25.6

The Mirror Equation and the Magnification Equation

CONCEPTS AT A GLANCE Ray diagrams drawn to scale are useful for determining the location and size of the image formed by a mirror. However, for an accurate description of the image, a more analytical technique is needed. We will now derive two equations, known as the mirror equation and the magnification equation, that will provide a complete description of the image. The Concepts-at-a-Glance chart in Figure 25-23, which is an extension of the chart in Figure 25-17, emphasizes that these equations are based on the law of reflection and provide relationships between:

f

the focal length of the mirror

do

the object distance, which is the distance between the object and the mirror

di

the image distance, which is the distance between the image and the mirror

m

the magnification of the mirror, which is the ratio of the height of the image to the height of the object.

Figure 25-23 CONCEPTS AT A GLANCE The law of reflection gives rise to the mirror equation and the magnification equation, which give the location and relative size of an image. The photograph shows an image of a person reading a newspaper, which has been formed because light has been reflected by the curved outer surface of the eye. (Ben Welsh/SUPERSTOCK)

| Concave Mirrors

We begin our derivation of the mirror equation by referring to Figure 25-24a, which shows a ray leaving the top of the object and striking a concave mirror at the point where the principal axis intersects the mirror. Since the principal axis is perpendicular to the mirror, it is also the normal at this point of incidence. Therefore, the ray reflects at an equal angle and passes through the image. The two colored triangles are similar triangles because they have equal angles, so

where ho is the height of the object and hi is the height of the image. The minus sign appears on the left in this equation because the image is inverted in Figure 25-24a. In part b another ray leaves the top of the object, this one passing through the focal point F, reflecting parallel to the principal axis, and then passing through the image. Provided the ray remains close to the axis, the two colored areas can be considered to be similar triangles, with the result that

Figure 25-24 These diagrams are used to derive the mirror equation and the magnification equation. (a) The two colored triangles are similar triangles. (b) If the ray is close to the principal axis, the two colored regions are almost similar triangles.

Setting the two equations above equal to each other yields . Rearranging this result gives the mirror equation:

| (25.3)

We have derived this equation for a real image formed in front of a concave mirror. In this case, the image distance is a positive quantity, as are the object distance and the focal length. However, we have seen in the last section that a concave mirror can also form a virtual image, if the object is located between the focal point and the mirror. Equation 25.3 can also be applied to such a situation, provided that we adopt the convention that di is negative for an image behind the mirror, as it is for a virtual image.

The reflections of these three friends in this fun-house mirror are distorted because the mirror is curved and not flat. It is not a spherical mirror, however. (Gail Mooney/Masterfile)

In deriving the magnification equation, we remember that the magnification m of a mirror is the ratio of the image height to the object height: . If the image height is less than the object height, the magnitude of m is less than one, and if the image is larger than the object, the magnitude of m is greater than one. We have already shown that , so it follows that

| (25.4)

As Examples 3 and 4 show, the value of m is negative if the image is inverted and positive if the image is upright.

| Example 3 | | | A Real Image Formed by a Concave Mirror

A 2.0-cm-high object is placed 7.10 cm from a concave mirror whose radius of curvature is 10.20 cm. Find (a) the location of the image and (b) its size. Reasoning Since , the object is located between the focal point F and the center of curvature C of the mirror, as in Figure 25-19a. Based on this figure, we expect that the image is real and that, relative to the object, it is farther away from the mirror, inverted, and larger.
Solution

a. | With and , the mirror equation (Equation 25.3) can be used to find the image distance:

In this calculation, f and do are positive numbers, indicating that the focal point and the object are in front of the mirror. The positive answer for di means that the image is also in front of the mirror, and the reflected rays actually pass through the image, as Figure 25-19a shows. In other words, the positive value for di indicates that the image is a real image.

Problem solving insight According to the mirror equation, the image distance di has a reciprocal given by . After combining the reciprocals and , do not forget to take the reciprocal of the result to find di.

b. | According to the magnification equation (Equation 25.4), the image height hi is related to the object height ho and the magnification m by , where . Thus, we find that

The negative value for hi indicates that the image is inverted with respect to the object, as in Figure 25-19a.

| Example 4 | | | A Virtual Image Formed by a Concave Mirror

An object is placed 6.00 cm in front of a concave mirror that has a 10.0-cm focal length, (a) Determine the location of the image. (b) The object is 1.2 cm high. Find the image height. Reasoning The object is located between the focal point and the mirror, as in Figure 25-20a. The setup is analogous to a person using a makeup or shaving mirror. Therefore, we expect that the image is virtual and that, relative to the object, it is upright and larger.
Solution

a. | Using the mirror equation with and , we have

The answer for di is negative, indicating that the image is behind the mirror. Thus, as expected, the image is virtual.
b. | The image height hi can be found from the magnification equation, which indicates that , where ho is the object height and . It follows, then, that

The image is larger than the object, and the positive value for hi indicates that the image is upright (see Figure 25-20a).

Need more practice?

Interactive LearningWare25.1
A gemstone is placed 20.0 cm in front of a concave mirror and is within the focal point. When the concave mirror is replaced by a plane mirror, the image moves 15.0 cm toward the mirror. Find the focal length of the concave mirror.

| Convex Mirrors

The mirror equation and the magnification equation can also be used with convex mirrors, provided the focal length f is taken to be a negative number, as indicated explicitly in Equation 25.2. One way to remember this is to recall that the focal point of a convex mirror lies behind the mirror. Example 5 deals with a convex mirror.

Problem solving insight When using the mirror equation, it is useful to construct a ray diagram to guide your thinking and to check your calculation.

| Example 5 | | | A Virtual Image Formed by a Convex Mirror

A convex mirror is used to reflect light from an object placed 66 cm in front of the mirror. The focal length of the mirror is (note the minus sign). Find (a) the location of the image and (b) the magnification. Reasoning We have seen that a convex mirror always forms a virtual image, as in Figure 25-22a, where the image is upright and smaller than the object. These characteristics should also be indicated by the results of our analysis here.
Solution

a. | With and , the mirror equation gives

The negative sign for di indicates that the image is behind the mirror and, therefore, is a virtual image.
b. | According to the magnification equation, the magnification is

The image is smaller (m is less than one) and upright (m is positive) with respect to the object.

Convex mirrors, like plane (flat) mirrors, always produce virtual images behind the mirror. However, the virtual image in a convex mirror is closer to the mirror than it would be if the mirror were planar, as Example 6 illustrates.

| Example 6 | | | A Convex Versus a Plane Mirror

An object is placed 9.00 cm in front of a mirror. The image is 3.00 cm closer to the mirror when the mirror is convex than when it is planar (see Figure 25-25). Find the focal length of the convex mirror.

Figure 25-25 The object distance (9.00 cm) is the same for the plane mirror (top part of drawing) as for the convex mirror (bottom part of drawing). However, as discussed in Example 6, the image formed by the convex mirror is 3.00 cm closer to the mirror.

Reasoning For a plane mirror, the image and the object are the same distance on either side of the mirror. Thus, the image would be 9.00 cm behind a plane mirror. If the image in a convex mirror is 3.00 cm closer than this, the image must be located 6.00 cm behind the convex mirror. In other words, when the object distance is , the image distance for the convex mirror is (negative because the image is virtual). The mirror equation can be used to find the focal length of the mirror.
Solution According to the mirror equation, the reciprocal of the focal length is

Contact lenses are worn to correct vision problems. Optometrists take advantage of the mirror equation and the magnification equation in providing lenses that fit the patient’s eyes properly, as the next example illustrates.

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

| Example 7 | | | Measuring the Curvature of the Cornea of the Eye

A contact lens rests against the cornea of the eye. Figure 25-26 shows an optometrist using a keratometer to measure the radius of curvature of the cornea, thereby ensuring that the prescribed lenses fit accurately. In the keratometer, light from an illuminated object reflects from the corneal surface, which acts like a convex mirror and forms an upright virtual image that is smaller than the object (see Figures 25-22a and 25-23). With the object placed 9.0 cm in front of the cornea, the magnification of the corneal surface is measured to be 0.046. Determine the radius of the cornea.

Figure 25-26 An optometrist is using a keratometer to measure the radius of curvature of the cornea of the eye, which is the surface against which a contact lens rests. (Andrew McLenaghan/SPL/Photo Researchers, Inc.).

The physics of keratometers.

Reasoning The radius of a convex mirror can be determined from the mirror’s focal length, since the two are related. The focal length is related to the distances of the object and its image from the mirror via the mirror equation. The magnification of the mirror is also related to the object and image distances according to the magnification equation. By using the mirror equation and the magnification equation, we will be able to determine the focal length and, hence, the radius.
Knowns and Unknowns The following table summarizes the available data:

Description | Symbol | Value | Comment
Object distance | do | 9.0 cm | Distance of object from cornea.
Magnification of corneal surface | m | 0.046 | Cornea acts like a convex mirror and forms a virtual image.
Unknown Variable |  |  |
Radius of cornea | R | ? |

Modeling the Problem

| Step 1 Relation between Radius and Focal Length
| The focal length f of a convex mirror is given by Equation 25.2 as

where R is the radius of the spherical surface. Solving this expression for the radius gives Equation 1 at the right. In Step 2, we determine the unknown focal length.

| (1)

| Step 2 The Mirror Equation
| The focal length is related to the object distance do and the image distance di via the mirror equation, which specifies that

| (25.3)

Solving this equation for f gives

which can be substituted into Equation 1 as shown in the right column. A value for do is given in the data table, and we turn to Step 3 to determine a value for di.

| Step 3 The Magnification Equation
| According to the magnification equation, the magnification m is given by

| (25.4)

Solving for di, we obtain

and can substitute the result into Equation 2, as shown at the right.

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

Rearranging terms gives

Thus, the radius is

Related Homework: Problems 24 and 37

The following Reasoning Strategy summarizes the sign conventions that are used with the mirror equation and the magnification equation. These conventions apply to both concave and convex mirrors.

| Reasoning Strategy

Summary of Sign Conventions for Spherical Mirrors
Focal length

f is + for a concave mirror.
f is − for a convex mirror.

Object distance

do is + if the object is in front of the mirror (real object).
do is − if the object is behind the mirror (virtual object).*

Image distance

di is + if the image is in front of the mirror (real image).
di is − if the image is behind the mirror (virtual image).

Magnification

m is + for an image that is upright with respect to the object.
m is − for an image that is inverted with respect to the object.

Check Your Understanding 4

An object is placed in front of a spherical mirror, and the magnification of the system is . What does this number tell you about the image? Select one or more of the choices below. (The answer is given at the end of the book.)

A. | The image is larger than the object.
B. | The image is smaller than the object.
C. | The image is upright relative to the object.
D. | The image is inverted relative to the object.
E. | The image is a real image.
F. | The image is a virtual image.

Background:The magnification equation is the key here; it contains a wealth of information about the image.
For similar questions (including calculational counterparts), consult Self-Assessment Test 25.2, which is described next.

Test your understanding of the material in Sections 25.4, 25.5 and 25.6 :

· | Spherical Mirrors
· | The Formation of Images by Spherical Mirrors
· | The Mirror Equation and the Magnification Equation

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