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Homework Help: Strange slit / wave interference question

  1. Jun 29, 2007 #1
    Let me start by saying this is just a problem posed to me. I had this prof before so some people thought I might could see what he was thinking.

    OK here is the problem

    two in phase sources of waves are separated by a distance of 4.00m these sources produce identical waves that have a wave length of 5.00 m. on the line between them there are two places at which the same type of interference occurs. a: is it constructive or destructive interference and b: where are the places located

    answer: destructive and at .75m and 3.25m

    the problem can be solved simply using a destructive interference equation where L1 and L2 are distance traveled by wave.

    L2-L1 = (m+1/2) lamba

    L2 = 4-x

    L1 = x

    That's not the problem , the professor claims this can easily be solved using just two wave equations.

    Asin((2pi/wavelength)x) = Asin((2pi/wavelength)(x+1)

    The claim is that if you graph this equation in a TI or Excel you get the solutions at the 0 x intercept

    However algebraic solutions to this doesn't work.

    So my first question is can this problem be solved just using wave equations. I figure it can if you know the phase difference.

    My second is why would this work graphically and not algebraically.

    What a great way to spend a Friday night!
  2. jcsd
  3. Jun 29, 2007 #2

    Doc Al

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    Not sure what this equation means. You can represent the amplitude of the waves as a function of position like this:
    [tex]y_1 = A\sin[(2\pi/\lambda)x + \omega t][/tex]
    [tex]y_2 = A\sin[(2\pi/\lambda)(4-x) + \omega t][/tex]

    where x is between 0 and 4 (between the two sources).

    For destructive interference, the phase difference must be +/- [itex]\pi[/itex]:
    [tex](2\pi/\lambda)(4-2x) = \pm \pi[/tex]

    Which is the same thing as your destructive interference equation.

    So I guess I don't know what your professor is talking about.

    I've done worse. :wink:
  4. Jun 29, 2007 #3


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    I don't quite follow. what is x?

    why a plus one there?
  5. Jun 29, 2007 #4
    Trust me I am asking the same questions. I had this guy for pchem years ago. Now my ex has him for physics and he is being really confusing. Like I said before the problem is quickly solved with an equation from the book but he tried to do if from wave equations and says they can't use the equation in the book.

    the x's above represent the two points of destructive interferences

    it occurs at x and at 4-x since 4 is the slit width.

    I myself asked where the +1 came from. It seems to me the way he has the equation written is when y on both wave functions equal 0 but what I think he should be looking for is when the sum of both waves equals 0.

    as far as I figure the phase difference can be related to the slit distance by taking the two points of destructive inteference, in this case x and 4-x and dividing it by wavelength since it is constant.

    so wave1 + wave 2 will equal 0 at each point.

    now i have not seen this done but he claims he graphed the this function and that y = 0 at the two points (those being the answers above)

    sin( ( 2pi / wavelength) x ) = sin( ( 2pi / wavelength) (x+1) )

    however no one in the class can tell me how he actually graphed it.

    and I don't know where the 1 come from.
  6. Jun 29, 2007 #5

    sorry to bother you with this but i am not following the manipulation from the first to questions to the next equation.
  7. Jun 29, 2007 #6


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    I think I figured it out. I am typing my explanation
  8. Jun 29, 2007 #7

    Doc Al

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    I'm just setting the phase difference between those two waves equal to [itex]\pm \pi[/itex]; that's the condition for destructive interference.
  9. Jun 29, 2007 #8


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    The two waves will cancel if

    [tex] sin (\frac{2 \pi}{\lambda} (4-x) - \pi) = sin (\frac{2 \pi}{\lambda} x) [/tex]

    i.e. they are out of phase by Pi (notice that sin(theta - pi) = - sin(theta)

    Now, I will manipulate the argument of the first sin and use the fact that the wavelength is 5 meters

    \frac{2 \pi}{\lambda} (4-x) + \pi = \frac{2 \pi}{\lambda} (5 -1 -x) + \pi

    = 2 \pi + \frac{2 \pi}{\lambda} (-1 -x) - \pi = \frac{2 \pi}{\lambda} (-1 -x) + \pi [/tex]
    where I have used that the wavelength is 5 meters.

    Now I use

    [tex] sin( \frac{2 \pi}{\lambda} (-1 -x) + \pi ) = - sin(\frac{2 \pi}{\lambda} (-1 -x) ) = sin \frac{2 \pi}{\lambda} (1+x) [/tex]


    Hope this makes sense.

  10. Jun 29, 2007 #9
    why do you think he claims

    sin( ( 2pi / wavelength) x ) = sin( ( 2pi / wavelength) (x+1) )

    gives the solutions graphically but he is unable to solve it algebraically?
  11. Jun 29, 2007 #10


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    It seems possible to me to solve this algebraically!!
  12. Jun 29, 2007 #11
    well , do you get the results at the beginning. also he claims this leads to 0=1
  13. Jun 29, 2007 #12


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    It's because he is not considering all the possibilities for the inverse of a sine.

    If one takes naively the inverse of the sin on boths ides, one obviously gets

    [itex] x = x + 1 \rightarrow 0 = 1 [/itex]

    However, there is another solution for the inverse sin.
    What I mean is the following: If one takes the inverse sin of [itex] sin(\theta) [/itex] one possibility is obviously [itex] \theta [/itex] modulo multiples of 2 pi.

    However, the inverse sin may also give [itex] \pi - \theta [/itex] modulo multiples of 2 pi. This is because [itex] sin(\pi -\theta) = sin (\theta) [/itex].

    Taking that second choice instead of the naive one (which is obviosuly incorrect in this case since it would give 0=1) does lead to the correct equation. Since it's past midnight here and I have been working on learning differential geometry for 5 hours straight, I will let you try it!

  14. Jun 30, 2007 #13
    hey thanks for the response,

    unfortunately , my mind has be consumed with teaching high school for three years so my math isn't up to par

    i am not sure what modulo means

    i just go on so i will try to figure out what is going on here.

    if you have anyway to help idiot proof this for me, by all means

    thanks again and also sorry that it doesn't come to me.
  15. Jul 1, 2007 #14
    figured it out, thanks
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