How to number nodal and anti nodal lines

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SUMMARY

This discussion focuses on the numbering of nodal and antinodal lines in the context of double-slit interference patterns. The participants clarify that for in-phase sources, antinodal lines are assigned integer values starting from zero, while nodal lines receive half-integer values. The confusion arises when determining the correct values for path difference and wavelength equations, particularly when sources are out of phase. The consensus is that the path difference equation should be used to derive values, with specific attention to the phase relationship of the sources.

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  • Understanding of double-slit interference principles
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Homework Statement



I'm having trouble understanding how you are supposed to decide which nodal and antinodal line numbers are supposed to be attached to the corresponding line.

Homework Equations



https://roncalliphysics.wikispaces.com/Year+13+Double+Slit+Interference

I've read this and other links, and also the information provided to complete the work.
In the school book it says anti nodal lines should be m and only m should be in the center with a value of 0.

I'm talking about these numbers here and how they apply to path difference and wavelength equations.

nodal_lines.gif

The Attempt at a Solution



From what I read at the link it sounds like P should be n=2...
From what I read in my school book it sounds like for P n=3...
It also gives integer values to the anti nodal lines instead of decimals.

The answer is really going to determine the grade so I need to know which one is right?

1.jpg
 
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Your diagram shows two sources in-phase. Your image seems to show two sources antiphase.
 
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As for the numbering, there seem to be two schemes (http://www.wikiphys.org/images/TwoSource.pdf).

Assuming the sources are in phase:

Scheme A:
Nodal lines are numbered as first , second etc. (1, 2, ...)
Antinodal lines are zeroth, first, second...

Scheme B:
The lines are assigned 'order numbers', usually denoted m. The antinodes get integer values 0, 1, 2 (so are equal to their antinodal numbers), while the nodes get the halfway numbers, m = 1/2, 3/2, ...
 
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Do you know when it is out of phase and you get a centre nodal line, is it numbered 1?

It sounds like only when it is in phase you get a centre anti nodal = 0.

(for the first scheme) I'm reading but It's hard to find an answer to this specific question.
 
julianwitkowski said:
when it is out of phase and you get a centre nodal line, is it numbered 1?
I have no idea. Maybe there is no standard numbering for anything except the in-phase case. How would one standardise it so that it works for any phase relationship?
 
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P must be 2... I know you can't measure, but does this look right to you for the image I posted.

They ask to measure the wavelength, source paths to P, and the distance between source, and to provide the completed wavelength equation for this situation. Is this the right equation they want me to provide.

Path Difference = |6λ - 8λ| = (n-½)λ = 2λ-½λ = 2λ = 1.2 = 12mm

It works with my measurements...

λ = 0.8cm
P = 2
PnS1 = 48mm = 6λ
PnS2 = 64mm = 8λ
 
julianwitkowski said:
P must be 2... I know you can't measure, but does this look right to you for the image I posted.

They ask to measure the wavelength, source paths to P, and the distance between source, and to provide the completed wavelength equation for this situation. Is this the right equation they want me to provide.

Path Difference = |6λ - 8λ| = (n-½)λ = 2λ-½λ = 2λ = 1.2 = 12mm

It works with my measurements...

λ = 0.8cm
P = 2
PnS1 = 48mm = 6λ
PnS2 = 64mm = 8λ
Can't make much sense of your algebra there (2λ-½λ = 2λ ?), but I agree that since the sources seem to be antiphase the point P should be two wavelengths further from one than from the other.
 
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haruspex said:
Can't make much sense of your algebra there (2λ-½λ = 2λ ?)

This is why ignorance = bliss. What am I doing...

Path Difference = |6λ - 8λ| = 2λ = 1.6cm but (n-½)λ = 2λ-½λ = 1.5λ = 12mm

This means that either Path Difference ≠ 2λ, or n ≠2, or λ ≠ = 0.8cm...

My problem is that the measurements are pretty accurate, which suggests n = 2.5...
Is there anyway this is possible?
 
julianwitkowski said:
Path Difference = |6λ - 8λ| = 2λ = 1.6cm but (n-½)λ = 2λ-½λ = 1.5λ = 12mm
OK, but I don't understand where you are getting n-½ from in this context. Isn't that for in-phase sources? As you say, the path length difference in the image is 2λ.
 
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haruspex said:
OK, but I don't understand where you are getting n-½ from in this context. Isn't that for in-phase sources? As you say, the path length difference in the image is 2λ.
What equation is used for sources that are out of phase?
 
  • #11
Do they just want Path Difference = nλ...
 
  • #12
julianwitkowski said:
Do they just want Path Difference = nλ...
Yes, except perhaps in the form λ = ...
 
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  • #13
haruspex said:
Yes, except perhaps in the form λ = ...
Thanks for all your help. You are a gentleman and a scholar.
 

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