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

Answer

A

Explanation

(a) 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//cos r )/(c//n)=(nd)/(c cos r)`
From Snell's law, `n=( sin theta)/(sin r )`
`implies sin r=(sin theta)/(n)`
` cos r =sqrt(1-sin^(2) r )= sqrt(1-( sin ^(2) theta)/(n^(2)))`

`Delta t =(nd)/(C(1-(sin ^(2) theta)/(n^(2)))^(1//2))=(n^(2)d)/(C)(1-(sin ^(2) theta)/(n^(2)))^(-1//2)`
Phase difference= `Delta phi =(2pi)/(T)xx Delta t =(2 pi nd)/(lambda)(1-(sin ^(2) theta)/(n^(2)))^(-1//2)`
So, net phase difference `=Delta phi + pi`
`=( 4 pi d)/(lambda)(1-(1)/(n^(2))sin^(2) theta)^(-1//2)+ pi`