When the object is placed on the focal point in front of a convex lens the image produced is

By the end of this section, you will be able to:

List the rules for ray tracking for thin lenses.
Illustrate the formation of images using the technique of ray tracking.
Determine power of a lens given the focal length.

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.

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.

Figure 1. Rays of light entering a converging lens parallel to its axis converge at its focal point F. (Ray 2 lies on the axis of the lens.) The distance from the center of the lens to the focal point is the lens’s focal length f. An expanded view of the path taken by ray 1 shows the perpendiculars and the angles of incidence and refraction at both surfaces.

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.

The point at which the light rays cross is called the focal point F of the lens.

The distance from the center of the lens to its focal point is called focal length f.

Figure 2. Sunlight focused by a converging magnifying glass can burn paper. Light rays from the sun are nearly parallel and cross at the focal point of the lens. The more powerful the lens, the closer to the lens the rays will cross.

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

$P=1fP=\frac{1}{f}\\$

The power P of a lens is defined to be the inverse of its focal length. In equation form, this is

$P=1fP=\frac{1}{f}\\$

, 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,

$1m−11\text{D}=\frac{1}{\text{m}}\text{, or }1\text{m}^{-1}\\$

. (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.

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? 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). The focal length of the lens is the distance from the center of the lens to the spot, given to be 8.00 cm. Thus,

f = 8.00 cm.

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

$DP=\frac{1}{f}=\frac{1}{0.0800\text{ m}}=12.5\text{ D}\\$

Discussion

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.

Figure 3. Rays of light entering a diverging lens parallel to its axis are diverged, and all appear to originate at its focal point F. The dashed lines are not rays—they indicate the directions from which the rays appear to come. The focal length f of a diverging lens is negative. An expanded view of the path taken by ray 1 shows the perpendiculars and the angles of incidence and refraction at both surfaces.

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.

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.

A lens that causes the light rays to bend away from its axis is called a diverging lens.

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.

Figure 4. A small light source, like a light bulb filament, placed at the focal point of a convex lens, results in parallel rays of light emerging from the other side. The paths are exactly the reverse of those shown in Figure 1. This technique is used in lighthouses and sometimes in traffic lights to produce a directional beam of light from a source that emits light in all directions.

Figure 6. The light ray through the center of a thin lens is deflected by a negligible amount and is assumed to emerge parallel to its original path (shown as a shaded line).

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.

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.)

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.

Figure 6. Thin lenses have the same focal length on either side. (a) Parallel light rays entering a converging lens from the right cross at its focal point on the left. (b) Parallel light rays entering a diverging lens from the right seem to come from the focal point on the right.

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

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

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:

A ray entering a converging lens parallel to its axis passes through the focal point F of the lens on the other side. (See rays 1 and 3 in Figure 1.)
A ray entering a diverging lens parallel to its axis seems to come from the focal point F. (See rays 1 and 3 in Figure 2.)
A ray passing through the center of either a converging or a diverging lens does not change direction. (See Figure 5, and see ray 2 in Figure 1 and Figure 2.)
A ray entering a converging lens through its focal point exits parallel to its axis. (The reverse of rays 1 and 3 in Figure 1.)
A ray that enters a diverging lens by heading toward the focal point on the opposite side exits parallel to the axis. (The reverse of rays 1 and 3 in Figure 2.)

A ray entering a converging lens parallel to its axis passes through the focal point F of the lens on the other side.
A ray entering a diverging lens parallel to its axis seems to come from the focal point F.
A ray passing through the center of either a converging or a diverging lens does not change direction.
A ray entering a converging lens through its focal point exits parallel to its axis.
A ray that enters a diverging lens by heading toward the focal point on the opposite side exits parallel to the axis.

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.

Figure 7. Ray tracing is used to locate the image formed by a lens. Rays originating from the same point on the object are traced—the three chosen rays each follow one of the rules for ray tracing, so that their paths are easy to determine. The image is located at the point where the rays cross. In this case, a real image—one that can be projected on a screen—is formed.

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.

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.

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.

Figure 8. Real images can be projected. (a) A real image of the person is projected onto film. (b) The converging nature of the multiple surfaces that make up the eye result in the projection of a real image on the retina.

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

$1do+1di=1f\frac{1}{d_\text{o}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\$

and

$hiho=dido=m\frac{h_{\text{i}}}{h_\text{o}}=\frac{d_{\text{i}}}{d_{\text{o}}}=m\\$

We define the ratio of image height to object height

$(hiho)\left(\frac{h_{\text{i}}}{h_\text{o}}\right)\\$

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.

The distance of the image from the center of the lens is called image distance.

$1do+1di=1f\displaystyle\frac{1}{d_\text{o}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\$

$hiho=dido=m\displaystyle\frac{h_{\text{i}}}{h_\text{o}}=\frac{d_{\text{i}}}{d_{\text{o}}}=m\\$

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.

Figure 9. A light bulb placed 0.750 m from a lens having a 0.500 m focal length produces a real image on a poster board as discussed in the example above. Ray tracing predicts the image location and size.

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. 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 thin lens equations can be used to find di from the given information:

$1do+1di=1f\frac{1}{d_\text{o}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\$

Rearranging to isolate di gives

$1di=1f−1do\frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_\text{o}}\\$

Entering known quantities gives a value for

$1di\frac{1}{d_{\text{i}}}\\$

$m=0.667m\frac{1}{d_{\text{i}}}=\frac{1}{0.500\text{ m}}-\frac{1}{0.750\text{ m}}=\frac{0.667}{\text{m}}\\$

This must be inverted to find di:

$md_{\text{i}}=\frac{\text{m}}{0.667}=1.50\text{ m}\\$

Note that another way to find di is to rearrange the equation:

$1di=1f−1do\frac{1}{d_\text{i}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\$

This yields the equation for the image distance as:

$di=fdodo−fd_{\text{i}}=\frac{fd_{\text{o}}}{d_{\text{o}}-f}\\$

Note that there is no inverting here.

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

$m=−2.00\displaystyle{m}=-\frac{d_{\text{i}}}{d_{\text{o}}}=-\frac{1.50\text{ m}}{0.750\text{ m}}=-2.00\\$

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 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.)

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 10. (a) When a converging lens is held farther away from the face than the lens’s focal length, an inverted image is formed. This is a case 1 image. Note that the image is in focus but the face is not, because the image is much closer to the camera taking this photograph than the face. (credit: DaMongMan, Flickr) (b) A magnified image of a face is produced by placing it closer to the converging lens than its focal length. This is a case 2 image. (credit: Casey Fleser, Flickr)

Figure 11. Ray tracing predicts the image location and size for an object held closer to a converging lens than its focal length. Ray 1 enters parallel to the axis and exits through the focal point on the opposite side, while ray 2 passes through the center of the lens without changing path. The two rays continue to diverge on the other side of the lens, but both appear to come from a common point, locating the upright, magnified, virtual image. This is a case 2 image.

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.

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.

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.

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? 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. To find the magnification m, we try to use magnification equation,

$m=−didom=-\frac{d_{\text{i}}}{d_{\text{o}}}\\$

. 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

$1di=1f−1do\frac{1}{d_\text{i}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\$

Entering known values, we obtain a value for

$1di\frac{1}{d_\text{i}}\\$

$cm=−0.0333cm\frac{1}{d_\text{i}}=\frac{1}{10.0\text{ cm}}-\frac{1}{7.50\text{ cm}}=\frac{-0.0333}{\text{cm}}\\$

This must be inverted to find di:

$cmd_{\text{i}}=-\frac{\text{cm}}{0.0333}=-30.0\text{ cm}\\$

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

$cm=3.00\displaystyle{m}=-\frac{d_{\text{i}}}{d_{\text{o}}}=-\frac{-30.0\text{ cm}}{10.0\text{ cm}}=3.00\\$

Discussion

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 12. A car viewed through a concave or diverging lens looks upright. This is a case 3 image. (credit: Daniel Oines, Flickr)

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.

Figure 13. Ray tracing predicts the image location and size for a concave or diverging lens. Ray 1 enters parallel to the axis and is bent so that it appears to originate from the focal point. Ray 2 passes through the center of the lens without changing path. The two rays appear to come from a common point, locating the upright image. This is a case 3 image, which is closer to the lens than the object and smaller in height.

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? 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. To find the magnification m, we must first find the image distance di using thin lens equation

$1di=1f−1do\frac{1}{d_\text{i}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\$

, or its alternative rearrangement

$di=fdodo−f{d_\text{i}}=\frac{fd_{\text{o}}}{d_{\text{o}}-f}\\$

We are given that f = –10.0 cm and do = 7.50 cm. Entering these yields a value for

$1di\frac{1}{d_{\text{i}}}\\$

$cm=−0.2333cm\displaystyle\frac{1}{d_{\text{i}}}=\frac{1}{-10.0\text{ cm}}-\frac{1}{7.50\text{ cm}}=\frac{-0.2333}{\text{cm}}\\$

This must be inverted to find di:

$cm\displaystyle{d_\text{i}}=-\frac{\text{cm}}{0.2333}=-4.29\text{ cm}\\$

$cm\displaystyle{d}_{\text{i}}=\frac{\left(7.5\right)\left(-10\right)}{\left(7.5-\left(-10\right)\right)}=-\frac{75}{17.5}=-4.29\text{ cm}\\$

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

$cm=0.571\displaystyle{m}=-\frac{d_{\text{i}}}{d_{\text{o}}}=-\frac{-4.29\text{ cm}}{7.50\text{ cm}}=0.571\\$

Discussion

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.

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.

Table 1. Three Types of Images Formed By Thin Lenses
Type	Formed when	Image type	di	m
Case 1	f positive, do>f	real	positive	negative
Case 2	f positive, do<f	virtual	negative	positive m > 1
Case 3	f negative	virtual	negative	positive m < 1

In Image Formation by Mirrors, we shall see that mirrors can form exactly the same types of images as lenses.

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.

Step 1. Examine the situation to determine that image formation by a lens is involved. 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. Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns). 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. Step 5. If ray tracing is required, use the ray tracing rules listed near the beginning 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 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.

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. 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.

Light rays entering a converging lens parallel to its axis cross one another at a single point on the opposite side.
For a converging lens, the focal point is the point at which converging light rays cross; for a diverging lens, the focal point is the point from which diverging light rays appear to originate.
The distance from the center of the lens to its focal point is called the focal length f.
Power P of a lens is defined to be the inverse of its focal length,
$P=1fP=\frac{1}{f}\\$
. A lens that causes the light rays to bend away from its axis is called a diverging lens. Ray tracing is the technique of graphically determining the paths that light rays take.
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.
Thin lens equations are

$1do+1di=1f\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}=\frac{1}{f}$

and

$hiho=−dido=m\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=-\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=m\\$

(magnification).

The distance of the image from the center of the lens is called image distance.
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.

It can be argued that a flat piece of glass, such as in a window, is like a lens with an infinite focal length. If so, where does it form an image? That is, how are di and do related?
You can often see a reflection when looking at a sheet of glass, particularly if it is darker on the other side. Explain why you can often see a double image in such circumstances.
When you focus a camera, you adjust the distance of the lens from the film. If the camera lens acts like a thin lens, why can it not be a fixed distance from the film for both near and distant objects?
A thin lens has two focal points, one on either side, at equal distances from its center, and should behave the same for light entering from either side. Look through your eyeglasses (or those of a friend) backward and forward and comment on whether they are thin lenses.
Will the focal length of a lens change when it is submerged in water? Explain.

What is the power in diopters of a camera lens that has a 50.0 mm focal length?
Your camera’s zoom lens has an adjustable focal length ranging from 80.0 to 200 mm. What is its range of powers?
What is the focal length of 1.75 D reading glasses found on the rack in a pharmacy?
You note that your prescription for new eyeglasses is –4.50 D. What will their focal length be?
How far from the lens must the film in a camera be, if the lens has a 35.0 mm focal length and is being used to photograph a flower 75.0 cm away? Explicitly show how you follow the steps in the Problem-Solving Strategy for lenses.
A certain slide projector has a 100 mm focal length lens. (a) How far away is the screen, if a slide is placed 103 mm from the lens and produces a sharp image? (b) If the slide is 24.0 by 36.0 mm, what are the dimensions of the image? Explicitly show how you follow the steps in the Problem-Solving Strategy for Lenses (above).
A doctor examines a mole with a 15.0 cm focal length magnifying glass held 13.5 cm from the mole (a) Where is the image? (b) What is its magnification? (c) How big is the image of a 5.00 mm diameter mole?
How far from a piece of paper must you hold your father’s 2.25 D reading glasses to try to burn a hole in the paper with sunlight?
A camera with a 50.0 mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens must the film be? (b) If the film is 36.0 mm high, what fraction of a 1.75 m tall person will fit on it? (c) Discuss how reasonable this seems, based on your experience in taking or posing for photographs.
A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is 33.0 mm. (a) What is the closest object that can be photographed? (b) What is the magnification of this closest object?
Suppose your 50.0 mm focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus? (b) What is the height of the object if its image is 2.00 cm high?
(a) What is the focal length of a magnifying glass that produces a magnification of 3.00 when held 5.00 cm from an object, such as a rare coin? (b) Calculate the power of the magnifier in diopters. (c) Discuss how this power compares to those for store-bought reading glasses (typically 1.0 to 4.0 D). Is the magnifier’s power greater, and should it be?
What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 25.0 cm away?
In Example 3, the magnification of a book held 7.50 cm from a 10.0 cm focal length lens was found to be 3.00. (a) Find the magnification for the book when it is held 8.50 cm from the magnifier. (b) Do the same for when it is held 9.50 cm from the magnifier. (c) Comment on the trend in m as the object distance increases as in these two calculations.
Suppose a 200 mm focal length telephoto lens is being used to photograph mountains 10.0 km away. (a) Where is the image? (b) What is the height of the image of a 1000 m high cliff on one of the mountains?
A camera with a 100 mm focal length lens is used to photograph the sun and moon. What is the height of the image of the sun on the film, given the sun is 1.40 × 106 km in diameter and is 1.50 × 108 km away?
Combine thin lens equations to show that the magnification for a thin lens is determined by its focal length and the object distance and is given by
$m=f(f−do)m=\frac{f}{\left(f-d_{\text{o}}\right)}\\$
.

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

diverging lens: a concave lens in which light rays that enter it parallel to its axis bend away (diverge) from its axis

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

focal length: distance from the center of a lens or curved mirror to its focal point

magnification: ratio of image height to object height

power: inverse of focal length

real image: image that can be projected

virtual image: image that cannot be projected

2. 5.00 to 12.5 D 4. –0.222 m 6. (a) 3.43 m; (b) 0.800 by 1.20 m 7. (a) −1.35 m (on the object side of the lens); (b) +10.0; (c) 5.00 cm 8. 44.4 cm 10. (a) 6.60 cm; (b) –0.333 12. (a) +7.50 cm; (b) 13.3 D; (c) Much greater 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. 16. −0.933 mm

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Page 2

By the end of this section, you will be able to:

Illustrate image formation in a flat mirror.
Explain with ray diagrams the formation of an image using spherical mirrors.
Determine focal length and magnification given radius of curvature, distance of object and image.

We only have to look as far as the nearest bathroom to find an example of an image formed by a mirror. Images in flat mirrors are the same size as the object and are located behind the mirror. Like lenses, mirrors can form a variety of images. For example, dental mirrors may produce a magnified image, just as makeup mirrors do. Security mirrors in shops, on the other hand, form images that are smaller than the object. We will use the law of reflection to understand how mirrors form images, and we will find that mirror images are analogous to those formed by lenses. Figure 1 helps illustrate how a flat mirror forms an image. Two rays are shown emerging from the same point, striking the mirror, and being reflected into the observer’s eye. The rays can diverge slightly, and both still get into the eye. If the rays are extrapolated backward, they seem to originate from a common point behind the mirror, locating the image. (The paths of the reflected rays into the eye are the same as if they had come directly from that point behind the mirror.) Using the law of reflection—the angle of reflection equals the angle of incidence—we can see that the image and object are the same distance from the mirror. This is a virtual image, since it cannot be projected—the rays only appear to originate from a common point behind the mirror. Obviously, if you walk behind the mirror, you cannot see the image, since the rays do not go there. But in front of the mirror, the rays behave exactly as if they had come from behind the mirror, so that is where the image is situated.

Figure 1. Two sets of rays from common points on an object are reflected by a flat mirror into the eye of an observer. The reflected rays seem to originate from behind the mirror, locating the virtual image.

Now let us consider the focal length of a mirror—for example, the concave spherical mirrors in Figure 2. Rays of light that strike the surface follow the law of reflection. For a mirror that is large compared with its radius of curvature, as in Figure 2a, we see that the reflected rays do not cross at the same point, and the mirror does not have a well-defined focal point. If the mirror had the shape of a parabola, the rays would all cross at a single point, and the mirror would have a well-defined focal point. But parabolic mirrors are much more expensive to make than spherical mirrors. The solution is to use a mirror that is small compared with its radius of curvature, as shown in Figure 2b. (This is the mirror equivalent of the thin lens approximation.) To a very good approximation, this mirror has a well-defined focal point at F that is the focal distance f from the center of the mirror. The focal length f of a concave mirror is positive, since it is a converging mirror.

Figure 2. (a) Parallel rays reflected from a large spherical mirror do not all cross at a common point. (b) If a spherical mirror is small compared with its radius of curvature, parallel rays are focused to a common point. The distance of the focal point from the center of the mirror is its focal length f. Since this mirror is converging, it has a positive focal length.

Just as for lenses, the shorter the focal length, the more powerful the mirror; thus,

$P=1fP=\frac{1}{f}\\$

for a mirror, too. A more strongly curved mirror has a shorter focal length and a greater power. Using the law of reflection and some simple trigonometry, it can be shown that the focal length is half the radius of curvature, or

$f=R2f=\frac{R}{2}\\$

, where R is the radius of curvature of a spherical mirror. The smaller the radius of curvature, the smaller the focal length and, thus, the more powerful the mirror

The convex mirror shown in Figure 3 also has a focal point. Parallel rays of light reflected from the mirror seem to originate from the point F at the focal distance f behind the mirror. The focal length and power of a convex mirror are negative, since it is a diverging mirror.

Ray tracing is as useful for mirrors as for lenses. The rules for ray tracing for mirrors are based on the illustrations just discussed:

Figure 3. Parallel rays of light reflected from a convex spherical mirror (small in size compared with its radius of curvature) seem to originate from a well-defined focal point at the focal distance f behind the mirror. Convex mirrors diverge light rays and, thus, have a negative focal length.

A ray approaching a concave converging mirror parallel to its axis is reflected through the focal point F of the mirror on the same side. (See rays 1 and 3 in Figure 2b.)
A ray approaching a convex diverging mirror parallel to its axis is reflected so that it seems to come from the focal point F behind the mirror. (See rays 1 and 3 in Figure 3.)
Any ray striking the center of a mirror is followed by applying the law of reflection; it makes the same angle with the axis when leaving as when approaching. (See ray 2 in Figure 4.)
A ray approaching a concave converging mirror through its focal point is reflected parallel to its axis. (The reverse of rays 1 and 3 in Figure 2.)
A ray approaching a convex diverging mirror by heading toward its focal point on the opposite side is reflected parallel to the axis. (The reverse of rays 1 and 3 in Figure 3.)

We will use ray tracing to illustrate how images are formed by mirrors, and we can use ray tracing quantitatively to obtain numerical information. But since we assume each mirror is small compared with its radius of curvature, we can use the thin lens equations for mirrors just as we did for lenses.

Consider the situation shown in Figure 4, concave spherical mirror reflection, in which an object is placed farther from a concave (converging) mirror than its focal length. That is, f is positive and do > f, so that we may expect an image similar to the case 1 real image formed by a converging lens. Ray tracing in Figure 4 shows that the rays from a common point on the object all cross at a point on the same side of the mirror as the object. Thus a real image can be projected onto a screen placed at this location. The image distance is positive, and the image is inverted, so its magnification is negative. This is a case 1 image for mirrors. It differs from the case 1 image for lenses only in that the image is on the same side of the mirror as the object. It is otherwise identical.

Figure 4. A case 1 image for a mirror. An object is farther from the converging mirror than its focal length. Rays from a common point on the object are traced using the rules in the text. Ray 1 approaches parallel to the axis, ray 2 strikes the center of the mirror, and ray 3 goes through the focal point on the way toward the mirror. All three rays cross at the same point after being reflected, locating the inverted real image. Although three rays are shown, only two of the three are needed to locate the image and determine its height.

Electric room heaters use a concave mirror to reflect infrared (IR) radiation from hot coils. Note that IR follows the same law of reflection as visible light. Given that the mirror has a radius of curvature of 50.0 cm and produces an image of the coils 3.00 m away from the mirror, where are the coils? We are given that the concave mirror projects a real image of the coils at an image distance di=3.00 m. The coils are the object, and we are asked to find their location—that is, to find the object distance do. We are also given the radius of curvature of the mirror, so that its focal length is

$cmf=\frac{R}{2}=25.0\text{ cm}\\$

(positive since the mirror is concave or converging). Assuming the mirror is small compared with its radius of curvature, we can use the thin lens equations, to solve this problem. Since di and f are known, thin lens equation can be used to find do:

$1do+1di=1f\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\$

Rearranging to isolate do gives

$1do=1f−1di\frac{1}{d_{\text{o}}}=\frac{1}{f}-\frac{1}{d_{\text{i}}}\\$

Entering known quantities gives a value for

$1do\frac{1}{d_{\text{o}}}\\$

$m=3.667m\frac{1}{d_{\text{o}}}=\frac{1}{0.250\text{ m}}-\frac{1}{3.00\text{ m}}=\frac{3.667}{\text{m}}\\$

This must be inverted to find do:

$cmd_{\text{o}}=\frac{1\text{ m}}{3.667}=27.3\text{ cm}\\$

. Note that the object (the filament) is farther from the mirror than the mirror’s focal length. This is a case 1 image (do > f and f positive), consistent with the fact that a real image is formed. You will get the most concentrated thermal energy directly in front of the mirror and 3.00 m away from it. Generally, this is not desirable, since it could cause burns. Usually, you want the rays to emerge parallel, and this is accomplished by having the filament at the focal point of the mirror. Note that the filament here is not much farther from the mirror than its focal length and that the image produced is considerably farther away. This is exactly analogous to a slide projector. Placing a slide only slightly farther away from the projector lens than its focal length produces an image significantly farther away. As the object gets closer to the focal distance, the image gets farther away. In fact, as the object distance approaches the focal length, the image distance approaches infinity and the rays are sent out parallel to one another.

One of the solar technologies used today for generating electricity is a device (called a parabolic trough or concentrating collector) that concentrates the sunlight onto a blackened pipe that contains a fluid. This heated fluid is pumped to a heat exchanger, where its heat energy is transferred to another system that is used to generate steam—and so generate electricity through a conventional steam cycle. Figure 5 shows such a working system in southern California. Concave mirrors are used to concentrate the sunlight onto the pipe. The mirror has the approximate shape of a section of a cylinder. For the problem, assume that the mirror is exactly one-quarter of a full cylinder.

If we wish to place the fluid-carrying pipe 40.0 cm from the concave mirror at the mirror’s focal point, what will be the radius of curvature of the mirror?
Per meter of pipe, what will be the amount of sunlight concentrated onto the pipe, assuming the insolation (incident solar radiation) is 0.900 k W/m2?
If the fluid-carrying pipe has a 2.00-cm diameter, what will be the temperature increase of the fluid per meter of pipe over a period of one minute? Assume all the solar radiation incident on the reflector is absorbed by the pipe, and that the fluid is mineral oil.

To solve an Integrated Concept Problem we must first identify the physical principles involved. Part 1 is related to the current topic. Part 2 involves a little math, primarily geometry. Part 3 requires an understanding of heat and density. To a good approximation for a concave or semi-spherical surface, the point where the parallel rays from the sun converge will be at the focal point, so R = 2f = 80.0 cm. The insolation is 900 W /m2. We must find the cross-sectional area A of the concave mirror, since the power delivered is 900 W /m2 × A. The mirror in this case is a quarter-section of a cylinder, so the area for a length L of the mirror is

$A=14(2πR)L\text{A}=\frac{1}{4}\left(2\pi\text{R}\right)\text{L}\\$

. The area for a length of 1.00 m is then

$m)=1.262\displaystyle\text{A}=\frac{\pi}{2}R\left(1.00\text{ m}\right)=\frac{\left(3.14\right)}{2}\left(0.800\text{ m}\right)\left(1.00\text{ m}\right)=1.26^2\\$

The insolation on the 1.00-m length of pipe is then

$W\displaystyle\left(9.00\times10^{2}\frac{\text{W}}{\text{m}^2}\right)\left(1.26\text{ m}^2\right)=1130\text{ W}\\$

Solution to Part 3

The increase in temperature is given by Q = mcΔT. The mass m of the mineral oil in the one-meter section of pipe is

$kg\begin{array}{lll}{m}&=&\rho\text{V}=\rho\pi\left(\frac{d}{2}\right)^2\left(1.00\text{ m}\right)\\\text{ }&=&\left(8.00\times10^2\text{ kg/m}^3\right)\left(3.14\right)\left(0.0100\text{ m}\right)^2\left(1.00\text{ m}\right)\\\text{ }&=&0.251\text{ kg}\end{array}\\$

Therefore, the increase in temperature in one minute is

$=162∘C\begin{array}{lll}\Delta{T}&=&\frac{Q}{mc}\\\text{ }&=&\frac{\left(1130\text{ W}\right)\left(60.0\text{ s}\right)}{\left(0.251\text{ kg}\right)\left(1670\text{ J}\cdot\text{ kg/}^{\circ}\text{ C}\right)}\\\text{ }&=&162^{\circ}\text{C}\end{array}\\$

Discussion for Part 3

Figure 5. Parabolic trough collectors are used to generate electricity in southern California. (credit: kjkolb, Wikimedia Commons)

An array of such pipes in the California desert can provide a thermal output of 250 MW on a sunny day, with fluids reaching temperatures as high as 400ºC. We are considering only one meter of pipe here, and ignoring heat losses along the pipe.

What happens if an object is closer to a concave mirror than its focal length? This is analogous to a case 2 image for lenses ( do <f and f positive), which is a magnifier. In fact, this is how makeup mirrors act as magnifiers. Figure 6a uses ray tracing to locate the image of an object placed close to a concave mirror. Rays from a common point on the object are reflected in such a manner that they appear to be coming from behind the mirror, meaning that the image is virtual and cannot be projected. As with a magnifying glass, the image is upright and larger than the object. This is a case 2 image for mirrors and is exactly analogous to that for lenses.

Figure 6. (a) Case 2 images for mirrors are formed when a converging mirror has an object closer to it than its focal length. Ray 1 approaches parallel to the axis, ray 2 strikes the center of the mirror, and ray 3 approaches the mirror as if it came from the focal point. (b) A magnifying mirror showing the reflection. (credit: Mike Melrose, Flickr)

All three rays appear to originate from the same point after being reflected, locating the upright virtual image behind the mirror and showing it to be larger than the object. (b) Makeup mirrors are perhaps the most common use of a concave mirror to produce a larger, upright image.

A convex mirror is a diverging mirror (f is negative) and forms only one type of image. It is a case 3 image—one that is upright and smaller than the object, just as for diverging lenses. Figure 7a uses ray tracing to illustrate the location and size of the case 3 image for mirrors. Since the image is behind the mirror, it cannot be projected and is thus a virtual image. It is also seen to be smaller than the object.

Figure 7. Case 3 images for mirrors are formed by any convex mirror. Ray 1 approaches parallel to the axis, ray 2 strikes the center of the mirror, and ray 3 approaches toward the focal point. All three rays appear to originate from the same point after being reflected, locating the upright virtual image behind the mirror and showing it to be smaller than the object. (b) Security mirrors are convex, producing a smaller, upright image. Because the image is smaller, a larger area is imaged compared to what would be observed for a flat mirror (and hence security is improved). (credit: Laura D’Alessandro, Flickr)

A keratometer is a device used to measure the curvature of the cornea, particularly for fitting contact lenses. Light is reflected from the cornea, which acts like a convex mirror, and the keratometer measures the magnification of the image. The smaller the magnification, the smaller the radius of curvature of the cornea. If the light source is 12.0 cm from the cornea and the image’s magnification is 0.0320, what is the cornea’s radius of curvature? If we can find the focal length of the convex mirror formed by the cornea, we can find its radius of curvature (the radius of curvature is twice the focal length of a spherical mirror). We are given that the object distance is do = 12.0 cm and that m = 0.0320. We first solve for the image distance di, and then for f.

$m=−didom=-\frac{d_{\text{i}}}{d_{\text{o}}}\\$

. Solving this expression for di gives di = −mdo.

Entering known values yields di = –(0.0320)(12.0 cm) = –0.384 cm.

$1f=1do+1di\frac{1}{f}=\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}\\$

Substituting known values,

$cm=−2.52cm\frac{1}{f}=\frac{1}{12.0\text{ cm}}+\frac{1}{-0.384\text{ cm}}=\frac{-2.52}{\text{cm}}\\$

This must be inverted to find f:

$cmf=\frac{\text{cm}}{-2.52}=-0.400\text{ cm}\\$

The radius of curvature is twice the focal length, so that R = 2|f| = 0.800 cm.

Although the focal length f of a convex mirror is defined to be negative, we take the absolute value to give us a positive value for R. The radius of curvature found here is reasonable for a cornea. The distance from cornea to retina in an adult eye is about 2.0 cm. In practice, many corneas are not spherical, complicating the job of fitting contact lenses. Note that the image distance here is negative, consistent with the fact that the image is behind the mirror, where it cannot be projected. In this section’s Problems and Exercises, you will show that for a fixed object distance, the smaller the radius of curvature, the smaller the magnification.

The three types of images formed by mirrors (cases 1, 2, and 3) are exactly analogous to those formed by lenses, as summarized in the table at the end of Image Formation by Lenses. It is easiest to concentrate on only three types of images—then remember that concave mirrors act like convex lenses, whereas convex mirrors act like concave lenses.

Find a flashlight and identify the curved mirror used in it. Find another flashlight and shine the first flashlight onto the second one, which is turned off. Estimate the focal length of the mirror. You might try shining a flashlight on the curved mirror behind the headlight of a car, keeping the headlight switched off, and determine its focal length.

Step 1. Examine the situation to determine that image formation by a mirror is involved.

Step 2. Refer to the Problem-Solving Strategies for Lenses. The same strategies are valid for mirrors as for lenses with one qualification—use the ray tracing rules for mirrors listed earlier in this section.

The characteristics of an image formed by a flat mirror are: (a) The image and object are the same distance from the mirror, (b) The image is a virtual image, and (c) The image is situated behind the mirror.
Image length is half the radius of curvature:
$f=R2f=\frac{R}{2}\\$
A convex mirror is a diverging mirror and forms only one type of image, namely a virtual image.

What are the differences between real and virtual images? How can you tell (by looking) whether an image formed by a single lens or mirror is real or virtual?
Can you see a virtual image? Can you photograph one? Can one be projected onto a screen with additional lenses or mirrors? Explain your responses.
Is it necessary to project a real image onto a screen for it to exist?
At what distance is an image always located—at do, di, or f?
Under what circumstances will an image be located at the focal point of a lens or mirror?
What is meant by a negative magnification? What is meant by a magnification that is less than 1 in magnitude?
Can a case 1 image be larger than the object even though its magnification is always negative? Explain.
Figure 8 shows a light bulb between two mirrors. One mirror produces a beam of light with parallel rays; the other keeps light from escaping without being put into the beam. Where is the filament of the light in relation to the focal point or radius of curvature of each mirror?
Figure 8. The two mirrors trap most of the bulb’s light and form a directional beam as in a headlight.
The two mirrors trap most of the bulb’s light and form a directional beam as in a headlight.
Two concave mirrors of different sizes are placed facing one another. A filament bulb is placed at the focus of the larger mirror. The rays after reflection from the larger mirror travel parallel to one another. The rays falling on the smaller mirror retrace their paths.
Devise an arrangement of mirrors allowing you to see the back of your head. What is the minimum number of mirrors needed for this task?
If you wish to see your entire body in a flat mirror (from head to toe), how tall should the mirror be? Does its size depend upon your distance away from the mirror? Provide a sketch.
It can be argued that a flat mirror has an infinite focal length. If so, where does it form an image? That is, how are di and do related?
Why are diverging mirrors often used for rear-view mirrors in vehicles? What is the main disadvantage of using such a mirror compared with a flat one?

What is the focal length of a makeup mirror that has a power of 1.50 D?
Some telephoto cameras use a mirror rather than a lens. What radius of curvature mirror is needed to replace a 800 mm focal length telephoto lens?
(a) Calculate the focal length of the mirror formed by the shiny back of a spoon that has a 3.00 cm radius of curvature. (b) What is its power in diopters?
Find the magnification of the heater element in Example 1. Note that its large magnitude helps spread out the reflected energy.
What is the focal length of a makeup mirror that produces a magnification of 1.50 when a person’s face is 12.0 cm away?
A shopper standing 3.00 m from a convex security mirror sees his image with a magnification of 0.250. (a) Where is his image? (b) What is the focal length of the mirror? (c) What is its radius of curvature?
An object 1.50 cm high is held 3.00 cm from a person’s cornea, and its reflected image is measured to be 0.167 cm high. (a) What is the magnification? (b) Where is the image? (c) Find the radius of curvature of the convex mirror formed by the cornea. (Note that this technique is used by optometrists to measure the curvature of the cornea for contact lens fitting. The instrument used is called a keratometer, or curve measurer.)
Ray tracing for a flat mirror shows that the image is located a distance behind the mirror equal to the distance of the object from the mirror. This is stated di = −do, since this is a negative image distance (it is a virtual image). (a) What is the focal length of a flat mirror? (b) What is its power?
Show that for a flat mirror hi = ho, knowing that the image is a distance behind the mirror equal in magnitude to the distance of the object from the mirror.
Use the law of reflection to prove that the focal length of a mirror is half its radius of curvature. That is, prove that
$f=R2f=\frac{R}{2}\\$
. Note this is true for a spherical mirror only if its diameter is small compared with its radius of curvature.
Referring to the electric room heater considered in the first example in this section, calculate the intensity of IR radiation in W/m2 projected by the concave mirror on a person 3.00 m away. Assume that the heating element radiates 1500 W and has an area of 100 cm2, and that half of the radiated power is reflected and focused by the mirror.
Consider a 250-W heat lamp fixed to the ceiling in a bathroom. If the filament in one light burns out then the remaining three still work. Construct a problem in which you determine the resistance of each filament in order to obtain a certain intensity projected on the bathroom floor. The ceiling is 3.0 m high. The problem will need to involve concave mirrors behind the filaments. Your instructor may wish to guide you on the level of complexity to consider in the electrical components.

converging mirror: a concave mirror in which light rays that strike it parallel to its axis converge at one or more points along the axis

diverging mirror: a convex mirror in which light rays that strike it parallel to its axis bend away (diverge) from its axis

law of reflection: angle of reflection equals the angle of incidence

1. +0.667 m

3. (a) −1.5 × 10−2 m; (b) −66.7 D

5. +0.360 m (concave) 7. (a) +0.111; (b) −0.334 cm (behind "mirror"); (c) 0.752cm

$m=hiho=−dido=−−dodo=dodo=1⇒hi=hom=\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=-\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=-\frac{-{d}_{\text{o}}}{{d}_{\text{o}}}=\frac{{d}_{\text{o}}}{{d}_{\text{o}}}=1\Rightarrow {h}_{\text{i}}={h}_{\text{o}}\\$