Image Formation by Thin Lenses
In some circumstances, a lens forms an obvious image, such as when a movie projector casts an image onto a screen. In other cases, the image is less obvious. Where, for example, is the image formed by eyeglasses? We will use ray tracing for thin lenses to illustrate how they form images, and we will develop equations to describe the image formation quantitatively.
Consider an object some distance away from a converging lens, as shown in Figure 8.26. To find the location and size of the image formed, we trace the paths of selected light rays originating from one point on the object, in this case the top of the person’s head. The figure shows three rays from the top of the object that can be traced using the ray tracing rules given above. Rays leave this point going in many directions, but we concentrate on only a few with paths that are easy to trace. The first ray is one that enters the lens parallel to its axis and passes through the focal point on the other side (rule 1). The second ray passes through the center of the lens without changing direction (rule 3). The third ray passes through the nearer focal point on its way into the lens and leaves the lens parallel to its axis (rule 4). The three rays cross at the same point on the other side of the lens. The image of the top of the person’s head is located at this point. All rays that come from the same point on the top of the person’s head are refracted in such a way as to cross at the point shown. Rays from another point on the object, such as her belt buckle, will also cross at another common point, forming a complete image, as shown. Although three rays are traced in Figure 8.26, only two are necessary to locate the image. It is best to trace rays for which there are simple ray tracing rules. Before applying ray tracing to other situations, let us consider the example shown in Figure 8.26 in more detail.
The image formed in Figure 8.26 is a real image, meaning that it can be projected. That is, light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye, for example. Figure 8.27 shows how such an image would be projected onto film by a camera lens. This figure also shows how a real image is projected onto the retina by the lens of an eye. Note that the image is there whether it is projected onto a screen or not.
Real Image
The image in which light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye is called a real image.
Several important distances appear in Figure 8.26. We define to be the object distance, the distance of an object from the center of a lens. Image distance is defined to be the distance of the image from the center of a lens. The height of the object and height of the image are given the symbols and respectively. Images that appear upright relative to the object have heights that are positive and those that are inverted have negative heights. Using the rules of ray tracing and making a scale drawing with paper and pencil, like that in Figure 8.26, we can accurately describe the location and size of an image. But the real benefit of ray tracing is in visualizing how images are formed in a variety of situations. To obtain numerical information, we use a pair of equations that can be derived from a geometric analysis of ray tracing for thin lenses. The thin lens equations are
8.24 and
8.25 We define the ratio of image height to object height to be the magnification The minus sign in the equation above will be discussed shortly. The thin lens equations are broadly applicable to all situations involving thin lenses, and thin mirrors, as we will see later. We will explore many features of image formation in the following worked examples.
Image Distance
The distance of the image from the center of the lens is called image distance.
Thin Lens Equations and Magnification
8.26 8.27
Example 8.6 Finding the Image of a Light Bulb Filament by Ray Tracing and by the Thin Lens Equations
A clear glass light bulb is placed 0.750 m from a convex lens having a 0.500 m focal length, as shown in Figure 8.28. Use ray tracing to get an approximate location for the image. Then use the thin lens equations to calculate (a) the location of the image and (b) its magnification. Verify that ray tracing and the thin lens equations produce consistent results.
Strategy and Concept
Since the object is placed farther away from a converging lens than the focal length of the lens, this situation is analogous to those illustrated in Figure 8.26 and Figure 8.27. Ray tracing to scale should produce similar results for Numerical solutions for and can be obtained using the thin lens equations, noting that
Solutions (Ray tracing)
The ray tracing to scale in Figure 8.28 shows two rays from a point on the bulb’s filament crossing about 1.50 m on the far side of the lens. Thus, the image distance is about 1.50 m. Similarly, the image height based on ray tracing is greater than the object height by about a factor of 2, and the image is inverted. Thus is about –2. The minus sign indicates that the image is inverted.
The thin lens equations can be used to find from the given information.
8.28 Rearranging to isolate gives
8.29 Entering known quantities gives a value for
8.30 This must be inverted to find
8.31 Note that another way to find is to rearrange the equation.
8.32 This yields the equation for the image distance as
8.33 Note that there is no inverting here.
The thin lens equations can be used to find the magnification since both and are known. Entering their values gives
8.34
Discussion
Note that the minus sign causes the magnification to be negative when the image is inverted. Ray tracing and the use of the thin lens equations produce consistent results. The thin lens equations give the most precise results, being limited only by the accuracy of the given information. Ray tracing is limited by the accuracy with which you can draw, but it is highly useful both conceptually and visually.
Real images, such as the one considered in the previous example, are formed by converging lenses whenever an object is farther from the lens than its focal length. This is true for movie projectors, cameras, and the eye. We shall refer to these as case 1 images. A case 1 image is formed when and is positive, as in Figure 8.29(a). A summary of the three cases or types of image formation appears at the end of this section.
A different type of image is formed when an object, such as a person's face, is held close to a convex lens. The image is upright and larger than the object, as seen in Figure 8.29(b), and so the lens is called a magnifier. If you slowly pull the magnifier away from the face, you will see that the magnification steadily increases until the image begins to blur. Pulling the magnifier even farther away produces an inverted image as seen in Figure 8.29(a). The distance at which the image blurs, and beyond which it inverts, is the focal length of the lens. To use a convex lens as a magnifier, the object must be closer to the converging lens than its focal length. This is called a case 2 image. A case 2 image is formed when and is positive.
Figure 8.30 uses ray tracing to show how an image is formed when an object is held closer to a converging lens than its focal length. Rays coming from a common point on the object continue to diverge after passing through the lens, but all appear to originate from a point at the location of the image. The image is on the same side of the lens as the object and is farther away from the lens than the object. This image, like all case 2 images, cannot be projected and, hence, is called a virtual image. Light rays only appear to originate at a virtual image; they do not actually pass through that location in space. A screen placed at the location of a virtual image will receive only diffuse light from the object, not focused rays from the lens. Additionally, a screen placed on the opposite side of the lens will receive rays that are still diverging, and so no image will be projected on it. We can see the magnified image with our eyes, because the lens of the eye converges the rays into a real image projected on our retina. Finally, we note that a virtual image is upright and larger than the object, meaning that the magnification is positive and greater than 1.
Virtual Image
An image that is on the same side of the lens as the object and cannot be projected on a screen is called a virtual image.
Example 8.7 Image Produced by a Magnifying Glass
Suppose the book page in Figure 8.30 (a) is held 7.50 cm from a convex lens of focal length 10.0 cm, such as a typical magnifying glass might have. What magnification is produced?
Strategy and Concept
We are given that and so we have a situation where the object is placed closer to the lens than its focal length. We therefore expect to get a case 2 virtual image with a positive magnification that is greater than 1. Ray tracing produces an image like that shown in Figure 8.30, but we will use the thin lens equations to get numerical solutions in this example.
Solution
To find the magnification
we try to use magnification equation,
We do not have a value for
so that we must first find the location of the image using lens equation. The procedure is the same as followed in the preceding example, where
and
were known. Rearranging the magnification equation to isolate
gives
8.35 Entering known values, we obtain a value for
8.36 This must be inverted to find
8.37 Now the thin lens equation can be used to find the magnification since both and are known. Entering their values gives
8.38
Discussion
A number of results in this example are true of all case 2 images, as well as being consistent with Figure 8.30. Magnification is indeed positive, as predicted, meaning the image is upright. The magnification is also greater than 1, meaning that the image is larger than the object—in this case, by a factor of 3. Note that the image distance is negative. This means the image is on the same side of the lens as the object. Thus the image cannot be projected and is virtual. Negative values of occur for virtual images. The image is farther from the lens than the object, since the image distance is greater in magnitude than the object distance. The location of the image is not obvious when you look through a magnifier. In fact, since the image is bigger than the object, you may think the image is closer than the object. But the image is farther away, a fact that is useful in correcting farsightedness, as we shall see in a later section.
A third type of image is formed by a diverging or concave lens. Try looking through eyeglasses meant to correct nearsightedness. (See Figure 8.31.) You will see an image that is upright but smaller than the object. This means that the magnification is positive but less than 1. The ray diagram in Figure 8.32 shows that the image is on the same side of the lens as the object and, hence, cannot be projected—it is a virtual image. Note that the image is closer to the lens than the object. This is a case 3 image, formed for any object by a negative focal length or diverging lens.
Example 8.8 Image Produced by a Concave Lens
Suppose an object such as a book page is held 7.50 cm from a concave lens of focal length –10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. What magnification is produced?
Strategy and Concept
This example is identical to the preceding one, except that the focal length is negative for a concave or diverging lens. The method of solution is thus the same, but the results are different in important ways.
Solution
To find the magnification we must first find the image distance using thin lens equation
8.39 or its alternative rearrangement
8.40 We are given that and
Entering these yields a value for
8.41 This must be inverted to find :
8.42 or
8.43 Now the magnification equation can be used to find the magnification since both and are known. Entering their values gives
8.44
Discussion
A number of results in this example are true of all case 3 images, as well as being consistent with Figure 8.32. Magnification is positive, as predicted, meaning the image is upright. The magnification is also less than 1, meaning the image is smaller than the object—in this case, a little over half its size. The image distance is negative, meaning the image is on the same side of the lens as the object. The image is virtual. The image is closer to the lens than the object, since the image distance is smaller in magnitude than the object distance. The location of the image is not obvious when you look through a concave lens. In fact, since the image is smaller than the object, you may think it is farther away. But the image is closer than the object, a fact that is useful in correcting nearsightedness, as we shall see in a later section.
Table 8.2 summarizes the three types of images formed by single thin lenses. These are referred to as case 1, 2, and 3 images. Convex, converging, lenses can form either real or virtual images (cases 1 and 2, respectively), whereas concave, diverging, lenses can form only virtual images (always case 3). Real images are always inverted, but they can be either larger or smaller than the object. For example, a slide projector forms an image larger than the slide, whereas a camera makes an image smaller than the object being photographed. Virtual images are always upright and cannot be projected. Virtual images are larger than the object only in case 2, where a convex lens is used. The virtual image produced by a concave lens is always smaller than the object—a case 3 image. We can see and photograph virtual images only by using an additional lens to form a real image.
Type |
Formed When |
Image Type |
di
|
m |
Case 1 |
positive, |
real |
positive |
negative |
Case 2 |
positive, |
virtual |
negative |
positive |
Case 3 |
negative |
virtual |
negative |
positive |
Table 8.2 Three Types of Images Formed By Thin Lenses
In Image Formation by Mirrors, we shall see that mirrors can form exactly the same types of images as lenses.
Take-Home Experiment: Concentrating Sunlight
Find several lenses and determine whether they are converging or diverging. In general those that are thicker near the edges are diverging and those that are thicker near the center are converging. On a bright sunny day take the converging lenses outside and try focusing the sunlight onto a piece of paper. Determine the focal lengths of the lenses. Be careful because the paper may start to burn, depending on the type of lens you have selected.