Critical angle for a medium is `sin^(-1) (0.6)`, then polarization angle for this medium is ......

Options

A. `sin^(-1)(0.8)`
B. `sin^(-1)(1.5)`
C. `tan^(-1)(1.6666)`
D. `tan^(-1)(0.6)`

Answer

C

Explanation

Critical angle `C= sin^(-1)(0.6)`
`:. sinC=0.6`
`:.(1)/(n)=0.6`
`:.n=(1)/(0.6)`
`:. "tan" theta_(p)=(1)/(0.6) " " [ :. n= tan theta_(p)]`
`:. theta_(p)=tan^(-1)((1)/(0.6))`
`:.theta_(p)tan^(-1)((1.6666)`