To understand interference, we need to focus on the phases of the two waves, which are

The difference between the two phases ${\varphi }_1$ and ${\varphi }_2$, called the phase difference $\text{Δ}\varphi$, is

You can see that there are two contributions to the phase difference. $\text{Δ}x=x_2-x_1$, the distance between the two sources, is called path-length difference. It is the extra distance traveled by wave 2 on the way to the point where the two waves are combined. $\text{Δ}{\varphi }_0={\varphi }_{20}-{\varphi }_{10}$ is the inherent phase difference between the sources.

The condition of being in phase, where crests are aligned with crests and troughs with troughs, is $\text{Δ}\varphi =0$, $2\pi$, $4\pi$, or any integer multiple of $2\pi \text{rad}$. Thus the condition for maximum constructive interference is

For identical sources, which have $\text{Δ}{\varphi }_0=0\text{rad,}$ maximum constructive interference occurs when $\text{Δ}x=m\text{λ}$. That is, two identical sources produce maximum constructive interference when the path-length difference is an integer number of wavelengths.

FIGURE 17.19 shows two identical sources (i.e., the two loudspeakers are doing the same thing at the same time), so $\text{Δ}{\varphi }_0=0\text{rad}$. The path-length difference $\text{Δ}x$ is the extra distance traveled by the wave from loudspeaker 2 before it combines with loudspeaker 1. In this case, $\text{Δ}x=\text{λ}$. Because a wave moves forward exactly one wavelength during one period, loudspeaker 1 emits a crest exactly as a crest of wave 2 passes by. The two waves are “in step,” with $\text{Δ}\varphi =2\pi \text{rad,}$ so the two waves interfere constructively to produce a wave of amplitude 2a.

FIGURE 17.19 Two identical sources one wavelength apart.

Maximum destructive interference, where the crests of one wave are aligned with the troughs of the other, occurs when two waves are out of phase, meaning that $\text{Δ}\varphi =\pi$, $3\pi$, $5\pi$, or any odd multiple of $\pi \text{rad}$. Thus the condition for maximum destructive interference is

For identical sources, which have $\text{Δ}{\varphi }_0=0\text{rad,}$ maximum destructive interference occurs when $\text{Δ}x=(m+\frac{1}{2})\text{λ}$. That is, two identical sources produce maximum destructive interference when the path-length difference is a half-integer number of wavelengths.

Two waves can be out of phase because the sources are located at different positions, because the sources themselves are out of phase, or because of a combination of these two. FIGURE 17.20 illustrates these ideas by showing three different ways in which two waves interfere destructively. Each of these three arrangements creates waves with $\text{Δ}\varphi =\pi \text{rad}$.

FIGURE 17.20 Destructive interference three ways.

Full Alternative Text

Full Alternative Text

EXAMPLE 17.7 Interference between two sound waves

You are standing in front of two side-by-side loudspeakers playing sounds of the same frequency. Initially there is almost no sound at all. Then one of the speakers is moved slowly away from you. The sound intensity increases as the separation between the speakers increases, reaching a maximum when the speakers are 0.75 m apart. Then, as the speaker continues to move, the intensity starts to decrease. What is the distance between the speakers when the sound intensity is again a minimum?

MODEL The changing sound intensity is due to the interference of two overlapped sound waves.

VISUALIZE Moving one speaker relative to the other changes the phase difference between the waves.

SOLVE A minimum sound intensity implies that the two sound waves are interfering destructively. Initially the loudspeakers are side by side, so the situation is as shown in Figure 17.20a with $\text{Δ}x=0$ and $\text{Δ}{\varphi }_0=\pi \text{rad}$. That is, the speakers themselves are out of phase. Moving one of the speakers does not change $\text{Δ}{\varphi }_0$, but it does change the path-length difference $\text{Δ}x$ and thus increases the overall phase difference $\text{Δ}\varphi$. Constructive interference, causing maximum intensity, is reached when

$$\text{Δ}\varphi =2\pi {\displaystyle \frac{\text{Δ}x}{\text{λ}}}+\text{Δ}{\varphi }_0=2\pi {\displaystyle \frac{\text{Δ}x}{\text{λ}}}+\pi =2\pi \text{rad}$$

where we used $m=1$ because this is the first separation giving constructive interference. The speaker separation at which this occurs is $\text{Δ}x=\text{λ}\text{/}2$. This is the situation shown in FIGURE 17.21.

FIGURE 17.21 Two out-of-phase sources generate waves that are in phase if the sources are one half-wavelength apart.

FIGURE 17.21 Full Alternative Text

Because $\text{Δ}x=0.75\text{m}$ is $\text{λ}\text{/}2$, the sound’s wavelength is $\text{λ}=1.50\text{m}$. The next point of destructive interference, with $m=1$, occurs when

$$\text{Δ}\varphi =2\pi {\displaystyle \frac{\text{Δ}x}{\text{λ}}}+\text{Δ}{\varphi }_0=2\pi {\displaystyle \frac{\text{Δ}x}{\text{λ}}}+\pi =3\pi \text{rad}$$

Thus the distance between the speakers when the sound intensity is again a minimum is

$$\text{Δ}x=\text{λ}=1.50\text{m}$$

REVIEW A separation of $\text{λ}$ gives constructive interference for two identical speakers $(\text{Δ}{\varphi }_0=0)$. Here the phase difference of $\pi \text{rad}$ between the speakers (one is pushing forward as the other pulls back) gives destructive interference at this separation.