The critical angle of a certain medium is `40.2^(@)`. The speed of light in that medium is

Options

A. `2.936 xx 10^(8)m//s`
B. `1.936 xx 10^(8)m//s`
C. `1.936 xx 10^(8)m//s`
D. `4.936 xx 10^(8)m//s`

Answer

C

Explanation

`mu = (v_(a))/(v_(m)) = (1)/(sin c)`
`(3 xx 10^(8))/(v_(m)) = (1)/(sin 40.2)`
`v_(m) = 1.936 xx 10^(8)m//s`