What is thin Film interference? What is the Condition for bright and dark fringe?
Inference by division of amplitude: Thin Film interference
In case of soap bubbles the interference of light waves reflecting off the top surface of a film with the waves reflecting from the bottom surface. To obtain a nice colored pattern due to constructive and destructive interference of light waves and the thickness of the film has to be on the order of the wavelength of light.
So, the thinfilm interference takes place when incident light waves reflected by the upper and lower surface of a thin film interfere with one another to form a new wave. Thin films commercial applications: optical filters, mirrors, Antireflection coatings etc.
Path Difference in thin film interference
If we consider a uniform thin film with µ refractive index and thickness of the thin film is t. There are two surfaces present in the thin film is PQ and ST. Suppose that the ray AB incident on surface PQ with incidence angle i; there are two possibilities one is the ray AB is partially reflected back in same medium i.e. BC and another possibility is partially refracted from the surface PQ at angle r i.e. ray BD and again partially reflected beck along DE from the surface ST and finally emerges out along EF from the surface PQ. In this condition the reflected ray BC and refracted ray EF interfere and form an interference pattern.
According to diagram;
Angle of incidence <ABN = i and <BEG = i {< = Angle}
Angle of refraction <HBD = r
Thickness of thin film = t
Refractive index of thin film = µ
So, the path difference between two rays i.e. BC and EF is given as
Ð = Path in film (BD + DE) – Path in air (BG) {Ð = Delta}
Ð = µ (BD + DE) –BG (refractive index µ=1 for air) (1)
According to Snell’s law
because a ray reflected from the surface of a denser medium suffers an abrupt phase change of p which is equivalent to λ/2 path difference.

Condition for bright fringe or maxima or constructive interference
For the constructive inference the path difference should be nλ
So
Ð = 2µt cos r λ/2 = nλ (n=0,1,2,3…)
2µt cos r = (2n1)λ/2

Condition for dark fringe or minima or destructive interference
For the destructive inference the path difference should be (2n1)λ/2
Ð = 2µt cos r λ/2 = (2n1)λ/2 (n=0,1,2,3…)
2µt cos r = nλ
Problem 1: Light of wavelength 5893A^{0} is reflected at normal incidence from a soap film of refractive index 1.45. What is the least thickness of the film that will appear dark or bright?
Solution: Given that l = 5893A^{0} = 5.893 X 10^{7}m ; i=r=0; m = 1.45 ; consider thickness t=1
For bright fringe (constructive inference) condition
2mt cos r = (2n1)l/2
The thickness of film is
t = (2n1)l/4m cos r
t = (21) X 5.893 X 10^{7}/4 X 1.45 X 1
t = 1.017 X 10^{4} mm
For dark fringe (destructive inference) condition
2mt cos r = nl
The thickness of film is
t = nl/2m cos r
t = 1 X 5.893 X 10^{7} / 2 X 1.45 X 1
t = 2.032 X 10^{4} mm
Problem 2 : What is the thickness of film of m=1.45 that will result in constructive interference in the reflected light; the wavelength of incidence light is 6100A^{0}.
Solution: Given that: l = 6.1 X 10^{7}m; m=1.45
For constructive inference condition in thin film
2mt cos r = (2n1)l/2
The thickness of film is
t = (2n1)l/4m cos r
t = (21) X 6.1 X 10^{7}/4 X 1.45 X 1
t = 1.052 X 10^{4}mm
Leave a Reply