The shimmering colors of soap bubbles and oil slicks, as seen in the photo at the beginning of the chapter, are due to the interference of light waves. In fact, the idea of light-wave interference in one dimension has an important application in the optics industry, namely the use of thin-film optical coatings. These films, less than $1\mu \text{m }({10}^{-6}\text{m})$ thick, are placed on glass surfaces, such as lenses, to control reflections from the glass. Antireflection coatings on the lenses in cameras, microscopes, and other optical equipment are examples of thin-film coatings.

FIGURE 17.23 shows a light wave of wavelength $\text{λ}$ approaching a piece of glass that has been coated with a transparent film of thickness d whose index of refraction is n. The air-film boundary is a discontinuity at which the wave speed suddenly decreases, and you saw earlier, in Figure 17.7, that a discontinuity causes a reflection. Most of the light is transmitted into the film, but a little bit is reflected.

FIGURE 17.23 The two reflections, one from the coating and one from the glass, interfere.

Furthermore, you saw in Figure 17.7 that the wave reflected from a discontinuity at which the speed decreases is inverted with respect to the incident wave. For a sinusoidal wave, which we’re now assuming, the inversion is represented mathematically as a phase shift of $\pi \text{rad}$. The speed of a light wave decreases when it enters a material with a larger index of refraction. Thus a light wave that reflects from a boundary at which the index of refraction increases has a phase shift of $\mathit{\pi }$ rad. There is no phase shift for the reflection from a boundary at which the index of refraction decreases. The reflection in Figure 17.23 is from a boundary between air $(n_{\text{air}}=1.00)$ and a transparent film with $n_{\text{film}}>n_{\text{air}}$, so the reflected wave is inverted due to the phase shift of $\pi \text{rad}$.

When the transmitted wave reaches the glass, most of it continues on into the glass but a portion is reflected back to the left. We’ll assume that the index of refraction of the glass is larger than that of the film, $n_{\text{glass}}>n_{\text{film}}$, so this reflection also has a phase shift of $\pi \text{rad}$. This second reflection, after traveling back through the film, passes back into the air. There are now two equal-frequency reflected waves traveling to the left, and these waves will interfere. If the two reflected waves are in phase, they will interfere constructively to cause a strong reflection. If the two reflected waves are out of phase, they will interfere destructively to cause a weak reflection or, if their amplitudes are equal, no reflection at all.

This suggests practical uses for thin-film optical coatings. The reflections from glass surfaces, even if weak, are often undesirable. For example, reflections degrade the performance of optical equipment. These reflections can be eliminated by coating the glass with a film whose thickness is chosen to cause destructive interference of the two reflected waves. This is an antireflection coating.

The amplitude of the reflected light depends on the phase difference between the two reflected waves. This phase difference is

where we explicitly included the reflection phase shift of each wave. In this case, because both waves had a phase shift of $\pi \text{rad,}$ the reflection phase shifts cancel.

The wavelength ${\text{λ}}_{\text{f}}$ is the wavelength in the film because that’s where the path-length difference $\text{Δ}x$ occurs. You learned in Chapter 16 that the wavelength in a transparent material with index of refraction n is ${\text{λ}}_{\text{f}}=\text{λ}\text{/}n$, where the unsubscripted $\text{λ}$ is the wavelength in vacuum or air. That is, $\text{λ}$ is the wavelength that we measure on “our” side of the air-film boundary.

The path-length difference between the two waves is $\text{Δ}x=2d$ because wave 2 travels through the film twice before rejoining wave 1. The two waves have a common origin—the initial division of the incident wave at the front surface of the film—so the inherent phase difference is $\text{Δ}{\varphi }_0=0$. Thus the phase difference of the two reflected waves is

The interference is constructive, causing a strong reflection, when $\text{Δ}\varphi =m\cdot 2\pi \text{rad}$. So when both reflected waves have a phase shift of $\pi \text{rad,}$ constructive interference occurs for wavelengths

You will notice that m starts with 1, rather than 0, in order to give meaningful results. Destructive interference, with minimum reflection, requires $\text{Δ}\varphi =(m-\frac{1}{2})\cdot 2\pi \text{rad}$. This—again, when both waves have a phase shift of $\pi \text{rad}$—occurs for wavelengths

We’ve used $m-\frac{1}{2}$, rather than $m+\frac{1}{2}$, so that m can start with 1 to match the condition for constructive interference.