27.6

Resolving Power

Figure 27-26 shows three photographs of an automobile’s headlights taken at progressively greater distances from the camera. In parts a and b, the two separate headlights can be seen clearly. In part c, however, the car is so far away that the headlights are barely distinguishable and appear almost as a single light. The resolving power of an optical instrument, such as a camera, is its ability to distinguish between two closely spaced objects. If a camera with a higher resolving power had taken these pictures, the photograph in part c would have shown two distinct and separate headlights. Any instrument used for viewing objects that are close together must have a high resolving power. This is true, for example, for a telescope used to view distant stars or for a microscope used to view tiny organisms. We will now see that diffraction occurs when light passes through the circular, or nearly circular, openings that admit light into cameras, telescopes, microscopes, and human eyes. The resulting diffraction pattern places a limit on the resolving power of these instruments.

Figure 27-26 These automobile headlights were photographed at various distances from the camera, closest in part (a) and farthest in part (c). In part (c), the headlights are so far away that they are barely distinguishable. (© Truax/The Image Finders).

Figure 27-27 shows the diffraction pattern created by a small circular opening when the viewing screen is far from the opening. The pattern consists of a central bright circular region, surrounded by alternating bright and dark circular fringes. These fringes are analogous to the rectangular fringes that a single slit produces. The angle in the picture locates the first circular dark fringe relative to the central bright region and is given by

| (27.5)

where is the wavelength of the light and D is the diameter of the opening. This expression is similar to Equation 27.4 for a slit (, when ) and is valid when the distance to the screen is much larger than the diameter D.

Figure 27-27 When light passes through a small circular opening, a circular diffraction pattern is formed on a screen. The angle locates the first dark fringe relative to the central bright region. The intensities of the bright fringes and the diameter of the opening have been exaggerated for clarity.

An optical instrument with the ability to resolve two closely spaced objects can produce images of them that can be identified separately. Think about the images on the film when light from two widely separated point objects passes through the circular aperture of a camera. As Figure 27-28 illustrates, each image is a circular diffraction pattern, but the two patterns do not overlap and are completely resolved. On the other hand, if the objects are sufficiently close together, the intensity patterns created by the diffraction overlap, as Figure 27-29a suggests. In fact, if the overlap is extensive, it may no longer be possible to distinguish the patterns separately. In such a case, the picture from a camera would show a single blurred object instead of two separate objects. In Figure 27-29b the diffraction patterns overlap, but not enough to prevent us from seeing that two objects are present. Ultimately, then, diffraction limits the ability of an optical instrument to produce distinguishable images of objects that are close together.

Figure 27-28 When light from two point objects passes through the circular aperture of a camera, two circular diffraction patterns are formed as images on the film. The images here are completely separated or resolved because the objects are widely separated.

Figure 27-29 (a) According to the Rayleigh criterion, two point objects are just resolved when the first dark fringe (zero intensity) of one image falls on the central bright fringe (maximum intensity) of the other image. (b) This photograph shows two overlapping but still resolvable diffraction patterns. (From Michel Cagnet et al., Atlas of Optical Phenomena, Springer-Verlag, Berlin, 1962).

It is useful to have a criterion for judging whether two closely spaced objects will be resolved by an optical instrument. Figure 27-29a presents the Rayleigh criterion for resolution, first proposed by Lord Rayleigh (1842–1919):

> Two point objects are just resolved when the first dark fringe in the diffraction pattern of one falls directly on the central bright fringe in the diffraction pattern of the other.

The minimum angle between the two objects in the drawing is that given by Equation 27.5. If is small (less than about ) and is expressed in radians, . Then, Equation 27.5 can be rewritten as

| (27.6)

For a given wavelength and aperture diameter D, this result specifies the smallest angle that two point objects can subtend at the aperture and still be resolved. According to Equation 27.6, optical instruments designed to resolve closely spaced objects (small values of ) must utilize the smallest possible wavelength and the largest possible aperture diameter. For example, when short-wavelength ultraviolet light is collected by its large 2.4-m-diameter mirror, the Hubble Space Telescope is capable of resolving two closely spaced stars that have an angular separation of about . This angle is equivalent to resolving two objects only 1 cm apart when they are (about 62 miles) from the telescope. Example 7 deals with the resolving power of the human eye.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 7 | | | The Human Eye Versus the Eagle’s Eye

(a) A hang glider is flying at an altitude of 120 m. Green light (wavelength = 555 nm in vacuum) enters the pilot’s eye through a pupil that has a diameter of 2.5 mm. Determine how far apart two point objects must be on the ground if the pilot is to have any hope of distinguishing between them (see Figure 27-30). (b) An eagle’s eye has a pupil with a diameter of 6.2 mm. Repeat part (a) for an eagle flying at the same altitude as the glider.

Figure 27-30 The Rayleigh criterion can be used to estimate the smallest distance s that can separate two objects on the ground, if a person on a hang glider is to be able to distinguish between them.

Reasoning A greater distance s of separation between the objects on the ground makes it easier for the eye of the observer (the pilot or the eagle; see Figure 27-30) to resolve them as separate objects. This is because the angle that the two objects subtend at the pupil of the eye is greater when the separation distance is greater. This angle must be at least as large as the angle specified by the Rayleigh criterion for resolution. In applying this criterion, we will use the concept of the radian to express the angle as an arc length (approximately the separation distance) divided by a radius (the altitude), as discussed in Section 8.1.
Knowns and Unknowns The following table summarizes the data that are given:

Description | Symbol | Value | Comment
Altitude | H | 120 m | Same for pilot and eagle.
Wavelength of light in vacuum |  | 555 nm |
Diameter of pupil of eye | D | 2.5 mm or 6.2 mm | Smaller value is for pilot. Larger value is for eagle.
Unknown Variable |  |  |
Separation between objects on ground | s | ? |

Modeling the Problem

| Step 1 Radian Measure
| The angle that the two objects subtend at the pupil of the eye must be at least as large as the angle specified by the Rayleigh criterion for resolution. Using radian measure as discussed in Section 8.1, we refer to Figure 27-30 and express this angle as

This is an approximate application of Equation 8.1, which states that an angle in radians is the arc length divided by the radius. Here, the arc length is approximately the separation distance s, assuming that the altitude H is much greater than s. Solving for s gives Equation 1 at the right. In this result, the altitude is known, and we proceed to Step 2 to evaluate the angle .

| (1)

| Step 2 The Rayleigh Criterion
| The Rayleigh criterion specifies the angle in

| (27.6)

where is the wavelength of the light in vacuum* and D is the diameter of the pupil of the eye. The substitution of this expression into Equation 1 is shown at the right.

Solution Combining the results of each step algebraically, we find that

Problem solving insight The minimum angle between two objects that are just resolved must be expressed in radians, not degrees, when using Equation 27.6 ().

The separation distance between the objects on the ground can now be obtained.

a. | For the pilot to have any hope of distinguishing between the objects, the separation distance must be at least

b. | For the eagle, we find that

The physics of comparing human eyes and eagle eyes.

Since the pupil of the eagle’s eye is larger than that of a human eye, diffraction creates less of a limitation for the eagle; the two objects can be closer together and still be resolved by the eagle’s eye.
Related Homework: Problems 29, 31, 34
Many optical instruments have a resolving power that exceeds that of the human eye. The typical camera does, for instance. Conceptual Example 8 compares the abilities of the human eye and a camera to resolve two closely spaced objects.

| Conceptual Example 8 | | | What You See is Not What You Get

The French postimpressionist artist Georges Seurat developed a technique of painting in which dots of color are placed close together on the canvas. From sufficiently far away the individual dots are not distinguishable, and the images in the picture take on a more normal appearance. Figure 27-31 shows a person in an art museum, looking at one of Seurat’s paintings. Suppose that this person stands close to the painting, then backs up until the dots just become indistinguishable to his eyes and takes a picture from this position. When he gets home and has the film developed, however, he can see the individual dots in the enlarged photograph. Why does the camera resolve the individual dots, while his eyes do not? Assume that light enters his eyes through pupils that have diameters of 2.5 mm and enters the camera through an aperture, or opening, with a diameter of 25 mm.

Figure 27-31 This person is about to take a photograph of a famous painting by Georges Seurat, who developed the technique of using tiny dots of color to construct his images. Conceptual Example 8 discusses what the person sees when the film is developed. (Georges Seurat, A Sunday on La Grande Jatte. © 2005. All rights reserved. Reproduced with permission of The Art Institute of Chicago).

Reasoning and Solution To answer this question, we turn to the Rayleigh criterion for resolving two point objects—namely, . According to the Rayleigh criterion, a larger value of the diameter D of the opening means that the value for is smaller, so that an optical instrument can be farther away and still resolve the objects. In other words, a larger value of D means a greater resolving power. For the eye and the camera the diameters are and , so the diameter for the camera is ten times larger than that for the eye. Thus, at the distance at which the eye loses its ability to resolve the individual dots in the painting, the camera can easily resolve the dots. As discussed in the footnote to Example 7, we can ignore the effect on the wavelength of the index of refraction of the material from which the eye is made. We conclude, then, that the camera succeeds in resolving the dots in the painting while the eye fails because the camera gathers light through a much larger opening.
Related Homework: Conceptual Question 14, Problem 32

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