Suppose that a thin film of acetone (index of refraction n = 1.25) of thickness

Question

Suppose that a thin film of acetone (index of refraction n = 1.25) of thickness d is coating a thick plate of glass (index of refraction = 1.50). Take the magnitude of the amplitude for reflection of a photon from the top or the bottom surface of the acetone at normal incidence to be r and assume that there is an additional phase change of in the reflection from the top and the bottom surface of the acetone, since at each of these surfaces light is passing from a medium with a lower index of refraction to one with a higher index of refraction. Calculate the probability that a photon of wavelength is reflected. Assume that amplitudes that involve multiple reflections at the bottom surface of the film can be neglected in your calculation. Express your answer in terms and r as well as the thickness d and the index of refraction n of the acetone. What is the minimum thickness of the coating necessary to produce zero reflection? Note: For the air-acetone and acetone-glass surfaces r 0.1.

Explanation / Answer

A)
Thickness of acetone = d
Refractive index of acetone na = 1.25
Refractive index of glass ng = 1.50  
Let the wavelength of light in vacuum =
The wavelength light in acetone a =/na
Here due to superposition of two reflected rays interference phenomena takes place .

The phase difference between these two reflected rays is 1800
Therefore condition for constructive interference and for normal incidence is
2d = ( 2n-1)a/2
2d = (2n-1) /2na
= 4nad /(2n-1), n=1,2,3,……
Condition for destructive interference and for normal reaction is
2d = n a
2d = n /na
= 2nad/n, n=1,2,3…..

B)
The reflected light is minimum or zero when reflected rays meet condition of destructive interference.
Both rays undergo a phase change upon reflection.
The net change in phase due to reflection is zero.
Therefore 2dmin = a /2
2dmin = / 2na,
dmin = / 4na
It is the minimum thickness of coating required to produce zero reflection

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