Correct Answer - B
After introducing the film, path difference at point O is
`Delta = t(mu-1)=11 xx10^(-6)xx1.1 = 12.1 xx10^(-6)m`
Phase difference at point O is
`phi=(2 pi)/(lamda)xx Delta`
`=(2pi)/(6 xx 10^(-7))12.1 xx 10^(-6)`
`=(12.1pi)/(3)=(pi)/(3)`
Thus intensity at point O is
`I=I_(0) "cos"^(2)(phi)/(2)=I_(0)"cos"^(2)(pi)/(6)= 3 (I_(0))/(4)`
Now path difference at point in terms of `lamda` is
`Delta _(0)=(12.1 xx10^(-6))/(6xx 10^(-7))lamda = 20.167 lamda`
Thus due to placement of thin film 20 bright fringes cross the point O.