27.2

Young’s Double-Slit Experiment

In 1801 the English scientist Thomas Young (1773–1829) performed a historic experiment that demonstrated the wave nature of light by showing that two overlapping light waves interfered with each other. His experiment was particularly important because he was also able to determine the wavelength of the light from his measurements, the first such determination of this important property. Figure 27-4 shows one arrangement of Young’s experiment, in which light of a single wavelength (monochromatic light) passes through a single narrow slit and falls on two closely spaced, narrow slits S1 and S2. These two slits act as coherent sources of light waves that interfere constructively and destructively at different points on the screen to produce a pattern of alternating bright and dark fringes. The purpose of the single slit is to ensure that only light from one direction falls on the double slit. Without it, light coming from different points on the light source would strike the double slit from different directions and cause the pattern on the screen to be washed out. The slits S1 and S2 act as coherent sources of light waves because the light from each originates from the same primary source—namely, the single slit.

Figure 27-4 In Young’s double-slit experiment, two slits S1 and S2 act as coherent sources of light. Light waves from these slits interfere constructively and destructively on the screen to produce, respectively, the bright and dark fringes. The slit widths and the distance between the slits have been exaggerated for clarity.

To help explain the origin of the bright and dark fringes, Figure 27-5 presents three top views of the double slit and the screen. Part a illustrates how a bright fringe arises directly opposite the midpoint between the two slits. In this part of the drawing the waves (identical) from each slit travel to the midpoint on the screen. At this location, the distances and to the slits are equal, each containing the same number of wavelengths. Therefore, constructive interference results, leading to the bright fringe. Part b indicates that constructive interference produces another bright fringe on one side of the midpoint when the distance is larger than by exactly one wavelength. A bright fringe also occurs symmetrically on the other side of the midpoint when the distance exceeds by one wavelength; for clarity, however, this bright fringe is not shown. Constructive interference produces additional bright fringes (not shown) on both sides of the middle wherever the difference between and is an integer number of wavelengths: , , , and so on. Part c shows how the first dark fringe arises. Here the distance is larger than by exactly one-half a wavelength, so the waves interfere destructively, giving rise to the dark fringe. Destructive interference creates additional dark fringes on both sides of the center wherever the difference between and equals an odd integer number of half-wavelengths: and so on.

Figure 27-5 The waves that originate from slits S1 and S2 interfere constructively (parts a and b) or destructively (part c) on the screen, depending on the difference in distances between the slits and the screen. Note that the slit widths and the distance between the slits have been exaggerated for clarity.

The brightness of the fringes in Young’s experiment varies, as the photograph in Figure 27-6 shows. Below the photograph is a graph to suggest the way in which the intensity varies for the fringe pattern. The central fringe is labeled with a zero, and the other bright fringes are numbered in ascending order on either side of the center. It can be seen that the central fringe has the greatest intensity. To either side of the center, the intensities of the other fringes decrease symmetrically in a way that depends on how small the slit widths are relative to the wavelength of the light.

Figure 27-6 The results of Young’s double-slit experiment, showing a photograph of the bright and dark fringes formed on the screen and a graph of the light intensity. The central or zeroth fringe is the brightest fringe (greatest intensity). (From Michel Cagnet, et al., Atlas of Optical Phenomena, Springer-Verlag, Berlin, 1962).

The position of the fringes observed on the screen in Young’s experiment can be calculated with the aid of Figure 27-7. If the screen is located far away compared with the separation d of the slits, then the lines labeled and in part a are nearly parallel. Being nearly parallel, these lines make approximately equal angles with the horizontal. The distances and differ by an amount , which is the length of the short side of the colored triangle in part b of the drawing. Since the triangle is a right triangle, it follows that . Constructive interference occurs when the distances differ by an integer number m of wavelengths , or . Therefore, the angle for the interference maxima can be determined from the following expression:

| (27.1)

The value of m specifies the order of the fringe. Thus, identifies the “second-order” bright fringe. Part c of the drawing stresses that the angle given by Equation 27.1 locates bright fringes on either side of the midpoint between the slits. A similar line of reasoning leads to the conclusion that the dark fringes, which lie between the bright fringes, are located according to

| (27.2)

Example 1 illustrates how to determine the distance of a higher-order bright fringe from the central bright fringe with the aid of these expressions.

Figure 27-7 (a) Rays from slits S1 and S2, which make approximately the same angle with the horizontal, strike a distant screen at the same spot. (b) The difference in the path lengths of the two rays is . (c) The angle is the angle at which a bright fringe (, here) occurs on either side of the central bright fringe ().

| Example 1 | | | Young’s Double-Slit Experiment

Red light ( nm in vacuum) is used in Young’s experiment with the slits separated by a distance . The screen in Figure 27-8 is located at a distance of from the slits. Find the distance y on the screen between the central bright fringe and the third-order bright fringe.

Figure 27-8 The third-order bright fringe () is observed on the screen at a distance y from the central bright fringe ().

Reasoning This problem can be solved by first using Equation 27.1 to determine the value of that locates the third-order bright fringe. Then trigonometry can be used to obtain the distance y.
Solution According to Equation 27.1, we find

According to Figure 27-8, the distance y can be calculated from :

Need more practice?

Interactive LearningWare27.1
In Young’s double-slit experiment, blue light ( nm) gives a second-order bright fringe at a certain location on a flat screen. What wavelength of visible light would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.
Related Homework: Problem 50

In the preceding version of Young’s experiment, monochromatic light has been used. Light that contains a mixture of wavelengths can also be used. Conceptual Example 2 deals with some of the interesting features of the resulting interference pattern.

| Conceptual Example 2 | | | White Light and Young’s Experiment

Figure 27-9 shows a photograph that illustrates the kind of interference fringes that can result when white light, which is a mixture of all colors, is used in Young’s experiment. Except for the central fringe, which is white, the bright fringes are a rainbow of colors. Why does Young’s experiment separate white light into its constituent colors? In any group of colored fringes, such as the two singled out in Figure 27-9, why is red farther out from the central fringe than green is? And finally, why is the central fringe white rather than colored?

Figure 27-9 This photograph shows the results observed on the screen in one version of Young’s experiment in which white light (a mixture of all colors) is used. (Andy Washnik)

Reasoning and Solution To understand how the color separation arises, we need to remember that each color corresponds to a different wavelength and that constructive and destructive interference depend on the wavelength. According to Equation 27.1 , there is a different angle that locates a bright fringe for each value of , and thus for each color. These different angles lead to the separation of colors on the observation screen. In fact, on either side of the central fringe, there is one group of colored fringes for and another for each value of m.
Now, consider what it means that, within any single group of colored fringes, red is farther out from the central fringe than green is. It means that, in the equation , red light has a larger angle than green light does. Does this make sense? Yes, because red has the larger wavelength (see Table 26-2, where nm and nm).
In Figure 27-9, the central fringe is distinguished from all the other colored fringes by being white. In Equation 27.1, the central fringe is different from the other fringes because it is the only one for which . In Equation 27.1, a value of means that , which reveals that , no matter what the wavelength is. In other words, all wavelengths have a zeroth-order bright fringe located at the same place on the screen, so that all colors strike the screen there and mix together to produce the white central fringe.
Related Homework: Problem 52

Historically, Young’s experiment provided strong evidence that light has a wave-like character. If light behaved only as a stream of “tiny particles,” as others believed at the time,* then the two slits would deliver the light energy into only two bright fringes located directly opposite the slits on the screen. Instead, Young’s experiment shows that wave interference redistributes the energy from the two slits into many bright fringes.

Check Your Understanding 2

The drawing shows two double-slits that have slit separations of d1 and d2. Light whose wavelength is either or passes through the slits. For comparison, the wavelengths are also illustrated in the drawing. For which combination of slit separation and wavelength would the pattern of bright and dark fringes on the observation screen be (a) the most spread out and (b) the least spread out? (The answers are given at the end of the book.)

Background: The amount by which the fringe pattern on the observation screen would be spread out depends on how large the wavelength is relative to the slit separation.
For similar questions (including calculational counterparts), consult Self-Assessment Test 27.1, which is described at the end of Section 27.4.

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.