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. `(2 pi n 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

In Fig. a ray of light `AB` is incident from air onto glass slab of width `d` at angle `theta`. It is reflected partially at `B` and refracted partially at `B` along `BC` at `/_r`. At `C`, the ray is partially reflected along `CD` and partially refracted (not shown). To calculate phase difference between rays reflected from `B` and `C`, we find
time difference, `Delta T =` time taken to travel `BC` in glass
`= (BC)/(v) = ((d//cos r))/(c//n) = (n d)/(c cos r)`
From Snell's law, `n = (sin theta)/(sin r)`,
`sin r = (sin theta)/(n)`
`cos r = sqrt(1 - sin^(2) r) = (1 - (sin^(2) theta)/(n^(2)))^(-(1)/(2)`
`:. Delta T = (n d)/(c (1 - (sin^(2) theta)/(n^(2)))^((1)/(2))) = (n xx T d)/(lambda)(1 - (sin^(2)theta)/(n^(2)))^(-(1)/(2)` ....`(c = lambda//T)`
Phase diff. `= Delta phi = (2 pi Delta T)/(T) = (2pi n d)/(T)(1 - (sin^(2)theta)/(n^(2)))^(-(1)/(2)`
As reflection at `C` is from medium of higher refractive index, additional phase diff. of `pi` is introduced.
Hence required phase difference `= (2 pi n d)/(lambda) (1 - (sin^(2) theta)/(n^(2)))^(-(1)/(2)) + pi`
Choice (a) is correct.