1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interference - two wave source

  1. May 4, 2013 #1
    I do not know how to approach this. I am trying to use r1 - r2 = mλ but I cant seem to find the answer! please help.


    I figured it out that it is constructive, but i just dont know how to mathematically prove it.

    r1 - r2 = mλ.
     

    Attached Files:

  2. jcsd
  3. May 4, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    I suggest giving us the entire problem verbatim. For example, what is the frequency of the source?
     
  4. May 4, 2013 #3
    Sorry, I totally forgot about this. Both sources are on phase and emit radiation of 15 MHz in every direction. Radiation is detected at point P. They also say that at P, source 1 delivers an average of 4*10^-4 W on a small square of 1cm^2 .

    I also have an extra question. Imagine that the building is not there (the 10m long building). How can you calculate the phase difference at P from both sources? I feel like that can help out. Thank you.
     
  5. May 4, 2013 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Given f = 15MHz, what is the formula for wavelength?
     
  6. May 4, 2013 #5
    lets see, c = f*lambda, so then, c/f = wavelength. 3.0*10^8 / 15*10^6 = lambda. So, lambda= 20m.
     
  7. May 4, 2013 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    (Still assuming that the building is not there:)

    How far is it from each source to point P ?

    How many wave lengths does each of those distances represent?
     
  8. May 4, 2013 #7
    ok there is a right triangle there, so the distance from source 1 to P would be obviously 40m, and the distance from source 2 to P is 50m. The 40m represent 2 wavelengths and the 50m represent 2.5 wavelengths.
     
  9. May 4, 2013 #8

    rude man

    User Avatar
    Homework Helper
    Gold Member

    That's only true in vacuum (n = 1). What is is when n = 2 like in the building?
    (The first question asked how many wavelengths there were in the building).
     
  10. May 4, 2013 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That should tell you something regarding interference for the case of no building there.
     
  11. May 4, 2013 #10

    I believe λ changes with respect to n right? so λ = λnot/n, so λ= 20m/2 so λ=10m, is this the wavelength when n=2? I dont know if there is 1 wavelength in the 10m building, or 1/2 of the original one. I am assuming is 1wavelength of the new wavelength. is this correct?

    well, one arrives 0.5 wavelengths later, so is this interference destructive? if i wanna use the formula r2-r1 = mλ, then i would do 50-40=m20, where m is .5, so is destructive? How can I apply this for the problem with the building?
     
  12. May 4, 2013 #11

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes, destructive.

    rude man is well on the way to getting you to that answer.
     
  13. May 4, 2013 #12
    Thank you! you two are awesome, I am working on the one with the building, any help will be appreciated.
     
  14. May 4, 2013 #13

    rude man

    User Avatar
    Homework Helper
    Gold Member

    The actual formula for lambda is v/f.

    Now then, when n = 1, v = c. So the wavelength is only 1.5e8/15e6 = 10m, as you correctly decided. And since the building is 10m in length, how many of those new wavelengths can you fit in the building? And in the absence of the building, how many wavelengths can you fit in the same 10m?

    I'll let SammyS take it the rest of the way.
     
  15. May 5, 2013 #14
    In the presence of the building, you can fit 1 wavelength, in the absence you can fit only half. Let me see if I get this, assume we move the building to the very end of the 40m length. So up to 30m, there are 1.5 wavelengths, but when the wave enters the building, then it can 'fit' one wavelength, so at P, it arrives at 2.5 wavelengths in total? how can I do this using the r2-r1=mλ formula? thank you.
     
  16. May 5, 2013 #15

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Formulas, formulas, formulas.

    Can you do it by considering the physics of the situation? (Then we can discuss the formula with some confidence.)

    In terms of waves, both distances are 2.5 wavelengths. What is the nature of the resulting interference?
     
  17. May 5, 2013 #16
    Im sorry I didnt mean to get you upset or anything, but I wish to understand this fully for my upcoming test. They both arrive at 2.5 wavelengths so the interference is constructive I believe.
     
  18. May 5, 2013 #17

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes.

    So in some sense, r1 becomes equivalent to the distance occupied by 2.5 of these wavelengths.

    There is a quantity referred to as "optical path length". Yes, these are radio waves, a form of Electro-Magnetic wave, but so are light waves.

    The optical path length is the product of the index of refraction times the actual path length.

    The optical path length from Source 1 to point P consists of 30 meters through air and 10 meters through the building. that's
    1∙(30 m) + 2∙(10 m) = 30 m + 20 m = 50 m​

    Of course the optical path from Source 2 to point P remains 50 m .

    So if you use the respective optical path lengths for r1 and r2, you can use the equation r2 - r1 = mλ .
     
  19. May 5, 2013 #18
    Thank you!
     
  20. May 6, 2013 #19
    I would add a slight word of caution: that formula is a condition for constructive interference, but you shouldn't really think of applying it in the strict normal sense. What it is saying is, "does a whole number of wavelengths fit in the difference?" And if the answer is yes, then you have constructive interference. But if you just plug in r2 and r1 and subtract, you can get any number of course!!
     
  21. May 6, 2013 #20
    Thankss for replying, can you explain in more detail what you meant? are you saying that the difference in wavelengths must be a whole number otherwise the formula won't work?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Interference - two wave source
  1. Interference and Waves (Replies: 4)

  2. Interference waves (Replies: 6)

  3. Wave interference? (Replies: 9)

Loading...