Chapter 22: Mirrors and Lenses - West Windsor … · Chapter 22: Mirrors and Lenses ... x The ray...

Etkina/Gentile/Van Heuvelen Process Physics 1/e Ch 22 22-1

Chapter 22: Mirrors and Lenses x How do you see sunspots? x When you look in a mirror, where is the face you see? x What is a burning glass? Make sure you know how to: 1. Apply the properties of similar triangles; 2. Draw ray diagrams and normal lines; 3. Use the laws of reflection and refraction.

It was a cold spring day in 1612; he felt a little chilly. Despite the coolness of the room, he started sweating when he saw all the pictures together. He knew that what he was seeing was dangerous. He was looking at pictures of the Sun that he meticulously drew during the last 28 days using a telescope to project the Sun’s image on paper and therefore observe it when high in the sky without burning his eyes. He was busy that month and drawing the pictures of the Sun was relaxing. However, he never put them all together; now that he did, it scared him. Those dark spots on the Sun were first seen by Chinese before the birth of Christ and then studied by the Greek Anaxagoras. His acquaintance Kepler observed them with his new camera obscura. None of them said that the spots belonged to the Sun. What started his drawings was a new book by Jesuit Christoph Scheiner that he received for review. Scheiner argued that sunspots were little satellites circling the Sun, like the Moon circles the Earth. So he decided to check. And now he found that Scheiner was wrong – the spots were a part of the Sun! They changed their position on the Sun every day and after 28 days came back to where they started. The Sun had turned in a full circle. That was scary. The Sun should not have any spots. And it should not rotate. It is a heavenly fire, as Aristotle had said many years earlier. This is what everyone believed. The man with the drawings was scared because of his discovery. And he was right, although then he could not predict that almost 30 years later these very drawings would put him into prison. His name was Galileo Galilei.

Lead In Chapter 21 we learned the laws of reflection and refraction and applied them for situations mostly involving laser beams. In this chapter we learn how these laws can be used to explain how mirrors and lenses work, and will consider mirror and lens applications including mirrors for makeup and shaving, store surveillance mirrors, cameras, eyes and eyeglasses, telescopes and microscopes.


22.1 Plane mirrors We start this mirror-lens study with the simplest case – a plane mirror. A plane mirror

is made of flat glass with one side covered with a metal film that reflects light. When you stand in front of a mirror, you see a reflection of yourself. How does the second “you” appear?

Before answering this question, remember the model that we created to describe how extended objects emit light. Each point on a luminous object sends light rays in all directions. Some of these rays enter our eyes and we see the object at the place where the rays reaching our eyes originate (Fig. 22.1). Now, suppose that a small shining object is in front of a mirror. What happens to the rays emitted by the object when they reach the mirror? Consider Observational Experiment Table 22.1.

Figure 22.1 Seeing light from an object

Observational Experiment Table 22.1 Seeing a point object in a plane mirror.

Observational experiment Analysis A tiny light bulb is held 20 cm in front of a plane mirror. Observers A, B, and C look at the reflected light and point a ruler at the image. The dashed lines indicate the orientations of their rulers.

For friend A to see the image, one or more rays of light reflected from the mirror must reach her/his eyes. Rays 1 and 2 after reflection reach observer A’s eyes.

Patterns/Hypotheses x Rays 1 and 2 reaching observer 1 do not originate at the same spot

but the observer might think of them as coming behind the mirror at the place where their continuations intersect—where they seem to originate. This is the location of the perceived image of the bulb produced by the mirror.

x The ray diagram shows that the perceived image of the bulb is produced at the same distance behind the mirror as the bulb is in front of it.

To explain the experiment in the Observational Experiment Table 22.1, we devise

two hypotheses. (1) The image of the real shining object in the mirror that a person sees is the

ALG 22.1.1-22.1.2


place where the rays sent by the object and reflected off the mirror to the eye of the observer seem to originate. This image is behind the mirror and not on the surface of the mirror. (2) The image (the place where rays reflected off the mirror seem to originate) is exactly the same distance behind the plane mirror as the object is in front of it. Let us test these hypotheses. We assume that a person “sees” an image at the location where the extensions behind the mirror of the reflected rays intersect. Testing Experiment Table 24.2 Testing the image location of a plane mirror.

Testing experiment Prediction Outcome (a) Repeat the Table 22.1 experiment for person A but this time cover the part of the mirror directly in front of the bulb.

If the image is due to the reflected rays, then even if we cover part of the mirror, there are still reflected rays reaching our eyes. The location of the image should not change. It might be less bright.

We see the image of the bulb exactly where it was before.

(b) Replace the mirror with a clean sheet of glass (it reflects some of the light and allows some of it to pass through). Place a lit bulb 20 cm in front of the glass. Where can you place an identical bulb 2 behind the glass so the person looking in the mirror sees one bulb?

According to the second hypothesis, the image of the first bulb is exactly the same distance behind the mirror as the bulb is in front. Thus, bulb 2 should be 20 cm behind the glass. If you remove the glass, the person looking at it should not see any difference.

The light from bulb 2 appears to come from the same place as the image of the first bulb. Removing the glass does not change the number of bulbs the person sees.

Conclusions x Experiment (a) Our reasoning about image formation has not been disproved by this experiment.

However, it disproved the idea that the image forms on the surface of the mirror. x Experiment (b) gives us confidence that the image is the same distance behind the mirror as the

object is in front of it.

We now know where the image produced by the mirror is, but you probably still feel uneasy about the image being somewhere behind a wall when you are looking in a bathroom mirror. If you live in a dorm, the image might be in your neighbor’s room! It is important to understand that the image seen in a plane mirror is not real – there is no real light coming from behind the mirror. The reflected light appears to originate from a point behind the mirror. Our mind thinks that the image is there. This is called a virtual image.

Plane mirror virtual image A plane mirror produces a virtual image at the same distance behind the mirror as the object is in front. A virtual image is at the position where the paths of the reflected rays seem to diverge from behind the mirror. The light does not actually pass through that image point.


Conceptual Exercise 22.1 Where is the lamp? You place a tilted lamp in front of a mirror at the position shown in 22.2a. Where do you see the lamp when looking at the mirror?

Figure 22.2(a) Finding the image of an object seen in plane mirror

Sketch and Translate We need to find the virtual image of the lamp in the mirror. Remember that the virtual image of each point of an object is where the extensions of the reflected rays from that point intersect (that is, from where the observer thinks the reflected rays originate). Simplify and Diagram Assume that all points of the lamp (including the base) send out multiple rays. We can then consider each point on the lamp as a source of light rays that diverge outward in all directions. Consider the top and the bottom of the lamp and locate the images of these two points. Assume that the images of points on the lamp between its end points are formed between the end point images. Draw rays from each end point – the image of that point will be where the extensions of the reflected rays intersect behind the mirror. As you can see from the ray diagram (Fig. 22.2b), the image of each point is behind the mirror and each point is at the same distance behind the mirror as the point on the lamp is in front of the mirror.

Figure 22.2(b)

Try It Yourself: What happens to the distance between an observer and her image in a plane mirror, if the observer moves backward doubling her distance from the mirror? Answer: The distance between the observer and the image quadruples.

ALG 22.2.1-22.2.3


Conceptual Exercise 22.2 Can you save money by buying a shorter mirror? You want to save money when buying a mirror. Your only requirement for that mirror is that you can see yourself from head to toe. How long should the mirror be? Sketch and Translate Draw a sketch of the situation (Fig. 22.3a). What height H mirror do you need to buy? The problem says nothing about your height; thus assume that your height is h and then express the mirror height H as a fraction or a multiple of your height. There is also no information about your distance from the mirror; assume that you stand a distance d from the mirror. It is important to understand what the problem asks: you need to see your entire body in the mirror. Each point of your body emits light in all directions. Rays from the top of your head and from the bottom of your feet should reflect off the mirror into your eyes.

Figure 22.3(a) Seeing self in mirror

Simplify and Diagram Assume that the mirror is vertical on a wall. Also assume that you are a shining object of a particular height with the eyes located some distance below the top of your head. Use the sketch in Fig. 22.3a to help draw a ray diagram. The vertical distance between the feet and eyes is h1 ; the vertical distance between the top of the head and the eyes

is h2 ; and h1 � h2 h (Fig. 22.3b). You will see a toe if a ray from your toe reaches your eye

after reflection. This is ray 1 on the diagram. It hits the mirror at the height equal to h1 / 2

above your toes. Ray 2 comes from the top of your head. It hits the mirror at a distance h2 / 2

below a horizontal line from the top of your head to the mirror. From the diagram we see that we can cut the bottom h1 / 2 off of the mirror and the top h2 / 2 off the mirror and still see

the whole body. Thus the length of the smallest mirror is h1 / 2 � h2 / 2 h / 2 . Note that the

distance d between you and the mirror does not enter our reasoning – you see yourself in this size mirror at any distance from the mirror.

Figure 22.3(b)


Try It Yourself: You are 1.6 m tall and stand 2.0 m from a plane mirror. How high is your image? How far from the mirror on the other side is it? What is your image height if you double your distance form the mirror? Answer: 1.6 m; 2.0 m; the same size. Review Question 22.1 A mirror is hanging on a vertical wall. Will you see more of your body or less if you come closer to the mirror?

22.2 Qualitative analysis of curved mirrors In the previous section we learned about flat mirrors. Flat mirrors produce images of

objects that have the exact same size as the objects themselves. You have probably seen other mirrors that make your face look bigger – women use them to apply makeup; men use them for shaving. How are these mirrors different from plane mirrors? Instead of being cut from a flat piece of glass with one side covered with a shinny metal film, they are cut from a segment of a spherically shaped piece of glass and have a metal film either on the outside of the spherical segment (convex mirrors) or on the inside (concave mirrors). Several terms are important here (Fig. 22.4): x The center of the sphere of radius R from which the mirror is cut is called the center of

curvature of the mirror. x A horizontal line connecting the center of curvature with the center of the mirror’s

surface is called the main axis. x The point where the main axis intersects the mirror is called the center of the mirror.

Figure 22.4 Concave and convex mirrors

Concave mirrors

We start our analysis by cutting a concave mirror in a narrow band about 1 or 2 cm wide and laying the flat cut side on a piece of paper (Fig. 22.5). Then shine a laser beam parallel t the plane of page on the mirror and trace the path of the incident and reflected light (see the examples shown in Table 22.3).


Figure 22.5 A concave mirror reflects laser beam

Observational Experiment Table 22.3 Laser beams shining on a concave mirror.

Observational experiment Analysis (a) The paths of laser light incident and reflected from a concave mirror.

The line from the center of curvature is a normal line perpendicular to the surface of the mirror. The law of reflection accounts for the path of the reflected ray.

(b) This time send several beams parallel to the main axis. Reflected beams all pass through the same point on the main axis.

Using the law of reflection (the normal line here is the radius of the mirror) we find that all reflected rays should pass through the same point on the main axis (called the focal point of the mirror).

(c) This time send several beams parallel to each other but not parallel to the main axis. One ray passes though the center of curvature. All reflected rays pass through the same point.

According to the law of reflection, the ray passing through the center of curvature reflects back on itself. All other rays reflect according to the law of reflection. They pass through the same point on a line perpendicular to the main axis and the same distance from the center of the mirror as the focal point (in the focal plane).

ALG 22.1.4; 22.1.6 – 22.1.8


Patterns 1. The mirror reflects light according to the law of reflection; the radius line drawn from the center of curvature of mirror O to the point where any incident ray hits the mirror is the normal line. 2. All incident rays traveling parallel to the main axis after reflection pass through the same point on the main axis—the focal point of the lens. 3. All incident rays traveling parallel to each other but not parallel to the main axis after reflection intersect above or below the main axis and at the same distance from the lens as the focal point.

We found that a curved mirror reflects light in agreement with the law of reflection. But unlike a flat mirror that leaves incident parallel rays parallel after the reflection, the concave mirror causes parallel incident rays parallel after refection to converge through a single point. If the incident rays are parallel to the main axis, this point is called the focal point F of the mirror (see (b) in Table 22.3). That is why a concave mirror is sometimes called a converging mirror. The distance from the focal point to the mirror’s surface is called the focal length f . If a mirror has a large radius of curvature compared to its size and the

rays are close to the main axis, the focal point is half way between the center of the mirror

and the center of curvature ( f R2

). See the geometric proof in Fig. 22.6.

Figure 22.6 The focal length f is half the radius of curvature R

Finally, a concave mirror converges not only rays parallel to the main axis, but any

rays parallel to each other – after reflection they all go through the same point that is in a plane that passes through the focal point—this point belongs to a focal plane and sometimes is called a secondary focal point. This point is where the parallel ray that passes though the center of curvature crosses the focal plane; this ray reflects back on itself since it travels along a line that is perpendicular to the mirror (the normal line). See (c) in Table 22.3. Ray diagram for locating the image of an object for a concave mirror


We can use rays to locate the images produced by objects in front of concave mirrors. The method is illustrated in Skill Box 22.1. The object in this case is a light bulb that is placed beyond the center of curvature of a concave mirror.

Skill Box: Constructing a ray diagram to locate the image of an object produced by a concave mirror

The image formed in the skill box is called a real inverted image. The reflected rays

converge at a point on the same side of the mirror as the object and below the main axis of the mirror. You could actually see an inverted light bulb at the image position. What happens if the object is moved closer to the mirror—between positions O and F? Try the next conceptual exercise.

Tip! We choose one of three convenient rays to draw an image, however you must remember that each points of an object emits an infinite number of rays that reflect off the mirror according to the law of reflection. Rays 1, 2, and 3 are used for simplicity, as one does not


need a protractor to find their reflections. You can also use any ray that is incident to the mirror (ray 4). To find how it travels after reflection you draw another ray parallel to it passing through center f curvature O – this ray reflects on itself. Both this ray and ray 4 pass through the same point in the focal plane after reflection. Conceptual Exercise 22.3 Where is the image? Suppose the light bulb in Skill Box 22.1 is moved forward between the center of curvature O and the focal point F. Use the ray diagram technique to locate the image of the bulb. Sketch and Translate Consider the bulb as a shiny arrow with its tail on the main axis and oriented perpendicular to the axis. Use the ray diagram method to estimate the new image position and orientation. Simplify and Diagram We can use rays 1 and 2 to estimate the image location and orientation (see Fig. 22.7a). We cannot use ray 3 as light from the tip of the arrow that passes through O does not hit the mirror and will not contribute to the reflected image. Rays 1 and 2 have a similar behavior as that used in Skill Box 22.1, except ray 2 now has a larger angle of incidence, as the bulb is closer to the mirror. Thus, the reflected ray 2 also has at a larger angle of reflection. Rays 1 and 2 converge below the main axis again producing an inverted image. The image is now farther from the mirror than point O and is also bigger. If you place a screen at the location of the image you would see a larger inverted light bulb there (Fig. 22.7a).

Figure 22.7(a) Object close to concave mirror (F < S < 2F)

Try It Yourself: Draw the image of the same object using ray 1 and an arbitrary ray. Answer: See Fig. 22.7b.

Figure 22.7(b)

ALG 22.2.4-22.2.6; 22.2.8-22.2.10


Tip! If ray 2 does not hit the mirror as the angle becomes too large, you can just extend the mirror. Those rays are the ones we choose for simplicity, in real life there are many more rays that actually hit the mirror and reflect back to form the image. Virtual image In Skill Box that helped you learn how to construct a ray diagram and in Conceptual Exercise 22.3, the rays from the object after reflection converged on the same side of the mirror as the object. Due to this convergence an observer sees a real image. Many important applications of mirrors and lenses involve situations in which the rays do not converge—for example in make-up and shaving mirrors. Consider the situation in Fig. 22.08a. There your face is closer to a concave mirror than the focal point F.

Figure 22.8 Object close to concave mirror

We will use rays 1 and 2 (see Fig. 22.08b). To draw ray 2, draw a line from the top of the object that passes through the focal point. This is the direction of ray 2. When this line touches the mirror, the reflected ray is parallel to the main axis. Reflected rays 1 and 2 diverge after reflection and never converge to form a real image on the same side of the mirror as the object. If your eye looks at this reflected light, it seems to be coming from a point behind the mirror. This point is the location of the virtual image of the object. The image is said to be virtual because the light does not actually come from that point—it just appears to come from that point. Thus, for this situation, a virtual upright image is formed. The image is magnified (bigger than the object) if you are close to the mirror. Hence, you can use the mirror for makeup and shaving.

You can check the reasoning for the Skill Box and Conceptual Exercise situations using a tablespoon as a mirror. For the virtual image situation, you need a big tablespoon. If you hold it horizontally really close to your mouth, your eyes will see a large upright image of your mouth.

So far drawing images of objects we skipped drawing the image of the bottom of the object. The bottom of the object was always on the main axis. We assumed that the image of the object-representing arrow perpendicular to the main axis will be the arrow which is also perpendicular to the main axis. To validate this assumption we need to learn how to draw the

ALG 22.2.4-22.2.6; 22.2.8-22.2.10


image of the bottom of the arrow. To do this, use the main axis as ray 1. Then use the method from the Try It Yourself 22.3. Draw another ray 2 in any direction (arbitrary ray) from that object point, as long as it hits the mirror. Now, draw a ray 3 parallel to ray 2 originating from the center of curvature. This ray reflects on itself. Ray 2 after reflection and ray 3 go through the same point in the focal plane perpendicular to the main axis (Fig. 22.09). Continue ray 2 through that plane to find the point where it intersects ray 1. That is the location of the base of the object. Now you can try on your own to find the image of the top of the arrow and find out whether the shape of the image is the same as the shape of the object (in this case perpendicular to the main axis).

Figure 22.9 Locating base of image

Convex mirrors

We now repeat the approach used to study the properties of concave mirrors. Do convex mirrors have a focal point, as concave mirrors do? Consider Observational Experiment Table 22.4. Observational Experiment Table 22.4 Where is the focal point for a convex mirror?

Observational experiment Analysis Repeat the procedure used with a concave mirror but this time send only two laser beams parallel to the main axis. We observe that the reflected beams diverge.

Using the normal lines passing though the center of curvature, we find that although reflected rays do not pass through the same point but if we extend them behind the mirror, their extensions do intersect.

Pattern x Light rays moving parallel to the main axis of a convex mirror reflect and diverge away from

each other. x Lines drawn backward along the direction of the reflected light cross the axis at a focal point

behind the mirror.

ALG 22.1.5


Recall that for a concave mirror, incident rays parallel to the main axis after reflection converged at a focal point in front of the mirror. On the other hand, rays parallel to the main axis incident on a convex mirror diverge after reflection. They appear to diverge from what is called a virtual focal point F behind the mirror. The virtual focal point of a convex mirror is a

focal distance f equal to half the radius of curvature R behind the mirror ( f R2 ), as

shown in Fig. 22.10. Where does a convex mirror form images of an object and what is the nature of these images? We use ray diagrams.

Figure 22.10 A convex mirror

Ray diagram for locating the image of an object in front of a convex mirror A ray diagram method for qualitatively locating the image of an object that is produced by a convex mirror is described in Skill Box below.


Skill Box: Constructing a ray diagram to locate the image of an object produced by a convex mirror

In the convex mirror ray diagram in Skill Box for the convex mirrors, the object was about 1.6R from the mirror. An upright virtual image was formed about 0.4R behind the mirror. Let us apply this ray diagram method to see how the image distance and the nature of the image changes when the object is somewhat closer to the mirror.

Conceptual Exercise 22.4 Looking into a convex mirror You hold a convex mirror 0.7R in front of your face. Approximately where is the image of your face and what is the nature of that image.

ALG 22.2.1

The special rays whose directions after reflection are known 1. Ray 1 travels from the tip of the object toward the mirror parallel to the main axis.

After reflection, it seems to diverge form a focal point F behind the mirror. 2. Ray 2 passes toward the center of curvature O. It hits the mirror perpendicular to

its surface and reflects back in the same direction.

1. Place a vertical arrow on the main axis to represent the object. Each point on the object emits light in all directions. We find the image location of only the tip of the object arrow.

2. Choose two special rays from the tip of the arrow (described below). You know the directions of these rays after reflection from the mirror. The rays diverge after reflection.

3. Extend the rays behind the mirror (the dashed lines). A virtual image is formed where these two or three rays seem to diverge from behind the mirror.

4. The eye sees the image at the place where the reflected rays seem to diverge—at the image point.

O F

Ray 1

Ray 2

Object

Image


Sketch and Translate Represent your face by an arrow. Instead of the image of every point on you face (every point of the arrow), we locate only the image of the tip of the arrow and use this tip location to decide the image location. Use the ray diagram method described in Skill Box for convex mirrors. Simplify and Diagram Draw a ray diagram (see Fig. 22.11). We use rays 1 and 2 described in the Skill Box for convex mirrors, to locate the image. The image is virtual and upright about 0.3R on the backside of the mirror. It is the same type as in the skill box but is now taller and closer to the mirror. You can check the nature of the image by observing your face when looking at the backside of a shiny tablespoon. The images for convex lenses are always upright and virtual on the backsides of the mirrors.

Figure 22.11 Image due to convex mirror

Try It Yourself: Where do you need to place an object relative to a concave or convex mirror to have an image that is: (a) real, bigger than the object; (b) real, smaller than the object; (c) virtual, bigger than the object; and (d) virtual, smaller than the object. Answers: (a) Concave 2R ! object ! f ; (b) Concave object ! 2R ; (c) Concave

object < f ; (d) Convex with object at any location.

Review Question 22.2 What are the differences between real and virtual images?

22.3 Quantitative analysis of curved mirrors Our ray diagrams for curved mirrors indicated that the distance to the image from the

mirror depends on where the object is located and on the mirror’s type and focal length. We will use the ray diagram in Fig. 22.12 to help derive a mathematical relationship between the following three quantities: x the distance of the object from the mirror s , x the distance of the image from the mirror s ' , and x the focal length of the mirror f .

ALG 22.3.1 – 22.3.2


Figure 22.12 Ray diagram to help develop mirror equation

The following notation in Fig. 22.12 is used to help develop the desired relationship: AB

is the object; and A1B1 is the image of the object. M, N, and C are points on the mirror where light rays strike it. We assume that the mirror is curved very little and the rays are close to the main axis. Now, using this notation, we make the following steps to complete the derivation. x AM | BC s . x A1N | B1C s ' .

x CM | BA and CN | B1A1 .

x ABF and CNF are similar triangles. Thus,

CNCF A1B1

CF ABBF or AB

A1B1 BF

CF (s – f )f .

x A1B1F and MCF are similar triangles. Thus,

A1B1B1F

MCCF AB

CF or ABA1B1

CFB1F

fs '– f .

x Setting the previous two ABA1B1

ratios equal to each other, we find that:

s� ff

f

s'� f. (22.1)

After some algebra (see the tip below), we get a relationship that is called the mirror equation: 1s�

1s'

1f

(22.2)

Tip! To get from Eq. (22.1) to (22.2), multiply both sides by the product of f (s '– f ) . Then

carry out the multiplication on the left side of the equation and add the quantities. You should have ss '– sf – fs ' 0 . Now divide both sides of the equation by the product of ss ' f . You

should arrive to Eq. (22.2).

Equation (22.2) allows us to predict the distance s ' of the image from the mirror when the distance s of the object from the mirror and the mirror’s focal length f are known.

Before we test the equation experimentally, let us check to see if it is consistent with an extreme case. We know that a concave mirror focuses parallel incident rays at the focal point.


If an object is infinitely far away, we can assume that rays from the object reaching the mirror are parallel to each other. If Eq. (22.2) is correct, and we use infinity for the object distance, then we should find that the image distance for that object equals the focal length distance. Using Eq. (22.2) we get:

1f�

1s ' 0 �

1s '

1f

.

Therefore the image is at the focal distance from the mirror. The equation survived this extreme case analysis. However, we also need to test the equation experimentally. Example 22.5 Were should the screen be? You have a concave mirror whose radius of curvature of 0.60 m. You place a candle 0.80 m from the mirror. Where should you put a screen to see the image of the candle? Sketch and Translate Draw a sketch of the situation (Fig. 22.13a). Assemble all parts of the experiment on a meter stick so the distances can be easily measured. The known quantities are: s 0.80 m and f R / 2 0.30 m . To find where to place the screen, we need to find s ' .

Figure 22.13(a) Candle in front of concave mirror

Simplify and Diagram Assume that the candle can be modeled as a shiny arrow. Each point radiates multiple rays; draw a ray diagram using two (or more) rays that travel in known directions after reflection (Fig. 22.13b).

Figure 22.14(b)

Represent Mathematically We see from the ray diagram that the image is real, inverted, and closer to the mirror than the candle. Rearrange Eq. (22.2) to find s ' :

1s '

1f�

1s

.

Solve and Evaluate


1s '

1f�

1s

10.3 m

�1

0.8 m

0.8 m � 0.3 m(0.3 m)(0.8m)

2.08 m�1 .

Thus, s ' 12.08 m–1 0.48 m . The image is closer to the mirror than the object, as predicted

by our ray diagram. Thus, the predictions using mathematics and the ray diagram are consistent with each other. We can now take the screen and place it where we predicted and there will be the inverted real image of the flame. Try It Yourself: Where is the candle’s image when the candle is 0.20 m from the mirror? Does the answer make sense to you? Answer: –0.48 m. The image is behind the mirror – it is a virtual image. Evidently, the negative sign must mean that the image is virtual.

In the Try It Yourself part of the last example, the mirror equation led to a negative value

for the distance between the mirror and the image. The image was virtual. We need to agree on some sign conventions when using Eq. (22.2). If the focal point of a mirror is on the backside of the mirror or the image of an object is virtual (behind the mirror), the focal distance and/or the image distance has a negative sign:

x The distance s ' for a virtual image on the backside of a mirror is negative. x The focal distance f for a convex mirror is negative.

Let us see if this works when we use the mirror equation for a convex mirror. Quantitative Exercise 22.6 Test the mirror equation for a convex mirror You have a convex 0.50-m radius mirror and hold your face 0.60 m form the mirror. Where is the virtual image that you see in the mirror? Sketch and Translate The situation is sketched in Fig. 22.14a. The known information is: s 0.60 m and f = –R/2 = –(0.50 m)/2 = –0.25 m . Note that for this convex mirror, the

focal distance is a negative number.

Figure 22.14(a) Image of face produced by convex mirror

Simplify and Diagram Draw a ray diagram representing your face as a shiny arrow (Fig. 22.14b). The image is upright, virtual, and closer behind the mirror than the object is in front of it. Let us see if the mathematics matches this prediction.


Figure 22.14(b)

Represent Mathematically Rearrange the mirror equation [Eq. (22.2)] to locate the image of your face.

1s '

1f�

1s

.

Solve and Evaluate Insert the know information: 1s '

1–0.25 m

�1

0.60 m �5.67 m–1 .

The image distance is s ' 1–5.67 m–1 = –0.17 m . We see that the image distance is negative

and its magnitude is less than the magnitude of the object distance—consistent with the ray diagram. Try It Yourself: You have a convex mirror. The image of your face is upright, virtual, and 0.30 m behind the mirror when you hold the mirror 1.0 m from your face. Determine the radius of the sphere from which the mirror was cut. Answer: The object distance s 1.0 m , the image distance s ' –0.30 m , and the focal length f –0.43 m . Consequently, R 0.86 m .

It appears that the mirror equation works equally well for concave and convex lenses.

Mirror equation The distance s of an object from the center (front surface) of a concave or convex mirror, the distance s ' of the image from the center of the mirror, and the focal length f of the mirror are related by the mirror equation:

1s�

1s'

1f

(22.2)

The following sign conventions apply for these mirrors: x The focal length is positive for concave mirrors and negative for convex mirrors. x The image distance is positive for real images in front of the mirror and negative for

virtual images behind the mirror. x The focal length is half the radius of curvature: f R / 2 .

Magnification

You probably noticed that image sizes were sometimes bigger and sometimes smaller than the size of the object. We are familiar with this effect – your face looks much bigger when you look into a special bathroom mirror that amplifies all the little details that you cannot see


without it. This change in size of the image compared to the object (magnification of the object either making it smaller or bigger) involves a new physical quantity called linear magnification m . This quantity is defined in terms of the image height h ' and the object height h , where the height of the image or object is the perpendicular distance of each relative to the axis of the mirror:

linear magnification image heightobject height

m = h 'h

, (22.3)

where a height is positive if the image or object is upright and negative if inverted. To find how magnification is related to the object/image locations and the focal

length of a particular mirror, consider Fig. 22.15.

Figure 22.15

We draw the image of the object using rays 1 and 2 and then draw a new ray that travels form the top of the object to the center of the mirror. Using a protractor you can find the reflected ray - it has the same angle T with the main axis as the incident angle. Thus,

tanT hs

h 's '

.

Therefore the absolute value of the magnification is m h 'h s '

s . If the image is real, it

will be inverted. The height of the inverted image is considered negative (just a convention). Hence h '/ h will be negative, whereas s ' will be positive—the ratios need to have the

opposite signs. To correct this difficulty, a negative sign is added. Thus, m h 'h – s '

s .

Linear magnification m is ratio of the image height h ' and object height h and can be determined using the negative ratio of the image distance s ' and object distance s :

m h 'h –

s 's

. (22.4)

Remember that the height is positive for an upright image or an upright object and negative for an inverted image or object.

ALG 22.4.1-2.4.4


Example 22.7 A close examination Use a concave mirror of radius of curvature 0.32 m to examine your face when doing makeup or shaving. What is the magnification of the mirror and how big is the image of a 0.0030-m diameter birthmark on your face when your face is 0.080 m from the mirror? Sketch and Translate Visualize your face in front of the mirror, including a ray diagram (Fig.

22.16). The givens are h 0.30 cm , s 0.080 m , and f R2 �0.16 m . We need to

determine the linear magnification m and then the diameter of the birthmark image h ' .

Figure 22.16 Image produced by makeup/shaving mirror

Simplify and Diagram The ray diagram is shown in Fig. 22.16. We represent the birthmark as an arrow. Note that although the tip sends multiple rays, we use two rays from the tip of the object arrow. Pay special attention on how Ray 3 travels. The reflected rays diverge. Follow the dashed lines back from them on the back side of the mirror to find where they intersect. This is the place where reflected light seems to come from and the location of the image point of the arrow’s tip. It appears that we get an enlarged upright virtual image behind the mirror. Its magnification will be the same as the magnification of the face. Represent Mathematically First use Eq. (22.2) to determine the image distance s ' . Then use Eq. (22.4) to find the magnification m and the image height h ' . Solve and Evaluate Rearranging Eq. (22.2), we find:

1s '

1f

–1s

1(0.16 m)

�1

(0.08 m) �6.25 m–1 .

or s ' 1

–6.25 m–1 –0.16 m . Use Eq. (22.4) to determine the linear magnification:

m –s 's –

(–0.16 m)(0.08 m)

�2.0 .

The magnification is also m h 'h

, so:

h ' mh (�2.0)(0.30 cm) = +0.60 cm .


The birthmark image is upright and two times bigger than the object. All of the features on your face will look about twice as big. Try It Yourself: You hold a 1.0-cm tall coin 0.20 cm from a concave mirror with focal length +0.60 m. Is the image of the coin upright or inverted and what is its magnification and height? Answer: The image is upright; m 1.5 and h ' = 1.5 cm . Tip! Notice that the mirror size does not enter into the mirror equation and does not affect the magnification. However, its radius of curvature and focal length do affect the image location and hence the magnification. If you cut the same mirror in half, the size of the image should not change. However, the brightness of the image will be less as less light will reflect off the smaller mirror. Review Question 22.3 Compare and contrast concave mirrors and plane mirrors in terms of image formation.

22.4 Lenses—qualitative analysis You are probably familiar with magnifying glasses. Maybe you held one above a piece

of paper in sunlight producing a small bright spot on the paper and maybe even smoke. In this section we learn why.

Lenses, like curved mirrors, can be concave and convex. A convex lens has at least one surface formed from the outside of a sphere as shown in Fig. 22. 17a. You can see the centers of the sphere O1 and O2 in the figure. Sometimes one of the surfaces is flat, but we will not consider such lenses. A concave lens is sometimes made of the segments of two spherically shaped surfaces that are curved inward as shown in Fig. 22.17b.

Figure 22.17 Lens are made from spherical surfaces

Convex lenses

Lenses form images by changing the direction of light. In Observational Experiment Table 22.5, we show the path of several parallel laser beams passing through a convex lens made of solid glass. Remember that the index of refraction of glass is greater than that of air, so when a ray of light moves from air to glass, it bends towards the normal line and when it moves from

ALG 22.1.10-22.1.11


glass to air – it bends away from the normal line. Both bendings cause the ray to move toward the axis of the lens. The normal line at each interface is a line drawn along a radius of the sphere from which the lens was cut. Observational Experiment Table 22.5 Laser beams passing through a glass lens.

Observational experiment Analysis 1. Shine three laser beams parallel to the main axis.

The path of each ray can be explained by refraction relative to the normal lines at the air-glass interfaces (see ray 1 below). The ray converges when entering the left side of thelens and converges even more to the focal point F when leaving the right side.

2. Shine three laser beams parallel to each other at an arbitrary angle relative to the axis of the lens.

The path of each ray can be explained by refraction relative to the normal lines to each air-glass interface. The rays converge at the same distance from the lens as the parallel rays in Experiment 1.

Patterns--due to the shape of the lens

1. The rays passing through the center of the lens do not bend. 2. The rays parallel to the main axis pass through the same point after the lens—the focal point. 3. The rays parallel to each other pass through the same point but not on the main axis, although in

the same plane with the focal point—in the focal plane.

We found that a convex lens made of glass is similar to a concave mirror where incident rays parallel to the main axis focus at a focal point after passing through the lens. Note in Fig. 22.18 that the rays converge more when passing through a thick lens as opposed to a thin lens. The focal point is closer to a thick lens than to a thin lens.

Figure 22.18 Thick and thin convex lenses

Tip! In all our experiments light shined from the left side of the lens. The focal point was on the right side of the lens. If you repeat the experiments shining the light from the right, the


rays converge on the left side of the lens. A lens has two focal points—at equal distances from the lens on each side.

Lenses have been in use for a long time. One use was to burn things. For example, in the chemical experiments that led to the idea of mass conservation discussed in Chapter 5, Lavoisie used a burning glass, a large convex lens that focused the Sun’s rays on iron chips in a closed flask and caused them to become so hot they caught fire!

In fact, people knew about burning glasses long before chemical experiments. In the ancient world people used vases with water as lenses to start fires. The vase with water was placed between the Sun and a material that would burn. The vase had a cylindrical convex shape that focused light on the material causing it to become very hot. In modern times Solar furnaces produce extremely high temperatures without fuel or electric energy. They use a large parabolically shaped set of mirrors to focus light at high intensity.

So far the lenses we have used were made of glass (or water) in air. Their material has a higher index of refraction that air. What will happen if you place a convex lens filled with air in a medium such as water? Draw a ray diagram to answer this question and you will find out that instead of converging parallel rays such lens diverges them. As in practical life people mostly use glass lenses in air, we will only consider such cases in this book.

Concave lenses

Now that we have an understanding of how a convex lens bends light, let us send parallel laser beams through a concave lens made of glass to see what happens (Fig. 22.19). Unlike a convex lens, a concave lens does not bend laser beams closer to the main axis, but on the contrary causes them to refract away from the axis. The laser beams that move parallel to the axis before reaching the lens, diverge after passing through the lens. The dashed lines drawn back through the lens indicate places where the light seems to originate if your eye looked at the rays after passing through the lens. All of these parallel rays seem to diverge from one point on the axis. This point is called the virtual focal point.

Figure 22.19 Focal point of concave lens

Ray diagrams for lenses

ALG 22.1.12 – 22.1.13


Ray diagrams were useful for curved mirrors in qualitatively locating images and in checking mathematical image location. They are even more useful for the same reasons when analyzing images produced by lenses. The method for constructing ray diagrams for lenses is summarized in Skill Box below. Skill Box: Constructing ray diagrams for single lens situations

F F Image

f f s s’

Ray 1

Ray 2

Ray 3Object

2. Place an object arrow at the object position with its base on the axis a distance s from the center of the lens.

O

1. Draw an axis for the lens situation and a vertical line through the axis representing the location of the lens. Place dots a distance f on each side of the lens representing the focal points F.

3. choose two or three special rays whose direction you know after the light rays pass through the lens. The rays are described below.

4. The place where the rays intersect on the right side at a distance s ' from the lens is the location of a real image of the object. If the rays diverge after passing through the lens, the place from which they seem to diverge on the left side at a distance s ' from the lens is the location of a virtual image (your eye looking at light passing through the lens thinks the object is at the virtual image location).

5. Description of special rays used to find image type and location. x Ray 1 leaves the object and moves parallel to the main axis. As the ray passes through the lens,

it passes through the focal point on the right side of the lens. If it passes through a concave lens, ray 1 is refracted away from the main axis and appears to come from the focal point on the left side of the lens.

x Ray 2 leaves the object and passes directly through the middle of the lens and for thin lenses, its direction does not change.

x Ray 3 leaves the object and passes through the focal point on the left. A convex lens refracts the ray so that it moves parallel to the axis on the right. For a concave lens, ray 3 leaves the object and moves toward the focal point on the right. Before reaching the focal point, the ray is refracted parallel to the axis on the right.


As you probably know from experience, lenses are used in magnifying glasses,

cameras, glasses that we wear, microscopes, telescopes, and in many other devices. Some of them require the image to be smaller than the object (camera) and some require that the image is larger than the object (magnifying glass). Ray diagrams help us understand how lenses form images of objects in order to explain how all these devices work.

Table 22.6 summarizes possible combinations of the locations of objects and images produced by different types of lenses. Cover all of the columns in the table except the very left one with a piece of paper. Redraw the situations in the left column in your notebook. Then apply your knowledge constructing ray diagrams to find an image location. Start by choosing the rays whose paths you know. Draw the paths of these rays before and after the lens. Find the place where the rays cross – this will be the image of the top of the object. Assume that the image is perpendicular to the main axis. After you do this – open the next column in the table and check your work. In the third column you need to decide if the image is real or virtual, upright or inverted, and enlarged or reduced in size. After you fill out that column for every case, open the last column in the table to check your work. How did you do? Table 22.6 Drawing images for lenses.

Situation Ray diagram Description of the image

s ! 2 f

Real, inverted, reduced

2 f ! s ! f

Real, inverted, enlarged

f ! s

Virtual, upright, enlarged

F F

object

F F

object

F F

object

F F

object image

F F

object

image

F F object

image


s ! f

Virtual, upright, reduced

f ! s

Virtual, upright, reduced

Now we can use Table 22.6 to think of different applications of lenses. For example, in a camera we need to have a reduced real image of an object – we probably need to use arrangement 1. If we want to project a computer image++ on a screen, we need arrangement 2. If we want to use a lens as a magnifying glass, we use arrangement 3. Image location of the base of the object

So far in all of our examples we found the image location of the top of the shining arrow and assumed that the image of the whole arrow remained perpendicular to the main axis. Let us check whether this is the case. But before we do this, recall the observations described in Table 22.5(b): rays parallel to each other but not parallel to the main axis after refraction pass through the same point. This point is on the focal plane of the lens – a plane perpendicular to the main axis and passing through the focal point (Fig. 22.20).

Figure 22.20 Parallel rays converge in focal plane

Now we use this idea to draw a ray diagram for the base of an object that lies on the main

axis. Consider Fig 22.21a. An infinite number of rays leave the base of the object. The ray parallel to the main axis goes through the center of the lens and does not bend (ray 1). This means that the image of the shining point should be on the main axis; but where? We do not know the paths of any of these other rays. Let us choose one of them – ray 2. We can draw an imaginary ray (ray 3 in Fig. 22.21b) parallel to ray 2 passing through the center of the lens. This ray passes

F F F F

object

image

F F

object

F F

object

image


through the lens unbent. After the lens, rays 2 and 3 pass through the same point in the focal plane.

Figure 22.21 Locating image of base of object

The image of the object is where bent ray 2 intersects with ray 1 on the main axis (Fig.

22.21b). Now we can draw the image of the arrow using the top and the bottom points (Fig. 22.21c). As you see, the image remains perpendicular to the main axis. Conceptual Exercise 22.8 Lens jeopardy In Fig 22.22a you see a shining point-like object and its image that is produced by a lens. The dashed line is the main axis of the lens. Where is the lens, what kind of lens is it and where are its focal points?

Figure 22.22(a) Finding the lens and its focal point

Sketch and Translate The situation is already sketched. We need to decide the location of the lens that produced the image of S’. We know that the shining point S sends light in all directions. Rays, which reach the lens, bend and pass through the image point S’. We need to find rays whose paths will help us to find that lens and its focal points. Simplify and Diagram Assume that the lens is thin. Also assume that the focal points are at the same distance from the lens on each side. Redraw Fig. 22.22a and think of a ray that goes through the center of the lens. It does not bend and it passes thought the image point. Thus, if we draw a line from the tip of the light source (the object) to the tip of the image, the point where the line crosses the main axis will be the location of the center of the lens (Fig. 22.22b). It looks like we are dealing with the convex lens as the image is on the other side of the lens.


We still do not know where the focal points are. We know that a ray from the tip of the object and parallel to the main axis refracts through the lens and passes through the focal point on the right side of the lens and then through the tip of the image. Thus, one focal point is where this ray passes through the axis after passing through the lens (Fig. 22.22c). Now we know the location of the focal point on the right. As we assumed that the lens is symmetrical, we can put the other focal point on the other side of the lens at the same distance from the lens.

Figure 22.22(b)(c)

Try It Yourself: A ray from the object that passes through the focal point on the left side of the lens should after passing through the lens travel parallel to the axis on the right side of the lens. Does this ray also pass through the image point? Answer: It should—draw it carefully to see if it does. Conceptual Exercise 22.9 A partially covered lens Imagine that you have an object, a lens, and a screen. You place an object at a position s ! 2 f from the lens and cover most of the

top half of the lens. What part of the image will you see – the top or the bottom? Sketch and Translate To answer the question, we need to draw a ray diagram and see what happens to the image with the top part of the lens covered (Fig. 22.23).

Figure 22.23 Image when part of lens covered

Simplify and Diagram Represent the object by an arrow and find the image by examining rays radiated from points A and B (the top and the bottom of the object). With the top of the lens covered, the rays moving toward the top of the lens do not pass through; but rays moving toward the bottom half do. Remember that all rays from one point on the object that pass though the lens converge at the image point (Fig. 22.21). Covering the top half of the lens does not change the size or location of the image. It changes the image brightness, as less light passes through the lens.


Try It Yourself: Imagine that you covered the central part of the lens. You have a shining object far away from the lens. How will the location of the image and its size change compared to the situation when not covered? Answer: The size and location will not change—only the brightness.

So far, our analysis has involved only the qualitative use of ray diagrams. To make

our work more quantitative, we need to develop a mathematical relationship between the focal length f of a lens, the location s of the object, and the location s ' of the image—the

subject of the next section.

Review Question 22.4 How do we know how many rays an object sends onto a lens?

22.5 Thin lens equation and quantitative analysis of lenses We wish to derive a relation between s, s ', and f that can be used to find one of these

quantities (for example the image location) if the other two are known. To do this we use the same technique that was used for curved mirrors. Place an object at a known distance from a convex lens, draw its image, and then examine the geometry of the situation. The distance of the object from a convex lens is s, the distance of the image from the lens is s ' , and the focal length of the lens is f . AB is an object and A1B1 is the image (Fig.22.24).

Figure 22.24 Develop math relationship between s, s ', and f

x Assume that the lens is very thin and all the rays are close to the main axis. Triangles BAO and

B1A1O are similar. Thus ABs

A1B1

s '’ or after rearranging we get:

A1B1

AB =

s 's

.

x Triangles NOF and B1A1F are similar. Thus NO

f =

A1B1

s '� f. Also, NO = AB. Thus,

F

F B

A

B1

A1

O

N

f f s s’

ALG 22.3.3 – 22.3.5


ABf

= A1B1

s '� f.

x Rearranging this last equation, we get A1B1

AB

s '– ff

x Setting the above two equations with A1B1

AB on the left equal to each other, we get

s's

s'� ff

(22.5)

After some algebra (described in the following Tip), we get a relationship that is called the thin-lens equation:

1s�

1s'

1f

(22.6)

Tip! To get from Eq. (22.5) to (22.6), multiply both sides by the product of f (s '– f ) . Then carry

out the multiplication on the left side of the equation and add the quantities. You should have ss '– sf – sf ' 0 . Now divide both sides of the equation by the product of ss ' f . You should get

Eq. (22.6).

Now that we have derived a new relation, we need to check whether it works for extreme cases. Consider a first case in which an object is infinitely far from a convex lens. As we know, rays parallel to the main axis pass through the focal point. Thus, if the equation that we derived is correct, and an object that is infinitely far away sends rays of light parallel to the main axis, then the image of that object should be at the focal point. According to Eq. (22.6) for an object infinitely far away, we get:

1f�

1s '

1f

or 0 �1s '

1f

.

Therefore s ' f as expected.

Another extreme case is when the object is at a focal point. As we know, rays which pass through the focal point on the left side of the lens after passing through the lens become parallel to the main axis on the other side. They do not converge or diverge and there is no object formed anywhere relative to the lens. Thus, if the equation that we derived is correct, and an object is at a focal point on the left side of the lens, then the image of that object should be at infinity. For an object at a focal point ( s f ):

1s�

1s '

1f

or 1f�

1s '

1f

.


Therefore 1s '

= 0 or s ' f . So far our lens equation survived these extreme case

analyses. We will test it more later. The formation of an image by a lens is summarized below, including sign conventions for lenses.

Thin lenses The distance s of an object from a lens, the distance s ' of the image from the lens, and the focal length f of the lens are related by the thin-lens equation:

1s�

1s'

1f

. (22.6)

Several sign conventions are important when using the thin-lens equation. These conventions apply for light rays moving from left to right. 1. The focal length f is positive for convex lenses and negative for concave.

2. The object distance s is positive if the object is to the left of the lens and negative if the object is to the right of the lens (important later when we analyze two or more lens systems). 3. The image distance s ' is positive for real images formed to the right of the lens and negative for virtual images formed to the left of the lens.

Tip! Remember that all of the above reasoning applies to convex and concave lenses made of glass when they are used in air (made of material of higher index of refraction than that of the medium).

How can we explain why a sharp, clear image of an object appears in only one place? If a

screen is placed closer to the lens than distance s ' (Fig. 22.25a), the rays leaving the tip of the arrow have not yet converged to a point and the image on the screen will be blurry and out of focus. If the screen is placed to the right of s ' (Fig. 20.25b), where rays from the tip are diverging after having converged at s ' , a blurry image is again the result. Only at distance s ' from the lens is a sharp, clear, focused image seen. This is important for all applications involving optical instruments, including computer projectors.

Figure 22.25 Get focused image only at s '

Computer projectors A computer has an LCD (light emitting diode) screen. The LCD output can also be transferred to three LCD panels in a projector, one each for red, green, and blue light. When the


output of each of these three LCD panels is combined, the computer screen can be projected onto an external screen by the projector. You will learn in Chapter 24 how LCD panels work. Here we analyze the lens for the computer projector that converts the combined LCD screen in the projector into the much larger image on the outside screen. Example 22.10 Computer projector The final combined output from the three LCD panels is the object inside the projector that is 0.20 m from a +0.19-m focal length projector lens. Where should we place the external screen in order to get a focused image of the computer liquid crystal display inside the projector? Sketch and Translate The situation is sketched in Fig. 22.26a, including the known information and the unknown quantity (the distance s ' of the screen from the projector lens).

Figure 22.26(a) Computer projector

Simplify and Diagram Draw a ray diagram for the situation (Fig. 22.26b). Notice that the image is inverted, enlarged and real. Thus, the internal combined LCD panel has to be inverted so that the image produced on the external screen is upright.

Figure 22.26(b)

Represent Mathematically Rearrange Eq. (22.6) to determine the unknown image distance s ' :

1s'

1f

� 1s

Solve and Evaluate Substitute the known quantities in the right side of the above to get: 1s '

1f�

1s

1(0.19 m)

� 1

(0.20 m)

0.20 m - 0.19 m(0.19 m)(0.20 m)

= 0.263 m–1 .

To get the image distance, we have to invert the above (don’t forget to invert!):

s ' 1

(0.263 m–1) 3.8 m .

The screen should be placed 3.8 m from the lens—a reasonable distance for a projector.

ALG 22.4.5


Try It Yourself: Suppose the screen needs to be 4.8 m from the lens. What distance should the final combined LCD screen in the projector be from the same lens to get a focused image on the more distant screen? Answer: s 0.198 m (the lens and LCD output inside the projector needs to be moved closer to each other). Linear magnification

As we saw in Table 22.7, a lens can produce images that are bigger or smaller in size than the objects. The same definition for magnification that we used for the curved mirrors applies also for lenses:


m h 'h

(22.3)

where the heights h ' and/or h are positive if the image or object is upright and negative if inverted. As with mirrors, one can determine the linear magnification if the image location s ' and the object location s are known. Using identical triangles in the figure for the derivation of the lens equation, we can write:


m h 'h –

s 's

. (22.4)

Example 22.11 Looking at a bug Use a lens of focal +10.0 cm to look at a tiny bug on a book page. The lens is 5.0 cm from the paper. Where is the image of the bug? If the bug is 1.0 cm big, how big is the image? Sketch and Translate Draw a sketch of the situation and label the givens and the unknowns (Fig. 22.27a).

Figure 22.27(a)

Simplify and Diagram Assume that the bug is a bright object and the lens is parallel to the paper looking down at the bug on the paper. To make the ray diagram, reorient the bug and the lens main axis so it is horizontal—our usual orientation (Fig.22.27b). From the ray diagram, the image of the bug appears to be virtual (on the same side of the lens as the real bug) but farther away, upright, and enlarged. This means that in the situation shown in (a), the image is under the desk on which the paper lies. This is where the light appears to originate.


Figure 22.27(b)

Represent Mathematically Rearrange the lens equation [Eq. (22.6)] to find the image distance s ' :

1s'

1f

� 1s

.

Then use the linear magnification Eq. (22.4) to determine the magnification and the height of the bug’s image:

m h 'h –

s 's

Solve and Evaluate Insert f 0.10 m and s 0.05 m into the lens equation to get:

1s '

1f�

1s

10.10 m

– 1

0.05 m –10 m�1 .

Therefore s ' 1

–10 m–1 –0.10 m . The minus sign indicates that the image is virtual,

consistent with the ray diagram. The linear magnification is:

m –s 's –

(–0.10 m)(�0.05 m)

�2.0 .

The bug image is upright (the plus sign) and twice the object size or 2.0(1.0 cm) = 2.0 cm ,

consistent with the ray diagram. Try It Yourself: An image seen through a 10-cm convex lens is exactly the same size as the object but inverted and on the opposite side of the lens. Where are the object and the image located? Answer: The object is at a distance s 2 f 20 cm on one side of the lens and the image is on

the other side of the lens at a distance s ' 2 f 20 cm from the lens.

Example 22.12 Find image You place an object 20 cm to the left of a concave lens whose focal length is –10 cm (the negative sign is used for a concave lens). Where is the image of this object and is the image real or virtual? Sketch and Translate A sketch of the situation is shown in Fig. 22.28a along with the known information and the desired unknown.


Figure 22.28 Find image of concave lens

Simplify and Diagram A ray diagram for this system is shown in Fig. 22.28b. In the drawing, the image is less than half the distance between the object and the lens, or about 7 cm left of the lens, that is, s ' | –7 cm . The negative sign indicates that the image is virtual and to the left of the lens. The image appears smaller than the object. If our eye looked through the lens from its right side, the light would appear to come from this virtual image. Represent Mathematically We can use the thin-lens equation [Eq. (22.7)] to determine the image location quantiatively:

1s'

1f

� 1s

.

Solve and Evaluate Insert f –0.10 m and s 0.20 m into the above to get:

1s '

1f�

1s

1–0.10 m

-1

0.20 m �15 m�1 .

or s ' 1

–15 m–1 –0.067 m . This is very close to the prediction using the ray diagram!

Try It Yourself: Where will the image be if you place an object exactly at the focal point of a concave lens?

Answer: The image will be at a distance f2

on the same side of the lens as the object.

Tip! Forgetting to invert is a common error in solving optics problems! Review Question 22.5 Where should you place an object with respect to a convex lens to have an image at exactly the same distance from the lens? What size is the image?

22.6 Skills for solving mirror and lens problems The strategies for solving problems involving mirrors and lenses are illustrated in the

next example.

ALG 22.4.7 -22.4.8; 22.4.11


Example 20.13 A camera A simple 1880s one lens-camera (Fig. 22.29) consisted of a lens and a light sensitive film placed at the image location. To focus light on the image, the photographer would change the lens-film image distance. Imagine that the maximum image distance is 20 cm and that the film is (16 cm)(16 cm). A 1.9-m tall person stands 8.0-m from the camera. What focal-length lens should the camera have? Will you be able to see the head and the feet of the person?

Figure 22.29 An old-fashioned camera

Sketch and Translate x Sketch the situation in the problem

statement. x Include the known information and the

desired unknown(s) in the sketch.

The situation is sketched below:

• Known quantities:

maximum image distance s 'max �0.20 m , object size h 1.90 m .

• Unknowns: focal length f of the lens and the image height h ' compared to the 0.16-m film size,

Simplify and Diagram x Assume that distortions in the image

produced by small mirrors and lenses are small when using the mirror equation or the lens equation to find images.

x Draw a ray diagram representing the situation in the problem.

x A ray diagram with the person represented as a shining arrow is included in the sketch.

x The object distance s ! 2 f so the image is real, inverted and

smaller than the object.

Represent Mathematically x Use the picture and ray diagram to help

construct a mathematical description of the situation.

x Use lens equation to find the focal length: 1f

1s�

1s '

.

x Use magnification equation to find image height:

h ' mh – s 's

§©̈

·¹̧ h .

Solve and Evaluate x Solve the equations for an unknown

quantity. x Evaluate the results to see if they are

reasonable (the magnitude of the answer, its units, how the answer changes in limiting cases, and so forth).

1f

1s�

1s '

1

8.0 m�

10.20 m

0.125 m–1 + 5.0 m–1 = 5.125 m–1

f 1

5.125 m–1 0.20 m .

The image size is: h ' – s '

s§©̈

·¹̧ h – 0.20 m

8.0 m§©̈

·¹̧ (1.9 m) –0.048 m .

The inverted image size is 5 cm and will easily fit on the film. Try It Yourself: Suppose the 1.9-m tall person stands 4.0 m from the +0.20-m focal length lens. Now where is the image formed relative to the lens and what is the image height?


Answer: s ' 0.21 m from the lens, m –0.053 (the image is inverted), and h ' 0.10 m . Example 20.14 Shaving/makeup mirror You wish to order a mirror from a scientific supply company to use for shaving and makeup. The mirror should produce an upright image that is magnified by a factor of 2.0 when your face is 15 cm from the mirror. What type and focal length mirror should you order? Sketch and Translate A convex mirror always produces a smaller upright image between the object and the mirror—this will not meet our needs. A concave mirror produces inverted real images on the same side of the mirror if the object distance is greater than the mirror’s focal length. The only mirror that meets our needs is a concave mirror when our face is closer than the mirror’s focal length ( s � f )—see Fig. 22.30a. We then get an upright enlarged virtual image

behind the mirror. We want m �2.0 (the plus is because the image and object are both upright) and s 15 cm . We need to determine the mirror’s focal length f . We know it has to be

concave and should have a positive sign.

Figure 22.30(a) Image of face in make-up/shaving mirror

Simplify and Diagram Represent your face by an arrow. To construct a ray diagram that will show you where the focal point is you can trace one of the rays back. If you sent a ray to the mirror so that it would reflect back parallel to the main axis, the continuation of this ray behind the mirror will meet the top of the image. See Fig. 22.30b. Now extend the ray that you sent towards the mirror back to find its intersection with the main axis – this will be the focal point. Notice that the focal point is farther from the mirror than the object. This makes sense as only when the object is between the mirror and the focal point, the concave mirror produces a virtual image.

Figure 22.30(b)

Represent Mathematically Now that we understand the situation conceptually, we can use the mirror equation and the magnification equation to solve the problem mathematically. First


determine the image distance using the magnification equation: m h 'h – s '

s or s ' –ms .

Then use the mirror equation to determine the focal length of the mirror: 1f

1s�

1s '

.

Solve and Evaluate The image distance will be: s ' –ms –(�2.0)(15 cm) = –30 cm .

The minus sign means that the image is virtual and as desired behind the mirror. The focal length of the mirror is now determined from the mirror equation:

1f

1s�

1s '

1(15 cm)

�1

(–30 cm) 0.033 cm–1 .

Hence, the focal length of the mirror is: f 1(0.033 cm–1) �30 cm . A concave mirror has a

positive focal length. The result is consistent with the diagram and with our qualitative thinking at the start of the problem solution. Try It Yourself: You stand 3.0 m from a store’s –1.7-m focal length convex security mirror. Where is your image, and what is the magnification of your height? Answer: 1.1 m behind the mirror; 0.36 times your height. Review Question 22.6 If we have a mathematical lens equation, what is the purpose of drawing ray diagrams when solving lens and mirror problems?

22.7 Putting it all together: Single lens optical systems

Figure 22.31 A camera

There are many applications for mirrors and lenses, such as: cameras, the eye and

eyeglasses, home mirrors, security mirrors, microscopes, telescopes, computer projectors, and magnifying glasses. We consider the first two of these applications in this section.

ALG 22.4.9 – 22.4.10; 22.4.13


Photography and cameras A sketch of a simple camera is shown in Fig. 22.31. It has a lens of fixed focal length.

Light from an object enters the camera through the lens that focuses light on a surface that has light sensitive properties. To have sharp images of objects at different distances from the lens, cameras have a built-in mechanism that can move the lens with respect to the light sensitive surface. Twenty years ago this surface was exclusively a piece of photographic film. Now, most cameras are digital with a surface covered with a huge number (millions) of photosensitive elements that record the light incident on them when the camera shutter is open for a fraction of a second.

The work of any camera can be broken into two parts: the optical part that produces the image of an object and the chemical or electrical part of recording or saving the image. Recording the image: The first cameras had no lenses. They were camera-obscuras with a small hole and an image projected on a wall opposite the hole (we discussed them in Chapter 21). In the 5th and 4th centuries B.C., Chinese and Greek philosophers described the basic principles of these cameras. In 16th century Europe camera obscuras were used by artists to draw portraits. Unfortunately, the image of a sitting person remained as long as the person was in front of the camera; the artist had to draw over the image to preserve it.

In 1727 Johann Heinrich Schulze discovered that silver nitrate darkened upon light exposure and in 1814 Joseph Nicéphore Niépce achieved the first photographic image with a camera obscura by using paper that was coated with a chemical. However, the image required eight hours of light exposure and later faded. In 1829, Niepce partnered with another Frenchman Louis-Jacques-Mandé Daguerre to create permanent photographs. Joseph Niépce died in 1833 but Daguerre continued the work and made the first permanent photograph is 1833. The photograph was called daguerreotype. It is interesting that the word photography was coined by a British astronomer Sir John Hershel (famous for his studies of the Galaxy) in 1839. The word photography comes from two Greek words: photos” meaning light and “graphein” meaning draw.

Camera optics: After the process of capturing the image was invented, it took science almost 30 years to make the first camera with a lens. In 1840 the first American photography patent was issued to Alexander Wolcott. To make the image focused when the object approached the lens (decreasing object distance s ), the film was moved farther from the lens (increasing

image distance s ' ). For a camera, then, the focal length f is constant, and the image distance

s ' is varied by adjusting the camera's distance setting to accommodate for different object distances s . Example 22.15 Camera image A digital camera has a convex lens with a focal length of 5.0 cm. A dog 1.0 m tall stands 3.0 m from the lens. Where should the light sensitive detection surface in the camera be located relative to the lens so that a sharp image is produced and how large is the image?


Sketch and Translate A sketch of the situation is shown in Fig. 22.32a. The known information

is f �5.0 cm , s 300 cm , and h 1.0 m .

Figure 22.32(a)

Simplify and Diagram A ray diagram representing the process is shown in Fig. 22.32b.

Represent Mathematically Use thin-lens Eq. (22.6) to determine the image distance s ' :

1s '

1f

–1s

.

Use the linear magnification Eq. (22.4) to determine the image height h ' :

m h 'h –

s 's

Solve and Evaluate The image distance s ' is: 1s '

1f

–1s

1(5.0 cm)

–1

(300 cm) 0.1967 cm–1

or s ' 1(0.1967 cm–1) 5.085 cm .

The linear magnification then is:

m –s 's –

(5.085 cm)(300 cm)

–0.017 .

The negative sign means that the image is inverted. The height of the image is: h ' mh (–0.017)(1.0 m) = –0.017 m = –1.7 cm ,

easily fitting on a modern camera photo sensitive recording surface.


Try It Yourself: Where should the recording surface be relative to the lens if the dog is twice as far from the lens? How much does the light recording surface distance from the lens change compared to the object’s distance from the lens? Answer: 5.04 cm. The object’s distance doubled – changed from 300 cm to 600 cm; the image distance changed by only 0.04 cm, from 5.08 cm to 5.04 cm! The human eye

The human eye resembles in many ways an expensive digital camera. The eye is equipped with a built-in cleaning and lubricating system, an exposure meter, an automatic field finder, and about 130 million photosensitive elements that send electric signals to the brain. The detection system of the eye is similar to but much more extensive than the light detectors in modern digital cameras. Light from an object enters the cornea (Fig.22.33), a transparent covering over the surface of the eye, and passes through a transparent lens held in place by ciliary muscles. An iris in front of the lens opens or closes, like the shutter on a camera to regulate the amount of light entering the eye. The cornea and lens together act as a lens of variable focal length that focuses light from an object to form a real inverted image on the back surface of the eye, called the retina.

Figure 22.33 The human eye

The retina acts like the digital recording surface of a camera. It contains the 130 million light-

sensitive receptor cells called cones and rods. Light absorbed by these cells initiates photochemical reactions that cause electrical impulses in nerves attached to the cones and rods. The signals from individual cones and rods are combined in a complicated network of nerve cells and transferred from the eye to the brain via the optic nerve. What we see depends on which cones and rods are excited by absorbing light and on the way in which the electrical signals from different cones and rods are combined and interpreted by the brain.

The cones are concentrated in one part of the retina called the fovea. The fovea is about 0.3 mm in diameter and contains 10,000 cones and no rods. Each cone in this region has a separate nerve fiber that leads to the brain along the optic nerve. Because of the large number of nerves coming from this small area, the fovea is the best part of the retina for resolving the fine details of a bright object.


Besides providing a region of high visual acuity, the cones in the fovea and in other parts of the retina are specialized for detecting different colors of light.

The concentration of cones decreases outside the fovea. In these peripheral regions, the rods

predominate. Their density in the retina (about 150,000 / mm2 ) is about the same as that of the

cones in the fovea region. However, the light signals from perhaps 100 adjacent rods are brought together into a single nerve cell that leads to the brain. This combining of the rod signals reduces our ability to see the fine details of an object but helps us see dimly lit objects since many small signals are combined to produce a larger signal. Optics of the eye

A normal eye can focus on objects located anywhere from about 25 cm to hundreds of miles away. This ability to focus on objects at different distances is called accommodation. Unlike the camera, which uses a fixed focal length lens and a variable image distance to accommodate different object distances, the eye has a fixed image distance of about 2.1 cm (roughly the distance from the cornea and lens to the retina) and a variable focal length lens system.

The changing focal length of the eye's lens is illustrated in Fig. 22.34. When the eye looks at distant objects, the ciliary muscle attached to the lens of the eye relaxes, and the lens becomes less curved (Fig. 22.34a). When less curved, the focal length increases and an image is formed at the retina. If the lens remains flattened and the object moves closer to the lens, the image will then move back behind the retina, causing a blurred pattern of light on the retina. To avoid this, the ciliary muscles contract and cause an increase in the curvature of the lens, reducing its focal length (Fig. 22.34b). With reduced focal length, the image moves forward and again forms a sharp, focused image on the retina. If your eyes become tired after reading for many hours, it is because the ciliary muscles have been tensed to keep the lenses of your eyes curved.

Figure 22.34 Eye lens changes shape

The far point of the eye is the greatest distance to an object on which the relaxed eye can

focus. The near point of the eye is the closest distance of an object on which the tensed eye can focus. For the normal eye, the far point is effectively infinity (we can focus on the Moon and on distant stars) and the near point is about 25 to 50 cm. We will discuss later the changes in the eye that cause nearsightedness and farsightedness but before we do this, we will explain a phenomenon that you probably encountered if you tried to swim with your eyes open underwater.

It is important to understand that the eye’s lens is not the only part of the eye that participates in making the images. When you open your eyes underwater you see everything

ALG 22.4.14 - 22.4.15


blurry. If you wear a mask, you can see clearly. Why can't the eye focus underwater and what is the role of the mask? The answer here is that the eye’s lens is not the only optical element in the eye. Another element is the surface of the eye (the cornea) that is moist, curved and forms a lens. In normal conditions, this lens works with the rest of the eye to form an image on the retina. Without the cornea, your eye can't bring the light passing through it to a focus on your retina.

When the eye is in water, the eye's curved outer surface stops acting as a lens because its refractive index is the same as the refractive index of water. Therefore light does not focus on the retina anymore and everything looks blurry. When you wear a mask, there is air between the eye and the water; therefore you gain back the optical system and can focus underwater objects on the retina again. Nearsightedness

A nearsighted person's eyes can see clearly objects that are close to the eye but not those that are distant. The far point of a nearsighted person may be only a few meters rather than infinity.

Nearsightedness occurs if a person's eyeball is larger than the usual diameter (variations in diameter of 1 mm can easily cause a person to be nearsighted) or the cornea is too curved. In such cases, the image of a distant object is formed in front of the retina (Fig. 22.35a), even when the eye lens is relaxed. Tensing the ciliary muscles increases the curvature of the lens and causes the image to move even more in front of the retina.

To correct nearsighted vision, one can place a concave eyeglass lens in front of the cornea. The rays from an object diverge slightly while passing through the eyeglass lens so that when they pass through the lens of the eye, an image is formed farther back in the eye (Fig. 22.35b). You can think of the eyeglass lens as performing a trick on the eye. If an object is very distant (for example, s f ), then the focal length of the eyeglass lens is chosen so that its virtual image is formed at the far point of the eye (Fig. 22.35c). Light passing through the eyeglass lens appears to come from the image at the far point, not from the more distant object. To calculate the desired focal length of the eyeglass lens, we set s f and s ' –(far point distance) . The

negative sign accounts for the fact that the image of the object as seen through the lens of the eyeglass is virtual and to the left of the lens. We can then calculate the value of the desired focal length f of the eyeglass lens using the thin-lens equation.

ALG 22.4.17; 22.4.18; 22.4.20


Figure 22.35 Glasses for near-sighted people Example 22.16 Glasses for a nearsighted person Alex is nearsighted and his far point is 2.0 m from his eyes. What focal length lens should he use in order to focus on a very distant object ( s f )? Sketch and Translate A sketch of the eyeglass lens is show in Fig. 22.35b. The image of an object at infinity is to be formed at the far point 2.0 m in front of the lens. Simplify and Diagram The ray diagrams in Fig. 22.35c represents the optics of the eyeglass lens. Without glasses the image of a distant object is formed before reaching his retina—the distant object is blurred. When Alex wears glasses, the object at infinity ( s f ) has an image 2.0 m to the left of the eyeglass lens ( s ' –2.0 m ). Represent Mathematically We can use the given values of s and s ' in the lens Eq. (22.6) to determine the focal length of the eyeglass lenses:

1f

1s

+ 1s '

Solve and Evaluate Insert the object and image distances in the above equation: 1f

1f

+ 1

(–2.0 m)

Therefore f –2.0 m . The number is negative. This means that the lens should be concave.

The focal length of the lens for a nearsighted person equals the negative value of the person's far point. Try it yourself If Alex is able to see very distant objects with glasses of focal length –1.5 m, then what is his far point? Answer 1.5 m. Power of eyeglass lenses

An optometrist prescribes glasses using a different physical quantity – the optic power P measured in diopters:

P 1f (22.7)

where f is measured in meters. For example, the power of a concave lens of focal length

f –50 cm is P 1(–0.50 m) –2.0 diopters . A lens of high power has a short focal

length and causes rays to converge rapidly after passing through the lens. Tip! To calculate the focal length or the power of eyeglasses, imagine that your eye is looking at the image of some object produced by the glasses. Farsightedness

ALG 22.4.19; 22.4.21 – 22.4.22


Unlike nearsighted people, farsighted persons are able to see distant objects but cannot focus on nearby objects—like a book or newspaper. Whereas the normal eye has a near point of 25-50 cm, a farsighted person may have a near point several meters from the eye. If the person's vision is uncorrected by glasses, a farsighted person will have to hold a book or newspaper at the eye's near point several meters from the eye in order to focus on the print. If the book is held closer to the eye, the image of the book is formed behind the retina, as shown in Fig. 22.36a. Light striking the retina is blurred and out of focus. Farsightedness may occur if the diameter of a person's eyeball is smaller than usual or if the lens is unable to curve enough when the ciliary muscles contract.

Optometrists use convex eyeglass lenses to correct farsighted vision. The eyeglass lens bends light from an object toward the axis of the lens. The rays converge more than without the glasses, and the image is formed on the retina (Fig. 22.36b). Without the bending caused by the eyeglass lens, the image is formed behind the retina.

The convex lens used to correct farsightedness must have a focal length that is greater than the near point of a normal eye, which by convention is said to be 25 cm. When a person holds an object such as a book at the normal near point and views it through the eyeglass lens, she sees a virtual image of the object that is formed by the eyeglass lens behind the object (Fig. 22.36c). The eye looking through the eyeglass lens perceives light to be coming from the virtual image. An eyeglass lens of the correct focal length will cause the virtual image to be formed at the actual near point of the farsighted person's eye.

Figure 22.36 Glasses for a far-sighted person

To determine the focal length of the eyeglass lens, we need to solve the thin-lens

equation for f , given that s 25 cm s = 25 cm, the actual distance of the book or nearby

object from the eyeglass lens, and s ' –(Near point distance) , the position of the virtual

image produced by the eyeglass lens. Light entering the eye through the glasses appears to come from s ' . The negative sign means that the virtual image is on the same side of the lens as the object.


Example 22.17 Glasses for a farsighted person Eugenia is farsighted with a near point of 1.5 m. If you were an optometrist, what glasses would you prescribe for her so she can read a book held 25 cm from her eyes? Sketch and Translate Eugenia holds a book 0.25 m away (see Fig. 22.36b) and wants her glasses to make her eyes think it is 1.5 m away at her nearpoint. Thus s 0.25 m and s ' –1.5 m for her glasses (on the same side of the lens as the book). We need to find their focal length f and power P .

Simplify and Diagram A ray diagram for the optics of the eyeglass system is shown in Fig. 22.36c. Represent Mathematically Use Eq. (22.6) to find the focal length of the lenses:

1f

1s

+ 1s '

Solve and Evaluate Insert the known object and image distances in the above to determine the focal length of the desired glasses:

1f

1

0.25 m +

1–1.50 m

4.0 m–1 – 0.67 m–1 = 3.33 m–1

Thus,

f 13.33 m–1 0.30 m

and the power of the lenses is P 1f

1(0.30 m) 3.3 diopters . Eugenia needs strong

glasses! Try It Yourself: Eugenia went to a drug store and bought glasses. They only had 4.0 diopter glasses. Where will Eugenia needs to hold the book so its image is at her 1.5-m near point? Answer: 30 cm.

The history of glasses is quite interesting. The first person on record to use a corrective lens was the emperor Nero in the 1st century. He watched the games of gladiators through an emerald. Abbas Ibn Firnas in the 9th century invented a way to sand finish glass and made the first corrective glass lenses. An Italian Salvino D’Armate is considered the inventor of the first wearable eye glasses in 1284. However, physicists credit Roger Bacon who wrote a note about the magnifying properties of lenses in 1262. The early spectacles had convex lenses that could correct farsightedness; later people figured out how to treat nearsightedness. In 1604 after Kepler published his book on astronomy and optics, the first explanation was provided for how these glasses corrected vision problems. Benjamin Franklin had both near and far sighted vision problems and invented bifocals in 1784 so he did not need to switch between glasses. Review Question 22.7


What is the main difference between how a camera produces focused images of objects at different distances and how the human eye does it?

22.8 Angular magnification and magnifying glasses The linear magnification of an optical system compares only the heights of the image and

the object, but the apparent size of an object as judged by the eye depends not only on its height but also on its distance from the eye. For example, a pencil held 25 cm from your eye appears longer than one held 100 cm away (Fig. 22.37). In fact, the pencil may appear longer than a 100-story building that is several miles away, even though the pencil is much shorter.

Figure 22.37 Angular size of an object

A person's impression of an object's size depends on its so-called angular size T , shown in Fig. 22.37. The angular size depends on an object's height h and its distance r from the eye:

T hr

(22.8)

Remember that an angle T , in radians, is the ratio of an arc length ( h in Fig. 22.37) and the radius of a circle ( r in Fig. 22.37). The expression for T is exact only if h is the curved length of the arc of the circle and not the straight line between the corners of the triangle. For small angles, Eq. (22.7) is a good approximation.

If a person looks at an object through one or more lenses, he or she sees light that appears to come from the final image of the system of lenses. If the person's eye is close to the last lens, then the angular size T ' as seen through the lens is:

T ' h 's '

(22.9)

where h ' is the height of the final image and s ' is the distance of the final image from the last lens. The angular magnification M of an optical system is defined as the ratio of T ' and T :

M T 'T

Angular size of the final image of optical systemAngular size of object as seen by unaided eye

(22.10)


A magnifying glass is the simplest optical device that provides angular magnification. The

original inventor of the magnifying glass was Roger Bacon (attributed by physicists as the inventor of glasses). The famous detective Sherlock Holmes made a magnifying glass a cool thing to carry in a pocket – it was always with him.

A magnifying glass consists of a single convex lens. To use the lens as a magnifying glass, hold the object between its focal point and the lens, as shown in Fig. 22.38. A magnified virtual image is formed behind the object. When looking through the magnifying glass, the viewer thinks the light is coming from this enlarged image.

Figure 22.38 Magnifying glass

Now look carefully at Fig. 22.38. Note that both the object and the image make the same

angle T ' with the main optical axis (T ' h 's '

hs ). The object has the same angular size when

we look directly at it when a distance s from our eye or when we look through the magnifying glass at the image a distance s ' from our eye. So why use the magnifying glass? Unfortunately, we cannot focus on objects brought closer to our eyes than our near point. A magnifying glass, however, allows us to bring the object closer. Instead of trying to look directly at the object held real close to the eye (closer than we can focus), we look at the enlarged image, which is at our near point or farther. We can focus on the image because it is at or beyond our eye’s near point.

To calculate the angular magnification of the magnifying glass, compare the angular size T ' of the image seen through the magnifying glass and the angular size T of the object seen with

the unaided eye. The angular size of the image viewed through the magnifying glass is T ' h 's ' .

By considering ray 2 in Fig. 22.38, we find that h 's '

hs . Thus,

T ' hs

where h is the actual height of the object and s is its distance from the magnifying glass. Next consider the angular size of the object as seen with the unaided eye. The angular size

T of an object of height h that we hold a distance r from the eye is T hr (Fig.22.39a). As

we bring the object closer to the eye (Fig. 22.39b and c), its angular size increases. If we bring the


object closer than the near point, the image on the retina is blurred (Fig. 22.39d). The maximum angular size Tmax of an object of height h , when viewed by the unaided eye and when focused on

the retina, is

Tmax h

Near point distance (22.11)

The angular magnification of the magnifying glass (called magnifying power) is the ratio of the angular size T ' as seen through the magnifying glass and the maximum angular size Tmax as seen

with the unaided eye. Thus,

M TTmax

h / s

h / (Near point distance)

Near point distances

. (22.12)

Figure 22.39 Maximum angular size

Example 22.18 Examining a rare stamp You are interested in a rare stamp. You hold a magnifying glass 4.2 cm from a stamp that is lying on the desk. The focal length of the glass is +5.0 cm. Where is the virtual image of the stamp and what is the angular magnification of the magnifying glass? Sketch and Translate Draw a sketch of the situation (Fig. 22.40). The givens are s 4.2 cm and f = +5.0 cm . Since s � f , we should have an enlarged virtual image of the stamp. We need to

find s ' . If larger than the normal near point distance, we can immediately determine the angular magnification of the image.

Figure 22.40 Looking at stamp with magnifying glass

ALG 22.4. 6; 22.4.16


Simplify and Diagram Assume that the angles involved in the problem are small, the lens is parallel to the stamp, and the near point distance is 0.25 m. The situation is analyzed with a ray diagram, included in Fig. 22.40. Note that the image is enlarged and virtual and is located on the same side of the lens as the object—your eye looks at the enlarged virtual image created by the magnifying glass. Represent Mathematically Rearrange the thin lens Eq. (22.6) to find the image location:

1s '

1f�

1s

.

Solve and Evaluate Insert the known information in the above to find s ' : 1s '

15.0 cm

�1

4.2 cm �0.038 cm�1 .

Thus s ' 1(–0.038 cm–1) –26 cm . The image distance is negative and larger than the near

point distance – you will be able to see the stamp. The angular magnification of the magnifying glass is:

M Near point distance

s

25 cm4.2 cm

6.0 .

Try It Yourself: Suppose you want to use the +5.0-cm focal length magnifying glass for a farsighted person whose near point is 1.0 m = 100 cm. Where should the stamp be located relative to the lens? What then is its magnification? Answer: s ' 4.76 cm; M = 21. Review Question 22.8 If a person with normal vision and a farsighted person use the same magnifying glass, who will get the greater magnification? Why?

22.9 Optical instruments Many optical instruments have more than one lens (such as microscopes and telescopes)

and form an image of an object that is magnified more than is possible using a single lens. The technique for locating the final image and calculating its magnification requires careful attention to several details. Combinations of Lenses

Consider the two-lens system shown in Fig. 22.41a. The lenses are separated by a distance d and have focal lengths f1 and f2 . The object is a distance s from lens 1. Where is

the final image? To determine the location of the final image of the two-lens system we use the following approach:


1. Use Eq. (22.6) to determine the location s1 ' of the image formed by the first lens (Fig.

22.41b). If s1 ' is positive, the image is to the right of lens 1. If s1 ' is negative, the image is

to the left of lens 1. In our example, s1 ' is positive.

2. The image of lens 1 is now the object for lens 2. The object distance s2 (Fig. 22.41c) is

s2 d – s1 ' (22.12)

where d , the separation of lenses, is a positive number and s1 ' is either positive or negative,

depending on the value calculated in step 1. If s1 ' is a larger positive number than d , the

image of lens 1 is formed to the right of lens 2; s2 will then be a negative number. Be sure

to retain this negative sign in subsequent calculations—the “object” at s2 does not seem like

an object for the second lens but is. The second lens changes the direction of light passing through it so that the image of the second lens is at a different location than at s2 ).

3. We now use the thin-lens equation to calculate the image distance s2 ' of lens 2 (Fig.

22.41d): 1s2 '

1f�

1s2

This is the location of the final image relative to the second lens. If s2 ' is positive, the final

image is real and located to the right of lens 2. If s2 ' is negative, the final image is virtual

and to the left of lens 2. In our example, the final image is real and to the right of lens 2. 4. The total linear magnification m equals the product of the linear magnifications of each lens:

m m1m2 . If m is positive, the final image has the same orientation as the original object.

If m is negative, the final image is inverted relative to the original object. Techniques for calculating the total angular magnification M are illustrated in the discussions and examples that follow.

Figure 22.41 Analyzing two-lens optics system


The order of the steps used in the problem-solving procedure just outlined is often altered, depending on the known and desired information. You should find the procedure simple to use; just be sure to follow the appropriate sign conventions. Telescopes

It is not clear who was the first person to invent a telescope. One legend says that it was a Dutch spectacle maker Hans Lippershey in about 1600. He thought of a telescope while his two children were playing with his lenses and put two of them together and looked through them at a building far away – and saw it magnified. Lippershey checked what the children were doing and quickly mounted two lenses together to create his own "looker." He tried to market it to the Dutch army but did not arouse any interest in the military. One reason was that another Dutch spectacle maker Zaccharias Janssen claimed that he built the same device in 1590. Thus Lippeshey’s invention was not an invention at all. But the news of Lippershey’s work spread through Europe and reached Venice, where Galileo thought of pointing the looker into the heavens creating the first telescope. The instrument was not popular. Most thought that it created optical illusions until 1610 when Galileo published his famous book The Starry Messenger (Sidereus Nuncius). The book was only 24 pages but contained the results of Galileo’s night observations. It described mountains and craters on the Moon, moons of Jupiter, and stars inside the Milky Way. His observations supported the heliocentric system proposed earlier by Copernicus, which was rejected because of the absence of experimental support. And of course, Galileo observed sunspots that got him into serious trouble later. But this is the subject for science history books. The name telescope was coined by Federico Ceci in 1611, after Galileo published his book.

Galileo used a telescope that consisted of two lenses – one convex and the other concave. However, professional astronomers prefer a different version described below. This telescope has two convex lenses separated by a distance approximately equal to the sum of their focal lengths (Fig. 22.42a). When you observe a distant object, the first lens produces a real image just beyond its own focal length, as shown in part (b). The second lens is located so that this image of the first lens is just inside the focal point of the second lens. The image of the second lens is a magnified inverted virtual image, as is shown in part (c). Our eye perceives that light comes from this enlarged virtual image. The magnification of the telescope is the product of the magnification of each lens. Example 22.19 Looking at a lion with telescope A 1.2 m tall lion stands 50 m from the first lens of the telescope shown in Fig. 22.42a. Locate the final image of the lion. Determine the linear magnification of the telescope and the height of the final image and the angular magnification of the telescope.


Sketch and Translate We already drew a picture of the situation in 22.42a. The givens: f1 �20.0 cm , f2 �5.0 cm , d = 24.5 cm , s1 = 50 m = 5000 cm , and h 1.2 m . We

need to find s2 ' , m , and M .

Simplify and Diagram The ray diagrams in Fig. 22.42b and c help us locate the first and second images. Assume that the eye is positioned next to the second lens. Represent Mathematically Use the familiar procedure to determine the locations of the first image, the second object, and the second image s1 ', s2 and s2 ' —we’ll leave it to you. Then use

the values of s1 ', s2 and s2 ' to determine the linear magnification m m1m2 of the two lens

system, where m1 – s1 's1

and m2 – s2 's2

. Next determine the height of the final image of

the lion h ' mh . To find the angular magnification, compare the angular size of the lion as seen

through the optical system T ' h '

2

s2 ' and its angular size as seen with the unaided eye T

hr1

where r1 s1 � d . Then take the ratio of the angular sizes in order to determine the angular

magnification: M T 'T

.

Solve and Evaluate Use the mathematical procedure outlined earlier to determine s1 ' , s2 , and

s2 ' . You should find that s1 ' 20.08 cm , s2 4.42 cm , and s2 ' –38.1 cm . Notice that the

final distance came out negative in agreement with the ray diagram; we are viewing a virtual image. Now determine the linear magnification:

m m1 m2 s1 's1

s2 's2

(20.08 cm)(-38.1cm)(5000cm)(4.42cm)

�0.035

Notice that the linear magnification is less than 1; the final image is smaller than the original object. The image height is h2 ' mh (–0.035)(1.2 m) = –0.042 m = –4.2 cm .

Finally, determine the angular magnification M :

T ' h2 's2 '

(–4.2 cm)(–38.1 cm)

0.11 rad ;

T hs1

(1.2 m)(50 m)

0.024 rad ; and

M T 'T

0.11 rad0.024 rad

4.6 .

The details of the lion appear almost 5 times bigger when looking through the telescope. Try It Yourself: How can the lion appear larger when its final image is smaller? Answer: The final image was much closer than the lion and thus looked bigger through the telescope, even though its final image was smaller.


Figure 22.42 A telescope

There are a couple of reasons that telescopes are used to look at stars. Consider a ray

diagram for an infinitely far away object as seen through a telescope (Fig.22.43). As you can see, the telescope turns parallel rays into parallel rays, only now they are much closer together. This means that the same amount of light is concentrated over a small surface area, making the object appear blighter. This is how faint stars that cannot be seen with an eye can be seen through a telescope. Another advantage that the telescope offers for the stars is though their angular magnification. Although the telescope cannot increase the size of stars to help see surface details, it can increase the angular distance between two nearby stars helping to see them as two stars and not one.

Figure 22.43 Telescope concentrates light

The Compound Microscope

If you search for the inventor of the first microscope, you will probably stumble upon the name of Anton van Leeuwenhoek (1632-1723) from Holland (Dutch scientists invented both the telescope and the microscope). Anton van Leeuwenhoek was the best glass-maker of his time. He polished tiny lenses of great curvature. This work led to his invention of a microscope through which he observed bacteria, yeast plants, the rich life of the inhabitants of a drop of water, and the


circulation of blood particles in capillaries. Another scientist, the Englishman Robert Hooke (1635-1703), continued the work of van

Leeuwenhoek. He replicated van Leeuwenhoek’s microscope and in 1665 looked at a small piece of cork though a microscope and notice pores or “cells” in it. He believed that only plants had these cells. Hooke published a book Micrographia that contained detailed drawings of insects, snowflakes, and other small objects. How could he see all this tiny things? To answer this question we need to examine how a microscope works.

A microscope, like a telescope, has two convex lenses. But the purpose of a microscope is very different from that of a telescope; instead of looking at large and very distant objects, it looks at tiny and very close objects. Therefore the arrangement of the lenses is different (Fig. 22.44a). Both lenses have relatively short focal distances and are separated by 10-20 cm. The object is placed just outside and very close to the focal point of the first lens (called the objective lens). The real, enlarged inverted image of the first lens is closer to the second lens (called an eyepiece) than the focal distance of the second lens, as shown in Fig. 22.44b. Thus when an observer looks through the eyepiece, she sees an inverted, enlarged virtual image of the object (Fig. 22.44c).

Figure 22.44 A microscope

Consider the magnification of such a system. The image produced by the objective lens

has the linear magnification m1 – s1 's1

. The magnification is negative because the image is

inverted. The image of the first lens is the object for the second. Since this object is located just inside the focal point of the second lens, a final virtual image is formed outside the microscope (Fig. 22.44c). The second lens of the microscope acts as a magnifying glass used to view the real


image of the first lens. The angular magnification of this magnifying glass is, according to Eq.

(22.11), M 2 Near point distance

s2. The total angular magnification of the microscope is:

M m1M 2 –s1 's1

Near point distances2

.

This expression for angular magnification can be rewritten in terms of the focal lengths of the lenses and their separation d . Consider Fig. 22.45. Notice that s1 | f1 ; s1 ' | d – f2 ; and

s2 | f2 . By substituting these values in the previous equation, we have

�M ; (�

d � f2

f1)(

Near point distancef2

) . (22.13)

The negative sign indicates that the final image is inverted. This expression is easy to use because the magnification depends only on the focal distances of the lenses, their separation, and the near point distance of the eye.

Figure 22.45 Developing alternative expression for angular magnification

The microscope that we examined above is similar to the one that R. Hooke used. His magnification was about 100. Today, the very best light microscopes provide angular magnifications of about 1000. With these instruments, scientists can distinguish objects separated

by 2 u10–7 m (about the length of 700 atoms in a row). Although modern microscopes may contain as many as ten or more lenses, they work in principle much like the two-lens microscope used by Robert Hooke. Example 22.20 Seeing a cell with a microscope A compound microscope has an objective lens of focal length 0.80 cm and an eyepiece of focal length 1.25 cm. The lenses are separated by 18.0 cm. If a red blood cell is located 0.84 cm in front of the objective lens, where is the final image of the cell, and what is its angular magnification? The viewer's near point is 25 cm. Sketch and Translate We have the following information for the problem situation: f1 �0.80 cm ; f2 �1.25 cm , d = 18.00 cm , and s1 = +0.84 cm . We need to find s2 '

and the angular magnification M of the microscope.


Simplify and Diagram Assume that the red blood cell is brightly illuminated and can be considered a shining arrow. We use the microscope ray diagrams in Fig. 22.44. Represent Mathematically First find the image produced by the first lens:

1s '1

1f1�

1s1

The image of the first lens is the object of the second. The object distance for the second lens is s2 d – s1 ' . Use this distance to find the location of the image produced by the second lens (the

final image): 1

s '2

1f2

�1s2

To find the angular magnification M , first determine the linear magnification of the first lens m1

to determine how big the object is that is viewed by the second lens. Then determine the angular magnification of the second lens M 2 . The total final angular magnification is the product of these

two magnifications:

M m1M 2 –s1 's1


.

Solve and Evaluate 1s1 '

1f1�

1s1

1

0.80 cm-

10.84 cm

= 0.0595 cm�1

Thus, s1 ' 1(0.0595 cm–1) �16.8 cm . The object distance for the second lens is

s2 d – s1 ' 1.2 cm . The final image distance is:

1s2 '

1f2

�1s2

1

1.25 cm–

11.20 cm

= –0.033 cm-1

or s2 ' 1(0.033 cm–1) –30 cm . The total angular magnification is:

M m1M 2 –s1 's1


–16.8 cm0.84 cm

25 cm1.2 cm

–(20)(21) –420 .

The negative sign means that the final image is inverted. Try It Yourself: How does the magnification in this example using the magnifications for each lens compare to the approximate magnification equation [Eq. (22.13)] for a microscope? Answer: You get the same result –420. Review question 22.9 Why is saying that a telescope magnifies is simultaneously a correct and an incorrect statement?


Summary Words Sketches and/or diagrams Mathematical

representations Plane mirror: A plane mirror produces a virtual image at the same distance behind the mirror as the object is in front. A virtual image is at the position where the paths of the reflected rays seem to diverge from behind the mirror.

s s '

Concave and convex mirrors: The distance s of an object from the surface of a concave or convex mirror, the distance s ' of the image from the surface of the mirror, and the focal length f of the mirror are related by the mirror equation. The focal point is the location through which all incident rays parallel to the main axis pass after reflection (concave mirrors) or appear to pass after reflection (convex mirrors). The sign conventions for these quantities are: x The focal length is positive for concave

lenses and negative for convex lenses. x The image distance is positive for real images

in front of the lens and negative for virtual images behind the mirror.

Concave mirror

Convex mirror

1s�

1s'

1f

Convex and concave lens: The distance s of an object from a lens, the distance s ' of the image from the lens, and the focal length f of the lens are related by the thin-lens equation. Sign conventions for light moving from left to right are: x The focal length f is positive for convex

lenses and negative for concave. x The object distance s is positive if the object

is to the left of the lens and negative if the object is to the right of the lens.

x The image distance s ' is positive for real images formed to the right of the lens and negative for virtual images formed to the left of the lens.

Convex lens

Concave lens

1s�

1s'

1f

Linear magnification m is the ratio of the height h ' of the image and the height h of the object. The heights are positive if upright and negative if inverted. The linear magnification can be determined using the negative ratio of the image distance s ' and object distance s .

m = h'h

= – s's

.

Angular magnification M is the ratio of the angular size T ' of an image as seen through a lens or lenses and the maximum angular size of the object Tmax as seen with the unaided eye.

M = T 'Tmax

= near point distances

Multiple lens systems: The strategy for locating the final image of a multiple lens system is outlined at the beginning of Section 22.9—see Fig. 22.41.

Chapter 22: Mirrors and Lenses - West Windsor … · Chapter 22: Mirrors and Lenses ... x The ray...

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