Determining the number of antinodal and nodal lines

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Homework Help Overview

The discussion revolves around determining the number of antinodal and nodal lines produced by two coherent sources vibrating in phase, separated by a distance of 4.5 wavelengths. Participants are exploring the implications of the separation distance on the resulting interference pattern.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of nodal and antinodal lines based on the given separation and question whether endpoints should be included in the count. There is also a consideration of how changes in the separation distance might affect the results.

Discussion Status

The discussion is active, with participants questioning the assumptions regarding the counting of lines and the implications of the separation distance. Some guidance has been offered regarding the interpretation of the formula, but no consensus has been reached on the specifics of counting lines or the impact of slight variations in separation.

Contextual Notes

Participants note the ambiguity in the problem statement regarding the counting of lines at the endpoints and the potential for infinite lines if not constrained to a specific region. The discussion also touches on the nuances of the formula used for calculating nodal lines.

superconduct
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Homework Statement


For two coherent sources vibrating in phase separated by 4.5(lambda), what are the numbers of antinodal and nodal lines?

Homework Equations


No. of nodal lines equals twice the floor function of separation over wavelength.

The Attempt at a Solution


No. of antinodal lines= 2(4)+1=9
No. of nodal lines= 2(4)=8

Is this correct?
 
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superconduct said:

Homework Statement


For two coherent sources vibrating in phase separated by 4.5(lambda), what are the numbers of antinodal and nodal lines?

Homework Equations


No. of nodal lines equals twice the floor function of separation over wavelength.

The Attempt at a Solution


No. of antinodal lines= 2(4)+1=9
No. of nodal lines= 2(4)=8

Is this correct?
Are you supposed to count nodes at the endpoints? The question as you've posted it does not even limit it to the space between the sources, so you could argue there's an infinity of each.
 
haruspex said:
Are you supposed to count nodes at the endpoints? The question as you've posted it does not even limit it to the space between the sources, so you could argue there's an infinity of each.
What does it mean by endpoints here? How are there an infinity of nodal and antinodal lines? Under what conditions will the number be limited?
 
superconduct said:
What does it mean by endpoints here? How are there an infinity of nodal and antinodal lines? Under what conditions will the number be limited?
No, sorry, forget the bit about infinity.
But for the endpoints, I mean nodal lines straight out to the side from the two sources. In this set-up, if you move from one source directly away from the other source, they will cancel all the way along that line, no? Same for the other source. Are these two lines counted? The formula you quote implies not.
 
haruspex said:
No, sorry, forget the bit about infinity.
But for the endpoints, I mean nodal lines straight out to the side from the two sources. In this set-up, if you move from one source directly away from the other source, they will cancel all the way along that line, no? Same for the other source. Are these two lines counted? The formula you quote implies not.
Do you mean the two lines on the plane joining the sources and with a path difference of 4.5 lambda ?
 
superconduct said:
Do you mean the two lines on the plane joining the sources and with a path difference of 4.5 lambda ?
yes
 
haruspex said:
yes
If the separation becomes 4.4999999lamba instead of 4.5, then is my answer above correct?
 
superconduct said:
If the separation becomes 4.4999999lamba instead of 4.5, then is my answer above correct?
Yes I think so, it's just the special case of n+1/2 I was asking about. It seems to me the formula you quote doesn't quite work there.
 

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