Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle `theta` . The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

Options

A. `(4pid)/(lambda)(1-(1)/(n^(2))sin^(2)theta)^(1//2)+ pi`
B. `(4pid)/(lambda)(1-(1)/(n^(2))sin^(2)theta)^(1//2)`
C. `(4pid)/(lambda)(1-(1)/(n^(2))sin^(2)theta)^(1//2)+ pi//2`
D. `(4pid)/(lambda)(1-(1)/(n^(2))sin^(2)theta)^(1//2)+ 2pi`

Answer

A

Explanation

Consider the diagram, the ray (P) is incident at an angle `theta` and gets reflected in the direction P . and refracted in the direction P" . Due to reflection from the glass medium, There is a phase change of `pi` .
time taken to travel along OP"
`Deltat= (OP")/(v)= (d//cosr)/(c//n)= (nd)/(ccosr)`
From Snell.s law, `n=(sintheta)/(sinr)`
`implies sinr= (sintheta)/(n)`
`cosr= sqrt(1-sin^(2)r)= sqrt(1-(sin^(2)theta)/(n^(2)))`

`:.Deltat= (nd)/(c(1-(sin^(2)theta)/(n^(2)))^(1//2))= (nd)/(c)(1-(sin^(2)theta)/(n^(2)))^(-1//2))`
Phase difference
`Deltaphi= (2pi)/(T) xx Deltat= (2pind)/(lambda) (1-(sin^(2)theta)/(n^(2)))^(-1//2)`
So, net phase difference = `Deltaphi + pi = (4pid)/(lambda)(1-(1)/(n^(2))sin^(2)theta)^(-1//2)+ pi`