We’ll illustrate this general approach in a number of examples throughout this chapter. Most of these examples involve lenses, so Figure 26.2 shows our sign con- ventions (from Fig. 24.42) for applying the thin-lens equation to a single lens. A positive image distance si corresponds to an image on the right of the lens, whereas a negative value of si describes an image to the left of the lens (a “virtual image”).
Notice also that a positive object distance so corresponds to an object to the left of the lens. In all examples in Chapter 24, the object distance was positive. Our sign conventions also allow for a negative value of so, which means that the object is to the right of the lens (a “virtual object”). We’ll encounter such negative values of so in systems with two lenses, when the image produced the fi rst lens is “beyond” the second lens.
Rather than dwell on the rules for analyzing systems of multiple lenses, we’ll illustrate the key concepts by working through a series of examples.
Designing a Pair of Contact Lenses
As you age, the lenses in your eyes usually become less fl exible and are not able to change focal length over the range found in Equations 26.4 and 26.5. The focal length is instead confi ned to a narrower range of values, reducing your ability to focus. In some cases, the near-point distance increases and is greater than the near- point distance for a normal eye; objects located closer than this near-point distance cannot be focused on the retina (Fig. 26.3A), and you must move things (such as a newspaper or this book) to a greater distance to see them clearly. You are then farsighted. To compensate, you can place a lens (a contact lens) immediately in front each eye. Let’s now see how the required focal length of this lens depends on a person’s near-point distance.
Figure 26.3 A For a farsighted person, the near point is farther away than is convenient for read- ing or other close work. Rays from an object within the near point focus behind the retina. B When a converging lens (such as a contact lens) is placed in front of the eye, a virtual image is produced by the lens. C The image produced by the contact lens acts as the object for the eye. If this image is at or outside the near-point distance, the eye can form an image on the retina.
Object
RAY DIAGRAM FOR A FAR-SIGHTED PERSON
Near point
Object Image
Converging lens (contact lens)
Contact lens
Final image so
si
B
C A
Image produced by contact lens is the object for the eye.
Image so
sN
Without glasses, the image is behind the retina.
The image produced by this lens is outside the near point.
Figure 26.2 Sign conventions for working with the thin-lens equation.
Principal axis
CONVERGING LENS:
f positive
si positive so positive so negative si negative
o and i positive
o and i negative
Principal axis Other conventions as for converging lens
DIVERGING LENS:
f negative
Suppose the near-point distance for a certain individual is sN⫽ 75 cm, which is much greater than the “normal” near-point distance of 25 cm. We wish to design a contact lens that enables this person to focus on objects at the normal near-point distance of 25 cm. The contact lens is placed just in front of the eye as in Figure 26.3B; it forms the fi rst lens of our “system,” and the eye acts as the second lens.
We want this system of two lenses to be able to take an object at so⫽ 25 cm from the contact lens and produce a fi nal image on the retina. We follow our general rules for analyzing optical systems with two lenses and analyze the problem one lens at a time. The contact lens produces an image (Fig. 26.3B) that then acts as the object for the second “lens,” the eye, but where should the image formed by the contact lens be located? We are given that the near-point distance for the eye is 75 cm, which means that the eye alone can focus an object that is 75 cm away to produce an image on the retina. So, if the image produced by the contact lens is at this distance (or greater, Fig. 26.3C), the eye can produce a fi nal image at the retina. We thus want the contact lens to take an object at so⫽ 25 cm and produce an image that is 75 cm to the left of the eye. Since the contact lens is touching the eye, the image distance is si⫽ ⫺75 cm; this image distance is negative because it is to the left of the contact lens (and is thus a virtual image; see the sign conventions in Fig. 26.2).
We now apply the thin-lens equation (Eq. 26.1) to the contact lens; we substitute so⫽ 25 cm and si⫽⫺75 cm to fi nd
1 flens5 1
so
1 1 si
5 1
125 cm1 1
275 cm (26.6) and solve for flens:
flens⫽ 38 cm (26.7)
which is the focal length required for the contact lens. Since flens is positive, this lens is a converging lens.
If the person’s near-point distance is larger than 75 cm, Equation 26.6 shows that a lens with a shorter focal length is needed. In words, we say that a person with a large near-point distance needs a “stronger” lens than a person with a short sN. The strength of a lens is sometimes measured in terms of its refractive power, defi ned as
refractive power5 1
flens (26.8) The unit of refractive power is m⫺1, which is called a diopter, and
1 diopter ⫽ 1 m⫺1
The contact lens with the focal length in Equation 26.7 thus has a refractive power of 1/flens⫽ 1/(0.37 m) ⫽ 2.7 diopters.
Let’s now consider what type of contact lens is needed by a nearsighted person.
Such a person is unable to focus light from distant objects on the retina as shown in the ray diagram in Figure 26.4A. The incoming rays from an object very far away are approximately parallel to the axis (an object at “infi nity”); a nearsighted eye produces an image in front of the retina.
A nearsighted person is able to properly focus objects that are within a certain distance. Suppose a person can focus objects within 2.0 m on the retina. What kind of contact lens does this person need? We follow the approach we took for a farsighted person in Figure 26.3. We want an object that is very far away (so⫽⬁) to lead to a fi nal image at the retina. It is also given that the eye alone can focus an object that is 2.0 m away on the retina. So, the contact lens must take an object at so⫽⬁ and give an image at si⫽⫺2.0 m (Fig. 26.4B).
Applying the thin-lens equation to the contact lens, we have so⫽ ⬁ and si ⫽
⫺2.0 m, and we get
1 flens5 1
so 11 si5 1
` 1 1 22.0 m
26.1 | APPLICATIONS OF A SINGLE LENS: CONTACT LENSES, EYEGLASSES, AND THE MAGNIFYING GLASS 883 Figure 26.4 A When a person is nearsighted, rays from an object very far away (at infi nity) focus in front of the retina. B Placing a diverging lens in front of the eye produces a virtual image to the left of the lens. C The image formed by the lens acts as the object for the eye. The nearsighted eye is able to focus this image on the retina.
B A
C
Without glasses, the image is in front of the retina.
Diverging lens (contact lens for a near-sighted person) Image produced
by contact lens Final image
The object for the eye gives the image on the retina.
so⫽⬁ Distant object
RAY DIAGRAM FOR A NEAR-SIGHTED PERSON
Image
Virtual image produced by diverging lens
si
Solving for flens, we have
flens5 22.0 m
which is the focal length required for the contact lens. Since flens is negative, this lens is a diverging lens.
CO N C E P T C H E C K 26.1 | What Kind of Image Is It?
Figures 26.3 and 26.4 show ray diagrams for different types of contact lenses.
Which of the following statements correctly describes the images formed by these contact lenses?
(a) A contact lens always forms a real image.
(b) A contact lens always forms a virtual image.
(c) A contact lens can form a real or virtual image, depending on the type of lens.
CO N C E P T C H E C K 26. 2 |
A farsighted person has a very long near-point distance. What should the focal length of this person’s contact lenses be so that he can read a newspaper at a normal distance of (a) 25 cm, (b) 400 cm, or (c) 4.0 diopters? Hint: Assume the near-point distance is extremely long so that sN⬇⬁.
Designing a Pair of Eyeglasses
The problem of selecting the proper lens for a set of eyeglasses is very similar to the contact lens problems considered in Figures 26.3 and 26.4. The main difference is that the eyeglass lens is a short distance in front of the eye, rather than touching it (Fig. 26.5). This distance must be accounted for when relating the image distance from the eyeglass lens and the object distance for the eye. The rest of the analysis is the same as with the contact lens problem and is illustrated in Example 26.1.
Figure 26.5 Image produced by the combination of an eyeglass lens plus the eye. A The eyeglass lens produces an image that then acts as the object for the eye. B When calculating the object distance for the eye, we must account for the distance dspacing between the eyeglass lens and the eye.
Object Image
Object for eye
so si
A
B
Eyeglass lens
dspacing
so(eye) ⫽ si ⫹ dspacing
Final image
E X A M P L E 2 6 . 1 What Kind of Eyeglass Lenses Do You Need?
Consider a farsighted person with a near-point distance of 75 cm. In Figure 26.3 (and Eqs. 26.6 and 26.7), we designed a contact lens that would enable this person to read a book located at the “normal” near-point distance of 25 cm. The same person now wants a set of eyeglasses that he can use to read this book at 25 cm. What is the required focal length of the eyeglass lenses? Assume the glasses are positioned about dspacing⫽ 2 cm in front of the person’s eye.
RECOGNIZE T HE PRINCIPLE
With an object 25 cm in front of the eye, we want the eyeglass lens to produce an image 75 cm in front of the eye. This image then forms the object for the eye of our farsighted person, and we know that for this distance he will then be able to focus the fi nal image on his retina.
SK E TCH T HE PROBLEM
The problem is described by the ray diagrams in Figure 26.5.
IDENT IF Y T HE REL AT IONSHIPS
Since the eyeglass lens is dspacing⫽ 2 cm in front of the eye and the object is 25 cm in front of the eye, the object and image distances for the eyeglass lens are so⫽ 25 ⫺ dspacing⫽ 23 cm and si⫽⫺73 cm (Fig. 26.5A). This image is a virtual image (si is negative) because light does not actually pass through this point.
SOLV E
Using the thin-lens equation with so⫽ 23 cm and si⫽⫺73 cm, we get 1
flens 5 1 so 1 1
si5 1
23 cm 1 1
273 cm Solving for flens gives
flens⫽ 34 cm Since flens is positive, this lens is a converging lens.
What does it mean?
Comparing this result for flens with the result in Equation 26.7 for a contact lens, we see that the eyeglass lenses have a focal length about 10% shorter. That is why your prescription for contact lenses is slightly different than for your eyeglasses.
The Magnifying Glass
The magnifying glass is one of the most basic of all optical devices and is a com- ponent in microscopes, telescopes, and other instruments. The simplest magnify- ing glass is a single lens, used to view things that are too small to examine with the naked eye (Fig. 26.6). To analyze this optical “instrument,” we must again consider a system consisting of two lenses—the magnifying glass lens and the eye behind it—so the problem is similar to our analysis of contact lenses and eyeglasses.
In those applications, the goal was to choose a lens that would allow the eye to produce an image at the retina. With a magnifying glass, the goal is to produce a greatly magnifi ed image at the retina. We thus want the image on the retina to be as large as possible.
Consider fi rst the case without a magnifying glass. To examine a small object, the natural thing to do is to place the object as close as possible to your eye. Figure 26.7A shows how the position of an object in front of the eye affects the size of its image on the retina. As the object is brought closer and closer to the eye from point A to B, the image on the retina grows larger and the apparent size of the object increases, but there is a limit to how close you can bring the object and still keep it in focus. The largest clearly focused image for the unaided eye results when the object is at the near point. The object’s apparent size when it is located at the near point can be measured using the angle u defi ned in Figure 26.7B; we’ll call it the image angle.
To make the object even larger on the retina than in Figure 26.7B, we need to move it closer to the eye and hence inside the near point. The eye by itself cannot then produce an image on the retina, but it can do so with the help of the magnify- ing glass lens.
Image Properties with a Magnifying Glass
A magnifying glass is a single converging lens placed in front of the eye as shown in Figure 26.7C. The object is positioned inside the focal length of this lens, that is, so that the focal point F of the lens is to the left of the object in Figure 26.7C.
We showed in Chapter 24 (Fig. 24.44) that this position leads to an upright virtual image at a point farther from the eye, on the far left in Figure 26.7C. As far as the eye is concerned, light from the object emanates from this virtual image, and if this image produced by the magnifying glass is at the near point, the eye is able to focus this light onto the retina. Comparing the ray diagrams in Figures 26.7B and C shows that the image angle with the magnifying glass uM (Fig. 26.7C) is greater than the image angle for the eye alone (u in Fig. 26.7B), so the image on the retina is enlarged by the magnifying glass.
26.1 | APPLICATIONS OF A SINGLE LENS: CONTACT LENSES, EYEGLASSES, AND THE MAGNIFYING GLASS 885 Figure 26.6 A magnifying glass produces an upright virtual image.
© Thinkstock Images/Jupiterimages
The enlargement of the image on the retina is given by the angular magnifi cation mu , defi ned by
mu5 uM
u (26.9) The angular magnifi cation is related to (but slightly different from) the linear magnifi cation m defi ned in Chapter 24. The linear magnifi ca- tion involves the height of the object and image, whereas the angular magnifi cation involves the angle the image makes on the retina.
We can calculate the angular magnifi cation mu by applying the thin- lens equation (Eq. 26.1) to the lens of the magnifying glass (Fig. 26.7C):
1 so1 1
si5 1
f (26.10) As always, we must be careful to follow the proper sign conventions.
Because the magnifying glass is a converging lens, its focal length f is positive. The object is on the left side of the lens, so the object distance so is positive; the image is also on the left side of the lens, so the image distance si is negative. For simplicity, let’s assume the magnifying glass is very close to the eye; the image is at the near point and hence si⫽
⫺sN in Equation 26.10. Inserting all this information leads to 1
so 2 1 sN 51
f 1
so 5 1 f 1 1
sN5 sN1f fsN Solving for so, we get
so5 sNf
sN1f (26.11) We can now fi nd the image angle with and without the magnifying glass in place. From the geometry in parts B and C of Figure 26.7, we have
tan u 5ho
sN and tan uM5 hi sN
These angles are usually small, so we can use the approximations tan u<u and tan uM<uM, leading to
u 5ho
sN and uM5 hi
sN (26.12) Inserting these angles into the defi nition for the angular magnifi cation (Eq. 26.9) gives
mu5 uM u 5 hi
ho
There are two right triangles in Figure 26.7C, one involving ho and so and another involving hi and sN. These triangles are similar (having the same interior angles), so hi/sN5ho/so. The angular magnifi cation is thus
mu5 hi ho 5sN
so
Inserting the result for so from Equation 26.11 gives our fi nal result, mu5 sN
so 5 sN 1sNf2/1sN1f2 mu5 sN1f
f 5 sN
f 11 (26.13)
Figure 26.7 A The size of an object as perceived by the eye depends on the size of the image on the retina. If an object is brought closer to the eye (from A to B), the rays from its tip make a larger angle with the axis and the image on the retina becomes larger. B The closest object dis- tance at which the eye can focus the image on the retina is the near point, which for a typical young person is approximately 25 cm from the eye. C A magnifying glass produces a magnifi ed (enlarged) image at the near point of the eye.
The eye focuses the light from this image on the retina.
A B
A B
Apparent size of object with magnifying glass
B
C A
Image Eye
The size of an image on the retina determines the object’s apparent size.
Object at near point
Near point
u
u
uM
uM so⫽ sN艐 25 cm (near-point distance)
sN⫽ 25 cm (near-point distance)
o
Image at near point
Magnifying glass
Near point
F (focal point of lens)
i o
so Image
Practical magnifying glass lenses have focal lengths that are smaller than the near-point distance by a factor of 10 or more, so the ratio sN/f in Equation 26.13 is much greater than 1. Hence, to a good approximation we can write
mu5sN
f (26.14) For a typical magnifying glass, this result is accurate to 10% or better (and we can always use the result for mu in Eq. 26.13 if greater accuracy is needed). This expression for mu is also found when a magnifying glass is used as the eyepiece of a microscope or telescope, which we’ll explore in Example 26.2.
The near-point distance sN varies from person to person, so the magnifi cation actually obtained with a particular magnifying glass (having a particular value of f) will depend on the person. Because the “typical” near-point distance for a young adult is 25 cm, this value is used as a convention when specifying magnify- ing glasses, so
mu5 sN
f 5 25 cm f
For a magnifying glass specifi ed by the manufacturer as providing mu⫽ 10, the focal length is thus
f5 25 cm mu
5 25 cm
10 52.5 cm
The angular magnifi cation of a typical magnifying glass or eyepiece is usually 10 or 20.
Magnifi cation of a magnifying glass
Magnifi cation of a magnifying glass
26.1 | APPLICATIONS OF A SINGLE LENS: CONTACT LENSES, EYEGLASSES, AND THE MAGNIFYING GLASS 887
E X A M P L E 2 6 . 2 Angular Magnifi cation for an Object at the Focal Point
Figure 26.8A shows an object at the focal point of a magnifying glass.
This arrangement is slightly different from the one we considered in Figure 26.7C, where the object was inside the focal point. Now the image from the magnifying glass is at infi nity instead of at the near point of the eye (Fig. 26.7C). Show that the angular magnifi cation in this case is given by mu⫽ sN/f (Eq. 26.14).
RECOGNIZE T HE PRINCIPLE
To fi nd the angular magnifi cation, we must compare the angle u that the object makes with the principal axis of the eye when the object is at the near point (without the magnifying glass; Fig. 26.7B), with the angle uf when the object is at the focal point of the magnifying glass (Fig. 26.8B).
The magnifi cation is the ratio of uf to u. (Compare with Eq. 26.9.) mu5
uf
u (1)
SK E TCH T HE PROBLEM
Figure 26.8B shows a ray diagram. The object is at the focal point of a converging lens. From Chapter 24, we know that the lens will form an
image at infi nity (si⫽⬁); hence, the rays between the magnifying glass and the eye are parallel as shown in Figure 26.8B.
IDENT IF Y T HE REL AT IONSHIPS A ND SOLV E
To fi nd uf, we consider the right triangle shaded in red in Figure 26.8B. We assume the magnifying glass is very close to the eye, so the red triangle has sides f and ho and contains the angle uf. From the geometry of this triangle, tan uf⫽ ho/f. Using the small-angle approximation (tan u<u) gives uf5ho/f. From Figure 26.7B, an object at
Figure 26.8 Example 26.2.
A An object at the focal point of a magnifying glass. B The lens forms a virtual image an infi nite distance (at “infi nity”) in front of the eye.
The light from this virtual image forms a parallel bundle of rays that are then focused by the eye to form an image on the retina.
Image on retina Virtual image formed by
magnifying glass at infinity Object
Object
Magnifying glass
F
f A
B
o
f
uf uf o