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Multiplayer Refraction, Incident Angle

  1. Mar 18, 2015 #1

    rlc

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    1. The problem statement, all variables and given/known data
    The figure below shows the path of a beam of light through several layers (n1 = 1.58, n2 = 1.42, n3 = 1.20 and n4 = 1.00) of different indices of refraction.
    https://loncapa2.physics.sc.edu/res/brookscole/serway/College_Physics_7ed/Chap22/graphics/serw2244.gif

    a) If θ1 = 30.3 deg, what is the angle, θ2, of the emerging beam? (5.29×10^1 deg)

    b) What must the incident angle, θ1, be in order to have total internal reflection at the surface between the n3 = 1.20 medium and the n4 = 1.00 medium?

    2. Relevant equations
    For part a)
    n1sin(theta1)=n2sin(alpha)
    n2sin(alpha)=n3sin(beta)
    n3sin(beta)=n4sin(theta2)

    For part b) I'm not too sure.

    3. The attempt at a solution
    For part a)
    n1sin(theta1)=n2sin(alpha)
    (1.58)sin(30.3)=(1.42)sin(alpha)
    alpha=34.15099

    n2sin(alpha)=n3sin(beta)
    (1.42)sin(34.15099)=(1.20)sin(beta)
    beta=41.628239

    n3sin(beta)=n4sin(theta2)
    (1.20)sin(41.628)=(1)sin(theta2)
    theta2=52.90 deg ---> this is the correct answer
    (the formulas can also be simplified to... theta2=arcsin(n1sin(theta1)/n4)

    For part b)
    This is the part I"m having difficulties with.
    (1.20)sin(theta)=(1.00)sin(90)
    sin(theta)=(1.00)/(1.20)
    theta=56.44269 deg
    I know that this isn't the answer. Am I right in connecting the total internal reflection with sin(90)? Is the critical angle equivalent to the incident angle...are they the same thing?
     
  2. jcsd
  3. Mar 18, 2015 #2

    rlc

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  4. Mar 18, 2015 #3
    That angle that you found is the angle of incidence in the n3 medium. I think the question asks you for θ1, which is the incidence angle in the n1 medium. If you use Snells law to find θ1, that might be what you're looking for.
     
  5. Mar 20, 2015 #4

    rlc

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    (n3)sin(beta)=(n4)sin(90)
    (1.20)sin(beta)=(1.00)sin90
    beta=56.44

    (n2)sin(alpha)=(n3)sin(beta)
    (1.42)sin(alpha)=(1.20)sin(56.44)
    alpha=44.7669

    (n1)sin(theta1)=(n2)sin(alpha)
    (1.58)sin(theta1)=(1.42)sin(44.7669)
    theta1=39.26 deg
     
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