Proof of angle in path difference formula for two slits

AI Thread Summary
The discussion focuses on proving the relationship between angles in the path difference formula for two slits, specifically that the angle theta between rays is equal to another angle in the setup. Participants clarify that the assumption of a right angle in the proof can be justified by drawing a perpendicular line from one slit to the other ray, especially when considering the limit where the screen is far away. The approximation used in the proof is valid in the Fraunhofer zone, where the angles are small and the path length differences can be easily related to theta. Errors in this approximation are minimal under these conditions. Overall, the conversation emphasizes the importance of understanding the geometric relationships in the context of wave diffraction.
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Homework Statement
Please see below
Relevant Equations
##r_2 - r_1 = d\sin\theta##
For this
1678321819957.png

I am trying to prove that angle theta between PQ and QO is equal to theta highlighted so that I know I can use theta is the path difference formula. I assume that the rays ##r_1## and ##r_2## are parallel since ##L >> d##
1678322078982.png

1678322148938.png

My proof gives that the two thetas are equal, however I am wondering whether my assumption that the right angle circled in black can be proved. Is there a way to prove this?

Many thanks!
 
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Callumnc1 said:
My proof gives that the two thetas are equal, however I am wondering whether my assumption that the right angle circled in black can be proved. Is there a way to prove this?
It's a right angle if you draw a line starting at S1 perpendicular to the ray from S2. That's not the point. The point is whether, when you draw this perpendicular, the segment labeled ##\delta## is the path length difference between the rays from S1 and S2. That is the case when the rays are nearly parallel.
 
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You are free to choose that to be exactly a right angle before taking the limit that the screen is far away. Only in the limit are the distances delta and d easilly related to theta, but it can be shown that the approximation (not really an assumption) leads to errors that are very small in the Fraunhoffer (radiation) zone far from the slits.
 
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kuruman said:
It's a right angle if you draw a line starting at S1 perpendicular to the ray from S2. That's not the point. The point is whether, when you draw this perpendicular, the segment labeled ##\delta## is the path length difference between the rays from S1 and S2. That is the case when the rays are nearly parallel.
Thank you for your reply @kuruman!
 
hutchphd said:
You are free to choose that to be exactly a right angle before taking the limit that the screen is far away. Only in the limit are the distances delta and d easilly related to theta, but it can be shown that the approximation (not really an assumption) leads to errors that are very small in the Fraunhoffer (radiation) zone far from the slits.
Thank you for your reply @hutchphd !

True, I guess it is an approximation not assumption that ## L >> d##. Sorry, what did you mean by errors very small in the Fraunhoffer zone?

Many thanks!
 
You can look up Fraunhoffer zone. It is really that the diffraction angles aren't too large and that the d<< L. So all the angles are small, and the results are simple. This approximation (Fraunhoffer) works for things other two slit diffraction
 
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Might get better results if you search on "Fraunhofer" instead of "Fraunhoffer", although Google is pretty good at catching "obvious" spelling errorz. :cool:
 
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hutchphd said:
You can look up Fraunhoffer zone. It is really that the diffraction angles aren't too large and that the d<< L. So all the angles are small, and the results are simple. This approximation (Fraunhoffer) works for things other two slit diffraction
Thank you for your help @hutchphd!
 
jtbell said:
Might get better results if you search on "Fraunhofer" instead of "Fraunhoffer", although Google is pretty good at catching "obvious" spelling errorz. :cool:
Thank you for your reply @jtbell!
 
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