Finding speed of light in a material from critical angle of 32.4°

[jaron]

Homework Statement

the critical angle for a ray of light leaving an unknown material (surrounded by air) is
32.4'(degrees celsius). the speed of light in the material is:

Homework Equations

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

The Attempt at a Solution

niSin32'/Sin32' = 1Sin90'/Sin32'
ni = 1/0.53
ni = 1.89x10^8 m/s
that's 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

 

[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?

 

[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.