A Nicol prism is placed in the way of partially polarized beam of light. When the prism is turned from the position of maximum transmission through an angle `varphi = 60^(@)`, the intensity of transmitted light decreased by a factor of `eta = 3.0`. Find the degree of polarization of incident light.

Explanation

If, as above,
`I_(1) =` intensity of natural component
`I_(2) =` intensity of plane polarized component
then `I_(max) = (1)/(2)I_(1)+I_(2)`
and `I = (I_(max))/(eta) = (1)/(2)I_(1) + I_(2) cos^(2) varphi`
so `I_(2) = I_(max) (1-(1)/(eta)) coses^(2)varphi`
`I_(1) = 2I_(max) [1-(1-(1)/(eta))coses^(2) varphi] = (2I_(max))/(sin^(2)varphi) [(1)/(eta)-cos^(2)varphi]`
Then `P = (I_(2))/(I_(1) + I_(2)) = (1-(1)/(eta))/(2((1)/(eta)-cos^(2)varphi)+1-(1)/(eta)) =(eta - 1)/(1-etacos2 varphi)`
On putting `eta = 3.0, varphi = 60^(@)`
we get `P = (2)/(1+3xx(1)/(2)) = (4)/(5)= 0.8`