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Finding critical angles and index of refractions

by mark9159
Tags: angles, critical, index, refractions
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mark9159
#1
Apr29-05, 08:01 AM
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 29.2. What is the index of refraction?

so...Sin(29.2)= 1/n?...

my answer for this question is 2.049, rounded to 2.05

thank you,

mark
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xanthym
#2
Apr29-05, 08:34 AM
Sci Advisor
P: 412
Quote 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 29.2. What is the index of refraction?

so...Sin(29.2)= 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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OlderDan
#3
Apr29-05, 04:20 PM
Sci Advisor
HW Helper
P: 3,033
Quote Quote by mark9159
im having trouble setting my TI-83 Plus into "degree mode"
Press [MODE] and cursor down and right to the word Degree and hit [ENTER]

mark9159
#4
Apr29-05, 06:44 PM
P: 13
Finding critical angles and index of refractions

thank you very much
haydenbaird
#5
Apr12-10, 04:38 PM
P: 1
hahahhahah. are you in the BYU online homeschool thing too?


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