Explaining Thin Films and the Reflection of Light in Air-Film Interface

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In summary, thin films are layers of material that are only a few nanometers to micrometers thick. When light passes through an air-film interface, it is partially reflected and partially transmitted due to differences in the refractive index between the air and the film. This results in interference patterns, where certain wavelengths of light are reinforced and others are cancelled out, leading to colorful reflections. Thin films play a crucial role in various technologies such as anti-reflective coatings, optical filters, and solar cells.
  • #1
dekoi
This is not a textbook question. It is strictly some concepts which I am confused about.


Suppose we have thin film situation.

|||||||||||||||||||AIR|||||||||||||||||||
-----------------------------------------
|||||||||||||||||||FILM|||||||||||||||||
-----------------------------------------
|||||||||||||||||||AIR|||||||||||||||||||

The ray enters the air from the top. Part of it will be reflected from the surface of the film (it has a higher index of refraction). The ray will reflect 180 degrees and thus be 180 degrees out of phase, assuming that the incident ray was normal to surface.

Part of it will also pass through the air/film barrier. According to my professor, the ray will reflect of the film/air barrier, and then be 0 degrees out of phase. This is beyond me! How come the ray reflects of the boundary between film and air? Should it not pass through it without reflection? Also, why would the phase be 0?


In another situation:
|||||||||||||||||||AIR|||||||||||||||||||
-----------------------------------------
|||||||||||||||||||FILM|||||||||||||||||
-----------------------------------------
||||||||||DIAMOND (highest n)|||||||||||

The second reflected ray is said to be 180 degrees out of phase. Once again, why?!
 
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  • #2
anyone?/0
 
  • #3
dekoi said:
The ray enters the air from the top. Part of it will be reflected from the surface of the film (it has a higher index of refraction). The ray will reflect 180 degrees and thus be 180 degrees out of phase, assuming that the incident ray was normal to surface.

This has a problem: the reflected ray will reflect according to the law of reflection and take off an an angle equal to the incoming ray. The 180 degree out-of-phase means that the reflected wave is simply inverted. Look at a regular sine wave; if you flip the wave upside-down, notice that the crest has become a trough, and the trough becomes a crest. A full wavelength is chopped up into 360 "degrees" just like a circle; so the "distance" from a crest to a trough, is half a wavelength, or 180 degrees. Thus, the wave has shifted its phase by 180 degrees.

Exactly why light waves invert when reflecting from a fast-to-slow interface but not when reflecting from a slow-to-fast interface is not an easy explanation. You can check out the mechanical wave analogy, but the real answer lies in quantum electrodynamics ("QED").

Part of it will also pass through the air/film barrier. According to my professor, the ray will reflect of the film/air barrier, and then be 0 degrees out of phase. This is beyond me! How come the ray reflects of the boundary between film and air? Should it not pass through it without reflection?

All waves will partially reflect and partially transmit (refract) when they hit an interface of two media. The exception is for monochromatic light sources in special situations, such as what you are learning now.

The "mechanical wave" explanation says that in thin film interference, the two rays that reflect (one internally and one externally) come back together out of phase and mutually destruct, causing no wave of that frequency to emerge. But the weird thing is even if you fire one photon at a time, not one single photon will reflect in that situation even if there is nothing to interfere with. Bizarre! but true. QED.
 
  • #4
dekoi said:
This is not a textbook question. It is strictly some concepts which I am confused about.
Suppose we have thin film situation.
|||||||||||||||||||AIR|||||||||||||||||||
-----------------------------------------
|||||||||||||||||||FILM|||||||||||||||||
-----------------------------------------
|||||||||||||||||||AIR|||||||||||||||||||
The ray enters the air from the top. Part of it will be reflected from the surface of the film (it has a higher index of refraction). The ray will reflect 180 degrees and thus be 180 degrees out of phase, assuming that the incident ray was normal to surface.
Part of it will also pass through the air/film barrier. According to my professor, the ray will reflect of the film/air barrier, and then be 0 degrees out of phase. This is beyond me! How come the ray reflects of the boundary between film and air? Should it not pass through it without reflection? Also, why would the phase be 0?
In another situation:
|||||||||||||||||||AIR|||||||||||||||||||
-----------------------------------------
|||||||||||||||||||FILM|||||||||||||||||
-----------------------------------------
||||||||||DIAMOND (highest n)|||||||||||
The second reflected ray is said to be 180 degrees out of phase. Once again, why?!
When the light incides at the boundary of a two different materials part of it is reflected and the other part enters into the new medium, (it is called transmitted light) but will travell in different direction (it is refracted).

The laws for light reflection and refraction at a boundary between different media can be derived from Maxwell's equations of electrodynamics, and these laws state that both the magnitude and phase of the reflected and transmitted waves are connected to the refractive indices of those media.

In introductory Physics, it is enough to know that the phase change is 180 degree if the light ray incides from a medium of lower refractive index onto the surface of a medium of higher refractive index, and it is zero in the opposite case. Sometimes people call the refractive index "optical density" and say that the light reflects from an optically denser medium "out of phase" and in phase from an optically less dense one. The phenomenon is similar you experience with a wave traveling along a string: The wave turns over from fixed end and comes back in-phase from a loose end.
Air has the lowest refractive index in your problems, and diamond has the highest. The refractive index of the film is between the other two. So the phase change of the reflected wave is 180 degrees when the light goes from air to film. The other part of the light which has not ben reflected goes through the film and reaches the boundary of the film with air. Here again, it is partly reflected and partly transmitted, and the reflected wave is in-phase now, as it is directed from a high-optical density medium to a low one.
On the same way, the reflected wave is out-of phase when the light enters to diamond from the film.

ehild
 

1. What is a thin film?

A thin film is a layer of material that has a thickness of a few nanometers to a few micrometers. It is typically deposited onto a substrate, such as a solid surface or another film, and can be made of a variety of materials including metals, semiconductors, and polymers.

2. How does light interact with a thin film?

When light strikes a thin film, some of it is reflected off the surface and some of it penetrates into the film. The reflected and transmitted light waves interfere with each other, resulting in a unique pattern of light and dark areas known as interference fringes. This phenomenon is known as thin film interference.

3. What factors affect the reflection of light in an air-film interface?

The reflection of light in an air-film interface is affected by the refractive index of the film, the angle of incidence of the light, and the thickness of the film. The refractive index is a measure of how much a material bends light, and the angle of incidence is the angle at which the light strikes the surface. The thickness of the film determines how much the light waves will interfere with each other.

4. How is the thickness of a thin film determined using reflection of light?

The thickness of a thin film can be determined by measuring the distance between the interference fringes produced by the reflected light. Each fringe corresponds to a specific change in the thickness of the film. By counting the number of fringes and knowing the wavelength of the light used, the thickness of the film can be calculated.

5. What applications use the principles of thin film interference?

The principles of thin film interference are used in a variety of applications, including anti-reflective coatings on eyeglasses and camera lenses, optical filters, and thin film solar cells. Thin film interference is also used in the production of holograms and in non-destructive testing techniques for detecting flaws in materials.

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