26.8

The Thin-Lens Equation and the Magnification Equation

CONCEPTS AT A GLANCE When an object is placed in front of a spherical mirror, we can determine the location, size, and nature of its image by using the technique of ray tracing or the mirror and magnification equations. Both options are based on the law of reflection. The mirror and magnification equations relate the distances do and di of the object and image from the mirror to the focal length f and magnification m. For an object placed in front of a lens, Snell’s law of refraction leads to the technique of ray tracing and to equations that are identical to the mirror and magnification equations. The similarity between mirrors and lenses can be seen by comparing the Concepts-at-a-Glance chart in Figure 26-30 with that for mirrors in Figure 26-23.

Figure 26-30 CONCEPTS AT A GLANCE The thin-lens equation and the magnification equation describe how thin lenses form images and are based on Snell’s law of refraction. They are used to design lenses that correct vision, such as the contact lens that the woman in the photograph is putting in her eye. (Laurence Dutton/Stone/Getty Images)

The chart in Figure 26-30 indicates that Snell’s law leads to equations that are referred to as the thin-lens equation and the magnification equation:

| (26.6)

| (26.7)

Figure 26-31 defines the symbols in these expressions with the aid of a thin converging lens, but the expressions also apply to a diverging lens, if it is thin. The derivations of these equations are presented at the end of this section.

Figure 26-31 The drawing shows the focal length f, the object distance do, and the image distance di for a converging lens. The object and image heights are, respectively, ho and hi.

Certain sign conventions accompany the use of the thin-lens and magnification equations, and the conventions are similar to those used with mirrors in Section 25.6. The issue of real versus virtual images, however, is slightly different with lenses than with mirrors. With a mirror, a real image is formed on the same side of the mirror as the object (see Figure 25-19), in which case the image distance di is a positive number. With a lens, a positive value for di also means the image is real. But, starting with an actual object, a real image is formed on the side of the lens opposite to the object (see Figure 26-31). The sign conventions listed in the following Reasoning Strategy apply to light rays traveling from left to right from a real object.

BOX 26.2 | Reasoning Strategy

Summary of Sign Conventions for Lenses

Focal length f is + for a converging lens. f is − for a diverging lens.
Object distance do is + if the object is to the left of the lens (real object), as is usual. do is − if the object is to the right of the lens (virtual object).*
Image distance di is + for an image (real) formed to the right of the lens by a real object. di is − for an image (virtual) formed to the left of the lens by a real object.
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.

Examples 9 and 10 illustrate the use of the thin-lens and magnification equations.

| Example 9 | | | The Real Image Formed by a Camera Lens

A person 1.70 m tall is standing 2.50 m in front of a camera. The camera uses a converging lens whose focal length is 0.0500 m. (a) Find the image distance (the distance between the lens and the film) and determine whether the image is real or virtual. (b) Find the magnification and the height of the image on the film.
Reasoning This optical arrangement is similar to that in Figure 26-26a, where the object distance is greater than twice the focal length of the lens. Therefore, we expect the image to be real, inverted, and smaller than the object.
Solution

a. | To find the image distance di we use the thin-lens equation with do = 2.50 m and f = 0.0500 m:

Since the image distance is a positive number, a is formed on the film.
b. | The magnification follows from the magnification equation:

The image is 0.0204 times as large as the object, and it is inverted since m is negative. Since the object height is ho = 1.70 m, the image height is

Problem solving insight In the thin-lens equation, the reciprocal of the image distance di is given by , where f is the focal length and do is the object distance. After combining the reciprocals and , do not forget to take the reciprocal of the result to find di.

| Example 10 | | | The Virtual Image Formed by a Diverging Lens

An object is placed 7.10 cm to the left of a diverging lens whose focal length is f = −5.08 cm (a diverging lens has a negative focal length). (a) Find the image distance and determine whether the image is real or virtual. (b) Obtain the magnification.
Reasoning This situation is similar to that in Figure 26-29a. The ray diagram shows that the image is virtual, erect, and smaller than the object.
Solution

a. | The thin-lens equation can be used to find the image distance di:

The image distance is negative, indicating that the image is and located to the left of the lens.
b. | Since di and do are known, the magnification equation shows that

The image is upright (m is +) and smaller (m < 1) than the object.

The thin-lens and magnification equations can be derived by considering rays 1 and 3 in Figure 26-32a. Ray 1 is shown separately in part b of the drawing, where the angle θ is the same in each of the two colored triangles. Thus, is the same for each triangle:

A minus sign has been inserted in the numerator of the ratio for the following reason. The angle θ in Figure 26-32b is assumed to be positive. Since the image is inverted relative to the object, the image height hi is a negative number. The insertion of the minus sign ensures that the term , and hence , is a positive quantity.

Figure 26-32 These ray diagrams are used for deriving the thin-lens and magnification equations.

Ray 3 is shown separately in part c of the drawing, where the angles are the same. Therefore,

A minus sign has been inserted in the numerator of the term hi/di for the same reason that a minus sign was inserted earlier—namely, to ensure that is a positive quantity. The first equation gives , while the second equation yields . Equating these two expressions for hi/ho and rearranging the result produces the thin-lens equation, . The magnification equation follows directly from the equation , if we recognize that hi/ho is the magnification m of the lens.

Check Your Understanding 4

An object is located at a distance do in front of a lens.The lens has a focal length f and produces an upright image that is twice as tall as the object. What kind of lens is it? What is the object distance? Express your answer as a fraction or multiple of the focal length. (The answer is given at the end of the book.)
Background: The characteristics of the possible images for converging and diverging lenses allow you to determine the type of lens here.The thin-lens equation and the magnification equation provide the basis for determining the object distance in terms of the focal length.
For similar questions (including calculational counterparts), consult Self-Assessment Test 26.2, which is described at the end of Section 26.14.

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