Thin Film Interference and phasing

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SUMMARY

This discussion focuses on thin film interference, specifically the conditions under which reflected rays experience phase changes. A ray reflecting off a medium with a higher index of refraction incurs a 180-degree phase change, while one reflecting off a lower index does not. The example from the Serway Jewett Physics Book (7th ed) illustrates the behavior of rays in a thin film, highlighting the importance of the first two reflected rays for intensity comparison. The discussion emphasizes the relationship between phase shifts and constructive or destructive interference, particularly in the context of observing colors in thin films like soap bubbles.

PREREQUISITES
  • Understanding of thin film interference principles
  • Familiarity with phase changes in wave reflections
  • Knowledge of constructive and destructive interference formulas
  • Basic concepts of light refraction and reflection
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  • Study the phase change rules for light reflecting off different media
  • Explore the mathematical derivation of constructive and destructive interference conditions
  • Examine real-world applications of thin film interference in optical devices
  • Learn about the impact of film thickness on interference patterns
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Physics students, optical engineers, and anyone interested in the principles of light behavior in thin films and their applications in technology and nature.

jaredogden
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I am looking over thin film interference and am having some difficulty understanding when to know if a reflected ray will be 180 degrees out of phase or in phase. I know that a ray reflecting off a medium with a higher index of refraction will have a 180 degree phase change; and that a ray reflecting off a medium with a lower index of refraction will have no phase change.

I am looking at an example in the serway jewett physics book (7th ed) on page 1060 Figure 37.11 if anyone has it. There is a single ray incident on a thin film. It is reflected off the film back into air and has a 180 degree phase change, I understand that part. Part of the wave is refracted out of the bottom of the thin film into air producing ray 3 out of the bottom. A part of that ray is reflected back into the thin film and reflects off the bottom surface of the thin film and is refracted off the top layer of the film back into air where there is no phase change making ray 2 out of the top of the thin film. At the same point that ray is refracted back into the air out of the top part of the ray is reflected back into the thin film and travels to the bottom and is refracted out of the bottom of the thin film creating ray 4.

I'm sure that was a confusing explanation. But anyways I am trying to better understand phase changes and how to use the constructive and destructive interference formulas for thin films. If anyone can help explain this it would be appreciated. Thanks
 
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With thin films, you look only at the first two reflected rays because their intensity should be comparable (they have each been reflected only once). For a bubble of soapy water, the first reflected ray has an intensity of 2,006 % of the original intensity at normal incidence while the second one has an intensity of 1.966 %. Any ray that has been reflected more than once will have an intensity of less than 2 % of the previous numbers and will therefore not be noticeable to the eye.

Thin-film.jpg


If the first reflection if hard, the ray is as you say 180 degrees out of phase with the orginal. In the example of the soapy bubble, the second reflection is soft, hence the ray if in phase with the first ray. This ray has been refracted or transmitted twice but transmission does not change phase. Since ray one and ray two reach your eyes together, you are interested to know if they are in sync or not. In the soapy bubble example case, they are not.

If you look at the bubble and see magenta, then you know that interference is constructive (\Delta \phi= 0) for both blue and red wavelength. The total is zero, but the phase shift due to reflection is equal to pi, hence the phase shift due to the path difference should be pi, which translates to 2 * thickness = (2m + 1/2) wavelength of blue in the thin film (which is different from the one you observe in the air).

You do the same for red wavelength and the two conditions need to be met simultaneously.

You will find multiple solutions.
 

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