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When a thin film of oil or soap bubble is illuminated with white light, multiple colors appear.
This is due to
a) Diffraction b) polarization c) total internal reflection d) interference
In Interferenc
Answers
Answer:
It's due to Interference
Explanation:
By the end of this section, you will be able to:
Describe the phase changes that occur upon reflection
Describe fringes established by reflected rays of a common source
Explain the appearance of colors in thin films
The bright colors seen in an oil slick floating on water or in a sunlit soap bubble are caused by interference. The brightest colors are those that interfere constructively. This interference is between light reflected from different surfaces of a thin film; thus, the effect is known as thin-film interference.
As we noted before, interference effects are most prominent when light interacts with something having a size similar to its wavelength. A thin film is one having a thickness t smaller than a few times the wavelength of light, \lambda. Since color is associated indirectly with \lambda and because all interference depends in some way on the ratio of \lambda to the size of the object involved, we should expect to see different colors for different thicknesses of a film, as in (Figure).
These soap bubbles exhibit brilliant colors when exposed to sunlight. (credit: Scott Robinson)
A picture of soap bubbles is shown.
What causes thin-film interference? (Figure) shows how light reflected from the top and bottom surfaces of a film can interfere. Incident light is only partially reflected from the top surface of the film (ray 1). The remainder enters the film and is itself partially reflected from the bottom surface. Part of the light reflected from the bottom surface can emerge from the top of the film (ray 2) and interfere with light reflected from the top (ray 1). The ray that enters the film travels a greater distance, so it may be in or out of phase with the ray reflected from the top. However, consider for a moment, again, the bubbles in (Figure). The bubbles are darkest where they are thinnest. Furthermore, if you observe a soap bubble carefully, you will note it gets dark at the point where it breaks. For very thin films, the difference in path lengths of rays 1 and 2 in (Figure) is negligible, so why should they interfere destructively and not constructively? The answer is that a phase change can occur upon reflection, as discussed next.
Light striking a thin film is partially reflected (ray 1) and partially refracted at the top surface. The refracted ray is partially reflected at the bottom surface and emerges as ray 2. These rays interfere in a way that depends on the thickness of the film and the indices of refraction of the various media.
Reflection at an interface for light traveling from a medium with index of refraction {n}_{1} to a medium with index of refraction {n}_{2}, {n}_{1}<{n}_{2}, causes the phase of the wave to change by \pi radians.
Picture is a schematic drawing of the light undergoing interference by a thin film. Wave reflected from the top of the film is inverted; wave reflected from the bottom of the film is not inverted; refracted waves are not inverted.
If the film in (Figure) is a soap bubble (essentially water with air on both sides), then a phase shift of \lambda \text{/}2 occurs for ray 1 but not for ray 2. Thus, when the film is very thin and the path length difference between the two rays is negligible, they are exactly out of phase, and destructive interference occurs at all wavelengths. Thus, the soap bubble is dark here. The thickness of the film relative to the wavelength of light is the other crucial factor in thin-film interference.
Calculating the Thickness of a Nonreflective Lens Coating Sophisticated cameras use a series of several lenses. Light can reflect from the surfaces of these various lenses and degrade image clarity. To limit these reflections, lenses are coated with a thin layer of magnesium fluoride, which causes destructive thin-film interference. What is the thinnest this film can be, if its index of refraction is 1.38 and it is designed to limit the reflection of 550-nm light, normally the most intense visible wavelength? Assume the index of refraction of the glass is 1.52.
Strategy Refer to (Figure) and use {n}_{1}=1.00 for air, {n}_{2}=1.38, and {n}_{3}=1.52. Both ray 1 and ray 2 have a \lambda \text{/}2 shift upon reflection. Thus, to obtain destructive interference, ray 2 needs to travel a half wavelength farther than ray 1. For rays incident perpendicularly, the path length difference is 2t.
Solution To obtain destructive interference here,
2t=\frac{{\lambda }_{n2}}{2}
where {\lambda }_{n2} is the wavelength in the film and is given by {\lambda }_{n2}=\lambda \text{/}{n}_{2}. Thus,
2t=\frac{\lambda \text{/}{n}_{2}}{2}.
Solving for t and entering known values yields
t=
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