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Interference Question

  1. Mar 5, 2006 #1
    Two slits are separated by 0.180 mm. An interference pattern is formed on a screen 80.0 cm away by 656.3-nm light. Caclulate the fraction of the maximum intensity 0.600 cm above the central maximum.
    So to my understanding
    we know the wave-length = 656.3nm
    We know the distance between the two slits = .180 mm
    and we know how far away we are from the screen .800 m

    I know Light intensity(maximum)=
    I=I_max*(cos^2(pi*d*sin(thetha)/lambda))
    I know the relatship between phi and lambda as well..
    but I don't know how to go from here. Maybe I am heading the wrong directon with is? Any suggestions would be helpful.
    Thank you.
     
  2. jcsd
  3. Mar 6, 2006 #2

    mukundpa

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    1. What is the distance between two consequetive maximas?
    2. What is the phase difference between two consequetive maximas?
    3.What is the phase at a point 0.600 cm above the central maximum?

    Try to answer these questions.

    MP
     
  4. Mar 6, 2006 #3
    1. The distance by two consequitive maximas is Lambda.
    lambda is 656.3nm
    2. If PHI is in phase then phi between two consecutive maxima should be:
    PHI=m(2pi), where m =0,(+-)1,(+-)2,...etc.
    If PHI is not in phase,
    PhI= 2pi(m+1/2) where m =0,(+-)1,(+-)2,...etc.
     
    Last edited: Mar 6, 2006
  5. Mar 6, 2006 #4

    mukundpa

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    Sorry,

    The distance between two consequetive maxima is not lembda, this is the path difference for the two maxima.

    The distance between them is called fringewidth and is D(lembda)/d, where D is the screen distance and d is the distance between slits..
     
  6. Mar 6, 2006 #5
    Oh, it doesn't mention that in the book. The only mention of fringes is the discussion of the dark and light bands that appear. But that isn't explicity given.
    Wait, in our book it is different.
    they define it as
    D_bright= Lambda * L(distance from the screen)/d
    where d is the distance between the slit.
    t
    D_dark = Lambda*L/d(m+1/2)
    which is what i presume was meant by fringe distance.
    All right, i still don't see how this would answer the issue about light intensity. Unless this is what is meant? i don't think it is.
     
    Last edited: Mar 6, 2006
  7. Mar 6, 2006 #6

    mukundpa

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    ok

    In distance [D(lembda)/d] the phase changes by 2*(pi) so what is change in phase in distance in x =0.600 cm

    This will guve you phase difference between the two waves at that point...
     
  8. Mar 6, 2006 #7
    Nevermind. I just realized the following:
    I~=I_max*cos((pi*d/lambdaL)y)^2
    we know all those constants. The reason why we can use this is because our thetha is relatively small.
     
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