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Three stages in the evolution of the eye. The flatworm has two eye pits. The nautilus's eyes are pinhole cameras. The human eye incorporates a lens.

# Chapter 31. Refraction

Economists normally consider free markets to be the natural way of judging the monetary value of something, but social scientists also use questionnaires to gauge the relative value of privileges, disadvantages, or possessions that cannot be bought or sold. They ask people to imagine that they could trade one thing for another and ask which they would choose. One interesting result is that the average light-skinned person in the U.S. would rather lose an arm than suffer the racist treatment routinely endured by African-Americans. Even more impressive is the value of sight. Many prospective parents can imagine without too much fear having a deaf child, but would have a far more difficult time coping with raising a blind one.

So great is the value attached to sight that some have imbued it with mystical aspects. Joan of Arc saw visions, and my college has a “vision statement.” Christian fundamentalists who perceive a conflict between evolution and their religion have claimed that the eye is such a perfect device that it could never have arisen through a process as helter-skelter as evolution, or that it could not have evolved because half of an eye would be useless. In fact, the structure of an eye is fundamentally dictated by physics, and it has arisen separately by evolution somewhere between eight and 40 times, depending on which biologist you ask. We humans have a version of the eye that can be traced back to the evolution of a light-sensitive “eye spot” on the head of an ancient invertebrate. A sunken pit then developed so that the eye would only receive light from one direction, allowing the organism to tell where the light was coming from. (Modern flatworms have this type of eye.) The top of the pit then became partially covered, leaving a hole, for even greater directionality (as in the nautilus). At some point the cavity became filled with jelly, and this jelly finally became a lens, resulting in the general type of eye that we share with the bony fishes and other vertebrates. Far from being a perfect device, the vertebrate eye is marred by a serious design flaw due to the lack of planning or intelligent design in evolution: the nerve cells of the retina and the blood vessels that serve them are all in front of the light-sensitive cells, blocking part of the light. Squids and other molluscs, whose eyes evolved on a separate branch of the evolutionary tree, have a more sensible arrangement, with the light-sensitive cells out in front.

## 31.1 Refraction

a / A human eye.

b / The anatomy of the eye.

c / A simplified optical diagram of the eye. Light rays are bent when they cross from the air into the eye. (A little of the incident rays' energy goes into the reflected rays rather than the ones transmitted into the eye.)

d / The incident, reflected, and transmitted (refracted) rays all lie in a plane that includes the normal (dashed line).

e / The angles $$\theta_1$$ and $$\theta_2$$ are related to each other, and also depend on the properties of the two media. Because refraction is time-reversal symmetric, there is no need to label the rays with arrowheads.

f / Refraction has time-reversal symmetry. Regardless of whether the light is going into or out of the water, the relationship between the two angles is the same, and the ray is closer to the normal while in the water.

g / The relationship between the angles in refraction.

h / Example 1.

i / A mechanical model of refraction.

k / Total internal reflection in a fiber-optic cable.

l / A simplified drawing of a surgical endoscope. The first lens forms a real image at one end of a bundle of optical fibers. The light is transmitted through the bundle, and is finally magnified by the eyepiece.

m / Endoscopic images of a duodenal ulcer.

### Refraction

The fundamental physical phenomenon at work in the eye is that when light crosses a boundary between two media (such as air and the eye's jelly), part of its energy is reflected, but part passes into the new medium. In the ray model of light, we describe the original ray as splitting into a reflected ray and a transmitted one (the one that gets through the boundary). Of course the reflected ray goes in a direction that is different from that of the original one, according to the rules of reflection we have already studied. More surprisingly --- and this is the crucial point for making your eye focus light --- the transmitted ray is bent somewhat as well. This bending phenomenon is called refraction. The origin of the word is the same as that of the word “fracture,” i.e., the ray is bent or “broken.” (Keep in mind, however, that light rays are not physical objects that can really be “broken.”) Refraction occurs with all waves, not just light waves.

The actual anatomy of the eye, b, is quite complex, but in essence it is very much like every other optical device based on refraction. The rays are bent when they pass through the front surface of the eye, c. Rays that enter farther from the central axis are bent more, with the result that an image is formed on the retina. There is only one slightly novel aspect of the situation. In most human-built optical devices, such as a movie projector, the light is bent as it passes into a lens, bent again as it reemerges, and then reaches a focus beyond the lens. In the eye, however, the “screen” is inside the eye, so the rays are only refracted once, on entering the jelly, and never emerge again.

A common misconception is that the “lens” of the eye is what does the focusing. All the transparent parts of the eye are made of fairly similar stuff, so the dramatic change in medium is when a ray crosses from the air into the eye (at the outside surface of the cornea). This is where nearly all the refraction takes place. The lens medium differs only slightly in its optical properties from the rest of the eye, so very little refraction occurs as light enters and exits the lens. The lens, whose shape is adjusted by muscles attached to it, is only meant for fine-tuning the focus to form images of near or far objects.

### Refractive properties of media

What are the rules governing refraction? The first thing to observe is that just as with reflection, the new, bent part of the ray lies in the same plane as the normal (perpendicular) and the incident ray, d.

If you try shooting a beam of light at the boundary between two substances, say water and air, you'll find that regardless of the angle at which you send in the beam, the part of the beam in the water is always closer to the normal line, e. It doesn't matter if the ray is entering the water or leaving, so refraction is symmetric with respect to time-reversal, f.

If, instead of water and air, you try another combination of substances, say plastic and gasoline, again you'll find that the ray's angle with respect to the normal is consistently smaller in one and larger in the other. Also, we find that if substance A has rays closer to normal than in B, and B has rays closer to normal than in C, then A has rays closer to normal than C. This means that we can rank-order all materials according to their refractive properties. Isaac Newton did so, including in his list many amusing substances, such as “Danzig vitriol” and “a pseudo-topazius, being a natural, pellucid, brittle, hairy stone, of a yellow color.” Several general rules can be inferred from such a list:

• Vacuum lies at one end of the list. In refraction across the interface between vacuum and any other medium, the other medium has rays closer to the normal.
• Among gases, the ray gets closer to the normal if you increase the density of the gas by pressurizing it more.
• The refractive properties of liquid mixtures and solutions vary in a smooth and systematic manner as the proportions of the mixture are changed.
• Denser substances usually, but not always, have rays closer to the normal.

The second and third rules provide us with a method for measuring the density of an unknown sample of gas, or the concentration of a solution. The latter technique is very commonly used, and the CRC Handbook of Physics and Chemistry, for instance, contains extensive tables of the refractive properties of sugar solutions, cat urine, and so on.

### Snell's law

The numerical rule governing refraction was discovered by Snell, who must have collected experimental data something like what is shown on this graph and then attempted by trial and error to find the right equation. The equation he came up with was

$\begin{equation*} \frac{\sin\theta_1}{\sin\theta_2} = \text{constant} . \end{equation*}$

The value of the constant would depend on the combination of media used. For instance, any one of the data points in the graph would have sufficed to show that the constant was 1.3 for an air-water interface (taking air to be substance 1 and water to be substance 2).

Snell further found that if media A and B gave a constant $$K_{AB}$$ and media B and C gave a constant $$K_{BC}$$, then refraction at an interface between A and C would be described by a constant equal to the product, $$K_{AC}=K_{AB}K_{BC}$$. This is exactly what one would expect if the constant depended on the ratio of some number characterizing one medium to the number characteristic of the second medium. This number is called the index of refraction of the medium, written as $$n$$ in equations. Since measuring the angles would only allow him to determine the ratio of the indices of refraction of two media, Snell had to pick some medium and define it as having $$n=1$$. He chose to define vacuum as having $$n=1$$. (The index of refraction of air at normal atmospheric pressure is 1.0003, so for most purposes it is a good approximation to assume that air has $$n=1$$.) He also had to decide which way to define the ratio, and he chose to define it so that media with their rays closer to the normal would have larger indices of refraction. This had the advantage that denser media would typically have higher indices of refraction, and for this reason the index of refraction is also referred to as the optical density. Written in terms of indices of refraction, Snell's equation becomes

$\begin{equation*} \frac{\sin\theta_1}{\sin\theta_2} = \frac{n_2}{n_1} , \end{equation*}$

but rewriting it in the form

$\begin{equation*} n_1 \sin \theta_1=n_2 \sin \theta_2 \end{equation*}$

[relationship between angles of rays at the interface between media with indices of refraction $$n_1$$ and $$n_2$$; angles are defined with respect to the normal] makes us less likely to get the 1's and 2's mixed up, so this the way most people remember Snell's law. A few indices of refraction are given in the back of the book.

self-check:

(1) What would the graph look like for two substances with the same index of refraction?

(2) Based on the graph, when does refraction at an air-water interface change the direction of a ray most strongly?

##### Example 1: Finding an angle using Snell's law
$$\triangleright$$ A submarine shines its searchlight up toward the surface of the water. What is the angle $$\alpha$$ shown in the figure?

$$\triangleright$$ The tricky part is that Snell's law refers to the angles with respect to the normal. Forgetting this is a very common mistake. The beam is at an angle of $$30°$$ with respect to the normal in the water. Let's refer to the air as medium 1 and the water as 2. Solving Snell's law for $$\theta_1$$, we find

$\begin{equation*} \theta_1 = \sin^{-1}\left(\frac{n_2}{n_1}\sin\theta_2\right) . \end{equation*}$

As mentioned above, air has an index of refraction very close to 1, and water's is about 1.3, so we find $$\theta_1=40°$$. The angle $$\alpha$$ is therefore $$50°$$.

### The index of refraction is related to the speed of light.

What neither Snell nor Newton knew was that there is a very simple interpretation of the index of refraction. This may come as a relief to the reader who is taken aback by the complex reasoning involving proportionalities that led to its definition. Later experiments showed that the index of refraction of a medium was inversely proportional to the speed of light in that medium. Since $$c$$ is defined as the speed of light in vacuum, and $$n=1$$ is defined as the index of refraction of vacuum, we have

$\begin{equation*} n=\frac{c}{v} . \end{equation*}$

[$$n=$$ medium's index of refraction, $$v=$$ speed of light in that medium, $$c=$$ speed of light in a vacuum]

Many textbooks start with this as the definition of the index of refraction, although that approach makes the quantity's name somewhat of a mystery, and leaves students wondering why $$c/v$$ was used rather than $$v/c$$. It should also be noted that measuring angles of refraction is a far more practical method for determining $$n$$ than direct measurement of the speed of light in the substance of interest.

### A mechanical model of Snell's law

Why should refraction be related to the speed of light? The mechanical model shown in the figure may help to make this more plausible. Suppose medium 2 is thick, sticky mud, which slows down the car. The car's right wheel hits the mud first, causing the right side of the car to slow down. This will cause the car to turn to the right until is moves far enough forward for the left wheel to cross into the mud. After that, the two sides of the car will once again be moving at the same speed, and the car will go straight.

Of course, light isn't a car. Why should a beam of light have anything resembling a “left wheel” and “right wheel?” After all, the mechanical model would predict that a motorcycle would go straight, and a motorcycle seems like a better approximation to a ray of light than a car. The whole thing is just a model, not a description of physical reality.

j / A derivation of Snell's law.

### A derivation of Snell's law

However intuitively appealing the mechanical model may be, light is a wave, and we should be using wave models to describe refraction. In fact Snell's law can be derived quite simply from wave concepts. Figure j shows the refraction of a water wave. The water in the upper left part of the tank is shallower, so the speed of the waves is slower there, and their wavelengths is shorter. The reflected part of the wave is also very faintly visible.

In the close-up view on the right, the dashed lines are normals to the interface. The two marked angles on the right side are both equal to $$\theta_1$$, and the two on the left to $$\theta_2$$.

Trigonometry gives

\begin{align*} \sin \theta_1 &= \lambda_1/h \text{and} \\ \sin \theta_2 &= \lambda_2/h . \end{align*}

Eliminating $$h$$ by dividing the equations, we find

$\begin{equation*} \frac{\sin\theta_1}{\sin\theta_2}=\frac{\lambda_1}{\lambda_2} . \end{equation*}$

The frequencies of the two waves must be equal or else they would get out of step, so by $$v=f\lambda$$ we know that their wavelengths are proportional to their velocities. Combining $$\lambda\propto v$$ with $$v\propto 1/n$$ gives $$\lambda\propto 1/n$$, so we find

$\begin{equation*} \frac{\sin\theta_1}{\sin\theta_2}=\frac{n_2}{n_1} , \end{equation*}$

which is one form of Snell's law.

##### Example 2: Ocean waves near and far from shore

Ocean waves are formed by winds, typically on the open sea, and the wavefronts are perpendicular to the direction of the wind that formed them. At the beach, however, you have undoubtedly observed that waves tend come in with their wavefronts very nearly (but not exactly) parallel to the shoreline. This is because the speed of water waves in shallow water depends on depth: the shallower the water, the slower the wave. Although the change from the fast-wave region to the slow-wave region is gradual rather than abrupt, there is still refraction, and the wave motion is nearly perpendicular to the normal in the slow region.

### Color and refraction

In general, the speed of light in a medium depends both on the medium and on the wavelength of the light. Another way of saying it is that a medium's index of refraction varies with wavelength. This is why a prism can be used to split up a beam of white light into a rainbow. Each wavelength of light is refracted through a different angle.

### How much light is reflected, and how much is transmitted?

In chapter 20 we developed an equation for the percentage of the wave energy that is transmitted and the percentage reflected at a boundary between media. This was only done in the case of waves in one dimension, however, and rather than discuss the full three dimensional generalization it will be more useful to go into some qualitative observations about what happens. First, reflection happens only at the interface between two media, and two media with the same index of refraction act as if they were a single medium. Thus, at the interface between media with the same index of refraction, there is no reflection, and the ray keeps going straight. Continuing this line of thought, it is not surprising that we observe very little reflection at an interface between media with similar indices of refraction.

The next thing to note is that it is possible to have situations where no possible angle for the refracted ray can satisfy Snell's law. Solving Snell's law for $$\theta_2$$, we find

$\begin{equation*} \theta_2 = \sin^{-1}\left(\frac{n_1}{n_2}\sin\theta_1\right) , \end{equation*}$

and if $$n_1$$ is greater than $$n_2$$, then there will be large values of $$\theta_1$$ for which the quantity $$(n_1/n_2)\sin\theta$$ is greater than one, meaning that your calculator will flash an error message at you when you try to take the inverse sine. What can happen physically in such a situation? The answer is that all the light is reflected, so there is no refracted ray. This phenomenon is known as total internal reflection, and is used in the fiber-optic cables that nowadays carry almost all long-distance telephone calls. The electrical signals from your phone travel to a switching center, where they are converted from electricity into light. From there, the light is sent across the country in a thin transparent fiber. The light is aimed straight into the end of the fiber, and as long as the fiber never goes through any turns that are too sharp, the light will always encounter the edge of the fiber at an angle sufficiently oblique to give total internal reflection. If the fiber-optic cable is thick enough, one can see an image at one end of whatever the other end is pointed at.

Alternatively, a bundle of cables can be used, since a single thick cable is too hard to bend. This technique for seeing around corners is useful for making surgery less traumatic. Instead of cutting a person wide open, a surgeon can make a small “keyhole” incision and insert a bundle of fiber-optic cable (known as an endoscope) into the body.

Since rays at sufficiently large angles with respect to the normal may be completely reflected, it is not surprising that the relative amount of reflection changes depending on the angle of incidence, and is greatest for large angles of incidence.

##### Discussion Questions

What index of refraction should a fish have in order to be invisible to other fish?

Does a surgeon using an endoscope need a source of light inside the body cavity? If so, how could this be done without inserting a light bulb through the incision?

A denser sample of a gas has a higher index of refraction than a less dense sample (i.e., a sample under lower pressure), but why would it not make sense for the index of refraction of a gas to be proportional to density?

The earth's atmosphere gets thinner and thinner as you go higher in altitude. If a ray of light comes from a star that is below the zenith, what will happen to it as it comes into the earth's atmosphere?

Does total internal reflection occur when light in a denser medium encounters a less dense medium, or the other way around? Or can it occur in either case?

## 31.2 Lenses

Figures n/1 and n/2 show examples of lenses forming images. There is essentially nothing for you to learn about imaging with lenses that is truly new. You already know how to construct and use ray diagrams, and you know about real and virtual images. The concept of the focal length of a lens is the same as for a curved mirror. The equations for locating images and determining magnifications are of the same form. It's really just a question of flexing your mental muscles on a few examples. The following self-checks and discussion questions will get you started.

n / 1. A converging lens forms an image of a candle flame. 2. A diverging lens.

self-check:

(1) In figures n/1 and n/2, classify the images as real or virtual.

(2) Glass has an index of refraction that is greater than that of air. Consider the topmost ray in figure n/1. Explain why the ray makes a slight left turn upon entering the lens, and another left turn when it exits.

(3) If the flame in figure n/2 was moved closer to the lens, what would happen to the location of the image?

##### Discussion Questions

In figures n/1 and n/2, the front and back surfaces are parallel to each other at the center of the lens. What will happen to a ray that enters near the center, but not necessarily along the axis of the lens? Draw a BIG ray diagram, and show a ray that comes from off axis.

In discussion questions B-F, don't draw ultra-detailed ray diagrams as in A.

Suppose you wanted to change the setup in figure n/1 so that the location of the actual flame in the figure would instead be occupied by an image of a flame. Where would you have to move the candle to achieve this? What about in n/2?

There are three qualitatively different types of image formation that can occur with lenses, of which figures n/1 and n/2 exhaust only two. Figure out what the third possibility is. Which of the three possibilities can result in a magnification greater than one? Cf. problem 4, p. 835.

Classify the examples shown in figure o according to the types of images delineated in discussion question C.

In figures n/1 and n/2, the only rays drawn were those that happened to enter the lenses. Discuss this in relation to figure o.

In the right-hand side of figure o, the image viewed through the lens is in focus, but the side of the rose that sticks out from behind the lens is not. Why?

o / Two images of a rose created by the same lens and recorded with the same camera.

## 31.3 The lensmaker's equation (optional)

p / The radii of curvature appearing in the lensmaker's equation.

The focal length of a spherical mirror is simply $$r/2$$, but we cannot expect the focal length of a lens to be given by pure geometry, since it also depends on the index of refraction of the lens. Suppose we have a lens whose front and back surfaces are both spherical. (This is no great loss of generality, since any surface with a sufficiently shallow curvature can be approximated with a sphere.) Then if the lens is immersed in a medium with an index of refraction of 1, its focal length is given approximately by

$\begin{equation*} f = \left[(n-1)\left|\frac{1}{r_1}\pm\frac{1}{r_2}\right|\right]^{-1} , \end{equation*}$

where $$n$$ is the index of refraction and $$r_1$$ and $$r_2$$ are the radii of curvature of the two surfaces of the lens. This is known as the lensmaker's equation. In my opinion it is not particularly worthy of memorization. The positive sign is used when both surfaces are curved outward or both are curved inward; otherwise a negative sign applies. The proof of this equation is left as an exercise to those readers who are sufficiently brave and motivated.

## 31.4 Dispersion

q / Dispersion of white light by a prism. White light is a mixture of all the wavelengths of the visible spectrum. Waves of different wavelengths undergo different amounts of refraction.

For most materials, we observe that the index of refraction depends slightly on wavelength, being highest at the blue end of the visible spectrum and lowest at the red. For example, white light disperses into a rainbow when it passes through a prism, q. Even when the waves involved aren't light waves, and even when refraction isn't of interest, the dependence of wave speed on wavelength is referred to as dispersion. Dispersion inside spherical raindrops is responsible for the creation of rainbows in the sky, and in an optical instrument such as the eye or a camera it is responsible for a type of aberration called chromatic aberration (section 30.3 and problem 2). As we'll see in section 35.2, dispersion causes a wave that is not a pure sine wave to have its shape distorted as it travels, and also causes the speed at which energy and information are transported by the wave to be different from what one might expect from a naive calculation.

## 31.5 The principle of least time for refraction (optional)

r / The principle of least time applied to refraction.

We have seen previously how the rules governing straight-line motion of light and reflection of light can be derived from the principle of least time. What about refraction? In the figure, it is indeed plausible that the bending of the ray serves to minimize the time required to get from a point A to point B. If the ray followed the unbent path shown with a dashed line, it would have to travel a longer distance in the medium in which its speed is slower. By bending the correct amount, it can reduce the distance it has to cover in the slower medium without going too far out of its way. It is true that Snell's law gives exactly the set of angles that minimizes the time required for light to get from one point to another. The proof of this fact is left as an exercise (problem 9, p. 875).

## 31.6 Case study: the eye of the jumping spider (optional)

Figure s shows an exceptionally cute jumping spider. The jumping spider does not build a web. It stalks its prey like a cat, so it needs excellent eyesight. In some ways, its visual system is more sophisticated and more functional than that of a human, illustrating how evolution does not progress systematically toward “higher” forms of life.

s / Top left: A female jumping spider, Phidippus mystaceus. Top right: Cross-section in a horizontal plane, viewed from above, of the jumping spider Metaphidippus aeneolus. The eight eyes are shown in white. Bottom: Close-up of one of the large principal eyes.

One way in which the spider outdoes us is that it has eight eyes to our two. (Each eye is simple, not compound like that of a fly.) The reason this works well has to do with the trade-off between magnification and field of view. The elongated principal eyes at the front of the head have a large value of $$d_i$$, resulting in a large magnification $$M=d_i/d_o$$. This high magnification is used for sophisticated visual tasks like distinguishing prey from a potential mate. (The pretty stripes on the legs in the photo are probably evolved to aid in making this distinction, which is a crucial one on a Saturday night.) As always with a high magnification, this results in a reduction in the field of view: making the image bigger means reducing the amount of the potential image that can actually fit on the retina. The animal has tunnel vision in these forward eyes. To allow it to glimpse prey from other angles, it has the additional eyes on the sides of its head. These are not elongated, and the smaller $$d_i$$ gives a smaller magnification but a larger field of view. When the spider sees something moving in these eyes, it turns its body so that it can take a look with the front eyes. The tiniest pair of eyes are too small to be useful. These vestigial organs, like the maladaptive human appendix, are an example of the tendency of evolution to produce unfortunate accidents due to the lack of intelligent design. The use of multiple eyes for these multiple purposes is far superior to the two-eye arrangement found in humans, octopi, etc., especially because of its compactness. If the spider had only two spherical eyes, they would have to have the same front-to-back dimension in order to produce the same acuity, but then the eyes would take up nearly all of the front of the head.

Another beautiful feature of these eyes is that they will never need bifocals. A human eye uses muscles to adjust for seeing near and far, varying $$f$$ in order to achive a fixed $$d_i$$ for differing values of $$d_o$$. On older models of H. sap., this poorly engineered feature is usually one of the first things to break down. The spider's front eyes have muscles, like a human's, that rotate the tube, but none that vary $$f$$, which is fixed. However, the retina consists of four separate layers at slightly different values of $$d_i$$. The figure only shows the detailed cellular structure of the rearmost layer, which is the most acute. Depending on $$d_o$$, the image may lie closest to any one of the four layers, and the spider can then use that layer to get a well-focused view. The layering is also believed to help eliminate problems caused by the variation of the index of refraction with wavelength (cf. problem 2, p. 874).

Although the spider's eye is different in many ways from a human's or an octopus's, it shares the same fundamental construction, being essentially a lens that forms a real image on a screen inside a darkened chamber. From this perspective, the main difference is simply the scale, which is miniaturized by about a factor of $$10^2$$ in the linear dimensions. How far down can this scaling go? Does an amoeba or a white blood cell lack an eye merely because it doesn't have a nervous system that could make sense of the signals? In fact there is an optical limit on the miniaturization of any eye or camera. The spider's eye is already so small that on the scale of the bottom panel in figure s, one wavelength of visible light would be easily distinguishable --- about the length of the comma in this sentence. Chapter 32 is about optical effects that occur when the wave nature of light is important, and problem 14 on p. 900 specifically addresses the effect on this spider's vision.

## Vocabulary

refraction — the change in direction that occurs when a wave encounters the interface between two media

index of refraction — an optical property of matter; the speed of light in a vacuum divided by the speed of light in the substance in question

## Notation

$$n$$ — the index of refraction

## Summary

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Refraction is a change in direction that occurs when a wave encounters the interface between two media. Together, refraction and reflection account for the basic principles behind nearly all optical devices.

Snell discovered the equation for refraction,

$\begin{equation*} n_1 \sin \theta_1=n_2 \sin \theta_2 , \end{equation*}$

[angles measured with respect to the normal] through experiments with light rays, long before light was proven to be a wave. Snell's law can be proven based on the geometrical behavior of waves. Here $$n$$ is the index of refraction. Snell invented this quantity to describe the refractive properties of various substances, but it was later found to be related to the speed of light in the substance,

$\begin{equation*} n = \frac{c}{v} , \end{equation*}$

where $$c$$ is the speed of light in a vacuum. In general a material's index of refraction is different for different wavelengths of light.

As discussed in chapter 20, any wave is partially transmitted and partially reflected at the boundary between two media in which its speeds are different. It is not particularly important to know the equation that tells what fraction is transmitted (and thus refracted), but important technologies such as fiber optics are based on the fact that this fraction becomes zero for sufficiently oblique angles. This phenomenon is referred to as total internal reflection. It occurs when there is no angle that satisfies Snell's law.

## Homework Problems

t / Problem 6.

u / Problem 8.

v / Problem 13.

x / Problem 20.

1. Suppose a converging lens is constructed of a type of plastic whose index of refraction is less than that of water. How will the lens's behavior be different if it is placed underwater?

2. There are two main types of telescopes, refracting (using lenses) and reflecting (using mirrors). (Some telescopes use a mixture of the two types of elements: the light first encounters a large curved mirror, and then goes through an eyepiece that is a lens.) What implications would the color-dependence of focal length have for the relative merits of the two types of telescopes? Describe the case where an image is formed of a white star. You may find it helpful to draw a ray diagram.

3. Based on Snell's law, explain why rays of light passing through the edges of a converging lens are bent more than rays passing through parts closer to the center. It might seem like it should be the other way around, since the rays at the edge pass through less glass --- shouldn't they be affected less? In your answer:

• Include a ray diagram showing a huge close-up view of the relevant part of the lens.
• Make use of the fact that the front and back surfaces aren't always parallel; a lens in which the front and back surfaces are always parallel doesn't focus light at all, so if your explanation doesn't make use of this fact, your argument must be incorrect.
• Make sure your argument still works even if the rays don't come in parallel to the axis.

4. When you take pictures with a camera, the distance between the lens and the film has to be adjusted, depending on the distance at which you want to focus. This is done by moving the lens. If you want to change your focus so that you can take a picture of something farther away, which way do you have to move the lens? Explain using ray diagrams. [Based on a problem by Eric Mazur.]

5. (a) Light is being reflected diffusely from an object 1.000 m underwater. The light that comes up to the surface is refracted at the water-air interface. If the refracted rays all appear to come from the same point, then there will be a virtual image of the object in the water, above the object's actual position, which will be visible to an observer above the water. Consider three rays, A, B and C, whose angles in the water with respect to the normal are $$\theta_i=0.000°$$, $$1.000°$$ and $$20.000°$$ respectively. Find the depth of the point at which the refracted parts of A and B appear to have intersected, and do the same for A and C. Show that the intersections are at nearly the same depth, but not quite. [Check: The difference in depth should be about 4 cm.]

(b) Since all the refracted rays do not quite appear to have come from the same point, this is technically not a virtual image. In practical terms, what effect would this have on what you see?

(c) In the case where the angles are all small, use algebra and trig to show that the refracted rays do appear to come from the same point, and find an equation for the depth of the virtual image. Do not put in any numerical values for the angles or for the indices of refraction --- just keep them as symbols. You will need the approximation $$\sin\theta\approx \tan\theta\approx \theta$$, which is valid for small angles measured in radians.

6. The drawing shows the anatomy of the human eye, at twice life size. Find the radius of curvature of the outer surface of the cornea by measurements on the figure, and then derive the focal length of the air-cornea interface, where almost all the focusing of light occurs. You will need to use physical reasoning to modify the lensmaker's equation for the case where there is only a single refracting surface. Assume that the index of refraction of the cornea is essentially that of water.

7. When swimming underwater, why is your vision made much clearer by wearing goggles with flat pieces of glass that trap air behind them? [Hint: You can simplify your reasoning by considering the special case where you are looking at an object far away, and along the optic axis of the eye.]

8. The figure shows four lenses. Lens 1 has two spherical surfaces. Lens 2 is the same as lens 1 but turned around. Lens 3 is made by cutting through lens 1 and turning the bottom around. Lens 4 is made by cutting a central circle out of lens 1 and recessing it.

(a) A parallel beam of light enters lens 1 from the left, parallel to its axis. Reasoning based on Snell's law, will the beam emerging from the lens be bent inward, or outward, or will it remain parallel to the axis? Explain your reasoning. As part of your answer, make a huge drawing of one small part of the lens, and apply Snell's law at both interfaces. Recall that rays are bent more if they come to the interface at a larger angle with respect to the normal.

(b) What will happen with lenses 2, 3, and 4? Explain. Drawings are not necessary.

9. Prove that the principle of least time leads to Snell's law.

10. (answer check available at lightandmatter.com) An object is more than one focal length from a converging lens. (a) Draw a ray diagram. (b) Using reasoning like that developed in chapter 30, determine the positive and negative signs in the equation $$1/f=\pm1/d_i\pm1/d_o$$. (c) The images of the rose in section 4.2 were made using a lens with a focal length of 23 cm. If the lens is placed 80 cm from the rose, locate the image.

11. (answer check available at lightandmatter.com) An object is less than one focal length from a converging lens. (a) Draw a ray diagram. (b) Using reasoning like that developed in chapter 30, determine the positive and negative signs in the equation $$1/f=\pm1/d_i\pm1/d_o$$. (c) The images of the rose in section 4.2 were made using a lens with a focal length of 23 cm. If the lens is placed 10 cm from the rose, locate the image.

12. (answer check available at lightandmatter.com) Nearsighted people wear glasses whose lenses are diverging. (a) Draw a ray diagram. For simplicity pretend that there is no eye behind the glasses. (b) Using reasoning like that developed in chapter 30, determine the positive and negative signs in the equation $$1/f=\pm1/d_i\pm1/d_o$$. (c) If the focal length of the lens is 50.0 cm, and the person is looking at an object at a distance of 80.0 cm, locate the image.

13. (solution in the pdf version of the book) Two standard focal lengths for camera lenses are 50 mm (standard) and 28 mm (wide-angle). To see how the focal lengths relate to the angular size of the field of view, it is helpful to visualize things as represented in the figure. Instead of showing many rays coming from the same point on the same object, as we normally do, the figure shows two rays from two different objects. Although the lens will intercept infinitely many rays from each of these points, we have shown only the ones that pass through the center of the lens, so that they suffer no angular deflection. (Any angular deflection at the front surface of the lens is canceled by an opposite deflection at the back, since the front and back surfaces are parallel at the lens's center.) What is special about these two rays is that they are aimed at the edges of one 35-mm-wide frame of film; that is, they show the limits of the field of view. Throughout this problem, we assume that $$d_o$$ is much greater than $$d_i$$. (a) Compute the angular width of the camera's field of view when these two lenses are used. (b) Use small-angle approximations to find a simplified equation for the angular width of the field of view, $$\theta$$, in terms of the focal length, $$f$$, and the width of the film, $$w$$. Your equation should not have any trig functions in it. Compare the results of this approximation with your answers from part a. (c) Suppose that we are holding constant the aperture (amount of surface area of the lens being used to collect light). When switching from a 50-mm lens to a 28-mm lens, how many times longer or shorter must the exposure be in order to make a properly developed picture, i.e., one that is not under- or overexposed? [Based on a problem by Arnold Arons.]

14. A nearsighted person is one whose eyes focus light too strongly, and who is therefore unable to relax the lens inside her eye sufficiently to form an image on her retina of an object that is too far away.

(a) Draw a ray diagram showing what happens when the person tries, with uncorrected vision, to focus at infinity.

(b) What type of lenses do her glasses have? Explain.

(c) Draw a ray diagram showing what happens when she wears glasses. Locate both the image formed by the glasses and the final image.

(d) Suppose she sometimes uses contact lenses instead of her glasses. Does the focal length of her contacts have to be less than, equal to, or greater than that of her glasses? Explain.

15. Diamond has an index of refraction of 2.42, and part of the reason diamonds sparkle is that this encourages a light ray to undergo many total internal reflections before it emerges. (a) Calculate the critical angle at which total internal reflection occurs in diamond. (answer check available at lightandmatter.com) (b) Explain the interpretation of your result: Is it measured from the normal, or from the surface? Is it a minimum, or a maximum? How would the critical angle have been different for a substance such as glass or plastic, with a lower index of refraction?

16. Fred's eyes are able to focus on things as close as 5.0 cm. Fred holds a magnifying glass with a focal length of 3.0 cm at a height of 2.0 cm above a flatworm. (a) Locate the image, and find the magnification. (b) Without the magnifying glass, from what distance would Fred want to view the flatworm to see its details as well as possible? With the magnifying glass? (c) Compute the angular magnification.

w / Problem 17.

17. Panel 1 of the figure shows the optics inside a pair of binoculars. They are essentially a pair of telescopes, one for each eye. But to make them more compact, and allow the eyepieces to be the right distance apart for a human face, they incorporate a set of eight prisms, which fold the light path. In addition, the prisms make the image upright. Panel 2 shows one of these prisms, known as a Porro prism. The light enters along a normal, undergoes two total internal reflections at angles of 45 degrees with respect to the back surfaces, and exits along a normal. The image of the letter R has been flipped across the horizontal. Panel 3 shows a pair of these prisms glued together. The image will be flipped across both the horizontal and the vertical, which makes it oriented the right way for the user of the binoculars.
(a) Find the minimum possible index of refraction for the glass used in the prisms.
(b) For a material of this minimal index of refraction, find the fraction of the incoming light that will be lost to reflection in the four Porro prisms on a each side of a pair of binoculars. (See ch. 20.) In real, high-quality binoculars, the optical surfaces of the prisms have antireflective coatings, but carry out your calculation for the case where there is no such coating.
(c) Discuss the reasons why a designer of binoculars might or might not want to use a material with exactly the index of refraction found in part a.

18. It would be annoying if your eyeglasses produced a magnified or reduced image. Prove that when the eye is very close to a lens, and the lens produces a virtual image, the angular magnification is always approximately equal to 1 (regardless of whether the lens is diverging or converging).

19. A typical mirror consists of a pane of glass of thickness $$t$$ and index of refraction $$n$$, “silvered” on the back with a reflective coating. Let $$d_o$$ and $$d_i$$ be measured from the back of the mirror. Show that $$d_i=d_o- 2(1-1/n)t$$. Use the result of, and make the approximation employed in, problem 5c. As a check on your result, consider separately the special values of $$n$$ and $$t$$ that would recover the case without any glass.

20. The figure shows a lens with surfaces that are curved, but whose thickness is constant along any horizontal line. Use the lensmaker's equation to prove that this “lens” is not really a lens at all.(solution in the pdf version of the book)

21. Estimate the radii of curvature of the two optical surfaces in the eye of the jumping spider in in figure s on p. 870. Use physical reasoning to modify the lensmaker's equation for a case like this one, in which there are three indices of refraction $$n_1$$ (air), $$n_2$$ (lens), and $$n_3$$ (the material behind the lens), and $$n_1 \ne n_3$$. Show that the interface between $$n_2$$ and $$n_3$$ contributes negligibly to focusing, and verify that the image is produced at approximately the right place in the eye when the object is far away. As a check on your result, direct optical measurements by M.F. Land in 1969 gave $$f=512\ \mu\text{m}$$.(answer check available at lightandmatter.com)

This exercise was created by Dan MacIsaac.

Equipment:

eyeglasses

diverging lenses for students who don't wear glasses, or who use converging glasses

rulers and metersticks

scratch paper

marking pens

Most people who wear glasses have glasses whose lenses are diverging, which allows them to focus on objects far away. Such a lens cannot form a real image, so its focal length cannot be measured as easily as that of an converging lens. In this exercise you will determine the focal length of your own glasses by taking them off, holding them at a distance from your face, and looking through them at a set of parallel lines on a piece of paper. The lines will be reduced (the lens's magnification is less than one), and by adjusting the distance between the lens and the paper, you can make the magnification equal 1/2 exactly, so that two spaces between lines as seen through the lens fit into one space as seen simultaneously to the side of the lens. This object distance can be used in order to find the focal length of the lens.

1. Does this technique really measure magnification or does it measure angular magnification? What can you do in your experiment in order to make these two quantities nearly the same, so the math is simpler?

2. Before taking any numerical data, use algebra to find the focal length of the lens in terms of $$d_o$$, the object distance that results in a magnification of 1/2.

3. Use a marker to draw three evenly spaced parallel lines on the paper. (A spacing of a few cm works well.) Measure the object distance that results in a magnification of 1/2, and determine the focal length of your lens.

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(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.