A beam of plane-polarized light falls on the surface of water at the Brewster angle. The polarization plane of the electric vector of the electromagnetic wave makes an angle `varphi = 45^(@)` with the incidence plane. Find the reflection coefficient.

Explanation

The amplitude of the incident component whose oscillation vector is perpendicualr to the plane of incidence is
`A_(bot) = A_(0)sin varphi`
and similarly `A_(||) = A_(0) cos varphi`
Then `I'_(bot) = I_(0) (sin^(2)(theta_(1) - theta_(2)))/(sin^(2)(theta_(1) - theta_(2))) sin^(2) varphi`
`=I_(0) [(sintheta_(1)cos theta_(2) - cos theta_(1)sin theta_(2))/(sin theta_(1)cos theta_(2) + cos theta_(1) sin theta_(2))]^(2)sin^(2) varphi`
`=I_(0) [(n^(2) - 1)/(n^(2) +1)]^(2)sin^(2) varphi`
Hence `rho = (I'_(bot))/(I_(0)) = [(n^(2) - 1)/(n^(2) +1)]^(2)sin^(2) varphi`
Putting `n = 1.33` for water we get `rho = 0.0386`