Geometric Optics, Vision and Optical Instruments

66 The Law of Refraction

Learning Objectives

  • Determine the index of refraction from the speed of light in a material.
  • Explain why light changes direction when it passes from one material to another.

Have you ever noticed that a straw looks bent when placed in a glass of water, or that a fish in an aquarium appears to be in a different location than expected? These familiar observations occur because light changes direction as it passes from one material to another. This phenomenon is called refraction.

Refraction occurs because light travels at different speeds in different materials. As light passes between air, water, glass, or biological tissues, its speed changes, causing the light ray to bend. Refraction is responsible for many everyday and medical technologies, including eyeglasses, contact lenses, microscopes, cameras, endoscopes, and optical fibers used in modern medical communication and imaging.

Refraction

The change in the direction of a light ray as it passes from one material to another because its speed changes.

Clinical Connection

Refraction is fundamental to many areas of healthcare. Corrective eyeglasses and contact lenses work by refracting light so that images focus correctly on the retina. Endoscopes use optical fibers that rely on refraction to transmit light into the body, while microscopes use carefully designed lenses to magnify cells and microorganisms. Even the apparent position of tissues viewed through fluids during surgery is influenced by refraction.

A person looking into a fish tank sees the same fish appearing in two different locations because light rays bend as they leave the water and enter the air.
Figure 66.1: Light from the fish changes direction as it passes from water into air. Because the light can follow different paths to the observer's eyes, the fish appears to occupy different positions. This bending of light is called refraction.

Why does light bend when it moves from one material to another? The answer is that light travels at different speeds in different media. Light travels fastest in a vacuum and more slowly in materials such as water, glass, or plastic because it interacts with the atoms of the material. Understanding how the speed of light changes is the key to understanding refraction.

Speed of Light

The speed of light in a vacuum, represented by [latex]c[/latex], is one of the fundamental constants of nature. While nothing travels faster than light in a vacuum, light slows down when it passes through matter because of its interactions with the atoms of the material. This change in speed produces the bending of light known as refraction.

The constant speed of light in a vacuum also plays a central role in Einstein's theory of special relativity, which will be introduced later in this textbook.

The Speed of Light and the Index of Refraction

Refraction occurs because light travels at different speeds in different materials. Light moves fastest in a vacuum and slows whenever it passes through matter, such as air, water, glass, or biological tissues. Whenever the speed changes as light crosses from one material to another, the light ray changes direction. This simple idea explains a wide range of optical phenomena, from the apparent bending of a straw in water to the operation of eyeglasses, microscopes, and endoscopes.

Scientists spent centuries trying to determine whether light traveled instantaneously or at a finite speed. In 1676, the Danish astronomer Ole Rømer provided the first convincing evidence that light requires time to travel by carefully studying the eclipses of Jupiter's moon Io. Two centuries later, Albert A. Michelson developed laboratory techniques that measured the speed of light with remarkable precision. Today, the speed of light in a vacuum is known so accurately that it is defined exactly and serves as one of the fundamental constants of physics.

Diagram of Michelson's rotating mirror experiment used to measure the speed of light.
Figure 66.2: Albert Michelson used rotating mirrors to make one of the first highly accurate measurements of the speed of light. Modern measurements use lasers and atomic clocks, achieving even greater precision.

The speed of light in a vacuum is represented by the symbol [latex]c[/latex] and has the value

[latex]c = 2.99792458\times10^8~\text{m/s} \approx 3.00\times10^8~\text{m/s}.[/latex]

Light always travels more slowly in matter than in a vacuum because it interacts with the atoms and molecules of the material. The amount by which light slows down is described by the index of refraction, defined as

[latex]n=\frac{c}{v},[/latex]

where [latex]v[/latex] is the speed of light in the material.

Index of Refraction

The index of refraction is the ratio of the speed of light in a vacuum to its speed in a material.

[latex]n=\frac{c}{v}[/latex]

Because light always travels more slowly in matter than in a vacuum, the index of refraction is always at least 1:

[latex]n\ge1.[/latex]

The index of refraction depends on the material and, to a small extent, on the wavelength (color) of the light. This wavelength dependence explains phenomena such as prisms separating white light into different colors.

Clinical Connection

Different tissues in the eye have slightly different indices of refraction. Together, the cornea, aqueous humor, crystalline lens, and vitreous humor bend incoming light so that it focuses on the retina. Small changes in these refractive properties can produce nearsightedness, farsightedness, or astigmatism, which are corrected using eyeglasses, contact lenses, or laser surgery.

Table 66.1 Representative Indices of Refraction
Material Index of Refraction (n)
Vacuum 1.000
Air 1.0003
Water 1.333
Ice 1.309
Cornea 1.376
Aqueous humor 1.336
Vitreous humor 1.336
Glass (crown) 1.52
Plexiglas 1.51
Fused quartz 1.458
Diamond 2.42

Example 66.1: Calculating the Speed of Light in a Material

Zircon is a transparent material sometimes used as a diamond substitute in jewelry. Its index of refraction is [latex]n=1.923[/latex]. Calculate the speed of light in zircon.

Strategy

Use the definition of the index of refraction and solve for the speed of light in the material.

[latex]n=\frac{c}{v} \qquad\Rightarrow\qquad v=\frac{c}{n}[/latex]

Solution

[latex]v=\frac{3.00\times10^8~\text{m/s}}{1.923} =1.56\times10^8~\text{m/s}.[/latex]

Discussion

Although light travels much more slowly in zircon than in a vacuum, it still moves at more than half the speed of light in empty space. Materials with larger indices of refraction bend light more strongly, making them useful for optical devices such as lenses and contributing to the brilliance of gemstones.

Snell's Law of Refraction

When light passes from one material to another, its speed changes. As a result, the light ray changes direction at the boundary between the two materials. This change in direction is called refraction.

Figure 66.3 shows what happens when light travels between two materials with different indices of refraction. The angles are measured relative to the normal, an imaginary line drawn perpendicular to the surface where the light crosses the boundary.

If light enters a material in which it travels more slowly (a material with a larger index of refraction), the refracted ray bends toward the normal. Conversely, when light enters a material in which it travels more quickly, it bends away from the normal. The path of the light is completely reversible.

Light refracting at the boundary between two materials. When light enters a material with a higher index of refraction, it bends toward the normal. When it enters a material with a lower index of refraction, it bends away from the normal.
Figure 66.3: A light ray bends whenever it changes speed while crossing from one material to another. Entering a material with a higher index of refraction causes the ray to bend toward the normal, while entering a material with a lower index of refraction causes it to bend away from the normal. The path is completely reversible.

Clinical Connection

The refraction of light at boundaries between different materials makes vision possible. Most of the bending of incoming light occurs at the curved surface of the cornea, while the crystalline lens fine-tunes the focus so that a sharp image forms on the retina. Corrective lenses, contact lenses, surgical microscopes, endoscopes, and optical coherence tomography (OCT) all rely on the same principles of refraction.

The amount by which a ray bends depends on two factors:

  • the angle at which the light strikes the boundary, called the incident angle, and
  • the difference between the indices of refraction of the two materials.

This relationship is described mathematically by Snell's law, also known as the law of refraction:

[latex]n_1\sin\theta_1=n_2\sin\theta_2.[/latex]

In this equation:

  • [latex]n_1[/latex] and [latex]n_2[/latex] are the indices of refraction of the two materials.
  • [latex]\theta_1[/latex] is the angle between the incident ray and the normal.
  • [latex]\theta_2[/latex] is the angle between the refracted ray and the normal.

The incoming ray is called the incident ray, and the transmitted ray is called the refracted ray. Although Willebrord Snell discovered this relationship experimentally in 1621, we now understand that it results from the change in the speed of light between different materials.

Snell's Law (Law of Refraction)

[latex]n_1\sin\theta_1=n_2\sin\theta_2[/latex]

Take-Home Experiment: A Bent Pencil

Place a pencil or drinking straw in a clear glass that is half filled with water. Observe the object from the side. The portion submerged in the water appears bent because light rays from the underwater section refract as they pass from water into air before reaching your eyes.

Sketch a simple ray diagram showing how refraction changes the apparent position of the submerged portion of the object.

Example 66.2: Determining the Index of Refraction

Light travels from air into an unknown transparent material, as shown in Figure 66.3(a). If the angle of incidence is [latex]30.0^\circ[/latex] and the angle of refraction is [latex]22.0^\circ[/latex], determine the index of refraction of the second material.

Strategy

For most calculations, the index of refraction of air can be taken as [latex]n_1=1.00[/latex]. Since the incident angle, refracted angle, and index of refraction of air are known, Snell's law can be solved for the unknown index of refraction [latex]n_2[/latex].

Solution

Begin with Snell's law:

[latex]n_1\sin\theta_1=n_2\sin\theta_2.[/latex]

Rearrange to solve for the unknown index of refraction:

[latex]n_2=n_1\frac{\sin\theta_1}{\sin\theta_2}.[/latex]

Substitute the known values:

[latex]n_2 = 1.00 \left( \frac{\sin30.0^\circ}{\sin22.0^\circ} \right) = \frac{0.500}{0.375} = 1.33.[/latex]

Discussion

The calculated index of refraction is [latex]n=1.33[/latex], which corresponds to water. Measurements like this allow scientists to identify materials based on how strongly they refract light. The same principle is used in laboratory instruments, ophthalmology, and medical imaging systems that measure the optical properties of biological tissues.

Example 66.3: Refraction in Diamond

Light travels from air into a diamond with an incident angle of [latex]30.0^\circ[/latex]. Using the index of refraction of diamond, determine the angle of refraction.

Strategy

The index of refraction of air is approximately [latex]n_1=1.00[/latex]. From Table 66.1, diamond has an index of refraction of [latex]n_2=2.419[/latex]. Snell's law can therefore be solved for the refracted angle [latex]\theta_2[/latex].

Solution

Rearrange Snell's law to solve for the refracted angle:

[latex]\sin\theta_2= \frac{n_1}{n_2}\sin\theta_1.[/latex]

Substitute the known values:

[latex]\sin\theta_2 = \frac{1.00}{2.419} \sin30.0^\circ = (0.413)(0.500) = 0.207.[/latex]

Take the inverse sine:

[latex]\theta_2=\sin^{-1}(0.207)=11.9^\circ.[/latex]

Discussion

Because diamond has a much larger index of refraction than water, the light ray bends much more strongly toward the normal. For the same incident angle, the refracted angle in diamond is only [latex]11.9^\circ[/latex], compared with [latex]22.0^\circ[/latex] in water. Materials with large indices of refraction produce greater bending of light, which contributes to the brilliance and sparkle of gemstones and is an important consideration in the design of optical instruments.

Conceptual Questions

  1. Diffusion by reflection from a rough surface is described in this chapter. Light can also be diffused by refraction. Describe how this occurs in a specific situation, such as light interacting with crushed ice.
  2. Why is the index of refraction always greater than or equal to 1?
  3. Does the fact that the flash from lightning reaches you before its sound prove that the speed of light is extremely large, or simply that it is greater than the speed of sound? Discuss how you could use this effect to estimate the speed of light.
  4. Will light bend toward or away from the normal when it travels?
    • from air to water?
    • from water to glass?
    • from glass to air?
  5. Explain why an object under water appears to be at a shallower depth than it actually is. How does this optical illusion contribute to diving injuries in unfamiliar lakes or pools?
  6. Explain why a person's legs appear shorter when standing in a swimming pool. Support your explanation with a ray diagram showing the path of light from the feet to the eye of an observer standing outside the water.
  7. Why is the front surface of the thermometer shown in Figure 66.4 curved?
    A transparent thermometer with a curved front surface that acts as a magnifying lens, making the liquid column easier to read.
    Figure 66.4: The curved front surface of a liquid-in-glass thermometer magnifies the liquid column, making it easier to read accurately.
  8. Suppose light travels from air into a material with a negative index of refraction, such as [latex]n=-1.3[/latex]. Based on Snell's law, predict the direction of the refracted ray.

Problems & Exercises

  1. What is the speed of light in water? In glycerin?
  2. What is the speed of light in air? In crown glass?
  3. Calculate the index of refraction for a medium in which the speed of light is
    [latex]2.012\times10^{8}~\text{m/s}[/latex],
    and identify the most likely substance using Table 66.1.
  4. In what substance listed in Table 66.1 is the speed of light
    [latex]2.290\times10^{8}~\text{m/s}[/latex]?
  5. There was a major collision between an asteroid and the Moon during medieval times. Monks at Canterbury Cathedral in England described seeing a red glow on and around the Moon. The Moon is
    [latex]3.84\times10^5~\text{km}[/latex]
    from Earth. How long after the asteroid struck the Moon would the light first reach Earth?
  6. A scuba diver training in a pool looks at an instructor standing above the water, as shown in Figure 66.5. The angle between the light ray in the water and the normal to the surface is
    [latex]25.0^\circ[/latex].
    Determine the angle the light ray makes with the normal in the air.

    A scuba diver underwater and an instructor standing beside the pool observe each other. Light rays refract at the water surface, making each person appear at a different position than their actual location.
    Figure 66.5: A scuba diver and an instructor observe each other through the air-water interface.
  7. Components of some computers communicate using optical fibers with an index of refraction
    [latex]n=1.55[/latex].
    How many nanoseconds are required for a signal to travel
    [latex]0.200~\text{m}[/latex]
    through the fiber?
  8. Figure 66.5 illustrates a scuba diver and an instructor.
    1. Given that the angle between the light ray in the water and the normal is [latex]25.0^\circ[/latex], determine the height of the instructor's head above the water surface. (You must first determine the angle of refraction.)
    2. Determine the apparent depth of the diver's head below the water as seen by the instructor. Assume the diver and the apparent image are the same horizontal distance from the normal.
  9. An unknown transparent material is immersed in water. A light ray enters the material with an incident angle of
    [latex]45.0^\circ[/latex],
    and the refracted angle is
    [latex]40.3^\circ[/latex].
    Determine the index of refraction of the material and identify the most likely substance using Table 66.1.
  10. Lunar astronauts placed a corner reflector on the Moon that is periodically illuminated by laser pulses from Earth. Assuming the Moon is
    [latex]3.84\times10^8~\text{m}[/latex]
    away and Earth's atmosphere is equivalent to a
    [latex]30.0~\text{km}[/latex]
    layer with index of refraction
    [latex]n=1.000293[/latex],
    calculate the percentage correction that must be made to account for the slowing of light in the atmosphere.
  11. Figure 66.6 shows a light ray passing from air through a sheet of crown glass into water. If the incident angle is
    [latex]40.0^\circ[/latex]
    and the glass is
    [latex]1.00~\text{cm}[/latex]
    thick, calculate the lateral displacement
    [latex]\Delta x[/latex]
    of the ray.
  12. Figure 66.6 shows a light ray traveling through three different media. Show that the final angle
    [latex]\theta_3[/latex]
    is the same as it would be if the middle medium were absent (provided total internal reflection does not occur).

    A light ray passes from one medium into a second and then into a third. Although the ray is displaced sideways while passing through the middle layer, it emerges parallel to its original direction.
    Figure 66.6: A ray passing through a parallel-sided slab emerges parallel to its original direction but is laterally displaced by [latex]\Delta x[/latex].
  13. Unreasonable Results.
    Light travels from water into another substance with an angle of incidence of
    [latex]10.0^\circ[/latex]
    and an angle of refraction of
    [latex]14.9^\circ[/latex].

    1. Determine the index of refraction of the second substance.
    2. Explain what is unreasonable about the result.
    3. Identify which assumption or assumptions must be incorrect.
  14. Construct Your Own Problem.Consider sunlight entering Earth's atmosphere at sunrise or sunset, where the incident angle is
    [latex]90^\circ[/latex].
    Assuming the atmosphere has a sharp boundary, calculate the angle of refraction. Then develop a more realistic model in which the atmosphere consists of several layers with different refractive indices and determine how the refracted angle changes.
  15. Unreasonable Results.Light travels from water into a gemstone with an incident angle of
    [latex]80.0^\circ[/latex]
    and a refracted angle of
    [latex]15.2^\circ[/latex].

    1. Determine the speed of light in the gemstone.
    2. Explain what is unreasonable about the result.
    3. Identify which assumption or assumptions are inconsistent.

Glossary

refraction
the change in direction of a light ray as it passes from one medium to another because its speed changes
index of refraction
the ratio of the speed of light in a vacuum to the speed of light in a material, expressed as [latex]n=c/v[/latex]
definition

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Introductory Physics for the Health and Life Sciences II Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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