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. `(2pind)/(lamda)(1-(sin^(2)theta)/(n^(2)))^(-1//2)+pi`
B. `(4pid)/(lamda)(1-(sin^(2)theta)/(n^(2)))^(-1//2)`
C. `(4pid)/(lamda)(1-(sin^(2)theta)/(n^(2)))^(-1//2)+(pi)/(2)`
D. `(4pid)/(lamda)(1-(sin^(2)theta)/(n^(2)))^(-1//2)+2pi`

Answer

A

Explanation

`(2pind)/(lamda)(1-(sin^(2)theta)/(n^(2)))^(-1//2)+pi`

Situation given in the statement is depicted in above figure. Suppose light ray `vecPQ` is made incident at point Q on surface t=0.
Here refracted ray `vecQS` gets reflected from point S after sometime, say after time t. Now, if speed of light in glass is v then,
`v=(QS)/(t)`
`:.t=(QS)/(v)`
`:.t=(((d)/(cosr)))/((c)/(n))" "(":.cosr=(d)/(QS)andn=(c)/(v))`
`:.t=(nd)/(cxxcosr)" "....(1)`
Now, applying Snell.s law at point Q,
`sintheta=(n)sinr`
`:.n=(sintheta)/(sinr)`
`:.sinr=(sintheta)/(n)`
`:.sin^(2)r=(sin^(2)theta)/(n^(2))`
`:.1-cos^(2)r=(sin^(2)theta)/(n^(2))`
`:.cosr=(1-(sin^(2)theta)/(n^(2)))^(1/2)" "....(2)`
From equation (1) and (2), `t=(nd)/(cxx(1-(sin^(2)theta)/(n^(2)))^(1/2))`
`:.t=(nd)/(c)xx(1-(sin^(2)theta)/(n^(2)))^(-1/(2))" "......(3)`
Now, at the end of time t, if phase of ray `vecST` is `phi_(1)` and phase of ray `vecQR` is `phi_(2)` then phase difference between them is `Deltaphi=phi_(2)-phi_(1)=omegat`
`:.Deltaphi=(kc)t" "("":.c=(omega)/(k)=("angular frequency")/("wave vector"))`
But since ray `vecQR` gets reflected from the surface of denser medium, net path difference would be
`Deltaphi.=Deltaphi+pi`
`:.Deltaphi.=(kc)t+pi`
`:.Deltaphi.=((2pi)/(lamda)c)((nd)/(c))(1-(sin^(2)theta)/(n^(2)))^(-1/2)+pi`
( `:.k=(2pi)/(lamda)` and from equation (3))
`:.Deltaphi.=(2pind)/(lamda)(1-(sin^(2)theta)/(n^(2)))^(-1/2)+pi`