27.10

Concepts & Calculations

The ability to exhibit interference effects is a fundamental characteristic of any kind of wave. Our understanding of these effects depends on the principle of linear superposition, which we first encountered in Chapter 17. Only by means of this principle can we understand the constructive and destructive interference of light waves that lie at the heart of every topic in this chapter. Therefore, it is fitting that we review the essence of the principle in the next example.

| Concepts & Calculations Example 11 | | | The Principle of Linear Superposition

A square is 3.5 m on a side, and point A is the midpoint of one of its sides. On the side opposite this spot, two in-phase loudspeakers are located at adjacent corners, as Figure 27-41 indicates. Standing at point A, you hear a loud sound because constructive interference occurs between the identical sound waves coming from the speakers. As you walk along the side of the square toward either empty corner, the loudness diminishes gradually but does not entirely disappear until you reach either empty corner, where you hear no sound at all. Thus, at each empty corner destructive interference occurs. Find the wavelength of the sound waves.

Figure 27-41 Two loudspeakers are located at the corners of a square and produce identical sound waves. As Example 11 discusses, constructive interference occurs at point A, which is at the midpoint of the opposite side of the square, and destructive interference occurs at either empty corner.

Concept Questions and Answers Why does constructive interference occur at point A?

Answer The loudspeakers are in-phase sources of identical sound waves. The waves from one speaker travel a distance in reaching point A and the waves from the second speaker travel a distance . The condition that leads to constructive interference is , where is the wavelength of the waves and . Thus, the two distances are the same or differ by an integer number of wavelengths. Point A is the midpoint of a side of the square, so that and are the same, and constructive interference occurs.

What is the general condition that leads to destructive interference?

Answer The waves start out in phase, so the general condition that leads to destructive interference is , where .

The general condition that leads to destructive interference entails a number of possibilities. Which one of them, if any, applies at either empty corner of the square?

Answer The general condition that leads to destructive interference is written as , where . There are many possible values for m, and we are being asked what the specific value is. As you walk from point A toward either empty corner, one of the distances decreases and the other increases; the loudness diminishes gradually and disappears at either empty corner. Therefore, destructive interference occurs for the first time at an empty corner, when the difference between the distances to the speakers has attained the smallest possible value of . This means that we are dealing with the specific case in which .

Solution Consider the destructive interference that occurs at either empty corner. Using L to denote the length of a side of the square and taking advantage of the Pythagorean theorem, we have

The specific condition for the destructive interference at the empty corner is

Solving for the wavelength of the waves gives

The next example deals with thin-film interference and serves as a review of the factors that must be considered in such cases.

| Concepts & Calculations Example 12 | | | A Soap Film

A soap film is 375 nm thick and coats a flat piece of glass . Thus, air is on one side of the film and glass is on the other side, as Figure 27-42 illustrates. Sunlight, whose wavelengths (in vacuum) extend from 380 to 750 nm, travels through the air and strikes the film nearly perpendicularly. For which wavelength(s) in this range does constructive interference cause the film to look bright in reflected light?

Figure 27-42 The constructive interference that occurs with this thin soap film is analyzed in Example 12.

Concept Questions and Answers What, if any, phase change occurs when light, traveling in air, reflects from the air–film interface?

Answer In air the index of refraction is , whereas in the film it is . A phase change occurs whenever light travels through a material with a smaller refractive index toward a material with a larger refractive index and reflects from the boundary between the two. The phase change is equivalent to , where is the wavelength in the film.

What, if any, phase change occurs when light, traveling in the film, reflects from the film–glass interface?

Answer In the film the index of refraction is , whereas in the glass it is . Thus, the light is again traveling through a material with a smaller value of n toward a material with a greater value of n, and reflection at the film-glass interface is accompanied by a phase change that is equivalent to .

Is the wavelength of the light in the film greater than, smaller than, or equal to the wavelength in a vacuum?

Answer The wavelength of the light in the film (refractive index = n) is given by Equation 27.3 as . Since the refractive index of the film is , the wavelength in the film is less than the wavelength in a vacuum.

Solution Figure 27-42 shows the soap film and the two rays of light that represent the interfering light waves. At nearly perpendicular incidence, ray 2 travels a distance of 2t farther than ray 1, where t is the thickness of the film. In addition, as discussed in the Concept Questions, ray 2 experiences a phase shift of upon reflection at the bottom film surface, while ray 1 experiences the same phase shift at the upper film surface. Therefore, there is no net phase change for the two reflected rays, and only the extra travel distance determines the type of interference that occurs. For constructive interference the extra travel distance must be an integer number of wavelengths in the film:

This result is equivalent to

The wavelength in the film is (Equation 27.3), so the condition for constructive interference becomes

Solving for the vacuum wavelength gives

For the first four values of m and the given values for n and t, we find that

The range of visible wavelengths (in vacuum) extends from 380 to 750 nm. Therefore, the only visible wavelength in which the film appears bright due to constructive interference is .

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

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