| Just as pulses along a rope when reflected off a fixed-end are inverted, or when reflected off a free-end, remain in phase,
light waves also exhibit these behaviors when they encounter an interface between two mediums. When a light ray is reflected off an interface which has a less optically-dense medium above a more optically-dense medium the reflected ray is inverted. For example, air over water, air over glass, water over glass.
This reflection produces a phase difference of ½λ from the original wave and is called a fixed-end reflection. Shown below is a waveform diagram of this type of reflection. The incident crest (blue C) is reflected as a trough (orange T). When a light ray is reflected off an interface which has a more optically-dense medium above a less optically-dense medium the reflected ray is not inverted. For example, water over air [soap bubbles], glass over air [glass wedge], glass over water. This is called a free-end reflection. This reflection produces no phase difference with the original wave. Shown below is a waveform diagram of this type of reflection. The incident crest (blue C) is reflected as a crest (orange C). REMEMBER: the refracted rays NEVER undergo a phase change! Oil and Soap Films These reflective properties are critical to our understanding of the colors in such thin films as soap bubbles, coatings on camera lenses, colors in a butterfly's wings or peacock's feathers, or oil spills. Below is a diagram of a thin film and the light rays associated with the reflections and refractions as light impinges on the film. - When ray 1 strikes the top interface, some of the light is partially reflected, ray 2, and the rest is refracted, ray 3.
- When ray 3 strikes the bottom interface, some of it is reflected, ray 4, and the remainder is refracted, ray 6.
- When ray 4 strikes the top interface from underneath, some is reflected (not shown) and some is refracted, ray 5.
It is the interference between rays 2 and 5 that produces a thin film's color when the film is viewed from above. The formula used for thin film interference is EPD = 2t + Φ where - t represents the thickness of the thin film
- Φ is the net phase inversion between rays 2 and 5
- 0 λ
- ½ λ
- EPD represents the extreme path difference
- constructive interference: mλ where m
 {0, 1, 2, 3…} - destructive interference: ½(2m -1)λ where m
 {1, 2, 3...} The net phase inversion Φ is the difference in the phases between the two reflected rays - one off the top interface (#2) and the second off the top of the bottom interface (#5). These are the rays that span the thickness of the film. Remember that since we are dealing with the behavior of the light in the thin film, you must ALWAYS use the light's wavelength IN THE FILM. Also remember that refracted rays remain in-phase with their initial rays. - If both reflections occur at interfaces of less/more, then both are inverted and their net phase difference is zero.
- If both reflections occur at interfaces of more/less, then neither is inverted and the net phase difference between them is zero.
- However, if one reflection occurs at less/more and the other at more/less, then there is a net phase difference between the reflections of ½ λ.
Solve the above equation for t by calculating its value in terms of the appropriate wavelength in the thin film; that is, if the film is made of air, use the wavelength in air; if the film is of water, use the wavelength in water. Remember that the wavelength in a medium whose refractive index is n can be calculated with the formula λn = λvacuum/n and that the refractive indices for air, water, and glass are: 1, 1.33, and 1.5 respectively. Since the wavelength DECREASES when the light travels through an optically-denser medium, its average speed decreases. Color Addition and Subtraction Consequently, a given thickness of film can cancel or reinforce specific wavelengths of light resulting in the thin film displaying unique colors. If the film thickness cancels red, then you see its complementary color cyan. If the film thickness cancels green, then you see its complementary color magenta. If the film thickness cancels blue, then you see its complementary color yellow. |
Refer to the following information for the next eight questions. A thin soap film is formed by dipping a plastic rectangular wand into a solution of soapy water which has a refractive index of 1.4. When viewed in daylight, one portion of the film reflects blue light of wavelength 475 nm. Estimate the minimum thickness of that section of the film. | Q) | Using the same numbering scheme as in the lesson, will ray 2 be in-phase or out-of-phase with ray 1? | | |  | The outer interface is between air and the soap film. Since the refractive index of soap is greater than that of air, this interface will act like a fixed-end reflector and ray 2 will be reflected out-of-phase. This makes rays 1 and 2 out-of-phase with a phase inversion of ½ λ. | | | | Q) | Using the same numbering scheme as in the lesson, will ray 5 be in-phase or out-of-phase with ray 1? | | | | The inner interface is also between air and the soap film. But, the light is incident from soap to air, so this interface will act like a free-end reflector and ray 2 will be reflected in phase. This makes rays 3 and 4, in-phase, which translates into rays 1 and 5 being in-phase since refracted light is never inverted. Thus rays 1 and 5 are in-phase with no phase inversion. | | | | Q | What is Φ, the net phase inversion, for this series of reflections? | | | | Since one reflection is ½ λ and the other is 0 λ, the net phase inversion equals ½ λ. | | | | Q) | To see blue, are you going to use the formula for constructive or destructive interference? | | | | constructive interference - a bright | | | | Q) | What expression will represent the EPD in this problem? | | | | EPD = mλ where m  {0, 1, 2, 3...} | | | | Q) | An average blue wavelength is 475 nm in air, what is its wavelength in the soap film? | | | | 339 nm | | |  | In the soap film the effective wavelength is λn = 475/1.4 = 339 nm | | | | Q) | Write the equation for thin film interference as it applies to this problem. | | | | EPD = 2t + Φrays 2 & 5
mλ = 2t + ½ λ | | | | Q) | What is the minimum, non-zero thickness of the film? | | | | 84.8 nm | | | | Since we are determining the thickness of the film, we must use the light's wavelength in the film. 2t = mλn - ½λn
t = ½(m - ½)λn
t = ½(1 - ½)(339)
t = ¼(339)
t = 84.8 nm | | | |
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1561
