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//cos r)/(c//n)=(nd)/(c cosr)`
From Snell's law `n=(sin theta)/(sin r)rArrsin r=(sin theta)/n`
`cos r =sqrt(1-sin^(2)r)=sqrt(1-(sin^(2)theta)/n^(2))`
`:. Deltat=(nd)/(c(1-(sin^(2)theta)/n^(2))^(1//2))=(n^(2)d)/c(1-(sin^(2)theta)/n^(2))^(-1//2)`

Phase difference `=Deltaphi=(2pi)/TxxDeltat=(2pi nd)/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`