# Critical angle

1. Jun 16, 2009

### jaron

1. The problem statement, all variables and given/known data
the critical angle for a ray of light leaving an unknown material (surrounded by air) is
32.4'(degrees celcius). the speed of light in the material is:

2. Relevant equations
niSin(theda)c = nRSin(theda)R
(i am not sure how to make theda symbols on the computer)

3. The attempt at a solution
niSin32'/Sin32' = 1Sin90'/Sin32'
ni = 1/0.53
ni = 1.89x10^8 m/s
thats the answer that makes the most sense to me. but the back of my book has a different answer:
1.61x10^8 m/s (i have no idea how they got this)

i have a test on this stuff tomorrow

2. Jun 16, 2009

### rock.freak667

Well you can find the refractive index with respect to air given the critical angle..

What are the ratios you can use to find the refractive index?

3. Jun 16, 2009

### glueball8

hmm ni =/= 1.89*10^8 ... I guessing that's just a typo.

1/0.53... is right. I'm not sure how you got your answer but I got 1.61*10^8.

n=c/v

It might just be a math error.