Consider a ray of light incident from air onto a slab of glass (refr active 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. `(4pi d)/(lambda) (1- (1)/(n^(2))sin^(2)theta)^(1//2) + pi`
B. `(4pi d)/(lambda) (1- (1)/(n^(2))sin^(2)theta)^(1//2)`
C. `(4pi d)/(lambda) (1- (1)/(n^(2))sin^(2)theta)^(1//2)+ (pi)/(2)`
D. `(4pi d)/(lambda) (1- (1)/(n^(2))sin^(2)theta)^(1//2)+2pi`

Answer

A

Explanation

" Consider diagram, the ray " `()P` " is incident at an angle " `theta` " and gets reflected in the direction P' and reflected in the direction P . Due to reflection from the glass medium, there is a phase change of " `pi`. " Time taken to travel along OP "
`Delta t = ( "OP ")/(v) = (d// " cos r ")/( "c//n ") = (nd)/("c cos r")`
From Snell's, law `n = (sin theta)/(sin r)`
Phase difference,
`Delta phi = (2 pi)/(T)xxDelta t implies Delta phi = (2pi nd)/(lambda) (1-(sin^(2)theta)/(n^(2)))^(-1//2)`
So, net phase difference `= Delta phi + pi`