27.3

Thin-Film Interference

Young’s double-slit experiment is one example of interference between light waves. Interference also occurs in more common circumstances. For instance, Figure 27-10 shows a thin film such as gasoline floating on water. To begin with, let us assume that the film has a constant thickness. Consider what happens when monochromatic light (a single wavelength) strikes the film nearly perpendicularly. At the top surface of the film reflection occurs and produces the light wave represented by ray 1. However, refraction also occurs, and some light enters the film. Part of this light reflects from the bottom surface of the film and passes back up through the film, eventually reentering the air. Thus, a second light wave, which is represented by ray 2, also exists. Moreover, this wave, having traversed the film twice, has traveled farther than wave 1. Because of the extra travel distance, there can be interference between the two waves. If constructive interference occurs, an observer whose eyes detect the superposition of waves 1 and 2 would see a uniformly bright film. If destructive interference occurs, an observer would see a uniformly dark film. The controlling factor is whether the extra distance for wave 2 is an integer number of whole wavelengths (for constructive interference) or an odd integer number of half-wavelengths (for destructive interference).

Figure 27-10 Because of reflection and refraction, two light waves, represented by rays 1 and 2, enter the eye when light shines on a thin film of gasoline floating on a thick layer of water.

In Figure 27-10 the difference in path lengths between waves 1 and 2 occurs inside the thin film. Therefore, we have the following Problem solving insight: the wavelength that is important for thin-film interference is the wavelength within the film, not the wavelength in vacuum. The wavelength within the film can be calculated from the wavelength in vacuum by using the index of refraction n for the film. With the aid of Equations 26.1 and 16.1, it can be shown that . In other words,

| (27.3)

In explaining the interference that can occur in Figure 27-10, we need to add one other important part to the story. Whenever waves reflect at a boundary, it is possible for them to change phase. Figure 27-11, for example, shows that a wave on a string is inverted when it reflects from the end that is tied to a wall (see also Figure 17-19). This inversion is equivalent to a half-cycle of the wave, as if the wave had traveled an additional distance of one-half of a wavelength. In contrast, a phase change does not occur when a wave on a string reflects from the end of a string that is hanging free. When light waves undergo reflection, similar phase changes occur as follows:

1. | When light travels through a material with a smaller refractive index toward a material with a larger refractive index (e.g., air to gasoline), reflection at the boundary occurs along with a phase change that is equivalent to one-half of a wavelength in the film.
2. | When light travels from a larger toward a smaller refractive index, there is no phase change upon reflection at the boundary.

The next example indicates how the phase change that can accompany reflection is taken into account when dealing with thin-film interference.

Figure 27-11 When a wave on a string reflects from a wall, the wave undergoes a phase change. Thus, an upward-pointing half-cycle of the wave becomes, after reflection, a downward-pointing half-cycle, and vice versa, as the numbered labels in the drawing indicate.

| Example 3 | | | A Colored Thin Film of Gasoline

A thin film of gasoline floats on a puddle of water. Sunlight falls almost perpendicularly on the film and reflects into your eyes. Although sunlight is white since it contains all colors, the film looks yellow because destructive interference eliminates the color of blue from the reflected light. The refractive indices of the blue light in gasoline and in water are 1.40 and 1.33, respectively. Determine the minimum nonzero thickness t of the film.
Reasoning To solve this problem, we must express the condition for destructive interference in terms of the film thickness t and the wavelength in the gasoline film. We must also take into account any phase changes that occur upon reflection.
In Figure 27-10, the phase change for wave 1 is equivalent to one-half of a wavelength, since this light travels from a smaller refractive index toward a larger refractive index . In contrast, there is no phase change when wave 2 reflects from the bottom surface of the film, since this light travels from a material with a larger refractive index toward a material with a smaller one . The net phase change between waves 1 and 2 due to reflection is, thus, equivalent to one-half of a wavelength, . This half-wavelength must be combined with the extra travel distance for wave 2, to determine the condition for destructive interference. For destructive interference, the combined total must be an odd integer number of half-wavelengths. Since wave 2 travels back and forth through the film and since light strikes the film nearly perpendicularly, the extra travel distance is twice the film thickness, or 2t. Thus, the condition for destructive interference is

After subtracting the term from the left-hand side of this equation and from each term on the right-hand side, we can solve for the thickness t of the film that yields destructive interference:

Solution In order to calculate t, we need to know the wavelength of the blue light in the film. Equation 27.3, with , gives this wavelength as

With this value for and , our result for t gives the minimum nonzero film thickness for which the blue color is missing in the reflected light as follows:

Problem solving insight When analyzing thin-film interference effects, remember to use the wavelength of the light in the film () instead of the wavelength in a vacuum ().

In Example 3 a half-wavelength net phase change occurs due to the reflections at the upper and lower surfaces of the thin film. Depending on the refractive indices of the materials above and below the film, it is also possible that these reflections yield a zero net phase change. See Example 12 in Section 27.10 for a situation in which this occurs.
The thin film in Example 3 has the same yellow color everywhere. In nature, such a uniformly colored thin film would be unusual, and the next example deals with a more realistic situation.

| Conceptual Example 4 | | | Multicolored Thin Films

Under natural conditions, thin films, like gasoline on water or like the soap bubble in Figure 27-12, have a multicolored appearance that often changes while you are watching them. Why are such films multicolored, and what can be inferred from the fact that the colors change in time?

Figure 27-12 A soap bubble is multicolored when viewed in sunlight because of the effects of thin-film interference. (Raymond Forbes/Age Fotostock America, Inc.)

Reasoning and Solution In Example 3 we have seen that a thin film can appear yellow if destructive interference removes blue light from the reflected sunlight. The thickness of the film is the key. If the thickness were different, so that destructive interference removed green light from the reflected sunlight, the film would appear magenta. Constructive interference can also cause certain colors to appear brighter than others in the reflected light and thereby give the film a colored appearance. The colors that are enhanced by constructive interference, like those removed by destructive interference, depend on the thickness of the film. Thus, we conclude that the different colors in a thin film of gasoline on water or in a soap bubble arise because the thickness is different in different places on the film. Moreover, the fact that the colors change as you watch them indicates that the thickness is changing. A number of factors can cause the thickness to change, including air currents, temperature fluctuations, and the pull of gravity, which tends to make a vertical film sag, leading to thicker regions at the bottom than at the top.

Problem solving insight

Related Homework: Problems 14, 56

Need more practice?

Interactive LearningWare27.2
A thin film of gasoline floats on a surface of glass . Sunlight falls almost perpendicularly on the film and reflects into your eyes. Although sunlight is white since it contains all colors, the film has a yellow hue because destructive interference eliminates the color of blue from the reflected light. Determine the minimum nonzero thickness t of the film.
Related Homework: Problems 13, 18

The colors that you see when sunlight is reflected from a thin film also depend on your viewing angle. At an oblique angle, the light corresponding to ray 2 in Figure 27-10 would travel a greater distance within the film than it does at nearly perpendicular incidence. The greater distance would lead to destructive interference for a different wavelength.
Thin-film interference can be beneficial in optical instruments. For example, some cameras contain six or more lenses. Reflections from all the lens surfaces can reduce considerably the amount of light directly reaching the film. In addition, multiple reflections from the lenses often reach the film indirectly and degrade the quality of the image. To minimize such unwanted reflections, high-quality lenses are often covered with a thin nonreflective coating of magnesium fluoride . The thickness of the coating is usually chosen to ensure that destructive interference eliminates the reflection of green light, which is in the middle of the visible spectrum. It should be pointed out that the absence of any reflected light does not mean that it has been destroyed by the nonreflective coating. Rather, the “missing” light has been transmitted into the coating and the lens.

The physics of nonreflecting lens coatings.

Another interesting illustration of thin-film interference is the air wedge. As Figure 27-13a shows, an air wedge is formed when two flat plates of glass are separated along one side, perhaps by a thin sheet of paper. The thickness of this film of air varies between zero, where the plates touch, and the thickness of the paper. When monochromatic light reflects from this arrangement, alternate bright and dark fringes are formed by constructive and destructive interference, as the drawing indicates and Example 5 discusses.

Figure 27-13 (a) The wedge of air formed between two flat glass plates causes an interference pattern of alternating dark and bright fringes to appear in reflected light. (b) A side view of the glass plates and the air wedge.

| Example 5 | | | An Air Wedge

(a) Assuming that green light strikes the glass plates nearly perpendicularly in Figure 27-13, determine the number of bright fringes that occur between the place where the plates touch and the edge of the sheet of paper . (b) Explain why there is a dark fringe where the plates touch.
Reasoning A bright fringe occurs wherever there is constructive interference, as determined by any phase changes due to reflection and by the thickness of the air wedge. There is no phase change upon reflection for wave 1, since this light travels from a larger (glass) toward a smaller (air) refractive index. In contrast, there is a half-wavelength phase change for wave 2, since the ordering of the refractive indices is reversed at the lower air/glass boundary where reflection occurs. The net phase change due to reflection for waves 1 and 2, then, is equivalent to a half wavelength. Now we combine any extra distance traveled by ray 2 with this half wavelength and determine the condition for the constructive interference that creates the bright fringes. Constructive interference occurs whenever the combination yields an integer number of wavelengths. At nearly perpendicular incidence, the extra travel distance for wave 2 is approximately twice the thickness t of the wedge at any point, so the condition for constructive interference is

Subtracting the term from the left-hand side of this equation and from each term on the right-hand side yields

Therefore,

In this expression, note that the “film” is a film of air. Since the refractive index of air is nearly one, is virtually the same as that in a vacuum, so .

Problem solving insight Reflected light will experience a phase change only if the light travels from a material with a smaller refractive index toward a material with a larger refractive index. Be sure to take such a phase change into account when analyzing thin-film interference.

Solution

a. | When t equals the thickness of the paper holding the plates apart, the corresponding value of m can be obtained from the equation above:

Since the first bright fringe occurs when , the number of bright fringes is .
b. | Where the plates touch, there is a dark fringe because of destructive interference between the light waves represented by rays 1 and 2. Destructive interference occurs because the thickness of the wedge is zero here and the only difference between the rays is the half-wavelength phase change due to reflection from the lower plate.

Another type of air wedge can also be used to determine the degree to which the surface of a lens or mirror is spherical. When an accurate spherical surface is put in contact with an optically flat plate, as in Figure 27-14a, the circular interference fringes shown in part b of the figure can be observed. The circular fringes are called Newton’s rings. They arise in the same way that the straight fringes arise in Figure 27-13a.

Figure 27-14 (a) The air wedge between a convex spherical glass surface and an optically flat plate leads to (b) a pattern of circular interference fringes that are known as Newton’s rings. (Courtesy Bausch & Lomb)

Check Your Understanding 3

The figure shows three situations in which light reflects almost perpendicularly from the top and bottom surfaces of a thin film, with the indices of refraction as shown. (a) For which situation(s) is there a net phase shift (due to reflection) between waves 1 and 2 that is equivalent to either zero wavelengths or one wavelength , where is the wavelength of the light in the film? (b) For which situation(s) will the film appear dark when the thickness of the film is equal to ? (The answers are given at the end of the book.)

Background: For a film to appear dark, the waves represented by rays 1 and 2 must be out of phase.This can arise from phase shifts associated with the reflection at the top and bottom surfaces of the film and the extra distance traveled by ray 2 as it moves through the film.
For similar questions (including calculational counterparts), consult Self-Assessment Test 27.1, which is described at the end of Section 27.4.

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