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Unformatted text preview: Lecture 10 Notes: 07 / 14 Reflections and thin-film interference Instead of sending light through slits to interfere with itself, we can get interference between a source of light and its reflection, or two different reflections. First, we must understand how the phase of electromagnetic waves changes when they are reflected from an interface. Recall that for waves on a string, when a wave was reflected from an interface with a string that had a slower wave speed, the amplitude of the reflected wave was inverted. This corresponds to a phase change of . When the incident wave encountered an interface with a string that had a higher wave speed, the amplitude of the reflected wave had the same sign as that of the incident wave. This corresponds to no phase change. Electromagnetic waves behave the same way. When they reflect from a material with a higher index of refraction (lower wave speed) than the medium the wave is traveling in, they change phase by , but when they reflect from a material with a lower index of refraction, they do not change phase. The part of the wave that is transmitted into the new material never changes phase. Example: A plane wave is incident on a thin film of material with n = 1.36, with air on both sides of the film. The film is 1.00 m thick. The direction of travel of the plane wave is normal to the film. For what wavelengths of the plane wave is the reflection enhanced by constructive interference? For what wavelengths is the reflection suppressed? Which of these wavelengths are visible light? Let the thickness of the film be d = 1.00 m . Light that is reflected off the front surface of the film undergoes a phase change of . Light that is reflected off the back surface of the film undergoes no phase change, but it has an additional distance of 2.00 m to travel. Over this distance, the light accumulates a phase at a rate of 2 per wavelength, except the wavelength in the material is / n, where is the wavelength of the incident wave in air. This because the wave number k = / v is inversely proportional to the wave speed, and thus proportional to the index of refraction (since is the same on both sides). = 2 / k is inversely proportional to k , and therefore is inversely proportional to the index of refraction. Thus the additional phase accumulated by the wave that reflects off the back surface, due to its greater traveling distance, is 4 d / = 4 nd / . This is a schematic diagram of the situation. The collinear rays have been separated a bit to show which one is which. The difference in phase between the two rays is 4 nd / 0 - . For constructive interference, we must have the difference in phase equal to an even multiple of , while for destructive interference, it must be an odd multiple of : Simplifying and solving for , we obtain Running through a few values of j , we get constructive interference for wavelengths of 5440 nm , 1813 nm , 1088 nm , 777 nm , 605...
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This note was uploaded on 09/19/2011 for the course PHYS 1C taught by Professor Smith during the Spring '07 term at UCSD.
- Spring '07