# Diffraction and Interference

## Understanding Diffraction and Interference


### Section Key Terms


### Diffraction and Interference

We know that visible light is the type of electromagnetic wave to which our eyes responds. As we have seen previously, light obeys the equation
where  
   m/s is the speed of light in vacuum, f is the frequency of the electromagnetic wave in Hz (or s–1), and  
   is its wavelength in m. The range of visible wavelengths is approximately 380 to 750 nm. As is true for all waves, light travels in straight lines and acts like a ray when it interacts with objects several times as large as its wavelength. However, when it interacts with smaller objects, it displays its wave characteristics prominently. Interference is the identifying behavior of a wave.
In , both the ray and wave characteristics of light can be seen. The laser beam emitted by the observatory represents ray behavior, as it travels in a straight line. Passing a pure, one-wavelength beam through vertical slits with a width close to the wavelength of the beam reveals the wave character of light. Here we see the beam spreading out horizontally into a pattern of bright and dark regions that are caused by systematic constructive and destructive interference. As it is characteristic of wave behavior, interference is observed for water waves, sound waves, and light waves.
That interference is a characteristic of energy propagation by waves is demonstrated more convincingly by water waves.  shows water waves passing through gaps between some rocks. You can easily see that the gaps are similar in width to the wavelength of the waves and that this causes an interference pattern as the waves pass beyond the gaps. A cross-section across the waves in the foreground would show the crests and troughs characteristic of an interference pattern.
Light has wave characteristics in various media as well as in a vacuum. When light goes from a vacuum to some medium, such as water, its speed and wavelength change, but its frequency, f, remains the same. The speed of light in a medium is  
 , where n is its index of refraction. If you divide both sides of the equation  
   by n, you get  
  . Therefore,  
 , where  
   is the wavelength in a medium, and 
where  
   is the wavelength in vacuum and n is the medium’s index of refraction. It follows that the wavelength of light is smaller in any medium than it is in vacuum. In water, for example, which has n = 1.333, the range of visible wavelengths is (380 nm)/1.333 to (760 nm)/1.333, or  
  285–570 nm. Although wavelengths change while traveling from one medium to another, colors do not, since colors are associated with frequency.
The Dutch scientist Christiaan Huygens (1629–1695) developed a useful technique for determining in detail how and where waves propagate. He used wavefronts, which are the points on a wave’s surface that share the same, constant phase (such as all the points that make up the crest of a water wave). Huygens’s principle states, “Every point on a wavefront is a source of wavelets that spread out in the forward direction at the same speed as the wave itself. The new wavefront is a line tangent to all of the wavelets.”
 shows how Huygens’s principle is applied. A wavefront is the long edge that moves; for example, the crest or the trough. Each point on the wavefront emits a semicircular wave that moves at the propagation speed v. These are drawn later at a time, t, so that they have moved a distance  
  . The new wavefront is a line tangent to the wavelets and is where the wave is located at time t. Huygens’s principle works for all types of waves, including water waves, sound waves, and light waves. It will be useful not only in describing how light waves propagate, but also in how they interfere.
What happens when a wave passes through an opening, such as light shining through an open door into a dark room? For light, you expect to see a sharp shadow of the doorway on the floor of the room, and you expect no light to bend around corners into other parts of the room. When sound passes through a door, you hear it everywhere in the room and, thus, you understand that sound spreads out when passing through such an opening. What is the difference between the behavior of sound waves and light waves in this case? The answer is that the wavelengths that make up the light are very short, so that the light acts like a ray. Sound has wavelengths on the order of the size of the door, and so it bends around corners.
If light passes through smaller openings, often called slits, you can use Huygens’s principle to show that light bends as sound does (see ). The bending of a wave around the edges of an opening or an obstacle is called diffraction. Diffraction is a wave characteristic that occurs for all types of waves. If diffraction is observed for a phenomenon, it is evidence that the phenomenon is produced by waves. Thus, the horizontal diffraction of the laser beam after it passes through slits in  is evidence that light has the properties of a wave.
Once again, water waves present a familiar example of a wave phenomenon that is easy to observe and understand, as shown in .
The fact that Huygens’s principle worked was not considered enough evidence to prove that light is a wave. People were also reluctant to accept light’s wave nature because it contradicted the ideas of Isaac Newton, who was still held in high esteem. The acceptance of the wave character of light came after 1801, when the English physicist and physician Thomas Young (1773–1829) did his now-classic double-slit experiment (see ).
When light passes through narrow slits, it is diffracted into semicircular waves, as shown in  (a). Pure constructive interference occurs where the waves line up crest to crest or trough to trough. Pure destructive interference occurs where they line up crest to trough. The light must fall on a screen and be scattered into our eyes for the pattern to be visible. An analogous pattern for water waves is shown in  (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. Those angles depend on wavelength and the distance between the slits, as you will see below.

### Calculations Involving Diffraction and Interference

The fact that the wavelength of light of one color, or monochromatic light, can be calculated from its two-slit diffraction pattern in Young’s experiments supports the conclusion that light has wave properties. To understand the basis of such calculations, consider how two waves travel from the slits to the screen. 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 will end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively. More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths  
 , then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths  
 , then constructive interference occurs.
 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  
   between the path and a line from the slits perpendicular to the screen (see the figure) is nearly the same for each path. That approximation and simple trigonometry show the length difference,  
 , to be  
 , 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
The number m is the order of the interference. For example, m = 4 is fourth-order interference.
 shows how the intensity of the bands of constructive interference decreases with increasing angle.
Light passing through a single slit forms a diffraction pattern somewhat different from that formed by double slits.  shows a single-slit diffraction pattern. Note that the central maximum is larger than those on either side, and that the intensity decreases rapidly on either side.
The analysis of single-slit diffraction is illustrated in . Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel. That approximation allows a series of trigonometric operations that result in the equations for the minima produced by destructive interference.
or
When rays travel straight ahead, they remain in phase and a central maximum is obtained. However, when rays travel at an angle  
   relative to the original direction of the beam, each ray travels a different distance to the screen, and they can arrive in or out of phase. Thus, a ray from the center travels a distance  
   farther than the ray from the top edge of the slit, they arrive out of phase, and they interfere destructively. Similarly, for every ray between the top and the center of the slit, there is a ray between the center and the bottom of the slit that travels a distance  
   farther to the common point on the screen, and so interferes destructively. Symmetrically, there will be another minimum at the same angle below the direct ray.
Below we summarize the equations needed for the calculations to follow.
The speed of light in a vacuum, c, the wavelength of the light,  
 , and its frequency, f, are related as follows.
The wavelength of light in a medium,  
 , compared to its wavelength in a vacuum,  
 , is given by
To calculate the positions of constructive interference for a double slit, the path-length difference must be an integral multiple, m, of the wavelength.  
  
where d is the distance between the slits and  
   is the angle between a line from the slits to the maximum and a line perpendicular to the barrier in which the slits are located. To calculate the positions of destructive interference for a double slit, the path-length difference must be a half-integral multiple of the wavelength:
For a single-slit diffraction pattern, the width of the slit, D, the distance of the first (m = 1) destructive interference minimum, y, the distance from the slit to the screen, L, and the wavelength,  
 , are given by
Also, for single-slit diffraction,
where  
   is the angle between a line from the slit to the minimum and a line perpendicular to the screen, and m is the order of the minimum.

### Practice Problems


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### Section Summary

1. The wavelength of light varies with the refractive index of the medium.
2. Slits produce a diffraction pattern if their width and separation are similar to the wavelength of light passing through them.
3. Interference bands of a single-slit diffraction pattern can be predicted.
4. Interference bands of a double-slit diffraction pattern can be predicted.

### Key Equations


### Concept Items


### Critical Thinking


### Problems


### Test Prep Multiple Choice


### Test Prep Short Answer


### Test Prep Extended Response
