Ray tracing is the technique of determining or following (tracing) the paths that light rays take. For rays passing through matter, the law of refraction is used to trace the paths. Here we use ray tracing to help us understand the action of lenses in situations ranging from forming images on film to magnifying small print to correcting nearsightedness. While ray tracing for complicated lenses, such as those found in sophisticated cameras, may require computer techniques, there is a set of simple rules for tracing rays through thin lenses.

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?

We define the ratio of image height to object height  to be the magnification m. (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.

Based on desktop screen resolution statistics from January 2023, Full HD (1920x1080) is the most popular 16:9 aspect ratio resolution, followed by the Wide XGA (1366x768), which used to be the most popular resolution.

The ray tracing to scale in Figure 9 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 di 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 m is about –2. The minus sign indicates that the image is inverted.

The lens in which light rays that enter it parallel to its axis cross one another at a single point on the opposite side with a converging effect is called converging lens.

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.

Which mirrorisused inmagnifying glass

Several important distances appear in Figure 7. We define do to be the object distance, the distance of an object from the center of a lens. Image distance di 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 ho and hi, 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 7, 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

The image formed in Figure 7 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 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.

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.

Using paper, pencil, and a straight edge, ray tracing can accurately describe the operation of a lens. The rules for ray tracing for thin lenses are based on the illustrations already discussed:

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.

Look through your eyeglasses (or those of a friend) backward and forward and comment on whether they act like thin lenses.

A third type of image is formed by a diverging or concave lens. Try looking through eyeglasses meant to correct nearsightedness. (See Figure 12.) 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 13 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.

14. (a) +6.67; (b) +20.0; (c) The magnification increases without limit (to infinity) as the object distance increases to the limit of the focal distance.

We generally feel the entire lens, or mirror, is needed to form an image. Actually, half a lens will form the same, though a fainter, image.

What type of lens is magnifying glassused for

Step 7. Check to see if the answer is reasonable: Does it make sense? If you have identified the type of image (case 1, 2, or 3), you should assess whether your answer is consistent with the type of image, magnification, and so on.

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 7 and Figure 8. Ray tracing to scale should produce similar results for di. Numerical solutions for di and m can be obtained using the thin lens equations, noting that do = 0.750 m and f = 0.500 m.

Consider an object some distance away from a converging lens, as shown in Figure 7. 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 7, 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 7 in more detail.

What type of lens is magnifying glassvs convex

Image Formation by Lenses Copyright © 2021 by Lumen Learning is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Suppose the book page in Figure 11a 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?

converging lens: a convex lens in which light rays that enter it parallel to its axis converge at a single point on the opposite side

Suppose you take a magnifying glass out on a sunny day and you find that it concentrates sunlight to a small spot 8.00 cm away from the lens. What are the focal length and power of the lens?

To find the power of the lens, we must first convert the focal length to meters; then, we substitute this value into the equation for power. This gives

What type of lens isused to make amagnifying glassconverging or diverging

To find the magnification m, we must first find the image distance di using thin lens equation , or its alternative rearrangement .

Lenses are found in a huge array of optical instruments, ranging from a simple magnifying glass to the eye to a camera’s zoom lens. In this section, we will use the law of refraction to explore the properties of lenses and how they form images.

A thin lens is defined to be one whose thickness allows rays to refract, as illustrated in Figure 1, but does not allow properties such as dispersion and aberrations. An ideal thin lens has two refracting surfaces but the lens is thin enough to assume that light rays bend only once. A thin symmetrical lens has two focal points, one on either side and both at the same distance from the lens. (See Figure 6.)

Enter here your desired width for the image or video to get the actual height needed to maintain the HD 16:9 aspect ratio when resizing it.

We are given that do = 7.50 cm and f = 10.0 cm, 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 11, but we will use the thin lens equations to get numerical solutions in this example.

Now the thin lens equation can be used to find the magnification m, since both di and do are known. Entering their values gives

Another important characteristic of a thin lens is that light rays through its center are deflected by a negligible amount, as seen in Figure 5.

Magnifying lensconcave or convex

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.

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.

For example, if the distance to F in Figure 3 is 5.00 cm, then the focal length is f = –5.00 cm and the power of the lens is P = –20 D. An expanded view of the path of one ray through the lens is shown in the Figure to illustrate how the shape of the lens, together with the law of refraction, causes the ray to follow its particular path and be diverged.

Since 2009, the 16:9 format has been an international standard format for HDTV, widescreen TVs, movies, and smartphones.

A thin lens is defined to be one whose thickness allows rays to refract but does not allow properties such as dispersion and aberrations.

Table 1 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.

10 usesof magnifying glass

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 10b, 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 10a. 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 do < f and f is positive.

Figure 3 shows a concave lens and the effect it has on rays of light that enter it parallel to its axis (the path taken by ray 2 in the Figure is the axis of the lens). The concave lens is a diverging lens, because it causes the light rays to bend away (diverge) from its axis. In this case, the lens has been shaped so that all light rays entering it parallel to its axis appear to originate from the same point, F, defined to be the focal point of a diverging lens. The distance from the center of the lens to the focal point is again called the focal length f of the lens. Note that the focal length and power of a diverging lens are defined to be negative.

Now the magnification equation can be used to find the magnification m, since both di and do are known. Entering their values gives

Step 4. Make alist of what is given or can be inferred from the problem as stated (identify the knowns). It is helpful to determine whether the situation involves a case 1, 2, or 3 image. While these are just names for types of images, they have certain characteristics (given in Table 1) that can be of great use in solving problems.

Whichlens isused inmagnifying glassand where should the object be placed

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.

This is a relatively powerful lens. The power of a lens in diopters should not be confused with the familiar concept of power in watts. It is an unfortunate fact that the word “power” is used for two completely different concepts. If you examine a prescription for eyeglasses, you will note lens powers given in diopters. If you examine the label on a motor, you will note energy consumption rate given as a power in watts.

The power P of a lens is defined to be the inverse of its focal length. In equation form, this is , where f is the focal length of the lens, which must be given in meters (and not cm or mm). The power of a lens P has the unit diopters (D), provided that the focal length is given in meters. That is, . (Note that this power (optical power, actually) is not the same as power in watts defined in the chapter Work, Energy, and Energy Resources. It is a concept related to the effect of optical devices on light.) Optometrists prescribe common spectacles and contact lenses in units of diopters.

A number of results in this example are true of all case 2 images, as well as being consistent with Figure 11. 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 di 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.

Figure 11 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.

Today, most of our phones and cameras can record videos in 16:9 HD format. The 16:9 aspect ratio gives us the ability to make beautiful videos that are perfectly fitting for any widescreen display and social media platform.

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.

We do not realize that light rays are coming from every part of the object, passing through every part of the lens, and all can be used to form the final image.

A number of results in this example are true of all case 3 images, as well as being consistent with Figure 13. 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.

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 do > f and f is positive, as in Figure 10a. (A summary of the three cases or types of image formation appears at the end of this section.)

Step 6. Most quantitative problems require the use of the thin lens equations. These are solved in the usual manner by substituting knowns and solving for unknowns. Several worked examples serve as guides.

Step 2. Determine whether ray tracing, the thin lens equations, or both are to be employed. A sketch is very useful even if ray tracing is not specifically required by the problem. Write symbols and values on the sketch.

The situation here is the same as those shown in Figure 1 and Figure 2. The Sun is so far away that the Sun’s rays are nearly parallel when they reach Earth. The magnifying glass is a convex (or converging) lens, focusing the nearly parallel rays of sunlight. Thus the focal length of the lens is the distance from the lens to the spot, and its power is the inverse of this distance (in m).

What type of lens is magnifying glassfor reading

The greater effect a lens has on light rays, the more powerful it is said to be. For example, a powerful converging lens will focus parallel light rays closer to itself and will have a smaller focal length than a weak lens. The light will also focus into a smaller and more intense spot for a more powerful lens. The power P of a lens is defined to be the inverse of its focal length. In equation form, this is .

The word lens derives from the Latin word for a lentil bean, the shape of which is similar to the convex lens in Figure 1. The convex lens shown has been shaped so that all light rays that enter it parallel to its axis cross one another at a single point on the opposite side of the lens. (The axis is defined to be a line normal to the lens at its center, as shown in Figure 1.) Such a lens is called a converging (or convex) lens for the converging effect it has on light rays. An expanded view of the path of one ray through the lens is shown, to illustrate how the ray changes direction both as it enters and as it leaves the lens. Since the index of refraction of the lens is greater than that of air, the ray moves towards the perpendicular as it enters and away from the perpendicular as it leaves. (This is in accordance with the law of refraction.) Due to the lens’s shape, light is thus bent toward the axis at both surfaces. The point at which the rays cross is defined to be the focal point F of the lens. The distance from the center of the lens to its focal point is defined to be the focal length f of the lens. Figure 2 shows how a converging lens, such as that in a magnifying glass, can converge the nearly parallel light rays from the sun to a small spot.

focal point: for a converging lens or mirror, the point at which converging light rays cross; for a diverging lens or mirror, the point from which diverging light rays appear to originate

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 9. Use ray tracing to get an approximate location for the image. Then use the thin lens equations to calculate both the location of the image and its magnification. Verify that ray tracing and the thin lens equations produce consistent results.

To find the magnification m, we try to use magnification equation, . We do not have a value for di, 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 do and f were known.) Rearranging the magnification equation to isolate di gives

The thin lens equations can be used to find the magnification m, since both di and do are known. Entering their values gives

As noted in the initial discussion of the law of refraction in The Law of Refraction, the paths of light rays are exactly reversible. This means that the direction of the arrows could be reversed for all of the rays in Figure 1 and Figure 3. For example, if a point light source is placed at the focal point of a convex lens, as shown in Figure 4, parallel light rays emerge from the other side.