation. Hi Murray. Does the formula |PS2−PS1|=(n−1/2)λ/d work?

Answer 9

Cause I used |PS2−PS1|=(n−1/2)λ/d, like the teacher directed, but my answer was off by 0.01.

Answer 10

There are quite a few distances in the world, so could you be more specific as to what d refers to?

Answer 11

I actually dunno. I guess between the 2 points.

Answer 12

Also I'm not sure I'm doing "path length difference" correctly ... I'm just figuring out the distance between the two point sources = how far they are apart from each other. I'm making them a half-wavelength apart, using (1/2)λ

Answer 13

Your guess is better than mine. Hey, Cap Jack! Over here. You're lookin at the wrong question.

Answer 14

Alright, look, the way this works is the following. You get a dark spot if and only if