Chapter 25 THE REFLECTION OF LIGHT: MIRRORSbeardphysics.weebly.com/uploads/7/6/0/1/7601300/... ·...

Chapter 25 The Reflection of Light: Mirrors

281

Chapter 25

THE REFLECTION OF LIGHT: MIRRORS

PREVIEW

The ray model of light states that light may be represented by a straight line along thedirection of motion, and ray optics is the study of light using the ray model. Light can bereflected from a surface such as a mirror. In this chapter, we’ll study reflection fromplane mirrors and spherical mirrors. As light rays are reflected, they may form an image,which can be real or virtual, depending on the distance from the mirror to the objectwhich is the source of the light rays. Ray diagrams can be drawn to show the bending ofa light ray or to locate an image formed by a mirror.

The content contained in all sections of chapter 25 of the textbook is included on the APPhysics B exam.

QUICK REFERENCE

Important Terms

angle of incidencethe angle between the normal line to a surface and the incident ray or wave

angle of reflectionthe angle between the normal line to a surface and the reflected ray or wave

converging mirrora mirror which converges light rays reflecting from it; also known asa concave mirror

diverging mirrora mirror which diverges light rays reflecting from it; also known as aconvex mirror

focal lengththe distance between the center of a lens or mirror to the point at which therays converge at the focal point

focal pointthe point at which light rays converge or appear to originate

imagereproduction of an object using lenses or mirrors

magnificationratio of the size of an optical image to the size of the object

object (optics)the source of diverging light rays

plane mirrorsmooth, flat surface that reflects light regularly


282

principal axisthe line connecting the center of curvature of a curved mirror with itsgeometrical vertex; the line perpendicular to the plane of a lens passing throughits center

radius of curvaturethe radius of a spherical mirror which is equal to twice the focal length of themirror

ray model of lightlight may be represented by a straight line along the direction of motion

ray opticsstudy of light using the ray model

real imagean image that can be projected onto a screen

virtual imagean image which cannot be projected onto a screen; point at whichdiverging light rays appear to originate

Equations and Symbols

o

i

io

i

o

i

o

ir

d

dm

fdd

d

d

h

h

Rf

111

2

1

where

θr = angle of reflection as measured fromthe normal line

θi = angle of incidence as measured fromthe normal line

f = focal length of a mirrorR = radius of curvature of a convex or

concave mirrorho = height of the objecthi = height of the image produced by a

mirrordo = distance from the center of the

mirror to the objectdi = distance from the center of the

mirror to the imagem = magnification

Ten Homework Problems

Chapter 25 Problems 2, 5, 10, 11, 14, 16, 17, 23, 33, 36


283

DISCUSSION OF SELECTED SECTIONS

25.3 The Formation of Images by a Plane Mirror

Any wave that bounces off of a barrier follows the law of reflection: the angle ofincidence is equal to the angle of reflection as measured from a line normal(perpendicular) to the barrier (Figure A). In the case of light, the barrier is often a mirror.This is why if someone can see you in a mirror, you can see him in the mirror as well.

A plane mirror is simply a flat mirror. From your everyday experience you know that aplane mirror always produces an image which is the same size as the object (which couldbe you in the morning), left-right reversed, and the same distance behind the surface ofthe mirror as the object is in front of the mirror. We say that the image formed by a planemirror is virtual, since we cannot place a screen behind the mirror and see the imageprojected on the screen. A virtual image is one that cannot be projected onto a screen.

We can locate the image formed by a plane mirror by tracing two reflected rays and thenextending them backward (Figure B). The point at which the two rays seem to meet iswhere the virtual image is formed.

25.4 - 25.6 Spherical Mirrors, The Formation of Images by SphericalMirrors, and The Mirror Equation and the Magnification Equation

We will discuss two types of spherical (curved) mirrors: diverging (convex) andconverging (concave), although the AP Physics B exam usually focuses on concavemirrors.

A diverging mirror is sometimes referred to as a convex mirror. You may have seen thistype of mirror in the corner of a convenience store. The mirror diverges the rays of lightwhich strike it, allowing the clerk at the store to see practically the entire store in one

Figure B

θi

θr

Normal

Figure A


284

mirror. Light rays coming into the mirror parallel to its principal axis will diverge, orspread apart:

Notice that the rays appear to originate from a point behind the mirror. This point iscalled the virtual focus, and the distance between the surface of the mirror at its centerand the focal point is called the focal length f.

A converging mirror is sometimes referred to as a concave mirror. If a spherically-shaped concave mirror is small compared to its radius of curvature R, then light rayscoming in parallel to the principal axis of the mirror will converge to a focal point.

For this kind of mirror, the focal length f and the radius of curvature R of the mirror arerelated by the equation f = ½ R.

If you look into a converging mirror, you will at first see an image of yourself which isinverted (upside-down), but then as you move closer to the mirror your will see yourimage turn upright as you pass the focal point of the mirror. Satellite dishes act asconverging mirrors for radio and TV waves, gathering them at a detector located at thefocal point of the dish. Most research telescopes also use converging mirrors rather thanlenses to initially focus incoming light and study images.

The image formed by a converging mirror depends on the location of the object inrelation to the focal length and its orientation, and can be real, which means it can beprojected onto a screen, or virtual. The image can also be upright or inverted (upside-down), and smaller, larger, or the same size as the object.

The object distance do is the distance from the center of the surface of the mirror to theobject, and the image distance di is the distance from the center of the surface of themirror to the image.

3f 2f f f 2f 3f

3f 2f =R f f 2f =R 3f


285

For the image formed, the ratio of the object height ho to the image height hi is equal tothe ratio of the object distance to the image distance:

i

o

i

o

d

d

h

h

and the magnification of the image is given by

o

i

o

i

d

d

h

hm .

Magnification tells us how many times larger or smaller an image is than the object.The negative sign is inserted as a convention. The object height ho is always taken aspositive, and the image height hi is positive if the image is upright and negative ifinverted. The object and image distances are positive if the image and object are on thereflecting side of the mirror. If either the object or the image is behind the mirror, thecorresponding object or image distance is negative. In the end, the magnification m ispositive for an upright image and negative for an inverted image.

The relationship between the object distance, image distance, and focal length is

io ddf

111

where f = ½ R.

Example 1A converging (concave) mirror has a focal length of 20.0 cm. A 5.0-cm tall candle isplaced at a distance of 50.0 cm in front of the mirror.(a) By drawing a ray diagram, find the location of the image formed by the convergingmirror. State whether the image is real or virtual, upright or inverted, and larger, smaller,or the same size as the object (candle).(b) Using the mirror and magnification equations, verify your results from part (a).

Solution(a) To find the image formed by the mirror, we will draw two rays: (1) one ray from theflame which strikes the mirror parallel to the principal axis and reflecting through thefocal point, and (2) another ray from the flame which goes through the focal point andreflects back parallel to the principal axis. The image is formed at the location of theintersection of these two rays:


286

Note that when the object is place at a distance greater than twice the focal length, theimage is inverted and smaller than the object. If we place a screen at the location of theimage, we would see that the image is real. For a concave mirror, if the image is formedon the same side of the mirror as the object the image is real. If the image is formed onthe opposite side of the mirror as the object, it is virtual. Recall that a plane mirror willonly produce an image on the opposite side to the object, and thus is always virtual.

If we’ve drawn our diagram to scale we can simply measure the image distance from thecenter of the mirror. Let’s say our measurement is di = 32.5 cm, and the height of theimage is 3.2 cm.

(b) Using the mirror and magnification equations, we can solve the image distance:

cmd

dcmcm

ddf

i

i

io

3.33

1

50

1

20

1

111

Note that our measurement of the image distance is very close to the calculated valuewithin reasonable experimental error.

The magnification of the candle can be found by

cm

cm

cmcm

d

dhh

d

d

h

hm

o

ioi

o

i

o

i

3.350

3.330.5

Again, the negative sign indicates the image is inverted.

3f 2f f f 2f 3f

1

2

do = 60cm

di


287

Example 2A converging (concave) mirror has a focal length of 20.0 cm. A 5.0-cm tall candle isplaced at a distance of 10.0 cm in front of the mirror.(a) By drawing a ray diagram, find the location of the image formed by the convergingmirror. State whether the image is real or virtual, upright or inverted, and larger, smaller,or the same size as the object (candle).(b) Using the mirror and magnification equations, verify your results from part (a).

SolutionIf we place the candle at a distance less than the focal length, the reflected rays diverge,and the image is formed at the point from which the rays seem to originate. Drawing ourtwo principal rays:

Note that this time our second ray is drawn as if it originates at the center of curvature C,which is located at 2f. A ray that passes through the center of curvature will reflect backon itself, and therefore if we extend it behind the mirror it continues in a straight line. Wesee that the image of the candle is upright, larger, and behind the mirror, and therefore isvirtual.

(b) Using the mirror and magnification equations, we can solve the image distance:

cmd

dcmcm

ddf

i

i

io

20

1

10

1

20

1

111

The image distance is negative, indicating it is formed behind the mirror. Note that theimage distance shown in the diagram is slightly less than 20 cm due to measurementerror. The height of the image is

cm

cm

cmcm

d

dhh

d

d

h

hm

o

ioi

o

i

o

i

0.100.10

0.200.5

The positive sign indicates that the image is upright.

3f 2f f f 2f 3f

C


288

The image formed by a concave mirror when the object is placed inside the focal lengthis virtual upright and enlarged. Thus if you want to see an upright image of your face in aconcave mirror, you must move the mirror to a distance less than one focal length fromyour face.

summary of the images formed by a concave mirror (and a convex lens) is listed in thetable below.

Object placedat:

Imagedistance di

real or virtual upright orinverted

larger orsmaller

do > 2f + real inverted smallerdo = 2f + real inverted same size

f < do < 2f + real inverted largerdo = f No image No image No image No imagedo < f - virtual upright larger

Note that all real images are inverted.

PracticeVerify the results in the table above by drawing the ray diagrams below.

3f 2f f f 2f 3f

3f 2f f f 2f 3f

3f 2f f f 2f 3f


289

CHAPTER 25 REVIEW QUESTIONSFor each of the multiple choice questions below, choose the best answer.

1. In the figure shown the angle ofincidence is . Which angle(s) is/are theangle(s) of reflection?(A) 1(B) 3 and 4(C) 3(D) 4(E) 1, 3, and 4.

2. A plane mirror will produce a virtualimage(A) when the object distance is greater

than the image distance.(B) when the object distance is less than

the image distance.(C) when the object is on the principal

axis of the mirror.(D) when the rays converge at the focal

point of the mirror(E) at all distances from the mirror.

3. Which of the following mirrorsdiverge parallel light rays?(A) plane(B) convex(C) concave(D) inverted(E) upright

4. A candle is placed on the principalaxis of a concave mirror at a distance of30 cm from the mirror. The focal lengthof the mirror is 10 cm. The imageformed will be(A) real, upright, and enlarged(B) real, inverted, and enlarged(C) real, inverted, and smaller(D) virtual, upright, and enlarged(E) virtual, upright, and smaller

5. A candle is placed on the principalaxis of a concave mirror at a distance of10 cm from the mirror. The focal lengthof the mirror is 20 cm. The imageformed will be(A) real, upright, and enlarged(B) real, inverted, and enlarged(C) real, inverted, and smaller(D)virtual, upright, and enlarged(E) virtual, upright, and smaller

6. A candle is placed on the principalaxis of a concave mirror at a distance of20 cm from the mirror. The imageformed is magnified 3 times. The imagedistance is(A) 7 cm(B) 20 cm(C) 60 cm(D) 90 cm(E) 120 cm

N

2 31 4


290

Free Response Question

Directions: Show all work in working the following question. The question is worth 15 points,and the suggested time for answering the question is about 15 minutes. The parts within aquestion may not have equal weight.

1. (15 points)

The concave mirror shown above has a focal length of 30.0 centimeters. You are given a candle5.0 centimeters high. You wish to produce an image on a screen which is 15.0 cm high.

(a) State whether you would place the candle on the left side of the mirror shown above or theright side. Explain your choice.

(b) State whether you would place the screen on the left side of the mirror shown above or theright side in order to see the image formed by the mirror. Explain your choice.

(c) Choose an appropriate object distance, and calculate the corresponding distance from themirror at which the screen should be placed in order to see the 15.0 cm image.

(d) Using an appropriate scale, draw a ray diagram on the figure above which verifies yourcalculation.

180 150 120 90 60 30 30 60 90 120 150 180

cm


291

ANSWERS AND EXPLANATIONS TO CHAPTER 25 REVIEW QUESTIONS

Multiple Choice

1. CThe angles of incidence and reflection are both measured from the normal line.

2. EA plane (flat) mirror will reflect light and produce an image at any distance from the object.

3. BA convex mirror is also called a diverging mirror.

4. CThe candle is placed at a distance greater than twice the focal length, and so the image formedwill be real, inverted, and smaller than the candle.

5. DThe candle is placed at a distance less than the focal length, and so the image formed will bevirtual, upright, and enlarged.

6. C

Sinceo

i

d

dionMagnificat , then di = (M)(do) = (3)(20 cm) = 60 cm.

Free Response Question Solution

(a) 2 pointsThe candle must be placed on the left side of the mirror, since the left side is the concave(converging) side of the mirror.

(b) 2 pointsIn order to see an image on a screen, the image must be real. For a concave mirror, real imagesare formed on the same side of the mirror as the object.

(c) 5 pointsIn order to produce an enlarged real image, the candle would need to be placed at an objectdistance between f and 2f. Choosing 40 cm as the object distance, we can find the correspondingimage distance.


292

cmd

dcmcm

ddf

i

i

io

120

1

40

1

30

1

111

The screen should be placed 120 cm to the left of the mirror.

(d) 6 pointsDrawing one principal ray parallel to the principal axis and reflecting through the focal point,and one principal ray passing through the center of curvature (other principal rays could havebeen used):

cm

180 150 120 90 60 30 30 60 90 120 150 180

Chapter 26 The Refraction of Light: Lenses and Optical Instruments

293

Chapter 26

THE REFRACTION OF LIGHT: LENSES AND OPTICALINSTRUMENTS

PREVIEW

Light can be refracted, or bent, through a transparent medium such as a lens. Therelationship between the speed of light in two different media and the angle of the lightrays can be found using Snell’s law of refraction. As light rays are refracted, they mayform an image, which can be real or virtual, depending on the distance from the lens tothe object which is the source of the light rays. As in the previous case of mirrors, raydiagrams can be drawn to show the bending of a light ray or to locate an image formedby a lens.

The content contained in sections 1 – 8, and 15 (Example 17) of chapter 26 of thetextbook is included on the AP Physics B exam.

QUICK REFERENCE

Important Terms

angle of incidencethe angle between the normal line to a surface and the incident ray or wave

angle of refractionthe angle between the normal line to a surface and the refracted rayor wave at the boundary between two media.

converging lensa lens which converges light rays to a focal point; also known as aconvex lens

critical anglethe minimum angle entering a different medium at which total internalreflection will occur

diverging lensa lens which diverges light rays passing through it; also known as aconcave lens

focal lengththe distance between the center of a lens or mirror to the point at which therays converge at the focal point

focal pointthe point at which light rays converge or appear to originate

index of refractionthe ratio of the speed of light in a vacuum to the speed of light inanother medium


294

lensa piece of transparent material that can bend light rays to converge or diverge

magnifying glassoptical instrument which results from an object being placed within the focallength of a convex lens, producing an enlarged virtual image

refractionthe change in speed, wavelength, and direction of a light ray due to a changein medium

Snell’s law of refractionwhen light passes from one material with an index of refraction n1 into a materialof a different index of refraction n2, the angle of incidence θ1 is related to theangle of refraction θ2 by the equation n1sinθ1 = n2sin2.

total internal reflectionthe complete reflection of light that strikes the boundarybetween two media at an angle greater than the critical angle

Equations and Symbols

o

i

io

i

o

i

o

c

d

dm

fdd

d

d

h

h

n

n

nnv

cn

fc

111

sin

sinsin

1

2

2211

where

c = speed of light = 3 x 108 m/sf = frequency of lightλ = wavelength of lightn = index of refractionθ1 = angle of incidenceθ2 = angle of refractionθc = critical anglef = focal length of a lens

ho = height of the objecthi = height of the image produced by alensdo = distance from the center of the lensto the objectdi = distance from the center of the lensto the imagem = magnification

Ten Homework Problems

Chapter 26 Problems 2, 9, 10, 11, 25, 41, 42, 49, 57, 100


295

DISCUSSION OF SELECTED SECTIONS

26.1 and 26.2 The Index of Refraction, and Snell’s Law and theRefraction of Light

If you put a pencil in a clear glass of water, the image of the pencil in the water appearsto be bent and distorted. The light passing from the air into the water is refracted,bending due to the fact that it’s passing from one medium to another. If we consider asingle beam of laser light, we can observe it as it passes from air into a piece of glass.

The angle i from the normal line at which the beam approaches the glass from the air iscalled the angle of incidence. The angle r from the normal line in the glass is the angleof refraction. As the light passes from the air, a less dense medium, into the glass, a moredense medium, the beam bends toward the normal line. When the beam of light exits theglass and passes back into the air, it bends away from the normal at the same angle itentered the glass from the air.

The light bends toward the normal in the glass because the beam slows down as it entersthe glass. Light travels more slowly in a more dense medium. Recall that sound travelsfaster in a more dense medium, but sound is a mechanical wave, while light is anelectromagnetic wave. The ratio of the speed of light in air (approximately a vacuum) tothe speed of light in the glass (or any other medium) is called the index of refraction n:

glassv

cn

The index of refraction for a vacuum or air is 1, since v = c. The index of refraction forcrown glass is about 1.6, which means that light travels 1.6 times faster in a vacuum thanin crown glass.

θi

θi

θr

glass

air

air


296

We can relate the index of refraction to the angles of incidence and refraction by usingSnell’s law of refraction:

2211 sinsin nn

where n1 and n2 are the indices of refraction of the first and second media, and and2 are the angles of incidence and refraction, respectively.

Example 1A beam of light enters the flat surface of a diamond at an angle of 30º from the normal.The angle of refraction in the diamond is measured to be 12º from the normal. Determinethe speed of light in the diamond.

SolutionThe angle of incidence θ1 = 30º and the angle of refraction θ2 = 12º. The index ofrefraction can be found by Snell’s law:

5.2

12sin30sin0.1

sinsin

2

2

2211

n

n

nn

The speed of light in diamond can be found by

smxsmx

n

cv

v

cn

diamond

diamond

/102.15.2

/103 88

2

2

26.3 Total Internal Reflection

Consider a water-proof laser which you can put under the water and shine a beam of lightup out of the water into the air. If you shine the light at a small angle relative to thenormal, the light will emerge from the water and bend away from the normal as it entersthe air.

θi

θi

θr

θc

θr = 90̊

θr

totalinternalreflection


297

As you increase the angle at which the laser is pointed at the surface of the water, therefracted angle also increases, eventually causing the refracted ray to bend parallel to thesurface of the water:

The angle c is called the critical angle. If the laser is pointed at an angle greater than thecritical angle, the beam will not emerge from the water, but will reflect back into thewater.

This phenomena is called total internal reflection. The inside surfaces of a glass prism ina pair of binoculars can become like mirrors, reflecting light inside the prism if the lightis pointed at the surface at an angle greater than the critical angle. Total internal reflectionis the also the principle behind the transmitting of light waves through transparent fiberoptic cable for communication purposes.

Example 2

The speed of light in a particular piece of glass is 2.0 x 108 m/s, and the speed of light inwater is 2.3 x 108 m/s.(a) Find the index of refraction for

i. the glassii. water

(b) A sheet of this glass is placed over a tank full of water. Laser light is incident on theglass from the air above the glass at an angle of 40º. Determine whether or not the lightpasses into the water, and, if it does, find the angle of refraction of the light in the water.

Solution

(a) i. 5.1/100.2

/100.38

8

smx

smx

v

cn

gg

ii. 3.1/103.2

/100.38

8

smx

smx

v

cn

ww

(b) First, let’s find the critical angle for the light traveling from the glass to the water.

60

5.1

3.1sinsin

g

wc n

n

If the light passes from the glass toward the water at an angle greater than 60º, it willtotally internally reflect inside the glass. The angle of refraction inside the glass can befound by

4.255.1

40sinsin

sin

sinsin

1

g

airg

ggairair

n

nn


298

Since the angle of the light in the glass less than 60º, the light will refract in the water.The angle of refraction in the water can be found by

7.293.1

4.25sin5.1sin

sinsin

sinsin

11

w

ggw

wwgg

n

n

nn

26.5 The Dispersion of Light: Prisms and Rainbows

Each color in the spectrum refracts just a little differently than every other color. This iswhy we can separate white light into its component colors by passing it through a prism.The shorter wavelengths slow down and bend more than the longer wavelengths, soviolet bends the most, and red the least:

Red

Violet

Glass

40º

glass

air

water

θg

θw


299

Example 3

The glass plate shown above has an index of refraction that depends on the wavelength ofthe light that enters it. The index of refraction is 1.54 for yellow light of wavelength 5.80x 10-9 m in the air and 1.62 for violet light of wavelength 4.20 x 10-9 m in the air. Boththe yellow and violet beams of light enter the glass from the left at the same angle of 30ºabove the normal, are refracted inside the glass, and exit the glass on the right.

(a) Determine the following for each color for the time the light is inside the glass.i. the speed of each color in the glassii. the wavelength of each color in the glassiii. the frequency of each color in the glass

(b) On the figure above, sketch the approximate paths of both the yellow and the violetrays as they pass through the glass and then exit into the air.(c) The figure below represents a hollow space in a large piece of the type of glassdescribed above. On this figure, sketch the approximate path of the yellow and the violetrays as they pass through the hollow space and back into the glass.

glass

air

Y

V

Yellow

Violet

Glass30˚

30˚


300

Solution(a) i.

smxsmx

n

cv

smxsmx

n

cv

VV

YY

/1085.162.1

/1000.3

/1095.154.1

/1000.3

88

88

ii. Since the frequency of the light does not change as the beam passes from one mediumto another, the wavelength is proportional to the speed.

mxmx

n

mxmx

n

V

airV

Y

airY

77

77

1059.262.1

1020.4

1076.354.1

1080.5

iii. The frequency of each color is the same in the glass and in the air.

Hzxmx

smxcf

Hzxmx

smxcf

airV

airY

147

8

147

8

1014.71020.4

/1000.3

1017.51080.5

/1000.3

(b) Violet light slows down more than yellow light, and bends its path more than yellowlight. Both beams bend toward the normal line inside the glass, and away from thenormal line (at 30º) when they exit the glass into the air again.

(c) Since each beam of light is going from a more dense medium to a less dense medium,they will bend away from the normal as they enter the air, and toward the normal as theyenter the glass again. Again, violet light will bend more than yellow light.

Yellow

Violet

Glass30˚

30˚


301

26.6 - 26.8 Lenses, The Formation of Images by Lenses, and The Thin-Lens Equation and Magnification Equation

Lenses operate on the principle of refraction. A diverging (concave) lens is a lens whichis thicker on the edges than it is in the middle, and it diverges the light rays that passthrough it:

The focal point of a diverging lens can be found by extending the diverging rays backbehind the lens until they seem to meet.

A converging (or convex) lens is a lens which is thicker in the middle than on the edges,and it converges parallel rays that pass through it:

When you read the words diverging and converging, or convex and concave, be sure youidentify whether the question is asking you about lenses or mirrors. The answersassociated with lenses might be quite different than those associated with mirrors!

glass

air

Y

V

f

ff


302

Just as in the case of the concave mirror, a converging lens can create an image whichcan be real or virtual, upright or inverted, larger or smaller in size, or the same size asthe object. It all depends on where the object is placed relative to the focal length of thelens. Since the AP Physics B exam typically focuses on the converging reather than thediverging lens, we will look at some examples involving the converging lens.

The ray diagrams for the converging lens are very similar to the ray diagrams for theconcave mirror, as are the equations listed above for relating the focal length of the lens,the object distance, image distance, and magnification.

Example 4A converging (convex) lens has a focal length of 15.0 cm. A 5.0-cm tall candle is placedat a distance of 40.0 cm to the left of the lens.(a) By drawing a ray diagram, find the location of the image formed by the converginglens. State whether the image is real or virtual, upright or inverted, and larger, smaller, orthe same size as the object (candle).(b) Using the lens and magnification equations, verify your results from part (a).

Solution(a) To find out what kind of image will be formed by the lens, we will draw two rays: (1)one ray from the flame entering the lens parallel to the principal axis and bending throughthe focal point, and (2) another ray from the flame which passes straight through thecenter of the lens without bending. The image is formed at the location of the intersectionof these two rays:

We see that in the case where the object (candle) distance from the lens is greater thantwice the focal length, the image is inverted and reduced in size. The image is also real,so if we placed a screen at the location of the image, we would see the projection of asmall inverted candle. The image formed by a converging lens is real if the objectdistance is greater than the focal length.

From the diagram, let’s say we measure the image distance at 23 cm from the center ofthe lens, and the image is measured to be 3.0 cm tall.

45 30 15 15 30 45

cm

1

2


303

(b) Using the lens and magnification equations, we can solve the image distance:

cmd

dcmcm

ddf

i

i

io

0.24

1

0.40

1

0.15

1

111

Note that our measurement of the image distance is very close to the calculated valuewithin reasonable experimental error.

The height of the candle can be found by

cm

cm

cmcm

d

dhh

d

d

h

hm

o

ioi

o

i

o

i

0.30.40

0.240.5

Again, the negative sign indicates the image is inverted.

If we place the candle at a distance from the lens equal to the focal length, our two rayswould emerge parallel to each other, and no image would be formed.

If we place the candle inside the focal length of the lens, we get a result similar to thesame case for a concave mirror.

Example 5A converging lens has a focal length of 30.0 cm. A 5.0-cm tall candle is placed at adistance of 10.0 cm in front of the lens.(a) By drawing a ray diagram, find the location of the image formed by the converginglens. State whether the image is real or virtual, upright or inverted, and larger, smaller, orthe same size as the object (candle).(b) Using the lens and magnification equations, verify your results from part (a).

Solution(a)

45 30 15 15 30 45

cm


304

With the candle placed inside the focal length, our two rays diverge as they emerge fromthe lens. No image is formed on the side opposite to the candle, but extending the raysbackward, we find that they seem to originate on the same side as the candle. The pointfrom which they seem to originate is where a virtual image of the candle is formed.

From the diagram, let’s say we measure the image distance at 17.0 cm from the center ofthe lens, and the image is measured to be 8.0 cm tall.

(b) Using the lens and magnification equations, we can solve the image distance:

cmd

dcmcm

ddf

i

i

io

0.15

1

0.10

1

0.30

1

111

The height of the candle can be found by

cm

cm

cmcm

d

dhh

d

d

h

hm

o

ioi

o

i

o

i

5.70.10

0.150.5

The positive sign indicates the image is upright.

A summary of the images formed by a converging lens and a converging mirror is listedin the table below, where do is the distance from the candle to the object and f is the focallength of the lens or mirror. For a converging lens, a positive (+) image distance di

implies that the image is formed on the opposite side of the lens as the object, and anegative (-) image distance implies that the image is formed on the same side as theobject. This sign convention is just the opposite for a converging (concave) mirror.

Object placedat:

Imagedistance di

real or virtual upright orinverted

enlarged orreduced

do > 2f + real inverted reduceddo = 2f + real inverted same size

f < do < 2f + real inverted enlargeddo = f No image No image No image No imagedo < f - virtual upright enlarged

PracticeVerify the results in the table above by drawing the ray diagrams below.

3f 2f f f 2f 3f


305

CHAPTER 26 REVIEW QUESTIONSFor each of the multiple choice questions below, choose the best answer.

1. A beam of light passes from the airthrough a thick piece of glass as shown.Which of the following angles is theangle of refraction?(A) 1(B) 2(C) 3(D) 4(E) 5

2. The speed of light in a piece of glassis 1.5 x 108 m/s. What is the index ofrefraction of the glass?(A) 2(B) 1.5(C) 0.67(D) 0.33(E) 0.2

glass

2

1

3

5

4

3f 2f f f 2f 3f

3f 2f f f 2f 3f


306

3. A beam of light passes from air intowater. Which of the followingstatements is true?(A) The angle of incidence is greater

than the angle of refraction in thewater.

(B) The angle of incidence is less thanthe angle of refraction in the water.

(C) The angle of incidence is equal tothe angle of refraction in the water.

(D) The frequency of the light decreases.(E) The frequency of the light increases.

4. Total internal reflection occurs when(A) light passes from air into water.(B) light refracts as it exits glass into air.(C) light reflects off of a mirror.(D) light passing through glass is

reflected inside the glass.(E) the angle of incidence is less than

the critical angle.

5. Which of the following is true of adiverging lens?(A) Incoming parallel rays passing

through the lens converge to a focalpoint.

(B) The lens is thinner in the center thanon the edges.

(C) The lens is thicker in the center thanon the edges.

(D) Light must enter it parallel to theprincipal axis.

(E) The lens must be flat on one side.

6. A candle is placed on the principalaxis of a convex lens at a distance of 30cm from the lens. The focal length of thelens is 10 cm. The image formed will be(A) real, upright, and enlarged(B) real, inverted, and enlarged(C) real, inverted, and smaller(D) virtual, upright, and enlarged(E) virtual, upright, and smaller

7. A candle is placed on the principalaxis of a convex lens at a distance of 10cm from the lens. The focal length of thelens is 20 cm. The image formed will be(A) real, upright, and enlarged(B) real, inverted, and enlarged(C) real, inverted, and smaller(D)virtual, upright, and enlarged(E) virtual, upright, and smaller

8. A candle is placed on the principalaxis of a convex lens at a distance of 20cm from the lens. The image formed ismagnified 3 times. The image distance is(A) 7 cm(B) 20 cm(C) 60 cm(D) 90 cm(E) 120 cm

9. A beam of green light and a separatebeam of blue light enter a converginglens parallel to the principal axis. Whichof the following statements is true?(A) The focal length of the green light is

greater than the focal length of theblue light.

(B) The focal length of the green light isless than the focal length of the bluelight.

(C) The focal length of the green light isequal to the focal length of the bluelight

(D) The green light will be absorbed inthe lens, and the blue light will betransmitted through the lens.

(E) The blue light will be absorbed in thelens, and the green light will betransmitted through the lens.


307

Free Response Question

Directions: Show all work in working the following question. The question is worth 15 points,and the suggested time for answering the question is about 15 minutes. The parts within aquestion may not have equal weight.

1. (15 points)

A beam of light is directed from the air toward a glass prism in the shape of an equilateraltriangle. The index of refraction of the glass is 1.47.

(a) On the diagram above, sketch the path of the beam of light in the prism, and as it exits theprism. Be sure to use a straight-edge to draw the path of the light in each region.

(b) Calculate the angle from a line normal to the surface of the right face of the prism at whichthe beam exits the prism.

The prism is actually the top of a convex lens of focal length 50.0 cm. In addition to this lens,you are given a lens holder, an optical bench on which the lens and holder can be placed, a 5.0-cm tall candle and matches, and a screen on a holder which can be mounted on the optical bench.Your teacher tells you that you are to produce an image on the screen which is twice as tall as thecandle itself.

(c) On the diagram below, draw and label the equipment so that it is assembled in such a way asto produce this image. Be sure and place each item at an accurate distance, and show anycalculations you used to help you find the accurate distances.

60˚ 60˚

N


308

(d) If the lens were placed under water, would the focal length of the lens increase, decrease, orremain the same as in air? Explain.

ANSWERS AND EXPLANATIONS TO CHAPTER 26 REVIEW QUESTIONS

Multiple Choice

1. EAngle 5 is the angle of refraction, as measured from the normal line in the glass.

2. A

2/105.1

/1038

8

smx

smx

v

cn

3. AWhen the light enters the water, the beam bends toward the normal line, causing the angle ofrefraction to be less than the angle of incidence The frequency remains constant.

4. DTotal internal reflection implies that no light exits the glass since it is reflected inside the glass.

5. BA diverging lens is also called a concave lens, and is thinner in the center than on the edges.

6. CThe candle is placed at a distance greater than twice the focal length, and so the image formedwill be real, inverted, and smaller than the candle.

0 25 50 75 100 125 150 175 200 250 275 300

LensandHolder

candle ScreenandHolder


309

7. DThe candle is placed at a distance less than the focal length, and so the image formed will bevirtual, upright, and enlarged. This is the case for a magnifying glass.

8. CIf the image is 3 times larger than the object, the image distance must be 3 times larger than theobject distance.

9. AGreen has a longer wavelength than blue, and longer wavelengths travel at a higher speed thanshorter wavelengths in glass. Thus, the green light undergoes a smaller change in speed in theglass, and therefore bends less than blue, producing a longer wavelength.

Free Response Question Solution

(a) 3 points

(b) 5 pointsThe angle of refraction θ2 in the glass can be found by Snell’s law:

9.19

47.1

30sinsin

sinsin 111

2glassn

Using this angle and some geometry gives the angle of incidence θ3 = 40.1º for the beam beforeit exits the prism. Then the refracted angle θ4 as the light exits the prism is

3.71

1

1.40sin47.1sin

sinsin 131

4air

glass

n

n

(c) 5 pointsFirst, let’s choose an object distance and calculate the corresponding image distance. If we are toproduce a real image twice as large as the object, we must place the candle between f and 2f suchthat the image distance is twice the object distance (di = 2do). Using the lens equation:

60˚ 60˚

N

30˚

θ2 θ4

θ3


310

cmd

cmd

dcm

df

ddf

ddf

i

o

o

o

oo

io

0.150

0.75

2

3

0.50

1

2

31

2

111

111

So we can place the candle at 0, the lens at 75, and the screen at 75 + 150 = 225.

(d) 2 pointsThe focal length of the lens in the water would be greater, since there would be less of a changein speed of the light between the glass and the water as compared to the glass and the air.

0 25 50 75 100 125 150 175 200 250 275 300

ff

Air

ff

Water

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