$$\require{cancel}$$

# 13.3: Images, Quantitatively

• • Contributed by Benjamin Crowell
• Professor (Physics) at Fullerton College

It sounds a bit odd when a scientist refers to a theory as “beautiful,” but to those in the know it makes perfect sense. One mark of a beautiful theory is that it surprises us by being simple. The mathematical theory of lenses and curved mirrors gives us just such a surprise. We expect the subject to be complex because there are so many cases: a converging mirror forming a real image, a diverging lens that makes a virtual image, and so on for a total of six possibilities. If we want to predict the location of the images in all these situations, we might expect to need six different equations, and six more for predicting magnifications. Instead, it turns out that we can use just one equation for the location of the image and one equation for its magnification, and these two equations work in all the different cases with no changes except for plus and minus signs. This is the kind of thing the physicist Eugene Wigner referred to as “the unreasonable effectiveness of mathematics.” Sometimes we can find a deeper reason for this kind of unexpected simplicity, but sometimes it almost seems as if God went out of Her way to make the secrets of universe susceptible to attack by the human thought-tool called math.

# 12.3.1 A real image formed by a converging mirror

Location of the image

We will now derive the equation for the location of a real image formed by a converging mirror. We assume for simplicity that the mirror is spherical, but actually this isn't a restrictive assumption, because any shallow, symmetric curve can be approximated by a sphere. The shape of the mirror can be specified by giving the location of its center, C. A deeply curved mirror is a sphere with a small radius, so C is close to it, while a weakly curved mirror has C farther away. Given the point O where the object is, we wish to find the point I where the image will be formed. a / The relationship between the object's position and the image's can be expressed in terms of the angles $$\theta_o$$ and $$\theta_i$$.

To locate an image, we need to track a minimum of two rays coming from the same point. Since we have proved in the previous chapter that this type of image is not distorted, we can use an on-axis point, O, on the object, as in figure a/1. The results we derive will also hold for off-axis points, since otherwise the image would have to be distorted, which we know is not true. We let one of the rays be the one that is emitted along the axis; this ray is especially easy to trace, because it bounces straight back along the axis again. As our second ray, we choose one that strikes the mirror at a distance of 1 from the axis. “One what?” asks the astute reader. The answer is that it doesn't really matter. When a mirror has shallow curvature, all the reflected rays hit the same point, so 1 could be expressed in any units you like. It could, for instance, be 1 cm, unless your mirror is smaller than 1 cm!

The only way to find out anything mathematical about the rays is to use the sole mathematical fact we possess concerning specular reflection: the incident and reflected rays form equal angles with respect to the normal, which is shown as a dashed line. Therefore the two angles shown in figure a/2 are the same, and skipping some straightforward geometry, this leads to the visually reasonable result that the two angles in figure a/3 are related as follows:

$\begin{equation*} \theta_i+\theta_o = \text{constant} \end{equation*}$

(Note that $$\theta_i$$ and $$\theta_o$$, which are measured from the image and the object, not from the eye like the angles we referred to in discussing angular magnification on page 754.) For example, move O farther from the mirror. The top angle in figure a/2 is increased, so the bottom angle must increase by the same amount, causing the image point, I, to move closer to the mirror. In terms of the angles shown in figure a/3, the more distant object has resulted in a smaller angle $$\theta_o$$, while the closer image corresponds to a larger $$\theta_i;$$ One angle increases by the same amount that the other decreases, so their sum remains constant. These changes are summarized in figure a/4. b / The geometrical interpretation of the focal angle.

The sum $$\theta_i+\theta_o$$ is a constant. What does this constant represent? Geometrically, we interpret it as double the angle made by the dashed radius line. Optically, it is a measure of the strength of the mirror, i.e., how strongly the mirror focuses light, and so we call it the focal angle, $$\theta_f$$,

$\begin{equation*} \theta_i+\theta_o = \theta_f . \end{equation*}$

Suppose, for example, that we wish to use a quick and dirty optical test to determine how strong a particular mirror is. We can lay it on the floor as shown in figure c, and use it to make an image of a lamp mounted on the ceiling overhead, which we assume is very far away compared to the radius of curvature of the mirror, so that the mirror intercepts only a very narrow cone of rays from the lamp. This cone is so narrow that its rays are nearly parallel, and $$\theta_o$$ is nearly zero. The real image can be observed on a piece of paper. By moving the paper nearer and farther, we can bring the image into focus, at which point we know the paper is located at the image point. Since $$\theta_o\approx 0$$, we have $$\theta_i\approx \theta_f$$, and we can then determine this mirror's focal angle either by measuring $$\theta_i$$ directly with a protractor, or indirectly via trigonometry. A strong mirror will bring the rays together to form an image close to the mirror, and these rays will form a blunt-angled cone with a large $$\theta_i$$ and $$\theta_f$$.

Example 4: An alternative optical test c / Example 4, an alternative test for finding the focal angle. The mirror is the same as in figure b.

$$\triangleright$$ Figure c shows an alternative optical test. Rather than placing the object at infinity as in figure b, we adjust it so that the image is right on top of the object. Points O and I coincide, and the rays are reflected right back on top of themselves. If we measure the angle $$\theta$$ shown in figure c, how can we find the focal angle?

$$\triangleright$$ The object and image angles are the same; the angle labeled $$\theta$$ in the figure equals both of them. We therefore have $$\theta_i+\theta_o=\theta =\theta_f$$. Comparing figures b and c, it is indeed plausible that the angles are related by a factor of two.

At this point, we could consider our work to be done. Typically, we know the strength of the mirror, and we want to find the image location for a given object location. Given the mirror's focal angle and the object location, we can determine $$\theta_o$$ by trigonometry, subtract to find $$\theta_i=\theta_f-\theta_o$$, and then do more trig to find the image location.

There is, however, a shortcut that can save us from doing so much work. Figure a/3 shows two right triangles whose legs of length 1 coincide and whose acute angles are $$\theta_o$$ and $$\theta_i$$. These can be related by trigonometry to the object and image distances shown in figure d:

$\begin{equation*} \tan \theta_o = 1/d_o \tan \theta_i = 1/d_i \end{equation*}$ d / The object and image distances.

Ever since chapter 2, we've been assuming small angles. For small angles, we can use the small-angle approximation $$\tan x\approx x$$ (for $$x$$ in radians), giving simply

$\begin{equation*} \theta_o = 1/d_o \theta_i = 1/d_i . \end{equation*}$

We likewise define a distance called the focal length, $$f$$ according to $$\theta_f=1/f$$. In figure b, $$f$$ is the distance from the mirror to the place where the rays cross. We can now reexpress the equation relating the object and image positions as

$\begin{equation*} \frac{1}{f} = \frac{1}{d_i}+\frac{1}{d_o} . \end{equation*}$

Figure e summarizes the interpretation of the focal length and focal angle.1 e / Mirror 1 is weaker than mirror 2. It has a shallower curvature, a longer focal length, and a smaller focal angle. It reflects rays at angles not much different than those that would be produced with a flat mirror.

Which form is better, $$\theta_f=\theta_i+\theta_o$$ or $$1/f=1/d_i+1/d_o?$$ The angular form has in its favor its simplicity and its straightforward visual interpretation, but there are two reasons why we might prefer the second version. First, the numerical values of the angles depend on what we mean by “one unit” for the distance shown as 1 in figure a/1. Second, it is usually easier to measure distances rather than angles, so the distance form is more convenient for number crunching. Neither form is superior overall, and we will often need to use both to solve any given problem.2

Example 5: A searchlight

Suppose we need to create a parallel beam of light, as in a searchlight. Where should we place the lightbulb? A parallel beam has zero angle between its rays, so $$\theta_i=0$$. To place the lightbulb correctly, however, we need to know a distance, not an angle: the distance $$d_o$$ between the bulb and the mirror. The problem involves a mixture of distances and angles, so we need to get everything in terms of one or the other in order to solve it. Since the goal is to find a distance, let's figure out the image distance corresponding to the given angle $$\theta_i=0$$. These are related by $$d_i=1/\theta_i$$, so we have $$d_i=\infty$$. (Yes, dividing by zero gives infinity. Don't be afraid of infinity. Infinity is a useful problem-solving device.) Solving the distance equation for $$d_o$$, we have

\begin{align*} d_o &= (1/f - 1/d_i)^{-1} \\ &= (1/f - 0)^{-1} \\ &= f \end{align*}

The bulb has to be placed at a distance from the mirror equal to its focal point.

Example 6: Diopters

An equation like $$d_i=1/\theta_i$$ really doesn't make sense in terms of units. Angles are unitless, since radians aren't really units, so the right-hand side is unitless. We can't have a left-hand side with units of distance if the right-hand side of the same equation is unitless. This is an artifact of my cavalier statement that the conical bundles of rays spread out to a distance of 1 from the axis where they strike the mirror, without specifying the units used to measure this 1. In real life, optometrists define the thing we're calling $$\theta_i=1/d_i$$ as the “dioptric strength” of a lens or mirror, and measure it in units of inverse meters ($$\text{m}^{-1}$$), also known as diopters (1 D=1 $$\text{m}^{-1}$$).

## Magnification

We have already discussed in the previous chapter how to find the magnification of a virtual image made by a curved mirror. The result is the same for a real image, and we omit the proof, which is very similar. In our new notation, the result is $$M=d_i/d_o$$. A numerical example is given in subsection 12.3.2.

# 12.3.2 Other cases with curved mirrors

The equation $$d_i=(1/f-1/d_o)^{-1}$$ can easily produce a negative result, but we have been thinking of $$d_i$$ as a distance, and distances can't be negative. A similar problem occurs with $$\theta_i=\theta_f-\theta_o$$ for $$\theta_o>\theta_f$$. What's going on here?

The interpretation of the angular equation is straightforward. As we bring the object closer and closer to the image, $$\theta_o$$ gets bigger and bigger, and eventually we reach a point where $$\theta_o=\theta_f$$ and $$\theta_i=0$$. This large object angle represents a bundle of rays forming a cone that is very broad, so broad that the mirror can no longer bend them back so that they reconverge on the axis. The image angle $$\theta_i=0$$ represents an outgoing bundle of rays that are parallel. The outgoing rays never cross, so this is not a real image, unless we want to be charitable and say that the rays cross at infinity. If we go on bringing the object even closer, we get a virtual image. f / A graph of the image distance $$d_i$$ as a function of the object distance $$d_o$$.

To analyze the distance equation, let's look at a graph of $$d_i$$ as a function of $$d_o$$. The branch on the upper right corresponds to the case of a real image. Strictly speaking, this is the only part of the graph that we've proven corresponds to reality, since we never did any geometry for other cases, such as virtual images. As discussed in the previous section, making $$d_o$$ bigger causes $$d_i$$ to become smaller, and vice-versa.

Letting $$d_o$$ be less than $$f$$ is equivalent to $$\theta_o>\theta_f:$$ a virtual image is produced on the far side of the mirror. This is the first example of Wigner's “unreasonable effectiveness of mathematics” that we have encountered in optics. Even though our proof depended on the assumption that the image was real, the equation we derived turns out to be applicable to virtual images, provided that we either interpret the positive and negative signs in a certain way, or else modify the equation to have different positive and negative signs.

self-check:

Interpret the three places where, in physically realistic parts of the graph, the graph approaches one of the dashed lines. [This will come more naturally if you have learned the concept of limits in a math class.]

Example 7: A flat mirror

We can even apply the equation to a flat mirror. As a sphere gets bigger and bigger, its surface is more and more gently curved. The planet Earth is so large, for example, that we cannot even perceive the curvature of its surface. To represent a flat mirror, we let the mirror's radius of curvature, and its focal length, become infinite. Dividing by infinity gives zero, so we have

\begin{align*} 1/d_o &= -1/d_i ,\\ \text{or}\\ d_o &= -d_i . \end{align*}

If we interpret the minus sign as indicating a virtual image on the far side of the mirror from the object, this makes sense.

It turns out that for any of the six possible combinations of real or virtual images formed by converging or diverging lenses or mirrors, we can apply equations of the form

$\begin{gather*} \theta_f = \theta_i+\theta_o \\ \text{and} \frac{1}{f} = \frac{1}{d_i}+\frac{1}{d_o} , \end{gather*}$

with only a modification of plus or minus signs. There are two possible approaches here. The approach we have been using so far is the more popular approach in American textbooks: leave the equation the same, but attach interpretations to the resulting negative or positive values of the variables. The trouble with this approach is that one is then forced to memorize tables of sign conventions, e.g., that the value of $$d_i$$ should be negative when the image is a virtual image formed by a converging mirror. Positive and negative signs also have to be memorized for focal lengths. Ugh! It's highly unlikely that any student has ever retained these lengthy tables in his or her mind for more than five minutes after handing in the final exam in a physics course. Of course one can always look such things up when they are needed, but the effect is to turn the whole thing into an exercise in blindly plugging numbers into formulas.

As you have gathered by now, there is another method which I think is better, and which I'll use throughout the rest of this book. In this method, all distances and angles are positive by definition, and we put in positive and negative signs in the equations depending on the situation. (I thought I was the first to invent this method, but I've been told that this is known as the European sign convention, and that it's fairly common in Europe.) Rather than memorizing these signs, we start with the generic equations

\begin{align*} \theta_f &= \pm \theta_i \pm \theta_o \\ \frac{1}{f} &= \pm\frac{1}{d_i}\pm\frac{1}{d_o} , \end{align*}

and then determine the signs by a two-step method that depends on ray diagrams. There are really only two signs to determine, not four; the signs in the two equations match up in the way you'd expect. The method is as follows:

1. Use ray diagrams to decide whether $$\theta_o$$ and $$\theta_i$$ vary in the same way or in opposite ways. (In other words, decide whether making $$\theta_o$$ greater results in a greater value of $$\theta_i$$ or a smaller one.) Based on this, decide whether the two signs in the angle equation are the same or opposite. If the signs are opposite, go on to step 2 to determine which is positive and which is negative.

2. If the signs are opposite, we need to decide which is the positive one and which is the negative. Since the focal angle is never negative, the smaller angle must be the one with a minus sign.

In step 1, many students have trouble drawing the ray diagram correctly. For simplicity, you should always do your diagram for a point on the object that is on the axis of the mirror, and let one of your rays be the one that is emitted along the axis and reflected straight back on itself, as in the figures in subsection 12.3.1. As shown in figure a/4 in subsection 12.3.1, there are four angles involved: two at the mirror, one at the object $$(\theta_o)$$, and one at the image $$(\theta_i)$$. Make sure to draw in the normal to the mirror so that you can see the two angles at the mirror. These two angles are equal, so as you change the object position, they fan out or fan in, like opening or closing a book. Once you've drawn this effect, you should easily be able to tell whether $$\theta_o$$ and $$\theta_i$$ change in the same way or in opposite ways.

Although focal lengths are always positive in the method used in this book, you should be aware that diverging mirrors and lenses are assigned negative focal lengths in the other method, so if you see a lens labeled $$f=-30$$ cm, you'll know what it means.

Example 8: An anti-shoplifting mirror

$$\triangleright$$ Convenience stores often install a diverging mirror so that the clerk has a view of the whole store and can catch shoplifters. Use a ray diagram to show that the image is reduced, bringing more into the clerk's field of view. If the focal length of the mirror is 3.0 m, and the mirror is 7.0 m from the farthest wall, how deep is the image of the store? g / Example 8.

$$\triangleright$$ As shown in ray diagram g/1, $$d_i$$ is less than $$d_o$$. The magnification, $$M= d_i/d_o$$, will be less than one, i.e., the image is actually reduced rather than magnified.

Apply the method outlined above for determining the plus and minus signs. Step 1: The object is the point on the opposite wall. As an experiment, g/2, move the object closer. I did these drawings using illustration software, but if you were doing them by hand, you'd want to make the scale much larger for greater accuracy. Also, although I split figure g into two separate drawings in order to make them easier to understand, you're less likely to make a mistake if you do them on top of each other.

The two angles at the mirror fan out from the normal. Increasing $$\theta_o$$ has clearly made $$\theta_i$$ larger as well. (All four angles got bigger.) There must be a cancellation of the effects of changing the two terms on the right in the same way, and the only way to get such a cancellation is if the two terms in the angle equation have opposite signs:

$\begin{equation*} \theta_f = + \theta_i - \theta_o \end{equation*}$

or

$\begin{equation*} \theta_f = - \theta_i + \theta_o . \end{equation*}$

Step 2: Now which is the positive term and which is negative? Since the image angle is bigger than the object angle, the angle equation must be

$\begin{equation*} \theta_f = \theta_i - \theta_o , \end{equation*}$

in order to give a positive result for the focal angle. The signs of the distance equation behave the same way:

$\begin{equation*} \frac{1}{f} = \frac{1}{d_i}-\frac{1}{d_o} . \end{equation*}$

Solving for $$d_i$$, we find

\begin{align*} d_i &= \left(\frac{1}{f}+\frac{1}{d_o}\right)^{-1}\\ &= 2.1\ \text{m} . \end{align*}

The image of the store is reduced by a factor of $$2.1/7.0=0.3$$, i.e., it is smaller by 70%. h / A diverging mirror in the shape of a sphere. The image is reduced ($$M\lt1$$). This is similar to example 8, but here the image is distorted because the mirror's curve is not shallow.

Example 9: A shortcut for real images

In the case of a real image, there is a shortcut for step 1, the determination of the signs. In a real image, the rays cross at both the object and the image. We can therefore time-reverse the ray diagram, so that all the rays are coming from the image and reconverging at the object. Object and image swap roles. Due to this time-reversal symmetry, the object and image cannot be treated differently in any of the equations, and they must therefore have the same signs. They are both positive, since they must add up to a positive result.

# Aberrations

An imperfection or distortion in an image is called an aberration. An aberration can be produced by a flaw in a lens or mirror, but even with a perfect optical surface some degree of aberration is unavoidable. To see why, consider the mathematical approximation we've been making, which is that the depth of the mirror's curve is small compared to $$d_o$$ and $$d_i$$. Since only a flat mirror can satisfy this shallow-mirror condition perfectly, any curved mirror will deviate somewhat from the mathematical behavior we derived by assuming that condition. There are two main types of aberration in curved mirrors, and these also occur with lenses.

(1) An object on the axis of the lens or mirror may be imaged correctly, but off-axis objects may be out of focus or distorted. In a camera, this type of aberration would show up as a fuzziness or warping near the sides of the picture when the center was perfectly focused. An example of this is shown in figure i, and in that particular example, the aberration is not a sign that the equipment was of low quality or wasn't right for the job but rather an inevitable result of trying to flatten a panoramic view; in the limit of a 360-degree panorama, the problem would be similar to the problem of representing the Earth's surface on a flat map, which can't be accomplished without distortion. i / This photo was taken using a “fish-eye lens,” which gives an extremely large field of view.

(2) The image may be sharp when the object is at certain distances and blurry when it is at other distances. The blurriness occurs because the rays do not all cross at exactly the same point. If we know in advance the distance of the objects with which the mirror or lens will be used, then we can optimize the shape of the optical surface to make in-focus images in that situation. For instance, a spherical mirror will produce a perfect image of an object that is at the center of the sphere, because each ray is reflected directly onto the radius along which it was emitted. For objects at greater distances, however, the focus will be somewhat blurry. In astronomy the objects being used are always at infinity, so a spherical mirror is a poor choice for a telescope. A different shape (a parabola) is better specialized for astronomy. j / Spherical mirrors are the cheapest to make, but parabolic mirrors are better for making images of objects at infinity. A sphere has equal curvature everywhere, but a parabola has tighter curvature at its center and gentler curvature at the sides.

One way of decreasing aberration is to use a small-diameter mirror or lens, or block most of the light with an opaque screen with a hole in it, so that only light that comes in close to the axis can get through. Either way, we are using a smaller portion of the lens or mirror whose curvature will be more shallow, thereby making the shallow-mirror (or thin-lens) approximation more accurate. Your eye does this by narrowing down the pupil to a smaller hole. In a camera, there is either an automatic or manual adjustment, and narrowing the opening is called “stopping down.” The disadvantage of stopping down is that light is wasted, so the image will be dimmer or a longer exposure must be used. k / Even though the spherical mirror (solid line) is not well adapted for viewing an object at infinity, we can improve its performance greatly by stopping it down. Now the only part of the mirror being used is the central portion, where its shape is virtually indistinguishable from a parabola (dashed line).

What I would suggest you take away from this discussion for the sake of your general scientific education is simply an understanding of what an aberration is, why it occurs, and how it can be reduced, not detailed facts about specific types of aberrations. l / The Hubble Space Telescope was placed into orbit with faulty optics in 1990. Its main mirror was supposed to have been nearly parabolic, since it is an astronomical telescope, meant for producing images of objects at infinity. However, contractor Perkin Elmer had delivered a faulty mirror, which produced aberrations. The large photo shows astronauts putting correcting mirrors in place in 1993. The two small photos show images produced by the telescope before and after the fix.

# Contributors

Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.