How to Calculate Wavelength in a Two-Dimensional Wave Interference Pattern

[julianwitkowski]

Homework Statement

Using the two-dimensional wave interference pattern shown and the two equations involving path difference, complete the following.

i) measure the wavelength of the waves
ii) measure the distance between sources
iii) measure the path distance from each of the sources to the nodal point shown.

* note that this is a diagram I've made... I can't get the image out of the book for you at this moment but it's basically like this with wave interference lines... The dotted lines are obviously my nodal lines and I drew the triangle over it.

Do they just want me to measure this with a protractor and ruler?
For the wave I measured crest to crest...
With S₁ being the left source at the bottom, is this what I'm supposed to measure?

iv) show the complete calculation for wavelength

I don't know if this is equation and I'm not sure about the variables.

| P_nS₁ - P_nS₂ | = (n - ½) λ

Is this just the absolute value of the line P_nS₁ - P_nS₂ or absolute value of P_n ⋅ S₁
- P_n ⋅ S₂... If it is the later, I understand it less because it seems like coordinates to me.

For the anti nodal point in b) I believe this is the equation.
d sin ϴ = (m + ½) λ = mλ

ϴ = Is this the angle at the point separating S₁ and S₂?

For...

n = nodal line
m = antinodal line

I don't understand how to get numbers for n and m... I know what nodal and antinodal lines are in the wave interference pattern, but I'm not sure what the variable is a measure of?

b) Choose a point on any anti nodal point and show the complete calculation for wavelength.

Homework Equations

a) | P_nS₁ - P_nS₂ | = (n - ½) λ

b) d sin ϴ = m λ = (m + ½) λ

The Attempt at a Solution

i. 8mm
ii. 23mm
iii.P_nS₁ = 46mm
P_nS₂ = 63mm

I'm looking for advice on how to tackle the complete wavelength equations.
I also would like to understand the values of nodal and anti nodal variables n and m, that would probably set me straight.

Thanks :)

 

[julianwitkowski]

I edited a bit from first post...

 

[haruspex]

I don't know what the P_n are, so I can't comment on that equation. Your other equation, involving sin(theta) will do it, except that you seem to have conflated two equations - one for nodes and one for antinodes - into one equation.

 

[julianwitkowski]

I don't know what the P_n are, so I can't comment on that equation. Your other equation, involving sin(theta) will do it, except that you seem to have conflated two equations - one for nodes and one for antinodes - into one equation.

Thank you for the reply :)

I believe you mean the variables P_n, S₁ and S₂.
I believe these are points to make lines. P_n is the black dot on the nodal line.
S₁ and S₂ are the sources of the interference.
Here is the image from the question I captured with a cellphone camera ...so not to scale.

Here is a labeled diagram with what I think I'm supposed to measure...

I measured this with a ruler.

(Crest to Crest) λ = 8mm
S₁S₂ = 23mm
P_nS₁ = 46mm
P_nS₂ = 63mm

I don't know where the m and n numbers are to come from unless they're given to me.
I sense a pattern with their arrangement in other examples.
If this is the case, it might be 2. Not confident.
For the complete wavelength equation I am not sure if it is based from this.

|P₁S₁ – P₁S₂ | = (n - ½) λ
|46mm - 63mm | = (n - ½) 8mm
17mm = 8mm ∙ n – 4mm
21mm = 8mm ∙ n
n = 21mm/8mm
.
I don't think mine is right because 12/8 is a fraction and not an integer.
Either that or maybe I'm measuring the wrong thing.

Lastly, I mentioned the anti-nodal equation because the second part b) asks to pick a point on an anti-nodal line and provide the complete wave length equation. That's a problem I'm confident will be easily solved once I know how to do this right.

Thanks for reading.

 

[julianwitkowski]

I'm so confused, but from research I've concluded n = 3 as the center nodal line is n = 1. In my diagram shown I also left out two nodal lines on either side at the edges almost horizontal to the source.

I'm seems slightly juvenile to be measuring this with a ruler.
That being said, I remeasured again...

For P_nS₁ I get 48mm from the center of the points in the image.
If I measure from the edge of each point I get 45mm.

For P_nS₂ I get 64mm from the center of the points in the image.
If I measure from the edge of each point I get 62mm.

It works like this, but I have to tamper with my measurements...

|45mm - 65mm| = (3 - ½)8mm

 

[julianwitkowski]

I can also measure P_nS₁ as 5λ since that's how many wave lengths it is from the S₁.
Unless I can't, but I saw it on hyper physics so I'd take it this is allowed.

I don't know how I can do this with S₂ since it is on a diagonal.

 

[SammyS]

I'm so confused, but from research I've concluded n = 3 as the center nodal line is n = 1. In my diagram shown I also left out two nodal lines on either side at the edges almost horizontal to the source.

I'm seems slightly juvenile to be measuring this with a ruler.
That being said, I remeasured again...

For P_nS₁ I get 48mm from the center of the points in the image.
If I measure from the edge of each point I get 45mm.

For P_nS₂ I get 64mm from the center of the points in the image.
If I measure from the edge of each point I get 62mm.

It works like this, but I have to tamper with my measurements...

|45mm - 65mm| = (3 - ½)8mm

Looking at that image it appears that the sources are approximately 180° out of phase. The "center" nodal line is close to lying on the perpendicular bisector of the line segment from one source to the other.

Using a ruler is a reasonable approach - not juvenile.

How did you measure wave length. Did you just measure just one single wave? Measure the length for several waves - let's say 8 consecutive - then divide to get a more accurate value.

Definitely measure from your best estimate of the location of the center of each source.

Rather than using a ruler, some of this can be done using a compass/divider . Doing that for your image, I observe that P is about 5.5 λ from one source and 7.5 λ from the other. Using your measured distances for P from the sources, indicates that λ is in the range of 8.5 to 8.7 mm.

(Any measurement made using this image on a computer monitor is subject to errors caused by using the incorrect aspect ratio.)

 

[julianwitkowski]

λλ

Looking at that image it appears that the sources are approximately 180° out of phase. The "center" nodal line is close to lying on the perpendicular bisector of the line segment from one source to the other.

Thanks for the reply :) Would you say that n = 3?

How did you measure wave length.

I did a eight measurements and came to 0.795 average for λ

Also for I came to 5.5λ for PnS1 and 8λ for PnS2 because it works with 3 as the nodal line number and 0.8 for λ

 

[julianwitkowski]

Do you think I'm wrong that n=3 and really n=2, because the middle I guess could be an antinodal, m=0.

Basically Is this right? ... m4 Pn3 m2 n2 m1 n1 m1 n2 m2 n3 m3 m4

Of this? ... m3 n3 m2 Pn2 m1 n1 m0 n1 m1 n2 m2 n3 m3

 

[SammyS]

λλ

Thanks for the reply :) Would you say that n = 3?

That depends as to what you mean for n to represent.

... but in general I would say no.

If the sources are out of phase (180°) then the center line is a node rather than an anti-node. It does occur where there is zero path difference from the two sources. I would use an index of zero there.

Point P is also on a nodal line. It's the second one over from the center line, so, I would give it an index of 2 .

I did a eight measurements and came to 0.795 average for λ

Also for I came to 5.5λ for PnS1 and 8λ for PnS2 because it works with 3 as the nodal line number and 0.8 for λ

If you're averaging eight separate measurements, each of a single wave's length, then I would consider that to be of questionable accuracy.

My suggestion is to start at some crest and measure the whole span to maybe the 8th or 9th crest. Then divide by the number of waves that represents. Do that along a few different anti-nodal lines & average.

 

[julianwitkowski]

If you're averaging eight separate measurements, each of a single wave's length, then I would consider that to be of questionable accuracy.

It's pretty inconsistent to say the least... Here I averaged 8.06mm but it's only estimation by eye out of 6 tries...

 

[julianwitkowski]

Everything else in this chapter is measuring in cm for units in the equations...
Really wish they didn't make this so hard just to measure properly.