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Find angle required for total internal reflection of a beam of light

  1. Oct 10, 2013 #1
    1. The problem statement, all variables and given/known data

    An optical fiber consists of a glass core (index of refraction n1) surrounded by a coating (index of refration n2<n1). Suppose a beam of light enters the fiber from air at an angle θ with the fiber axis as shown in attached. (a) Show that the greatest possible value of θ for which a ray can travel down the fiber is given by θ = arcsin(sqrt(n1^2 - n2^2)). (b) If the indices of refraction of the glass and coating are 1.58 and 1.53, respectively, what is this value of the incident angle θ?

    2. Relevant equations

    na*sin(θa) = n1*sin(θ1)

    Critical Angle: θc = arcsin(n1/n2)

    3. The attempt at a solution

    See attached for my drawing. Any angle greater than θc will cause total internal reflection so,

    θo > θc

    where θo = 90 - θ1

    Substituting for θc

    θo > arcsin(n2/n1)

    Substituting for θo

    90 - θ1 > arcsin(n2/n1)

    Substituting for θ1

    90 - arcsin(sin(θa)/n1) > arcsin(n2/n1)

    Taking sine of the whole equation

    1 - sin(θa/n1) > n2/n1

    Solving for θa

    θa < arcsin(n1-n2)

    According to my calculations, this would be the maximum angle θa for which a ray can travel down the fiber. Unfortunately it's not correct.

    Where do they get the sqrt(n1^2-n2^2) term?

    Thanks in advance, MrMoose
     

    Attached Files:

  2. jcsd
  3. Oct 10, 2013 #2

    ehild

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    The equation in red is wrong. sin(90-ψ) is not 1-sinψ.

    ehild
     
  4. Oct 12, 2013 #3
    Nice, Thanks ehild! I think I got it now. Taking up from just before the incorrect equation:

    90 - arcsin(sin(θa)/n1) > arcsin(n2/n1)

    sin[90 - arcsin(sin(θa)/n1)] > n2/n1

    using the following property for sine: sin(x-y) = sinx*cosy - cosx*siny

    sin(90)*cos[arcsin(sinθa/n1)] - sin(θa)/n1*cos(90) > n2/n1

    Simplifying:

    cos[arcsin(sinθa/n1)] > n2/n1

    using the following property for cosine: cos[arcsin(x)] = sqrt(1-x^2)

    sqrt{1-[sin(θa)/n1]^2} > n2/n1

    solving for θa.

    θa < arcsin[sqrt(n1^2-n2^2)]

    Thanks for your help!
     
  5. Oct 12, 2013 #4

    ehild

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    Gold Member

    Good job!

    It is correct. Even simpler: the cosine of an angle θ in a right triangle is just the sine of the other angle, 90-θ. :smile:

    ehild
     
  6. Apr 7, 2015 #5
    It may be painfully obvious, but how exactly was θ1 = arcsin(sin(θa)/n1) derived?
     
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