In: Physics
A uniform film of TiO2, 1036 nm thick and having index of refraction 2.62, is spread uniformly over the surface of crown glass of refractive index 1.52. Light of wavelength 490 nm falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels.
a) What is the minimum thickness of TiO2 that you must add so the reflected light cancels as desired?
b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in nanometers.
c) Express your answer in wavelengths of the light in the TiO2 film.
The condition for destructive interference of light reflected from the two surfaces of a thin film can be calculated using the thickness dd of the material, its refractive index nn and the wavelength λλ. This case is very common in antireflection coatings. If one assumes normal incidence, the condition for these coatings is
4nd=(2k+1)λ k=1,2,3,...
In this configuration, we have the rare case that a coating with high refractive index nn is deposited at a substrate with lower refractive index. Therefore, we have to take into account a phase jump of λ/2λ/2 that the first reflected wave experiences and the second one doesn't. The second wave has a phase delay of 2nd2nd, where dd is the thickness of the layer. The difference between the two phases for destructive interference must be an odd multiple of half the wavelength. In mathematical notation, this reads
2nd−λ/2=(2k+1)λ/2
2nd=(2k+2)λ/2 (j=k+1)
d=jλn
=j⋅490 nm/(2.62)
=j⋅187.0 nm
The smallest integer yielding a thickness larger than the already existing 1036 nm is j = 6 resulting in a required thickness of 1122 nm.
A): A layer of thickness 86 nm has to be added to achieve destructive interference.
B): The path difference between the two waves is the difference of the optical path of the back reflection and the phase jump of the first reflection, i.e.
Δd=2 x 2.62 x 1122 nm−(460/2) nm = 5634.28 nm .
C): This corresponds to 11.5 wavelengths.