Learning objectives
Section Key Terms
| angle of refraction | corner reflector | critical angle | dispersion | incident ray |
| index of refraction | refracted ray | Snell’s law | total internal reflection |
The Law of Refraction
You may have noticed some odd optical phenomena when looking into a fish tank. For example, you may see the same fish appear to be in two different places (Figure_16_02_Fish). This is because light coming to you from the fish changes direction when it leaves the tank and, in this case, light rays traveling along two different paths both reach our eyes. The changing of a light ray’s direction (loosely called bending) when it passes a boundary between materials of different composition, or between layers in single material where there are changes in temperature and density, is called refraction. Refraction is responsible for a tremendous range of optical phenomena, from the action of lenses to voice transmission through optical fibers.
View original figure (external; image not reproduced in this text edition)
Original figure description: A man is looking at the corner of a fish tank. Although the fish is directly in his line of sight, he sees the fish to the right and left of the corner of the fish tank.
Why does light change direction when passing from one material (medium) to another? It is because light changes speed when going from one material to another. This behavior is typical of all waves and is especially easy to apply to light because light waves have very small wavelengths, and so they can be treated as rays. Before we study the law of refraction, it is useful to discuss the speed of light and how it varies between different media.
The speed of light is now known to great precision. In fact, the speed of light in a vacuum, c, is so important, and is so precisely known, that it is accepted as one of the basic physical quantities, and has the fixed value
where the approximate value of 3.00 108 m/s is used whenever three-digit precision is sufficient. The speed of light through matter is less than it is in a vacuum, because light interacts with atoms in a material. The speed of light depends strongly on the type of material, given that its interaction with different atoms, crystal lattices, and other substructures varies. We define the index of refraction, n, of a material to be
where v is the observed speed of light in the material. Because the speed of light is always less than c in matter and equals c only in a vacuum, the index of refraction (plural: indices of refraction) is always greater than or equal to one.
Table_16_02_Refractive_Indices lists the indices of refraction in various common materials.
| Medium | n |
|---|---|
| Gases at 0 °C and 1 atm | |
| Air | 1.000293 |
| Carbon dioxide | 1.00045 |
| Hydrogen | 1.000139 |
| Oxygen | 1.000271 |
| Liquids at 20 °C | |
| Benzene | 1.501 |
| Carbon disulfide | 1.628 |
| Carbon tetrachloride | 1.461 |
| Ethanol | 1.361 |
| Glycerin | 1.473 |
| Water, fresh | 1.333 |
| Solids at 20 °C | |
| Diamond | 2.419 |
| Fluorite | 1.434 |
| Glass, crown | 1.52 |
| Glass, flint | 1.66 |
| Ice at 0 °C | 1.309 |
| Plexiglas | 1.51 |
| Polystyrene | 1.49 |
| Quartz, crystalline | 1.544 |
| Quartz, fused | 1.458 |
| Sodium chloride | 1.544 |
| Zircon | 1.923 |
Figure_16_02_Lawnmower provides an analogy for and a description of how a ray of light changes direction when it passes from one medium to another. As in the previous section, the angles are measured relative to a perpendicular to the surface at the point where the light ray crosses it. The change in direction of the light ray depends on how the speed of light changes. The change in the speed of light is related to the indices of refraction of the media involved. In the situations shown in Figure_16_02_Lawnmower, medium 2 has a greater index of refraction than medium 1. This difference in index of refraction means that the speed of light is less in medium 2 than in medium 1. Note that, in Figure_16_02_Lawnmower(a), the path of the ray moves closer to the perpendicular when the ray slows down. Conversely, in Figure_16_02_Lawnmower(b), the path of the ray moves away from the perpendicular when the ray speeds up. The path is exactly reversible. In both cases, you can imagine what happens by thinking about pushing a lawn mower from a footpath onto grass, and vice versa. Going from the footpath to grass, the right front wheel is slowed and pulled to the side as shown. This is the same change in direction for light when it goes from a fast medium to a slow one. When going from the grass to the footpath, the left front wheel moves faster than the others, and the mower changes direction as shown. This, too, is the same change in direction as light going from slow to fast.
View original figure (external; image not reproduced in this text edition)
Original figure description: In view (a), a lawnmower is on a sidewalk and is approaching a strip of grass, which is perpendicular to the sidewalk, at angle θ1. As the right front wheel of the lawnmower hits the grass, the path of the angle, θ1, moves closer to the perpendicular. In view (b), a lawnmower is on the grass and is approaching the sidewalk, which is perpendicular to the grass, at angle θ2. As the left front wheel of the lawnmower hits the sidewalk, the path of the angle, θ2, moves farther from the perpendicular.
Bending Light
The Bending Light simulation allows you to show light refracting as it crosses the boundaries between various media (download animation first to view). It also shows the reflected ray. You can move the protractor to the point where the light meets the boundary and measure the angle of incidence, the angle of refraction, and the angle of reflection. You can also insert a prism into the beam to view the spreading, or dispersion, of white light into colors, as discussed later in this section. Use the ray option at the upper left.
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Teacher Support
Have students try all the different tabs at the top of the simulation. Point out to students that, although the tools work in both Ray and Wave mode, some may be easier to use in Wave mode because the region where the tool is able to read is larger.
Teacher Support
[BL]Be sure students understand that if c is always greater than v, n must always be greater than one. Demonstrating division using numbers that can be divided easily can reinforce student understanding.
[OL]Explain that, unlike the law of reflection, the law of refraction is most easily expressed as an equation, rather than in words. Walk students through the lawnmower analogy in Figure_16_02_Lawnmower. Suggest other wheeled vehicles with which they may be more familiar, and other surfaces, such as sand.
[AL]Ask students to try to explain why a prism separates white light into a rainbow of colors, but a window pane does not. If they cannot explain it, show them a ray diagram of light transmitted through a flat sheet of glass.
The amount that a light ray changes direction depends both on the incident angle and the amount that the speed changes. For a ray at a given incident angle, a large change in speed causes a large change in direction, and thus a large change in the angle of refraction. The exact mathematical relationship is the law of refraction, or Snell’s law, which is stated in equation form as
In terms of speeds, Snell’s law becomes
Here, n1 and n2 are the indices of refraction for media 1 and 2, respectively, and θ1 and θ2 are the angles between the rays and the perpendicular in the respective media 1 and 2, as shown in Figure_16_02_Lawnmower. The incoming ray is called the incident ray and the outgoing ray is called the refracted ray. The associated angles are called the angle of incidence and the angle of refraction. Later, we apply Snell’s law to some practical situations.
Dispersion is defined as the spreading of white light into the wavelengths of which it is composed. This happens because the index of refraction varies slightly with wavelength. Figure_16_02_Prism shows how a prism disperses white light into the colors of the rainbow.
View original figure (external; image not reproduced in this text edition)
Original figure description: In view (a), a ray of pure light (denoted by an arrow) hits the edge of a glass prism at an angle. As demonstrated by the arrow, this light is refracted as it enters the prism and as it exits the prism. In view (b), white light (also denoted by an arrow) hits the edge of a glass prism at an angle. In this case the ray of light disperses, as shown by two arrows originating where the white light enters the prism. When the rays pass through the prism, the angle widens and a rainbow of light is seen exiting the prism.
Rainbows are produced by a combination of refraction and reflection. You may have noticed that you see a rainbow only when you turn your back to the Sun. Light enters a drop of water and is reflected from the back of the drop, as shown in Figure_16_02_Rainbow. The light is refracted both as it enters and as it leaves the drop. Because the index of refraction of water varies with wavelength, the light is dispersed and a rainbow is observed.
View original figure (external; image not reproduced in this text edition)
Original figure description: A water droplet is depicted in this figure. A ray of sunlight, denoted by an arrow, hits drop of water and disperses at its point of entry, denoted by the arrow splitting in two. These two rays are then reflected off the back of the droplet at two different angles and are dispersed again as they leave the droplet, creating a rainbow.
Dispersion
This video explains how refraction disperses white light into its composite colors.
Original figure description: no alt text
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Teacher Support
Have students note that the dependence of the index of refraction on the speed of light implies a dependence on wavelength, as discussed in the video. This is because for light in a medium. Different wavelengths of light travel at different speeds, and so refract differently at media boundaries. Also, have students note that the frequency of light is not affected by refraction; it remains constant.
A good-quality mirror reflects more than 90 percent of the light that falls on it; the mirror absorbs the rest. But, it would be useful to have a mirror that reflects all the light that falls on it. Interestingly, we can produce total reflection using an aspect of refraction. Consider what happens when a ray of light strikes the surface between two materials, such as is shown in Figure_16_02_Total_internal(a). Part of the light crosses the boundary and is refracted; the rest is reflected. If, as shown in the figure, the index of refraction for the second medium is less than the first, the ray bends away from the perpendicular. Because n1 > n2, the angle of refraction is greater than the angle of incidence—that is, > . Now, imagine what happens as the incident angle is increased. This causes to increase as well. The largest the angle of refraction, , can be is 90°, as shown in Figure_16_02_Total_internal(b). The critical angle, , for a combination of two materials is defined to be the incident angle, , which produces an angle of refraction of 90°. That is, is the incident angle for which = 90°. If the incident angle, , is greater than the critical angle, as shown in Figure_16_02_Total_internal(c), then all the light is reflected back into medium 1, a condition called total internal reflection.
View original figure (external; image not reproduced in this text edition)
Original figure description: Views (a) through (c) depict how light behaves when light strikes a boundary and is reflected and refracted.
Recall that Snell’s law states the relationship between angles and indices of refraction. It is given by
When the incident angle equals the critical angle ( = ), the angle of refraction is 90° ( = 90°). Noting that sin 90° = 1, Snell’s law in this case becomes
The critical angle, , for a given combination of materials is thus
for n1 > n2.
Teacher Support
[OL]The superscript in sin−1 is not a power. It indicates arcsine, which is an inverse trigonometric function. It means, “that angle whose sine equals (in this case) (n2/n1).”
Total internal reflection occurs for any incident angle greater than the critical angle, , and it can only occur when the second medium has an index of refraction less than the first. Note that the previous equation is written for a light ray that travels in medium 1 and reflects from medium 2, as shown in Figure_16_02_Total_internal.
There are several important applications of total internal reflection. Total internal reflection, coupled with a large index of refraction, explains why diamonds sparkle more than other materials. The critical angle for a diamond-to-air surface is only 24.4°; so, when light enters a diamond, it has trouble getting back out (Figure_16_02_Diamond). Although light freely enters the diamond at different angles, it can exit only if it makes an angle less than 24.4° with the normal to a given surface. Facets on diamonds are specifically intended to make this unlikely, so that the light can exit only in certain places. Diamonds with very few impurities are very clear, so the light makes many internal reflections and is concentrated at the few places it can exit—hence the sparkle.
View original figure (external; image not reproduced in this text edition)
Original figure description: A sketch of a diamond shows a ray of light entering the diamond and bouncing of various facets of the diamond at an angle less than the critical angle.
A light ray that strikes an object that consists of two mutually perpendicular reflecting surfaces is reflected back exactly parallel to the direction from which it came. This parallel reflection is true whenever the reflecting surfaces are perpendicular, and it is independent of the angle of incidence. Such an object is called a corner reflector because the light bounces from its inside corner. Many inexpensive reflector buttons on bicycles, cars, and warning signs have corner reflectors designed to return light in the direction from which it originates. Corner reflectors are perfectly efficient when the conditions for total internal reflection are satisfied. With common materials, it is easy to obtain a critical angle that is less than 45°. One use of these perfect mirrors is in binoculars, as shown in Figure_16_02_Binoculars. Another application is for periscopes used in submarines.
View original figure (external; image not reproduced in this text edition)
Original figure description: A pair of binoculars is shown. The right side is a cutout view of the internal workings of the binoculars. A light ray (depicted by an arrow) enters the binoculars, reflects off two prisms (corner reflectors) to transmit the image, which exits the eyepiece.
Fiber optics are one common application of total internal reflection. In communications, fiber optics are used to transmit telephone, internet, and cable TV signals, and they use the transmission of light down fibers of plastic or glass. Because the fibers are thin, light entering one is likely to strike the inside surface at an angle greater than the critical angle and, thus, be totally reflected (Figure_16_02_Fiber). The index of refraction outside the fiber must be smaller than inside, a condition that is satisfied easily by coating the outside of the fiber with a material that has an appropriate refractive index. In fact, most fibers have a varying refractive index to allow more light to be guided along the fiber through total internal reflection. Rays are reflected around corners as shown in the figure, making the fibers into tiny light pipes.
View original figure (external; image not reproduced in this text edition)
Original figure description: View (a) shows a cross-section of a fiber optic cable that consists of a core (n1) and thin cladding (n2). A light ray (depicted by an arrow) enters the cable/fiber and reflects off various surfaces. View (b) shows a cross-section of a bent thin fiber and a light ray that reflects off the sides and around corners.
Medicine: Endoscopes
A medical device called an endoscope is shown in Figure_16_02_Endoscope.
View original figure (external; image not reproduced in this text edition)
Original figure description: A sketch of an endoscope shows the eyepiece attached to an optic tube.
The word endoscope means looking inside. Doctors use endoscopes to look inside hollow organs in the human body and inside body cavities. These devices are used to diagnose internal physical problems. Images may be transmitted to an eyepiece or sent to a video screen. Another channel is sometimes included to allow the use of small surgical instruments. Such surgical procedures include collecting biopsies for later testing, and removing polyps and other growths.
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Calculations with the Law of Refraction
Teacher Support
[BL]If an equation has two variables and a constant, such as the value of only one variable is needed to find the other.
[OL]Explain the difference between sine and arcsine. Explain why sin 90° = 1. Note that the index of refraction is a dimensionless number.
[AL]Show why the values sin 0°, sin 30°, sin 45°, sin 60°, and sin 90° can be expressed in the form
Show that the numerical values of these expressions are 0, 0.5, 0.707, 0.866, and 1.00, respectively.
The calculation problems that follow require application of the following equations:
and
These are the equations for refractive index, the mathematical statement of the law of refraction (Snell’s law), and the equation for the critical angle.
Snell’s Law Example 1
This video leads you through calculations based on the application of the equation that represents Snell’s law.
Original figure description: no alt text
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Calculating Index of Refraction from Speed
Calculate the index of refraction for a solid medium in which the speed of light is 2.012 108 m/s, and identify the most likely substance, based on the previous table of indicies of refraction.
Strategy
We know the speed of light, c, is 3.00 108 m/s, and we are given v. We can simply plug these values into the equation for index of refraction, n.
Solution
This value matches that of polystyrene exactly, according to the table of indices of refraction (Table_16_02_Refractive_Indices).
Discussion
The three-digit approximation for c is used, which in this case is all that is needed. Many values in the table are only given to three significant figures. Note that the units for speed cancel to yield a dimensionless answer, which is correct.
Calculating Index of Refraction from Angles
Suppose you have an unknown, clear solid substance immersed in water and you wish to identify it by finding its index of refraction. You arrange to have a beam of light enter it at an angle of 45.00°, and you observe the angle of refraction to be 40.30°. What are the index of refraction of the substance and its likely identity?
Strategy
We must use the mathematical expression for the law of refraction to solve this problem because we are given angle data, not speed data.
The subscripts 1 and 2 refer to values for water and the unknown, respectively, where 1 represents the medium from which the light is coming and 2 is the new medium it is entering. We are given the angle values, and the table of indicies of refraction gives us n for water as 1.333. All we have to do before solving the problem is rearrange the equation
Solution
The best match from Table_16_02_Refractive_Indices is fused quartz, with n = 1.458.
Discussion
Note the relative sizes of the variables involved. For example, a larger angle has a larger sine value. This checks out for the two angles involved. Note that the smaller value of compared with indicates the ray has bent toward normal. This result is to be expected if the unknown substance has a greater n value than that of water. The result shows that this is the case.
Calculating Critical Angle
Verify that the critical angle for light going from water to air is 48.6°. (See Table_16_02_Refractive_Indices, the table of indices of refraction.)
Strategy
First, choose the equation for critical angle
Then, look up the n values for water, n1, and air, n2. Find the value of . Last, find the angle that has a sine equal to this value and it compare with the given angle of 48.6°.
Solution
For water, n1 = 1.333; for air, n2 = 1.0003. So,
Discussion
Remember, when we try to find a critical angle, we look for the angle at which light can no longer escape past a medium boundary by refraction. It is logical, then, to think of subscript 1 as referring to the medium the light is trying to leave, and subscript 2 as where it is trying (unsuccessfully) to go. So water is 1 and air is 2.
Practice Problems
The refractive index of ethanol is 1.36. What is the speed of light in ethanol?
- 2.25×108 m/s
- 2.21×107 m/s
- 2.25×109 m/s
- 2.21×108 m/s
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Check Your Understanding
Teacher Support
Use these questions to assess student achievement of the section’s learning objectives. If students are struggling with a specific objective, these questions help identify which one, and then you can direct students to the relevant content.
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Section Summary
- The index of refraction for a material is given by the speed of light in a vacuum divided by the speed of light in that material.
- Snell’s law states the relationship between indices of refraction, the incident angle, and the angle of refraction.
- The critical angle, , determines whether total internal refraction can take place, and can be calculated according to .
Key Equations
| index of Refraction | |
| Snell’s law | |
| >Snell’s law in terms of speed | |
| critical angle |
Concept Items
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Critical Thinking
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Problems
An optical fiber uses flint glass (n = 1.66) clad with crown glass (n = 1.52) . What is the critical angle?
- 33.2°
- 23.7°
- 0.92 rad
- 1.16 rad
Suppose this figure represents a ray of light going from air (n = 1.0003) through crown glass (n = 1.52) into water, similar to a beam of light going into a fish tank.
View original figure (external; image not reproduced in this text edition)
Original figure description: An image is shown with a ray going into a medium. The incident ray has an angle labeled theta 1 and it splits into two rays as it hits the first horizon labeled en 1. The ray on the bottom is represented by a solid line and starts with an angle labeled theta two and the ray continues through a medium labeled en 2. When it encounters a horizon labeled en 3, the angle of theta 2 is on top of the horizon and theta 3 is under the horizon. The ray on top is represented by a dashed line. It continues straight from the incident ray until it hits horizon en 3. After the ray enters en 3, the angle of theta 3 is below the horizon. A distance of delta x is marked between the two rays at the bottom of the second medium.
Calculate the amount the ray is displaced by the glass (Δx), given that the incident angle is 40.0° and the glass is 1.00 cm thick.
- 0.839 cm
- 0.619 cm
- 0.466 cm
- 0.373 cm
Test Prep Multiple Choice
What does c represent in the equation ?
- the critical angle
- the refractive index
- the speed of light in a vacuum
- the speed of light in a transparent material
What is the term for the minimum angle at which a light ray is reflected back into a material and cannot pass into the surrounding medium?
- critical angle
- incident angle
- angle of refraction
- angle of reflection
Consider these indices of refraction: glass: 1.52, air: 1.0003, water: 1.333. Put these materials in order from the one in which the speed of light is fastest to the one in which it is slowest.
- The speed of light in water > the speed of light in air > the speed of light in glass.
- The speed of light in glass > the speed of light in water > the speed of light in air.
- The speed of light in air > the speed of light in water > the speed of light in glass.
- The speed of light in glass > the speed of light in air > the speed of light in water.
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Test Prep Short Answer
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Water floats on a liquid called carbon tetrachloride. The two liquids do not mix. A light ray passing from water into carbon tetrachloride has an incident angle of 45.0° and an angle of refraction of 40.1°. If the index of refraction of water is 1.33, what is the index of refraction of carbon tetrachloride?
- 1.60
- 1.49
- 1.21
- 1.46
Describe what happens to a light ray when it is refracted. Include in your explanation comparison of angles, comparison of refractive indices, and the term normal.
- When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of interference. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is greater in the medium with the greater refractive index.
- When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of refraction. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is less in the medium with the greater refractive index.
- When a ray of light goes from one medium to another medium with a different refractive index, the ray does not change its path. The angle between the ray and the normal (the line parallel to the surfaces of the two media) is the same in both media.
- When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of refraction. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is less in the medium with the lower refractive index.
Test Prep Extended Response
A scuba diver training in a pool looks at his instructor, as shown in this figure. The angle between the ray in the water and the normal to the water is 25°.
View original figure (external; image not reproduced in this text edition)
Original figure description: The figure depicts an instructor looking down at a scuba diver within a body of water.
What angle does the ray make from the instructor’s face with the normal to the water (n = 1.33) at the point where the ray enters? Assume n = 1.00 for air.
- 68°
- 25°
- 19°
- 34°
Describe total internal reflection. Include a definition of the critical angle and how it is related to total internal reflection. Also, compare the indices of refraction of the interior material and the surrounding material.
- When the interior material has a smaller index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs.
- When the interior material has a smaller index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is less than 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs.
- When the interior material has the same index of refraction as the surrounding material, the incident ray approaches the boundary at an angle (called the critical angle) such that the refraction angle is less than 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs.
- When the interior material has a greater index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is 90° . The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs.
Glossary
- angle of refraction
- the angle between the normal and the refracted ray
- critical angle
- an incident angle that produces an angle of refraction of 90°
- dispersion
- separation of white light into its component wavelengths
- incident ray
- the incoming ray toward a medium boundary or a reflective surface
- index of refraction
- the speed of light in a vacuum divided by the speed of light in a given material
- refracted ray
- the light ray after it has been refracted
- Snell’s law
- the law of refraction expressed mathematically as
- total internal reflection
- reflection of light traveling through a medium with a large refractive index at a boundary of a medium with a low refractive index under conditions such that refraction cannot occur