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Homework Help: Incident angle

  1. Jul 26, 2009 #1
    1. The problem statement, all variables and given/known data
    If the apex angle of a prism is f = 74.3° (see figure below),


    what is the minimum incident angle for a ray if it is to emerge from the opposite side (i.e., not be totally internally reflected), given n = 1.55?

    2. Relevant equations

    sin(alpha + delta)/2 =nsin(alpha/2)

    delta = incindence angle + alpha/2

    3. The attempt at a solution

    delta = min angle of deviation
    alpha = apex angle

    I solved for delta and got 65.4 deg

    Then I used the second equation to solve for the incident angle and got 28.3 deg

    I'm not even sure if I should go so far as to solve for delta. I'm thinking there is a simpler way
  2. jcsd
  3. Jul 26, 2009 #2


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    Science Advisor
    Homework Helper

    Hi NikkiNik! :smile:

    (have an alpha: α and a delta: δ and a phi: φ :wink:)
    Sorry, I've no idea what you're doing. :redface:

    Where does "/2" come from?

    And what are your first incident angle, refracted angle, and second incident angle? :confused:
  4. Jul 26, 2009 #3
    I found in the equations in my book

    (1) (sin(α + δ))/2 = n*sin(α/2)

    All I'm given is the apex angle and n and I'm not sure where to go from there. Since I'm not supposed to find the critical angle I'm not sure how to go about finding the incident angle.
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