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Diffraction Understanding a solution

  1. Mar 9, 2016 #1
    1. The problem statement, all variables and given/known data
    (see attached problem)

    2. Relevant equations
    (see attached problem)

    3. The attempt at a solution
    So assuming the area of the hole is circular, halving its diameter, will decrease its area by a factor of 16. Therefore LI will be I/16? However in the solution (see attached solution) it says: 'Halving the diameter of the hole will decrease the power through it by 1/4'. How does one make this conclusion?
     

    Attached Files:

  2. jcsd
  3. Mar 9, 2016 #2

    blue_leaf77

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    Because ##P=IA##, therefore any change in the scaling of ##A## will result in the same amount of scaling in ##P##.
     
  4. Mar 9, 2016 #3

    perhaps some error is there- Area is proportional to (d/2)^2 !

    i think one should remember 'as given' that halving the diameter will decrease the area of hole by a factor of four - but at the same time the diffracted beam increases its width by a factor of 2 - so it leads to a further spread of the same power in four times the area (decrease by a factor of four )so one gets a net 1/16 of the power.
     
  5. Mar 9, 2016 #4
    Why does twice the width mean 4 times the area?
     
  6. Mar 9, 2016 #5

    blue_leaf77

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    ##A_1=\frac{1}{4}\pi D_1^2##. Now if you set ##D_2=2D_1##, what is ##A_2=\frac{1}{4}\pi D_2^2## in terms of ##A_1##?
     
  7. Mar 9, 2016 #6
    So the new area is 4 times the old one I see, so we are just assuming that the area of the beam incident on the screen will be circular right?
     
  8. Mar 9, 2016 #7

    blue_leaf77

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    You are supposed to assume that the illumination is of plane wave from the left.
     
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