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

`sin theta = n sin r`
`implies sin r = (sin theta)/n`
when light is reflected at point A, then it suffers a phase change of `pi` , and we shall include it later.

Additional phase difference between lights emerging from A and D is due to path difference . Light emerging from point D travels extra distance `AB + BD` which is equal to `2 AB`. Optical path difference can be written as `Delta x = n (2AB)`
`implies Delta x = 2n xx d/(cos r)`
`implies Deltax = (2nd)/([1 - (sin^2 theta)/(n^2)]^(1/2))`
Total phase difference can be written as follows:
`delta = (2pi)/lambda Delta x + pi`
`implies delta = (4 pi n d)/(lambda) [1 - (sin^2 theta)/(n^2)]^(-1//2) + pi`