Finding critical angles and index of refractions

  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

    and

    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,

    mark
     
  2. jcsd
  3. xanthym

    xanthym 412
    Science Advisor

    In the following discussion, sin-1 is the Inverse Sine function.
    SOLUTION HINTS:
    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. OlderDan

    OlderDan 3,030
    Science Advisor
    Homework Helper

    Press [MODE] and cursor down and right to the word Degree and hit [ENTER]
     
  5. thank you very much
     
  6. hahahhahah. are you in the BYU online homeschool thing too?
     
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