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Finding critical angles and index of refractions

  1. Apr 29, 2005 #1
    im having trouble setting my TI-83 Plus into "degree mode" so i cant seem to get the answers for these problems...im probably doing them wrong too

    A material has an index of refraction of 1.75. What is the critical angle for this material?

    Sin(critical angle)=1/n, n being the index of refraction
    how would i set up the formula? (critical angle)= Sin(n) ?? im so confused


    The critical angle for cubic zirconium is 29.2º. What is the index of refraction?

    so...Sin(29.2º)= 1/n?...

    my answer for this question is 2.049, rounded to 2.05

    thank you,

  2. jcsd
  3. Apr 29, 2005 #2


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    In the following discussion, sin-1 is the Inverse Sine function.
    a) sin(θcrit) = (1/n) :: ⇒ :: θcrit = sin-1(1/n) = sin-1{1/(1.75)} = (34.85 deg)
    b) sin(29.2 deg) = 1/n :: ⇒ :: n = 1/{sin(29.2 deg)] = (2.05)

    Last edited: Apr 29, 2005
  4. Apr 29, 2005 #3


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    Homework Helper

    Press [MODE] and cursor down and right to the word Degree and hit [ENTER]
  5. Apr 29, 2005 #4
    thank you very much
  6. Apr 12, 2010 #5
    hahahhahah. are you in the BYU online homeschool thing too?
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