# Finding critical angles and index of refractions

by mark9159
Tags: angles, critical, index, refractions
 P: 13 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
P: 412
 Quote by mark9159 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
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)

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