26.2

Snell’s Law and the Refraction of Light

| Snell’s Law

When light strikes the interface between two transparent materials, such as air and water, the light generally divides into two parts, as Figure 26-1a illustrates. Part of the light is reflected, with the angle of reflection equaling the angle of incidence. The remainder is transmitted across the interface. If the incident ray does not strike the interface at normal incidence, the transmitted ray has a different direction than the incident ray. The ray that enters the second material is said to be refracted and behaves in one of the following two ways:

1. | When light travels from a medium where the refractive index is smaller into a medium where it is larger, the refracted ray is bent toward the normal, as in Figure 26-1a.
2. | When light travels from a medium where the refractive index is larger into a medium where it is smaller, the refracted ray is bent away from the normal, as in Figure 26-1b.

These two possibilities illustrate that both the incident and refracted rays obey the principle of reversibility. Thus, the directions of the rays in part a of the drawing can be reversed to give the situation depicted in part b. In part b the reflected ray lies in the water rather than in the air.

Figure 26-1 (a) When a ray of light is directed from air into water, part of the light is reflected at the interface and the remainder is refracted into the water. The refracted ray is bent toward the normal . (b) When a ray of light is directed from water into air, the refracted ray in air is bent away from the normal .

In both parts of Figure 26-1 the angles of incidence, refraction, and reflection are measured relative to the normal. Note that the index of refraction of air is labeled n1 in part a, whereas it is n2 in part b, because we label all variables associated with the incident (and reflected) ray with subscript 1 and all variables associated with the refracted ray with subscript 2.
The angle of refraction depends on the angle of incidence and on the indices of refraction, n2 and n1 , of the two media. The relation between these quantities is known as Snell’s law of refraction, after the Dutch mathematician Willebrord Snell (1591–1626) who discovered it experimentally. At the end of this section is a proof of Snell’s law.

| Snell’s Law of Refraction
| When light travels from a material with refractive index n1 into a material with refractive index n2, the refracted ray, the incident ray, and the normal to the interface between the materials all lie in the same plane. The angle of refraction is related to the angle of incidence by

| (26.2)

Example 1 illustrates the use of Snell’s law.

| Example 1 | | | Determining the Angle of Refraction

A light ray strikes an air/water surface at an angle of 46° with respect to the normal. The refractive index for water is 1.33. Find the angle of refraction when the direction of the ray is (a) from air to water and (b) from water to air. Reasoning Snell’s law of refraction applies to both part (a) and part (b). However, in part (a) the incident ray is in air, whereas in part (b) it is in water. We keep track of this difference by always labeling the variables associated with the incident ray with a subscript 1 and the variables associated with the refracted ray with a subscript 2.

Problem solving insight The angle of incidence and the angle of refraction that appear in Snell’s law are measured with respect to the normal to the surface, and not with respect to the surface itself.

Solution

a. | The incident ray is in air, so and n1 = 1.00. The refracted ray is in water, so n2 = 1.33. Snell’s law can be used to find the angle of refraction :

| (26.2)

Since is less than , the refracted ray is bent toward the normal, as Figure 26-1a shows.
b. | With the incident ray in water, we find that

Since is greater than , the refracted ray is bent away from the normal, as in Figure 26-1b.

We have seen that reflection and refraction of light waves occur simultaneously at the interface between two transparent materials. It is important to keep in mind that light waves are composed of electric and magnetic fields, which carry energy. The principle of conservation of energy (see Chapter 6) indicates that the energy reflected plus the energy refracted must add up to equal the energy carried by the incident light, provided that none of the energy is absorbed by the materials. The percentage of incident energy that appears as reflected versus refracted light depends on the angle of incidence and the refractive indices of the materials on either side of the interface. For instance, when light travels from air toward water at perpendicular incidence, most of its energy is refracted and little is reflected. But when the angle of incidence is nearly 90° and the light barely grazes the water surface, most of its energy is reflected, with only a small amount refracted into the water. On a rainy night, you probably have experienced the annoying glare that results when light from an oncoming car just grazes the wet road. Under such conditions, most of the light energy reflects into your eyes.

The physics of rearview mirrors.

The simultaneous reflection and refraction of light have applications in a number of devices. For instance, interior rearview mirrors in cars often have adjustment levers. One position of the lever is for day driving, while the other is for night driving and reduces glare from the headlights of the car behind. As Figure 26-2a indicates, this kind of mirror is a glass wedge, the back side of which is silvered and highly reflecting. Part b of the picture shows the day setting. Light from the car behind follows path ABCD in reaching the driver’s eye. At points A and C, where the light strikes the air–glass surface, there are both reflected and refracted rays. The reflected rays are drawn as thin lines, the thinness denoting that only a small percentage (about 10%) of the light during the day is reflected at A and C. The weak reflected rays at A and C do not reach the driver’s eye. In contrast, almost all the light reaching the silvered back surface at B is reflected toward the driver. Since most of the light follows path ABCD, the driver sees a bright image of the car behind. During the night, the adjustment lever can be used to rotate the top of the mirror away from the driver (see part c of the drawing). Now, most of the light from the headlights behind follows path ABC and does not reach the driver. Only the light that is weakly reflected from the front surface along path AD is seen, and the result is significantly less glare.

Figure 26-2 A car’s interior rearview mirror with a day–night adjustment lever.

Check Your Understanding 1

The drawing shows three layers of liquids, A, B, and C, each with a different index of refraction. Light begins in liquid A, passes into B, and eventually into C, as the ray of light in the drawing shows.The dashed lines denote the normals to the interfaces between the layers. Which liquid has the smallest index of refraction? (The answer is given at the end of the book.)

Background: The index of refraction is the ratio of the speed of light in a vacuum to the speed of light in a material. As light travels from one material into another, its direction of travel can change, depending on the indices of refraction of the materials. Snell’s law of refraction describes the change in direction.
For similar questions (including calculational counterparts), consult Self-Assessment Test 26.1, which is described at the end of Section 26.5.

| Apparent Depth

One interesting consequence of refraction is that an object lying under water appears to be closer to the surface than it actually is. Example 2 sets the stage for explaining why, by showing what must be done to shine a light on such an object.

| Example 2 | | | Finding a Sunken Chest

A searchlight on a yacht is being used at night to illuminate a sunken chest, as in Figure 26-3. At what angle of incidence should the light be aimed?

Figure 26-3 The beam from the searchlight is refracted when it enters the water.

Reasoning The angle of incidence must be such that, after refraction, the light strikes the chest. The angle of incidence can be obtained from Snell’s law, once the angle of refraction is determined. This angle can be found using the data in Figure 26-3 and trigonometry. The light travels from a region of lower into a region of higher refractive index, so the light is bent toward the normal and we expect to be greater than .

Problem solving insight Remember that the refractive indices are written as n1 for the medium in which the incident light travels and n2 for the medium in which the refracted light travels.

Solution From the data in the drawing it follows that , so = 31°. With n1 = 1.00 for air and n2 = 1.33 for water, Snell’s law gives

As expected, is greater than .

When the sunken chest in Example 2 is viewed from the boat (Figure 26-4a), light rays from the chest pass upward through the water, refract away from the normal when they enter the air, and then travel to the observer. This picture is similar to Figure 26-3, except the direction of the rays is reversed and the searchlight is replaced by an observer. When the rays entering the air are extended back into the water (see the dashed lines), they indicate that the observer sees a virtual image of the chest at an apparent depth that is less than the actual depth. The image is virtual because light rays do not actually pass through it. For the situation shown in Figure 26-4a, it is difficult to determine the apparent depth. A much simpler case is shown in part b, where the observer is directly above the submerged object, and the apparent depth d′ is related to the actual depth d by

| (26.3)

In this result, n1 is the refractive index of the medium associated with the incident ray (the medium in which the object is located), and n2 refers to the medium associated with the refracted ray (the medium in which the observer is situated). The proof of Equation 26.3 is the focus of problem 17 at the end of the chapter. Example 3 illustrates that the effect of apparent depth is quite noticeable in water.

| Example 3 | | | The Apparent Depth of a Swimming Pool

A swimmer is treading water (with her head above the water) at the surface of a pool 3.00 m deep. She sees a coin on the bottom directly below. How deep does the coin appear to be? Reasoning Equation 26.3 may be used to find the apparent depth, provided we remember that the light rays travel from the coin to the swimmer. Therefore, the incident ray is coming from the coin under the water , while the refracted ray is in the air .
Solution The apparent depth of the coin is

| (26.3)

Figure 26-4 (a) Because light from the chest is refracted away from the normal when the light enters the air, the apparent depth of the image is less than the actual depth. (b) The observer is viewing the submerged object from directly overhead.

In Example 3, a person sees a coin on the bottom of a pool at an apparent depth that is less than the actual depth. Conceptual Example 4 considers the reverse situation—namely, a person looking from under the water at a coin in the air.

| Conceptual Example 4 | | | On the Inside Looking Out

A swimmer is under water and looking up at the surface. Someone holds a coin in the air, directly above the swimmer’s eyes. To the swimmer, the coin appears to be at a certain height above the water. Is the apparent height of the coin greater than, less than, or the same as its actual height?
Reasoning and Solution Figure 26-5 shows that the apparent height of the coin above the water surface is greater than the actual height. Consider two rays of light leaving a point P on the coin. When the rays enter the water, they are refracted toward the normal because water has a larger index of refraction than air has. By extending the refracted rays backward (see the dashed lines in the drawing), we find that the rays appear to the swimmer as if they originate from a point that is farther from the water than point P is. Thus, the underwater swimmer sees the coin at an apparent height that is greater than the actual height d. Equation 26.3 reveals the same result, because represents the medium (air) associated with the incident ray and represents the medium (water) associated with the refracted ray. Since for water is greater than for air, the ratio is greater than one and is larger than d. This situation is the opposite of that in Figure 26-4b, where an object beneath the water appears to be closer to the surface than it actually is.
Related Homework: Problems 20, 100

Figure 26-5 Rays from point P on a coin in the air above the water refract toward the normal as they enter the water. An underwater swimmer perceives the rays as originating from a point P′ that is farther above the surface than the actual point P.

| The Displacement of Light by a Transparent Slab of Material

A windowpane is an example of a transparent slab of material. It consists of a plate of glass with parallel surfaces. When a ray of light passes through the glass, the emergent ray is parallel to the incident ray but displaced from it, as Figure 26-6 shows. This result can be verified by applying Snell’s law to each of the two glass surfaces, with the result that . Since air surrounds the glass, n1 = n3, and it follows that . Therefore, , and the emergent and incident rays are parallel. However, as the drawing shows, the emergent ray is displaced laterally relative to the incident ray. The extent of the displacement depends on the angle of incidence, the thickness of the slab, and its refractive index.

Figure 26-6 When a ray of light passes through a pane of glass that has parallel surfaces and is surrounded by air, the emergent ray is parallel to the incident ray , but is displaced from it.

| Derivation of Snell’s Law

Snell’s law can be derived by considering what happens to the wave fronts when the light passes from one medium into another. Figure 26-7a shows light propagating from medium 1, where the speed is relatively large, into medium 2, where the speed is smaller; therefore, n1 is less than n2. The plane wave fronts in this picture are drawn perpendicular to the incident and refracted rays. Since the part of each wave front that penetrates medium 2 slows down, the wave fronts in medium 2 are rotated clockwise relative to those in medium 1. Correspondingly, the refracted ray in medium 2 is bent toward the normal, as the drawing shows.
Although the incident and refracted waves have different speeds, they have the same frequency f. The fact that the frequency does not change can be understood in terms of the atomic mechanism underlying the generation of the refracted wave. When the electromagnetic wave strikes the surface, the oscillating electric field forces the electrons in the molecules of medium 2 to oscillate at the same frequency as the wave. The accelerating electrons behave like atomic antennas that radiate “extra” electromagnetic waves, which combine with the original wave. The net electromagnetic wave within medium 2 is a superposition of the original wave plus the extra radiated waves, and it is this superposition that constitutes the refracted wave. Since the extra waves are radiated at the same frequency as the original wave, the refracted wave also has the same frequency as the original wave.

Figure 26-7 (a) The wave fronts are refracted as the light passes from medium 1 into medium 2. (b) An enlarged view of the incident and refracted wave fronts at the surface.

The distance between successive wave fronts in Figure 26-7a has been chosen to be the wavelength . Since the frequencies are the same in both media but the speeds are different, it follows from Equation 16.1 that the wavelengths are different: and . Since v1 is assumed to be larger than v2, is larger than , and the wave fronts are farther apart in medium 1.
Figure 26-7b shows an enlarged view of the incident and refracted wave fronts at the surface. The angles and within the colored right triangles are, respectively, the angles of incidence and refraction. In addition, the triangles share the same hypotenuse h. Therefore,

and

Combining these two equations into a single equation by eliminating the common term hf gives

By multiplying each side of this result by c, the speed of light in a vacuum, and recognizing that the ratio c/v is the index of refraction n, we arrive at Snell’s law of refraction: .

CONCEPTS AT A GLANCE A number of important phenomena are related to refraction and Snell’s law. The Concepts-at-a-Glance chart in Figure 26-8 lists four of these phenomena, each of which will be discussed in the remainder of this chapter. In the next section, we turn our attention to the first of them, total internal reflection.

Figure 26-8 CONCEPTS AT A GLANCE Four of the phenomena that we will discuss in this chapter are listed at the right of this chart. This spectacular photograph shows a reptile swimming under water and its image. The image is formed because light from the reptile is completely reflected at the surface of the water (which is at the narrow horizontal white streak), a phenomenon known as total internal reflection. (Joel Sartore/National Geographic Society).

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