5. The object height ho is positive if the object extends above the axis and is negative if the object extends below.
6. The image height hi is positive if the image extends above the axis and is negative if the image extends below.
Once again, remember that there are several different sign conventions for working with lenses, so be careful when comparing our equations with those from other books.
CO N C E P T C H E C K 2 4 . 8 | Object and Image Distances
Figure 24.41B shows an object and its image as formed by a diverging lens. What are the signs (positive or negative) of (a) the object distance so, (b) the image distance si, and (c) the focal length f?
The Thin-Lens Equation
Figure 24.43A shows a ray diagram for a converging lens. We can use the geometry of this sketch to fi nd a mathematical relation for locating the image produced by a converging lens. Within this fi gure, we identify two pairs of similar right triangles.
One pair is shaded in yellow in Figure 24.43A; one yellow triangle has side lengths ho and so, and the other yellow triangle has sides of length hi and si. The similarity of these triangles leads to
ho so 5 2hi
si (24.19) The negative sign here is due to our convention that hi is negative because the image in this case points downward (extends below the axis). Another pair of similar tri- angles is shaded in red in Figure 24.43B and leads to
ho
f 5 2hi
si2f (24.20) with the same sign convention for the image height as in Equation 24.19. Taking the ratio of Equations 24.19 and 24.20 gives
ho/so
ho/f 5 hi/si hi/1si2f2 Canceling the factors of ho and hi, we have
f
so 5 si2f si and cross multiplying, we get
sif5so1si2f2 We can now rearrange to obtain
sof1sif5sosi
3Notice, however, that when light passes through two or more lenses (as in a telescope or micro- scope), it is possible for the object distance so to be negative. We’ll see examples of that in Chapter 26.
Figure 24.43 Geometry for deriving the thin-lens equation.
B A
o Axis
si so
FR
i
The two yellow triangles are similar.
o Axis f
FR i
The two red triangles are similar.
si f
Dividing both sides by a factor of fsosi leads to our fi nal result, 1
so1 1 si5 1
f (24.21) Equation 24.21 is called the thin-lens equation. It can be used to calculate the image properties of both converging lenses and diverging lenses, provided we use the sign conventions described above.4 Notice that the thin-lens equation is actually identical to the mirror equation (Eq. 24.15).
The thin-lens equation (Eq. 24.21) gives a relation between the object and image distances so and si, and can thus be used to fi nd the image location. Equation 24.21 can also be used to obtain the magnifi cation. As with mirrors, the magnifi cation is the ratio of the image height to the object height,
m5 hi
ho 5 2 si
so (24.22)
Applying the Thin-Lens Equation
Let’s now see how to use the thin-lens equation together with a ray diagram to locate and describe the image formed by a lens. Consider a converging lens with f 0.50 m and an object at so 0.30 m. We have included the positive signs here to emphasize that f is positive (a converging lens) and that so is positive because the object is on the left side of the lens. Figure 24.44, which also shows our standard ray diagram, is sketched to scale. We can fi nd the image distance from the thin-lens equation,
1 so1 1
si51 f Solving for 1/si fi rst, we get
1 si5 1
f 2 1
so 5 so2f fso Rearranging to fi nd si gives
si5 fso
so2f (24.23) Inserting the given values of f and so, we obtain
si5 fso
so2f 5 10.50 m2 10.30 m2
0.30 m20.50 m 5 20.75 m
According to our sign convention, this negative value for si means that the image is on the left side of the lens, so it is a virtual image. This result is confi rmed by the ray diagram in Figure 24.44.
CO N C E P T C H E C K 2 4 .9 | Image Formed by a Converging Lens
Is the image in Figure 24.44 (a) real and upright, (b) virtual and upright, (c) real and inverted, or (d) virtual and inverted?
Thin-lens equation Thin-lens equation
24.5 | LENSES 821
4As the name “thin lens” implies, “thick” lenses also exist, and their image properties are described by a different relation. We will not consider thick lenses in this book.
Figure 24.44 Applying the thin-lens equation to fi nd the location of an image formed by a converging lens.
si
FL FR
Object so0.30 m
f 0.50 mf 0.50 m Image
Central ray
Parallel ray Focal ray
Axis
E X A M P L E 2 4 . 8 Image Formed by a Diverging Lens
Suppose the converging lens in Figure 24.44 is replaced by a diverging lens with a focal length f 0.50 m. (a) If the object is again at so0.30 m (i.e., on the left side of the lens), what is the location of the image? (b) Is this image real or virtual?
RECOGNIZE T HE PRINCIPLE
We begin with a ray diagram, which will give us a qualitative understanding of the image. We then proceed to apply the thin-lens equation to fi nd the actual value of si. SK E TCH T HE PROBLEM
Figure 24.45 shows a ray diagram constructed by following our suggested procedures for ray tracing with lenses. Step 1: Figure 24.45A shows the lens and the focal points on the left and right. Step 2: We add the object to our drawing. Step 3: In Figure 24.45B, we add three rays. The parallel ray is initially parallel to the principal axis;
after passing through the lens, this ray is refracted away from the principal axis since we have a diverging lens. Extrapolating this ray back to the left-hand side, we see that it passes through the focal point on the left, FL. The focal ray is directed from the object toward the focal point to the right of the lens, FR; this ray is refracted by the lens and like the parallel ray is defl ected away from the principal axis. On the right side of the lens, the focal ray travels parallel to the principal axis. The central ray passes through the center of the lens and is not defl ected by the lens. Step 4:
These rays do not intersect at any point on the right side of the lens. Extrapolating them back, however, we fi nd intersection at a common image point to the left of the lens. This ray diagram tells us much about the image and provides a check on results obtained with the thin-lens equation.
IDENT IF Y T HE REL AT IONSHIPS A ND SOLV E
(a) We apply the thin-lens equation (Eq. 24.21) and solve for si just as we did in deriv- ing Equation 24.23:
si5 fso so2f
Substituting the given values of the focal length (f 0.50 m) and object distance (so0.30 m), we get
si5 fso
so2f5 120.50 m2 10.30 m2
0.30 m2 120.50 m2 5 0.19 m
(b) According to the ray diagram, light does not pass through this image point, so the image is virtual. This conclusion is confi rmed by our analysis; since the image distance si is negative, the image is indeed on the left side of the lens and is virtual .
What does it mean?
Because the focal length f for a diverging lens is negative, when so is positive (an object on the left side of the lens), the thin-lens equation will always give a nega- tive value for si and thus a virtual image. This result is different from a converging lens, which can give either a real or a virtual image, depending on how far the object is from the lens.
Applications of the thin-lens equation are mathematically straightforward, pro- vided you carefully follow the sign conventions in Figure 24.42. Constructing the image using ray tracing is always advisable, however, because it provides a check on results from the thin-lens equation and gives an intuitive understanding of the image properties.
2 4 . 6 | H O W T H E E Y E W O R K S
Lenses and mirrors are used in many different kinds of optical instruments, includ- ing microscopes and telescopes, which we’ll discuss in Chapter 26. In a sense, the oldest optical instrument is the eye. A sketch of a vertebrate eye is shown in Figure 24.46A; it is a complicated structure with several different regions, but its basic Figure 24.45 Example 24.8.
B A
Object
Axis
FL FR
FL FR
so0.30 m so0.30 m
Object
Axis
Focal ray
Central ray Parallel ray
Image Diverging lens
function can be understood using the principles of refraction and the properties of lenses. Light enters through the cornea, passes through a liquid region called the aqueous humor, through a lens, into a gel-like region called the vitreous humor, and then to the retina at the back surface of the eye. The retina contains light-sensitive cells that convert light into electrical signals carried to the brain by the optic nerve.
A Simple Model of the Eye: Corneal Refraction
Different regions of the eye have different indices of refraction, so light rays are refracted and change direction as they pass from the air into the cornea and then from region to region within the eye. The indices of refraction of the different regions inside the eye are all in the range of approximately 1.33 to 1.40 (Table 24.1), so the biggest index change and largest refraction occur when light fi rst enters the cornea.
In the simplifi ed model in Figure 24.46B, we approximate the eye as spherical, and we also treat all the different parts of the eye as a single region with a single index of refraction. In this model, the only refraction occurs at the surface of the cornea, and we want to calculate how this refraction focuses light. We are especially interested in the focal length of this corneal “lens” and how close (or how far away) the focal point is to the retina. Although this illustration is certainly a simplifi ed version of the real eye, it will still help us understand how the eye works. We’ll then consider improvements to the model that make it closer to the real thing.
Figure 24.46C shows a ray diagram for refraction by the cornea. The incident angle is u1, the angle the incoming ray makes with the normal to the surface. Here we consider a single incident ray that initially travels parallel to the principal axis;
we want to fi nd where its refracted ray intersects the axis since that will give the location of the focal point F. Because we have assumed the surface of the eye is spherical, the normal direction (the dashed line through points P and C) passes through the center of the eye at C. For simplicity, we assume the incoming ray is near the axis, so the height h and the angle u1 are small. The refracted angle is given by Snell’s law (Eq. 24.7), n1 sin u15n2 sin u2. The outside region is air, so n1 nair
⬇ 1.00, and we denote the index of refraction inside the eye as simply n. We also assume the angles are small so that sin u1⬇ u1 (and so forth for u2). We now have
n1 sin u15nair sin u1< 11.002u15 u1 n2 sin u25n sin u2<nu2
24.6 | HOW THE EYE WORKS 823 F
f
A C
⬇ R P
Axis
Cornea
A
B
C
180 u1 u1
u1
u3
n2 n n1 1
u2 Retina
Optic nerve Iris
Lens Pupil Cornea
Aqueous humor Ciliary muscle
Vitreous humor
⬇ 2.5 cm
Cornea
This is where we want the focus to be.
This is where the focus actually is.
Figure 24.46 A Structure of the eye. B Corneal refraction model of the eye. The curved surface of the cornea refracts and thus focuses light near the retina (the back surface of the eye).
C The refraction at the cornea can be analyzed using Snell’s law. Here we ignore refraction when the light leaves the right surface of the spherical eye.
Inserting into Snell’s law gives
u15nu2
u25 u1
n (24.24)
To locate the focal point and fi nd the focal length f, we use the geometry in Figure 24.46C to fi nd the angle u3. The sum of the interior angles of triangle PFC is 180°, so we have
180o2 u11 u21 u35180°
and thus u35 u12 u2. Inserting the result for u2 from Equation 24.24 gives u35 u12 u25 u12u1
n 5 u1an21
n b (24.25) We next consider the two right triangles PFA and PCA. From the fi rst of these triangles, we fi nd5 tan u3 h/f. The angle u3 is small, so we can also use the approxi- mation tan u3⬇ u3 to get
u3<tan u35h f
h<fu3 (24.26)
Applying the same approach to triangle PCA leads to u1<tan u15 h
R
h<Ru1 (24.27)
Setting Equations 24.26 and 24.27 equal gives fu35Ru1. The focal length of the corneal “lens” is then
f5R u1 u3
and inserting the result for u3 from Equation 24.25, we fi nd f5R u1
u11n212/n f5 n
n21 R (24.28)
Properties of the Spherical Eye
Equation 24.28 gives the distance from the refracting surface (the cornea) to the focal point for our simplifi ed model of the eye in Figure 24.46C. Where does this focal point lie? Is it inside the eye? We can estimate this focal length using the aver- age index of refraction of the eye, which is about n 1.4; we fi nd
f5 n
n21 R5 1.4
1.421 R5 1.4
0.4 R53.5R (24.29)
In our simple model of a spherical eye, the retina is a distance 2R from the front surface of the cornea, so the focal point found in Equation 24.29 is about 1.5R behind the retina.
We have worked through this simplifi ed model of the eye for two reasons. First, it shows how to deal with refraction from a spherical surface. Second, the result for the focal length in Equation 24.29 shows that this simple spherical eye would not function very well because it does not form an image at the retina; indeed, that
5We assume point A is very close to the spherical refracting surface. This approximation is good when the rays are close to the axis (as we have assumed) and h is small.
explains why the eye actually deviates somewhat from a spherical shape. In the more realistic picture of Figure 24.46A, you will notice that the cornea protrudes slightly from the otherwise spherical profi le of the eye. This protrusion makes the radius of curvature of the cornea smaller than the radius of the main spherical por- tion of the eye. The quantity R in Equation 24.29 is the radius of curvature of the cornea, so allowing for this more realistic shape will bring the focal point closer to the retina. Hence, our simple model has given some insight into why the cornea is shaped the way it is.
Refining the Model: Adding the Lens
Let’s now add the eye’s lens to our model. In a real eye, the lens is inside the eye (Fig. 24.46A), and light passes through this lens after being refracted by the cornea.
We will instead consider a slightly simpler model in which the lens is in front of the cornea and hence just outside the eye. Our model will thus not apply strictly to a real eye, but it will yield some useful insights and can be a good way to understand the combination of an eye with a contact lens. So, in our simple model, the incom- ing light fi rst passes through and is partially focused by the lens and then enters the cornea, which fi nishes the focusing job. By placing the lens in front of the cornea, we can more easily apply our results for lenses, including the thin-lens equation.
We now ask two questions: “What kind of lens is needed for the eye to function properly (i.e., converging or diverging)?” and “What is the shape of the lens (i.e., its radius of curvature)?”
We begin with the ray diagrams for two different cases. In Figure 24.47A, the eye is focused on an object at infi nity (i.e., on an object that is very far away). The incoming rays are parallel to the axis, and we assume the cornea has the correct radius of curvature so that these rays are focused on the retina. Hence, we assume the focusing for distant objects is done solely by the cornea using the refraction focusing of our fi rst model in Figure 24.46B. The ray diagram in Figure 24.47B shows why we need the added lens. Here the object has been moved very close to the eye; the incoming rays are now diverging from the axis, and the refracted rays produced by the cornea alone intersect at a point behind the retina. We need the extra focusing of the added lens in Figure 24.47C to focus the image on the retina.
Now let’s consider what properties this lens must have to make the ray diagram in Figure 24.47C work. We place the object a distance so in front of the lens and sup- pose the lens produces parallel, outgoing rays (i.e., an image at infi nity) as shown in the fi gure. We know from Figure 24.47A that the cornea by itself can focus these parallel rays on the retina. Hence, the lens should be designed such that an object at so focuses at infi nity (so that si ), producing parallel rays for the cornea. Using the thin-lens equation (Eq. 24.21) with an assumed focal length of flens, we have
1 flens5 1
so 1 1 si5 1
so 1 1
` 5 1
so (24.30) and thus
flens5so (24.31)
The eyes of a person with normal vision are able to focus on objects as close as so 25 cm from each eye. The lens in Figure 24.47C must therefore have a focal length flens 25 cm. Since flens is positive, the lens is a converging lens.
How must this lens be shaped to give this value for the focal length? The focal length is related to the radii of curvature of the two surfaces of a lens by a relationship6 called the lens maker’s formula. For a converging lens with two convex surfaces with radii of curvature R1 and R2, this formula reads
1
f 5 1n212 a 1
0R10 1 1
0R20 b (24.32) Lens maker’s formulaLens maker’s formula
24.6 | HOW THE EYE WORKS 825 Figure 24.47 Function of the eye’s lens. A In this model, light from a very distant object is focused by the cornea alone on the retina. B When an object is very close to the eye, the refracted rays do not fall on the retina, but instead intersect behind it.
C If we place a lens in front of the eye, light from a nearby object is focused by the combination of the eye’s lens and the cornea so that an image is formed at the retina.
A
B
The cornea alone focuses these rays onto the retina.
Object is very far away (so ).
Object is very close to eye.
C
Parallel rays produced by lens Lens Cornea
so
Cornea Cornea
6The lens maker’s formula can be derived by considering the refraction from the two surfaces of the lens in succession, similar to the calculation we did for our model eye in Figure 24.46C.
where n is the index of refraction of the lens. Equation 24.32 applies only to the double convex lens shape in Figure 24.37 and assumes the lens is in air. More general versions of the lens maker’s formula apply to other lens shapes and to sit- uations such as a lens under water. The absolute value terms in Equation 24.32 emphasize that both these terms are positive for the double convex lens considered here. For other lens shapes, one or both of these terms may be negative and hence describe a diverging lens with a negative value for f. A few such cases are discussed in the end-of- chapter problems. (See Problems 81 and 82.)
Applying the Lens Maker’s Formula
Let’s now apply the lens maker’s formula to estimate the radius of curvature of the lens in our model of the eye in Figure 24.47C. The required focal length of this lens was estimated using Equation 24.31, where we found flens⬇ 25 cm. Although the lens maker’s formula involves the radii of curvature for both sides of the lens sepa- rately, in the spirit of our approximate treatment we assume these radii are equal and R1 R2 R. Using this assumption in Equation 24.32 gives
1
flens 5 1n212 a 1
0R10 1 1
0R20 b5 21n212 R or
flens5 R 21n212
where we have now dropped the absolute value signs and assume R is positive. Solv- ing for R, we fi nd
R521n212flens (24.33)
Inserting the value of flens 25 cm given above and using the value of the index of refraction for the eye’s lens n 1.41 (see Table 24.1), we have
R521n212flens5211.41212 10.25 m250.21 m (24.34)
The radius of curvature of the lens is thus several times larger than the diameter of the eye. (Recall that the diameter of the eye is about 2.5 cm 0.025 m.) This result for the radius of curvature is in accord with the realistic drawing of the eye in Figure 24.46A. The front surface of the lens within the eye, which is responsible for most of the focusing, has a rather fl at surface, so its radius of curvature is much larger than the radius of the eye.
In addition to allowing the eye to focus on a nearby object, the eye’s lens plays another crucial role. When the object is at a particular location, the focal length of the eye’s lens has a value such that the fi nal image from the cornea is properly focused on the retina. If the object is moved closer to or farther away from the eye, the focal length of the lens must change to keep the fi nal image in focus. This change of shape is accomplished using muscles attached to the edges of the lens (Fig. 24.46A) to pull on the edges and change the lens’ radius of curvature. The lens must be suffi ciently fl exible to allow its shape to change accordingly. As a person grows older, the lens becomes less fl exible and the eye can no longer focus properly for objects that are both far away and very close. That is why the author wears glasses.
2 4 . 7 | O P T I C S I N T H E AT M O S P H E R E
Rainbows
In Section 24.3, we showed that refraction of light by a prism causes rays to change direction and that the amount light is defl ected depends on the color of the light.