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Refractive Index question

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data
    I uploaded the question below.


    2. The attempt at a solution

    At first what I did was using the Snell's Law formula [tex]n_1 \sin{\theta _1}=n_2 \sin{\theta _2}[/tex]

    then I did this, [tex]n\sin{\ 45}=1 \sin{\ 90}[/tex]

    so this is the minimum value that I got n=1.41

    I'm not sure if my answer is correct or not. Also how do you justify that particular n value is minimum
     

    Attached Files:

  2. jcsd
  3. Jan 24, 2010 #2

    ehild

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    Gold Member

    Total reflectance occurs here when [itex]n\sin(45)\geq 1[/itex], so [itex]n\geq 1/\sqrt{2}[/itex]. Your value is minimum as total reflectance holds for higher refractive index values, too.

    ehild
     
  4. Jan 25, 2010 #3
    I still can't get it
     
  5. Jan 25, 2010 #4
    The vocabulary phrase you want to look up is "total internal reflection."
     
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