Finding critical angles and index of refractions

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To find the critical angle for a material with an index of refraction of 1.75, use the formula θ_critical = sin⁻¹(1/n), resulting in approximately 34.85 degrees. For cubic zirconium with a critical angle of 29.2 degrees, the index of refraction can be calculated as n = 1/sin(29.2º), yielding an index of about 2.05. Users are advised to set their TI-83 Plus calculator to degree mode by pressing [MODE] and selecting "Degree." The discussion also touches on shared experiences in online education. Understanding these calculations is essential for solving refraction problems effectively.
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im having trouble setting my TI-83 Plus into "degree mode" so i can't 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) ?? I am 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
 
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mark9159 said:
im having trouble setting my TI-83 Plus into "degree mode" so i can't 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) ?? I am 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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Last edited:
mark9159 said:
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]
 
thank you very much
 
hahahhahah. are you in the BYU online homeschool thing too?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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