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Snell's Law and Lasers

  1. Apr 17, 2012 #1
    1. The problem statement, all variables and given/known data
    A laser beam of diameter = 3.1mm in air has an incident angle = 26 degrees at a flat air-glass surface.If the index of refraction of the glass is n=1.5, determine the diameter of the beam after it enters the glass.


    2. Relevant equations
    n1sinθ1=n2sinθ2


    3. The attempt at a solution
    okay, well the initial application of snell's law is obvious. The laser is going from air, where n = 1 into glass where n = 1.5

    So, (1)sin(26) = (1.5)sinθ2

    θ2 = (approx) 17 degrees or 16.99

    I know how to find the diameter of the laser, the solutions have this equation:

    d1/cosθ1 = d2/cosθ2

    and simple plug and chug would yield the answer 3.3 mm

    But I just don't understand where this second equation came from. Can some one walk me through the sense behind it? My book doesn't even mention it, so perhaps its obvious (just not to me).
     
  2. jcsd
  3. Apr 18, 2012 #2

    ehild

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    A picture always helps.


    ehild
     

    Attached Files:

  4. Apr 18, 2012 #3
    I see, the two diameters are related through the common interface (the hypothenuse of the two triangles). Just some geometry from there.

    It makes sense, thank you :)
     
    Last edited: Apr 18, 2012
  5. Apr 18, 2012 #4

    ehild

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    You are welcome. And start drawing!:smile:

    ehild
     
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