No discussion of wave optics would be complete without mentioning holography, which has both scientific and artistic applications. The basic idea is a simple extension of interferometry.

FIGURE 33.26a shows how a hologram is made. A beam splitter divides a laser beam into two waves. One wave illuminates the object of interest. The light scattered by this object is a very complex wave, but it is the wave you would see if you looked at the object from the position of the film. The other wave, called the reference beam, is reflected directly toward the film. The scattered light and the reference beam meet at the film and interfere. The film records their interference pattern.

FIGURE 33.26 Holography is an important application of wave optics.

FIGURE 33.26 Full Alternative Text

The interference patterns we’ve looked at in this chapter have been simple patterns of stripes and circles because the light waves have been well-behaved plane waves and spherical waves. The light wave scattered by the object in Figure 33.26a is exceedingly complex. As a result, the interference pattern recorded on the film—the hologram—is a seemingly random pattern of whorls and blotches. FIGURE 33.26b is an enlarged photograph of a portion of a hologram. It’s certainly not obvious that information is stored in this pattern, but it is.

The hologram is “played” by sending just the reference beam through it, as seen in FIGURE 33.26c. The reference beam diffracts through the transparent parts of the hologram, just as it would through the slits of a diffraction grating. Amazingly, the diffracted wave is exactly the same as the light wave that had been scattered by the object! In other words, the diffracted reference beam reconstructs the original scattered wave. As you look at this diffracted wave, from the far side of the hologram, you “see” the object exactly as if it were there. The view is three dimensional because, by moving your head with respect to the hologram, you can see different portions of the wave front.

CHAPTER 33 CHALLENGE EXAMPLE Measuring the index of refraction of a gas

A Michelson interferometer uses a helium-neon laser with wavelength ${\text{λ}}_{\text{vac}}=633\text{nm}$. In one arm, the light passes through a 4.00-cm-thick glass cell. Initially the cell is evacuated, and the interferometer is adjusted so that the central spot is a bright fringe. The cell is then slowly filled to atmospheric pressure with a gas. As the cell fills, 43 bright-dark-bright fringe shifts are seen and counted. What is the index of refraction of the gas at this wavelength?

MODEL Adding one additional wavelength to the round trip causes one bright-dark-bright fringe shift. Changing the length of the arm is one way to add wavelengths, but not the only way. Increasing the index of refraction also adds wavelengths because light has a shorter wavelength when traveling through a material with a larger index of refraction.

VISUALIZE FIGURE 33.27 shows a Michelson interferometer with a cell of thickness $d$ in one arm.

FIGURE 33.27 Measuring the index of refraction.

FIGURE 33.27 Full Alternative Text

SOLVE To begin, all the air is pumped out of the cell. As light travels from the beam splitter to the mirror and back, the number of wavelengths inside the cell is

$$m_1={\displaystyle \frac{2d}{{\text{λ}}_{\text{vac}}}}$$

where the 2 appears because the light passes through the cell twice.

The cell is then filled with gas at 1 atm pressure. Light travels slower in the gas, $v=c/n$, and you learned in Chapter 16 that the reduction in speed decreases the wavelength to ${\text{λ}}_{\text{vac}}/n$. With the cell filled, the number of wavelengths spanning distance $d$ is

$$m_2={\displaystyle \frac{2d}{\text{λ}}}={\displaystyle \frac{2d}{{\text{λ}}_{\text{vac}}/n}}$$

The physical distance has not changed, but the number of wavelengths along the path has. Filling the cell has increased the path by

$$\text{Δ}m=m_2-m_1=(n-1){\displaystyle \frac{2d}{{\text{λ}}_{\text{vac}}}}$$

wavelengths. Each increase of one wavelength causes one bright-dark-bright fringe shift at the output. Solving for $n$, we find

$$n=1+{\displaystyle \frac{{\text{λ}}_{\text{vac}}\text{Δ}m}{2d}}=1+{\displaystyle \frac{(6.33\times {10}^{-7}\text{m})\left(43\right)}{2\left(0.0400\text{m}\right)}}=1.00034$$

REVIEW This may seem like a six-significant-figure result, but there are really only two. What we’re measuring is not $n$ but $n-1$. We know the fringe count to two significant figures, and that has allowed us to compute $n-1={\text{λ}}_{\text{vac}}\text{Δ}m/2d=3.4\times {10}^{-4}$.