deadonce
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Hi hi, first post ^^
Anyway, I encountered this little problem with the derivation of Young's double slit formula. The standard derivation found on the internet and my textbook is shown here: http://schools.matter.org.uk/content/interference/formula.html
When deriving, a few errors had cropped up. I'll make reference to the linked diagram cause my work on paper is kinda messy:
1) The assumption that triangle S2-Z-S1 is a right-angled triangle. This assumption is made for sinθ=λ/s.
However, if said triangle is right-angled, then ZY≠S2Y, due to Pythagoras' theorem and some value of S2Z existing.
Therefore, the path difference cannot be λ for the first fringe, which goes against what we learn.
2) The assumption that θ reappears again in the triangle Y-(center of S1S2)-X. I don't see any similar triangles after error 1) is corrected. So does doing this base a working on an error which magnifies everything?
3) Assuming θ is negligible at the final step. This one is slightly more straightforward. If θ is negligible, then there would be no path difference, then there would be no λ, no fringe, and only one central fringe.
The major problem here is error 1). Unless the idea that sine=opposite/hypotenuses can be applied to NON-right-angled triangles, path difference x=λD/s seems flawed, at least, in terms of this method of derivation.
So, can someone please enlighten me a little about this?
Also, I have worked out an alternate equation that I'm pretty sure is wrong. Anyway, here it is:
Fringe difference x
=(a/2)+((S2Y)^2-D^2)^(1/2)
=(a/2)+((S1Y-λ)^2-D^2)^(1/2)
=(a/2)+Dtanβ
All 3 are alternate derivations depending on whether you have
a)distance S2Y
b)distance S1Y AND λ
c)angle β, which is the angle Y-S2-horizontal parallel to D
My derivation comes mostly from a different right angle triangle w.r.t Pythagoras' theorem. If anyone is interested, I can scan my messy drawing to be scrutinized properly.
I'm just pretty much certain that the derivation approximates too many things. Granted, the results are close enough to warrant the use of ≈, but =?
Thoughts?
Anyway, I encountered this little problem with the derivation of Young's double slit formula. The standard derivation found on the internet and my textbook is shown here: http://schools.matter.org.uk/content/interference/formula.html
When deriving, a few errors had cropped up. I'll make reference to the linked diagram cause my work on paper is kinda messy:
1) The assumption that triangle S2-Z-S1 is a right-angled triangle. This assumption is made for sinθ=λ/s.
However, if said triangle is right-angled, then ZY≠S2Y, due to Pythagoras' theorem and some value of S2Z existing.
Therefore, the path difference cannot be λ for the first fringe, which goes against what we learn.
2) The assumption that θ reappears again in the triangle Y-(center of S1S2)-X. I don't see any similar triangles after error 1) is corrected. So does doing this base a working on an error which magnifies everything?
3) Assuming θ is negligible at the final step. This one is slightly more straightforward. If θ is negligible, then there would be no path difference, then there would be no λ, no fringe, and only one central fringe.
The major problem here is error 1). Unless the idea that sine=opposite/hypotenuses can be applied to NON-right-angled triangles, path difference x=λD/s seems flawed, at least, in terms of this method of derivation.
So, can someone please enlighten me a little about this?
Also, I have worked out an alternate equation that I'm pretty sure is wrong. Anyway, here it is:
Fringe difference x
=(a/2)+((S2Y)^2-D^2)^(1/2)
=(a/2)+((S1Y-λ)^2-D^2)^(1/2)
=(a/2)+Dtanβ
All 3 are alternate derivations depending on whether you have
a)distance S2Y
b)distance S1Y AND λ
c)angle β, which is the angle Y-S2-horizontal parallel to D
My derivation comes mostly from a different right angle triangle w.r.t Pythagoras' theorem. If anyone is interested, I can scan my messy drawing to be scrutinized properly.
I'm just pretty much certain that the derivation approximates too many things. Granted, the results are close enough to warrant the use of ≈, but =?
Thoughts?
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