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Heisenberg's uncertanty

  1. May 26, 2015 #1
    1. The problem statement, all variables and given/known data
    particle pass through a slit of width 0.200mm. the wavelength is 633nm. after it pass through the slit they spread out over a range of angles . use uncertainty principle to determine the min. range of angles.

    2. Relevant equations
    (delta)p*(delta)y = h/4(pi)

    3. The attempt at a solution
    I have done some calculations using heisenbergs uncertainty principle but nothing like this.
    I usually did it in y-direction but it seems this one needs to be done in x direction so I though;
    (delta)px*(delta)x = h/4(pi) but would (delta)x be the width of the slit or the uncertanty of the particles in x-direction after passing through the slit? Also, how would (lambda) be integrated into this? The only thing I can think of is to use (delta)E * (delta)t = h/4(pi) and set E = hc/(lambda), however that would just give me the time? or use h/(lambda) * (delta)x = h/4(pi)

    Need some hints here..

    R
     
    Last edited: May 26, 2015
  2. jcsd
  3. May 26, 2015 #2
    You are only going the need to consider y direction from the looks of you since you are only confining the position in the y-plane.

    The slit size is the uncertainty in the y position of the photon. So from this calculate the uncertainty in the y momentum, call it p.
    Then the range in y momenta is from +p/2 to -p/2 . Use the expression for momentum and wavelength and consider the geometry.
     
  4. May 26, 2015 #3
    Also p = h / lambda
     
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