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Refraction; beam width

  1. Feb 10, 2012 #1
    1. The problem statement, all variables and given/known data

    (Refraction)
    A parallel beam of width W is incident on a interface. The index of refractions are n1 on
    the incident side and n2 on the transmitted side.
    (a) Find an equation that gives the width of the transmitted beam. (Assume there is an
    exit beam. IE we are not above the angle of total internal reflection.)

    2. Relevant equations

    n1 sin theta1 = n2 sin theta2

    3. The attempt at a solution

    If I draw a diagram, it seems that the refracted width is the same as the incident width. Anyone care to explain?
     
  2. jcsd
  3. Feb 10, 2012 #2

    wukunlin

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    Gold Member

    haven't actually tried calculating the width of the exit beam?
     
  4. Feb 11, 2012 #3
    yes, and the width is unchanged (?). both top and bottom "rays" of the beam are refracted at the same angle and come out of the interface parallel to each other.
     
  5. Feb 11, 2012 #4
    Erm, unless the interface is not perpendicular to the incident beam...
     
  6. Feb 11, 2012 #5

    wukunlin

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    Gold Member

    yes, the question is asking for the general expression of the exit beam width in relation to the incident beam as a function of the incident angle (well, it wouldn't be wrong to write them as a function of the angle of refraction but it's trivial)
     
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