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Finding critical angles and index of refractions 
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#1
Apr2905, 08:01 AM

P: 13

im having trouble setting my TI83 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
Apr2905, 08:34 AM

Sci Advisor
P: 412

SOLUTION HINTS: a) sin(θ_{crit}) = (1/n) :: ⇒ :: θ_{crit} = 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) ~~ 


#3
Apr2905, 04:20 PM

Sci Advisor
HW Helper
P: 3,031




#4
Apr2905, 06:44 PM

P: 13

Finding critical angles and index of refractions
thank you very much



#5
Apr1210, 04:38 PM

P: 1

hahahhahah. are you in the BYU online homeschool thing too?



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