26.3

Total Internal Reflection

When light passes from a medium of larger refractive index into one of smaller refractive index—for example, from water to air—the refracted ray bends away from the normal, as in Figure 26-9a. As the angle of incidence increases, the angle of refraction also increases. When the angle of incidence reaches a certain value, called the critical angle the angle of refraction is 90°. Then the refracted ray points along the surface; part b illustrates what happens at the critical angle. When the angle of incidence exceeds the critical angle, as in part c of the drawing, there is no refracted light. All the incident light is reflected back into the medium from which it came, a phenomenon called total internal reflection. Total internal reflection occurs only when light travels from a higher-index medium toward a lower-index medium. It does not occur when light propagates in the reverse direction—for example, from air to water.

Figure 26-9 (a) When light travels from a higher-index medium (water) into a lower-index medium (air), the refracted ray is bent away from the normal. (b) When the angle of incidence is equal to the critical angle , the angle of refraction is 90°. (c) If is greater than , there is no refracted ray, and total internal reflection occurs.

An expression for the critical angle can be obtained from Snell’s law by setting and (see Figure 26-9b):

| (26.4)

For instance, the critical angle for light traveling from water (n1 = 1.33) to air (n2 = 1.00) is . For incident angles greater than 48.8°, Snell’s law predicts that is greater than unity, a value that is not possible. Thus, light rays with incident angles exceeding 48.8° yield no refracted light, and the light is totally reflected back into the water, as Figure 26-9c indicates.
The next example illustrates how the critical angle changes when the indices of refraction change.

| Example 5 | | | Total Internal Reflection

A beam of light is propagating through diamond (n1 = 2.42) and strikes a diamond–air interface at an angle of incidence of 28°. (a) Will part of the beam enter the air (n2 = 1.00) or will the beam be totally reflected at the interface? (b) Repeat part (a), assuming that the diamond is surrounded by water (n2 = 1.33).
Reasoning Total internal reflection occurs only when the beam of light has an angle of incidence that is greater than the critical angle . The critical angle is different in parts (a) and (b), since it depends on the ratio n2/n1 of the refractive indices of the incident (n1) and refracting (n2) media.
Solution

a. | The critical angle for total internal reflection at the diamond–air interface is given by Equation 26.4 as

Because the angle of incidence of 28° is greater than the critical angle, there is no refraction, and the light is totally reflected back into the diamond.
b. | If water, rather than air, surrounds the diamond, the critical angle for total internal reflection becomes larger:

Now a beam of light that has an angle of incidence of 28° (less than the critical angle of 33.3°) at the diamond–water interface is refracted into the water.

The critical angle plays an important role in why a diamond sparkles, as Conceptual Example 6 discusses.

| Conceptual Example 6 | | | The Sparkle of a Diamond

A diamond gemstone is famous for its sparkle because the light coming from it glitters as the diamond is moved about. Why does a diamond exhibit such brilliance? And why does it lose much of its brilliance when placed under water?

The physics of why a diamond sparkles.

Reasoning and Solution A diamond sparkles because when it is held a certain way, the intensity of the light coming from it is greatly enhanced. Figure 26-10 helps to explain that this enhancement is related to total internal reflection. Part a of the drawing shows a ray of light striking a bottom facet of the diamond at an angle of incidence that is greater than the critical angle. This ray, and all other rays whose angles of incidence exceed the critical angle, are totally reflected back into the diamond, eventually exiting the top surface to give the diamond its sparkle. Many of the rays striking a bottom facet behave in this fashion because diamond has a relatively small critical angle in air. As part (a) of Example 5 shows, the critical angle is 24.4°. The value is so small because the index of refraction of diamond (n = 2.42) is large compared to that of air (n = 1.00).

Figure 26-10 (a) Near the bottom of the diamond, light is totally internally reflected, because the incident angle exceeds the critical angle for diamond and air. (b) When the diamond is in water, the same light is partially reflected and partially refracted, since the incident angle is less than the critical angle for diamond and water.

Now consider what happens to the same ray of light within the diamond when the diamond is placed in water. Because water has a larger index of refraction than air, the critical angle increases to 33.3°, as part (b) of Example 5 shows. Therefore, this particular ray is no longer totally internally reflected. As Figure 26-10b illustrates, only some of the light is reflected back into the diamond, and the remainder escapes into the water. Consequently, less light exits from the top of the diamond, causing it to lose much of its brilliance.

Need more practice?

Interactive LearningWare26.1
The drawing shows a ray of light whose angle of incidence is , traveling through two solid materials and then undergoing total internal reflection at a solid–liquid interface. What is the largest possible index of refraction for the liquid?

Related Homework: Problem 30

Many optical instruments, such as binoculars, periscopes, and telescopes, use glass prisms and total internal reflection to turn a beam of light through 90° or 180°. Figure 26-11a shows a light ray entering a glass prism (n1 = 1.5) and striking the hypotenuse of the prism at an angle of incidence of . The critical angle for a glass–air interface is . Since the angle of incidence is greater than the critical angle, the light is totally reflected at the hypotenuse and is directed vertically upward in the drawing, having been turned through an angle of 90°. Part b of the picture shows how the same prism can turn the beam through 180° when total internal reflection occurs twice. Prisms can also be used in tandem to produce a lateral displacement of a light ray, while leaving its initial direction unaltered. Figure 26-11c illustrates such an application in binoculars.

Figure 26-11 Total internal reflection at a glass–air interface can be used to turn a ray of light through an angle of (a) 90° or (b) 180°. (c) Two prisms, each reflecting the light twice by total internal reflection, are sometimes used in binoculars to produce a lateral displacement of a light ray.

The physics of fiber optics.

An important application of total internal reflection occurs in fiber optics, where hair-thin threads of glass or plastic, called optical fibers, “pipe” light from one place to another. Figure 26-12a shows that an optical fiber consists of a cylindrical inner core that carries the light and an outer concentric shell, the cladding. The core is made from transparent glass or plastic that has a relatively high index of refraction. The cladding is also made of glass, but of a type that has a relatively low index of refraction. Light enters one end of the core, strikes the core/cladding interface at an angle of incidence greater than the critical angle, and, therefore, is reflected back into the core. Light thus travels inside the optical fiber along a zigzag path. In a well-designed fiber, little light is lost as a result of absorption by the core, so light can travel many kilometers before its intensity diminishes appreciably. Optical fibers are often bundled together to produce cables. Because the fibers themselves are so thin, the cables are relatively small and flexible and can fit into places inaccessible to larger metal wires.

Figure 26-12 (a) Light can travel with little loss in a curved optical fiber, because the light is totally reflected whenever it strikes the core–cladding interface and because the absorption of light by the core itself is small. (b) Light being transmitted by a bundle of optical fibers. (PhotoDisc, Inc./Getty Images)

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 | | | An Optical Fiber

Figure 26-13 shows an optical fiber that consists of a core made of flint glass (nflint = 1.667) surrounded by a cladding made of crown glass (ncrown = 1.523). A ray of light in air enters the fiber at an angle with respect to the normal. What is if this light also strikes the core–cladding interface at an angle that just barely exceeds the critical angle?

Figure 26-13 A ray of light enters the left end of an optical fiber and strikes the core–cladding interface at the critical angle .

Reasoning The angle of incidence is related to the angle of refraction (see Figure 26-13) by Snell’s law, where the refraction angle is part of the right triangle in the drawing. The critical angle for the core–cladding interface is also part of the same right triangle, so that . The critical angle can be determined from a knowledge of the indices of refraction of the core and the cladding.
Knowns and Unknowns The data used in this problem are:

Description | Symbol | Value | Comment
Index of refraction of core material (flint glass) | nflint | 1.667 |
Index of refraction of cladding material (crown glass) | ncrown | 1.523 |
Index of refraction of air | nair | 1.000 | See Table 26-1.
Unknown Variable |  |  |
Angle of incidence of light ray entering optical fiber |  | ? |

Modeling the Problem

| Step 1 Snell’s Law of Refraction
| The ray of light, initially traveling in air, strikes the left end of the optical fiber at an angle of incidence labeled in Figure 26-13. When the light enters the flint-glass core, its angle of refraction is . Snell’s law of refraction gives the relation between these angles as

| (26.2)

Solving this equation for yields Equation 1 at the right. Values for nair and nflint are known. The angle will be evaluated in the next step.

| (1)

| Step 2 The Critical Angle
| We know that the light ray inside the core strikes the core–cladding interface at an angle that just barely exceeds the critical angle . When the angle of incidence exceeds the critical angle, all the light is reflected back into the core. From the right triangle in Figure 26-13, we see that the critical angle is related to by . The critical angle depends on the indices of refraction of the core (flint glass) and cladding (crown glass) according to Equation 26.4:

Substituting this expression for into gives

This result for can be substituted into Equation 1, as indicated at the right.

Solution Combining the results of Steps 1 and 2 algebraically to produce a single equation gives a rather cumbersome result. Hence, we follow the simpler procedure of evaluating Equation 2 numerically and then substituting the result into Equation 1:

| (2)

Related Homework: Problem 30

Optical fiber cables are the medium of choice for high-quality telecommunications because the cables are relatively immune to external electrical interference and because a light beam can carry information through an optical fiber just as electricity carries information through copper wires. The information-carrying capacity of light, however, is thousands of times greater than that of electricity. A laser beam traveling through a single optical fiber can carry tens of thousands of telephone conversations and several TV programs simultaneously.

The physics of endoscopy.

In the field of medicine, optical fiber cables have had extraordinary impact. In the practice of endoscopy, for instance, a device called an endoscope is used to peer inside the body. Figure 26-14 shows a bronchoscope being used, which is a kind of endoscope that is inserted through the nose or mouth, down the bronchial tubes, and into the lungs. It consists of two optical fiber cables. One provides light to illuminate interior body parts, while the other sends back an image for viewing. A bronchoscope greatly simplifies the diagnosis of pulmonary disease. Tissue samples can even be collected with some bronchoscopes. A colonoscope is another kind of endoscope, and its design is similar to that bronchoscope. It is inserted through the rectum and used to examine the interior of colon (see Figure 26-15). The colonoscope currently offers the best hope for diagnosing colon cancer in its early stages, when it can be treated.

Figure 26-14 A doctor is using an endoscope to collect samples of tissue and fluid from the lung of a patient who has a history of asthma and allergies. Subsequently, the samples are examined under a microscope to obtain a diagnosis. (James King-Holmes/SPL/Photo Researchers, Inc.)

Figure 26-15 A colonoscope reveals a benign (noncancerous) polyp attached to the wall of a patient’s colon (large intestine). This is a lipoma, a tumor arising from fatty tissue. Polyps that turn cancerous or grow large enough to obstruct the colon are surgically removed. (David M. Martin/SPL/Photo Researchers, Inc.)

The use of optical fibers has also revolutionized surgical techniques. In arthroscopic surgery, a small surgical instrument, several millimeters in diameter, is mounted at the end of an optical fiber cable. The surgeon can insert the instrument and cable into a joint, such as the knee, with only a tiny incision and minimal damage to the surrounding tissue (see Figure 26-16). Consequently, recovery from the procedure is relatively rapid compared to traditional surgical techniques.

Figure 26-16 Optical fibers have made arthroscopic surgery possible, such as the repair of a damaged knee shown in this photograph. (Margaret Rose Orthopaedic Hospital/Photo Researchers)

The physics of arthroscopic surgery.

Check Your Understanding 2

The drawing shows a prism and two light rays, A and B, both of which strike the prism perpendicularly.The prism is surrounded by an unknown liquid. When ray A reaches the hypotenuse in the drawing, it is totally internally reflected. Which one of the following statements applies to ray B when it reaches the hypotenuse? (a) It may or may not be totally internally reflected, depending on what the surrounding liquid is. (b) It is not totally internally reflected, no matter what the surrounding liquid is. (c) It is totally internally reflected, no matter what the surrounding liquid is. (The answer is given at the end of the book.)

Background: When light travels from a higher toward a lower index of refraction, total internal reflection occurs when the light strikes the interface between the two materials at an angle greater than the critical angle.The critical angle depends on the values of the two refractive indices.
For similar questions (including calculational counterparts), consult Self-Assessment Test 26.1, which is described at the end of Section 26.5.

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