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Approximation question -- How does this approximation simplify the integral?

  1. Dec 4, 2014 #1
    how does this aproximation reduse the integral
    upload_2014-12-5_1-47-46.png
     
  2. jcsd
  3. Dec 4, 2014 #2

    mfb

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    Staff: Mentor

    Where are lambda and k_z in the integral?

    The approximation simply assumes the integrand to be constant, evaluates it at the midpoint (z'=0) and multiplies it with the range z' takes.
     
  4. Dec 5, 2014 #3
    so its because r'=0 ??
    then why we need
    l<<lambda kz<<pi
    ?
     
  5. Dec 5, 2014 #4

    mfb

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    The exponential function should not change its argument too much, so kzl/r should be small compared to 1.
     
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