Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see Figure 27.10).

Why do we not ordinarily observe wave behavior for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source (the Sun) through a single slit to make the light somewhat coherent. By coherent, we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships. Why did Young then pass the light through a double slit? The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single $\lambda $) light to clarify the effect. Figure 27.11 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 27.12(a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 27.12(b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in Figure 27.13. Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in Figure 27.13(a). If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively as shown in Figure 27.13(b). More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [$(1/2)\lambda $, $(3/2)\lambda $, $(5/2)\lambda $, etc.], then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths ($\lambda $, $2\lambda $, $3\lambda $, etc.), then constructive interference occurs.

### Take-Home Experiment: Using Fingers as Slits

Look at a light, such as a street lamp or incandescent bulb, through the narrow gap between two fingers held close together. What type of pattern do you see? How does it change when you allow the fingers to move a little farther apart? Is it more distinct for a monochromatic source, such as the yellow light from a sodium vapor lamp, than for an incandescent bulb?

Figure 27.14 shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle $\theta $ between the path and a line from the slits to the screen (see the figure) is nearly the same for each path. The difference between the paths is shown in the figure; simple trigonometry shows it to be $d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $, where $d$ is the distance between the slits. To obtain constructive interference for a double slit, the path length difference must be an integral multiple of the wavelength, or

Similarly, to obtain destructive interference for a double slit, the path length difference must be a half-integral multiple of the wavelength, or

where $\lambda $ is the wavelength of the light, $d$ is the distance between slits, and $\theta $ is the angle from the original direction of the beam as discussed above. We call $m$ the order of the interference. For example, $m=4$ is fourth-order interference.

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 27.15. The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation

For fixed $\lambda $ and $m$, the smaller $d$ is, the larger $\theta $ must be, since $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda}/d$. This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance $d$ apart) is small. Small $d$ gives large $\theta $, hence a large effect.

### Example 27.1 Finding a Wavelength from an Interference Pattern

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of $\text{10}\text{.}\text{95\xba}$ relative to the incident beam. What is the wavelength of the light?

Strategy

The third bright line is due to third-order constructive interference, which means that $m=3$. We are given $d=0\text{.}\text{0100}\phantom{\rule{0.25em}{0ex}}\text{mm}$ and $\theta =\text{10}\text{.}\text{95\xba}$. The wavelength can thus be found using the equation $d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda}$ for constructive interference.

**Solution**

The equation is $d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda}$. Solving for the wavelength $\lambda $ gives

Substituting known values yields

Discussion

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with $\lambda $, so that spectra (measurements of intensity versus wavelength) can be obtained.

### Example 27.2 Calculating Highest Order Possible

Interference patterns do not have an infinite number of lines, since there is a limit to how big $m$ can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation $d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda}\phantom{\rule{0.25em}{0ex}}\text{(for}\phantom{\rule{0.25em}{0ex}}m=\mathrm{0,}\phantom{\rule{0.25em}{0ex}}\mathrm{1,}\phantom{\rule{0.25em}{0ex}}-\mathrm{1,}\phantom{\rule{0.25em}{0ex}}\mathrm{2,}\phantom{\rule{0.25em}{0ex}}-\mathrm{2,}\phantom{\rule{0.25em}{0ex}}\dots )$ describes constructive interference. For fixed values of $d$ and $\lambda $, the larger $m$ is, the larger $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $ is. However, the maximum value that $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $ can have is 1, for an angle of $\text{90\xba}$. (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which $m$ corresponds to this maximum diffraction angle.

**Solution**

Solving the equation $d\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda}$ for $m$ gives

Taking $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1$ and substituting the values of $d$ and $\lambda $ from the preceding example gives

Therefore, the largest integer $m$ can be is 15, or

Discussion

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.