Finding critical angles and index of refractions

  • Thread starter Thread starter mark9159
  • Start date Start date
  • Tags Tags
    Angles Index
AI Thread Summary
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.
mark9159
Messages
13
Reaction score
0
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
 
Physics news on Phys.org
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)


~~
 
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?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top