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## Homework Statement

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:

## 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

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