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Homework Help: Please help: Simple Wave Interference Problem

  1. Aug 25, 2014 #1
    1. The problem statement, all variables and given/known data

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

    a) i) measure the wavelength of the waves, ii) the distance between sources, and the iii) path distance from each of the sources to the nodal point shown on the diagram. iv) Show the complete calculation for wavelength
    b) Show the complete calculation for wavelength of the antinodal point shown on the diagram

    2. Relevant equations

    http://www.flickr.com/photos/100153549@N06/9487690991/ (source- another student)
    lPnS1-PnS2l =(n-1/2)λ
    lPnS1-PnS2l = mλ

    3. The attempt at a solution

    I am having problems here.

    a) i) Wavelength of waves = 8 mm
    *I measured this crest to crest vertically from the source points- Is this right?

    ii) Distance between sources is 13mm

    iii) nodal point 1 (top):
    P1S1= 36.5 mm
    P1S2= 44.5 mm

    anodal point 2 (bottom):
    P2S1= 40 mm
    P2S2= 34 mm

    iv) nodal point 1, n=2
    l PnS1-PnS2 l =(n-1/2)λ
    l 40 mm-34 mm l= 1.5λ
    ∴ λ= 6/1.5mm= 4mm

    anodal point 2 (bottom):
    nodal point 2, m=1
    lPnS1-PnS2l =mλ
    l 40 mm-34 mm l= (1)λ
    ∴ λ= 8mm

    Okay, so I know that for iv) the answers are not correct. As any point on the second nodal line the path difference (in lamda) will always be n-0.5= 2-0.5= 1.5 λ, but my answer is 40mm-34mm/8 = 0.75 λ And... for any point on the first anoidal line, the path difference (in lamda) should be mλ= 1λ, where my answer is 6λ.

    Can someone please help explain where I am going wrong?
    Thank you for your time.
  2. jcsd
  3. Aug 25, 2014 #2
    Correction. I have added iv) and b) together under iv). Thanks!
  4. Aug 26, 2014 #3
  5. Aug 26, 2014 #4
    Hi amandinger,
    I am sorry for the confusion. I pasted the link as the photo had already been uploaded from another source. Should I reload with a mm scale? Or is this better...The photo (coloured part) measures 74mm W x 87mm L.
    I looked at this website, but I still don't know what I am doing wrong.
  6. Aug 26, 2014 #5
    Two things I can tell you right now, is that you're right on ii) and wrong on i).

    I've uploaded the image you attached to SketchUp (basic CAD) and scaled it so that I can digitally measure the picture.

    ii) Yes, it appears the distance between centers is 13mm.

    i) You're right on the idea of measuring crest to crest, but I'm finding a different result. I decided to measure valley to valley (I'm taking those as the dark lines) because they are easier to measure. As you can see from the picture where I measured, I'm getting about 4.8 mm as the wavelength. Again, this is measured vertically from the center so that there isn't any interference from the other source. Also, I measure a little far out, because the center is quite congested.

    Attached Files:

  7. Aug 26, 2014 #6

    I'm now taking the measurements to the points, and am getting (in mm), with P1 being the top nodal point and P2 being the bottom anodal point, as you said:

    P1S1: 36.6
    P1S2: 39.6

    P2S1: 44.6
    P2S2: 33.8

    Some consistent, some not.

    The wavelength I found should make the largest difference.

    I'm not sure if your choice of n and m are correct, but that's the last piece of the puzzle. I would recommend trying those equations again with the new measurements. Also, pay attention to algebraic errors, you've made one. 40-34=6, not 8.
  8. Aug 27, 2014 #7
    Ok. So I remeasured the sources to the points a little more accurately.
    I also have sketchup. Never thought to use it as I thought that a ruler would have been accurate enough. But I'll import it and measure accordingly just to see. Measuring w/ ruler I get...

    Source 1 to the top nodal point n=2 (P2S1- I renamed it more accurately)= 36.5mm, and for P2S2=44.5mm.
    Source 1 to bottom anodal point m=1 (P1S1, again naming it more accurately)=39.5mm, for source 2 to that point(P2S1)=33.8mm
    Amadinger, are you certain that your P1S2 isnt switched for your P2S1? Because my measurements would very similar to yours if this was the case.

    For n=2
    l 36.5- 44.5 l = 2- 1/2 (lambda)
    Lambda= 5.33 mm
    For m=1
    l 39.5 - 33.8 l = 1(lambda)
    Lambda= 5.70 mm
    Would it make sense that they should be closer in value? Can you advise if I`m on the right track?
  9. Aug 27, 2014 #8
    Good catch! I've just confirmed that your values are correct in post #7 above, ignore my reported values.

    I've found a great video on the matter, and I can confirm you've got it! Check it out here: https://www.youtube.com/watch?v=_Vn...d8&feature=iv&annotation_id=annotation_840322

    Yes, your numbers should be close in value. If done right, they should be the same value. Here is what you are calculating.

    For a node on the n=1 line, you know that one wave traveled half a wavelength longer than the other in order for them to cancel. On the n=2 line (as in this case) you know one traveled one and a half wavelengths longer than the other in order for them to cancel. The video does a great job explaining this.

    So! If you can measure the two distances, and calculate the difference (how much longer one is than the other), you can relate that to the wavelengths of the waves creating the pattern. In this case, both sources are creating waves with the same wavelengths, and that's what you are after. The answer is probably somewhere between 5.33 and 5.7. You'd need more accurate P dots (they're relatively large in the picture) in order to get a more accurate answer!

    So you've got the right answer in #7! I hope it makes sense. I highly recommend the video.


    PS: Now that I understand it again, it looks like your initial problems arose from using P2S1 in both equations instead of just one, and a math errors. If the concept makes sense to you, then you are golden.
    Last edited: Aug 27, 2014
  10. Aug 31, 2014 #9
    Great video! Thank you for the link. But the more I look at this, I am wondering something. As you've confirmed, both wavelengths should be close or/the same value, but should it mean also that these values also should be close to/the same as the wavelength measurement taken directly from trough-to-trough?
    Also, I know that you said that my answers are correct but, what is an acceptable deviation in value if comparing only those two values? For example, is 0.0-0.5 acceptable?
    Thank you!
  11. Aug 31, 2014 #10
    Interesting question. I'm not sure about the answer, but I would think 'yes' if there were no interference on the vertical line. I would think there would be some, but can't confirm. That's the question you'd want to answer.

    As for your error range, that is a much more complicated question than you may think. It involves determining the amount of error possible by the size of the black dots, and propagating that error through the equations used. I think this is beyond your scope of study right now. I learned it my senior year of engineering undergrad. For this case, a .5 difference in wavelength is acceptable.
  12. Sep 1, 2014 #11
    Thank you so much for all your help!
  13. Sep 1, 2014 #12
    You're very welcome! Are you all set?
  14. Sep 1, 2014 #13
    I believe so. I will present the info as is with a "disclaimer" stating the inaccuracies due large source and nodal/antinodal points, how they should be close or exact in value. This way it shows that I know understand what is going on. Thanks again!
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