A beam of natural light falls on the surface of glass at an angle of `45^(@)`. Using the Fresnel equations, find the degree of polarization of
(a) reflected light,
(b) refracted light.

Explanation

Here `theta_(1) = 45^(@)`
`sin theta_(2) = (1)/(sqrt(2)) xx (1)/(n) = (2)/(3sqrt(2)) = (sqrt(2))/(3) = 0.4714`
`theta_(2) = sin^(-1) 0.4714 = 28.1^(@)`
Hence `I'_(bot) = I_(bot) (sin^(2)(theta_(1) - theta_(2)))/(sin^(2)(theta_(1) + theta_(2)))`
`= (1)/(2)I_(0) ((sin 16.9^(@))/(sin73.1^(@)))^(2) = (1)/(2)I_(0)xx 0.0923`
Thus `I'_(||) = (1)/(2)I_(0) ((tan 16.9^(@))/(tan73.1^(@)))^(2) = (1)/(2)I_(0) xx 0.0085`
(a) Degree of polarization `P` of the reflected light
` = (0.0838)/(0.1008) = 0.831`
(b) By conservation of energy
`I''_(bot) = (1)/(2)I_(0) xx 0.9077`
`I''_(||) = (1)/(2)I_(0) xx 0.9915`
Thus `P =( 0.0838)/(1.8982) = 0.044`