Critical angle for certain medium is `sin^(-1)` (0.6). The polarizing angle of that medium is

Options

A. `tan^(-1)[1.5]`
B. `sin^(-1)[0.8]`
C. `tan^(-1)[1.6667]`
D. `tan^(-1)[0.6667]`

Answer

C

Explanation

Here, critical angle, `C=sin^(-1)(0.6)` As, `sinC=(1)/(n)` where n is the refractive index of the medium
`n=(1)/(sinC)=(1)/(sin(sin^(-1)(0.6)))=(1)/(0.6)`
According to Brewster.s law
`n=tani_(p)`, where `i_(p)` is the polarizing angle
`i_(p)=tan^(-1)(n)`
`=tan^(-1)((1)/(0.6))=tan^(-1)(1.6667)`