Physics Chapter 13

CHAPTER 13 Instruction and Intervention Support

1 Core Instruction

Chapter Resources

■■ The Teacher’s Edition wrap has extensive teaching support for every lesson, including Misconception Alerts, Teach from Visuals, Demonstrations, Teaching Tips, Differentiated Instruction, and Problem Solving. ■■ Several additional presentation tools are available online for every chapter, including Animated Physics features, Interactive Whiteboard activities, and Visual Concepts.

Labs and Demonstrations Section Instruction Go Online for Materials Lists and Solutions Lists.

■■ Textbook: Characteristics of Light ■■ Demonstrations: Infrared Light • Radio Waves • How Light Travels 13.1 Visual Concepts: Characteristics of a Wave • Electromagnetic ■■ Lab: Light and Mirrors Waves • Electromagnetic Spectrum • Brightness and Distance ■■ Lab: Brightness of Light (Core Skill) from a Light Source ■■ Lab: Brightness of Light (Probeware) Teaching Visuals: Components of an Electromagnetic Wave • The Electromagnetic Spectrum • Predicting Wave Front Position Using Huygens’s Principle PowerPresentations

■■ Textbook: Flat Mirrors ■■ Demonstrations: Diffuse Reflection • Specular Reflection • 13.2 Visual Concepts: Reflection • Comparing Specular and Flat Mirror Images Diffuse Reflections • Angle of Incidence and Angle of Reflection • Comparing Real and Virtual Images Teaching Visuals: Image Formation by a Flat Mirror PowerPresentations

■■ Textbook: Curved Mirrors ■■ Demonstrations: Image Formed by a Concave Mirror • Focal Point 13.3 Animated Physics: Curved Mirrors of a Concave Mirror • Beams Reflected from a Concave Mirror • Visual Concepts: Comparing Concave, Convex, and Plane Convex Mirror Mirrors • Focal Point • Sign Conventions for Magnification • ■■ Lab: Curved Mirrors (STEM) Rules for Drawing Reference Rays for Mirrors • Ray Tracing for a ■■ QuickLab: Curved Mirrors Concave Spherical Mirror • Ray Tracing for a Convex Spherical Mirror • Sign Conventions for Mirrors • Reflecting Telescope Teaching Visuals: Image Formation by a Concave Spherical Mirror • Images Created by Concave Mirrors • Image Formation by a Convex Spherical Mirror • Spherical Aberration and Parabolic Mirrors • Sign Conventions and Rules for Drawing Reference Rays • Sign Conventions for Mirrors PowerPresentations

■■ Textbook: Color and Polarization ■■ Demonstrations: Reflection and Absorption of Color • Polarizing 13.4 Visual Concepts: Additive Color Mixing • Pixel • Subtractive Light by Transmission • Transmitting Light with Crossed Polarizers • Color Mixing • Linearly Polarized Light • Polarization Through Polarizing Light by Reflection Transmission • Polarization by Reflecting and Scattering ■■ QuickLab: Polarization of Sunlight Teaching Visuals: Additive and Subtractive Primary Colors • ■■ Lab: Polarization of Light (Probeware) Aligned and Crossed Polarizing Filters • Polarization of Light by Transmission, Reflection, and Scattering PowerPresentations

440A Chapter 13 PREMIUM Content Find all of your resources online at HMDScience.com.

2 Support and Intervention 3 Specialized Support

■■ Study Guide ■■ Chapter Summary Audio Files Concept Maps ■■ Differentiated Instruction: Inclusion, Below Level, and ■■ Scientific Reasoning Skill Builder English Learners (TE wrap) Interactive Demos Sample Problem Set I Sample Problem Set II

Where do I find it? PRINT DVD ONLINE Enrichment and Challenge Animated Physics ■ ■ Demonstrations (TE wrap) ■ ■ ■■ Differentiated Instruction: Pre-AP (TE wrap) Labs ■ ■ Power Presentations ■ ■ QuickLabs ■ ■ ■ Teaching Visuals ■ ■ Textbook ■ ■ ■ Visual Concepts ■ ■

Interactive Demos ■ ■ Concept Maps ■ ■ Sample Problem Set I ■ ■ Sample Problem Set II ■ ■ Scientific Reasoning Skill Builder ■ ■ Assessment Study Guide ■ ■

■■ Section Quizzes Chapter Summary Audio Files ■ ■ ■■ Chapter Tests A and B Differentiated Instruction (TE wrap) ■ ■■ Alternative Assessment (SE) Online Assessment and Remediation ■■ ExamView Banks

Light and Reflection 440B 440 likeobjects, oursun. distant celestial aswell objects ascloser isused tolight. TheVLA studyboth wavelengths thanat those ofvisible emitmoreobjects strongly at radio of celestial bodies. A number of celestial astronomy isused to produce images Astronomy Observatory (NRAO). Radio a project calledtheNational Radio The Very Large Array of ispart (VLA) About theImage phenomenon ofpolarization. subtractive colors andexplores the 4 Section and explainssphericalaberration. diagrams to confirm calculated results, concave andconvex mirrors, uses ray and magnification are calculated for 3 Section to determine image location. to planemirrors anduses ray diagrams 2 Section distance for alightsource. relationship between brightness and speed oflight,andintroduces the their frequency andwavelength to the the electromagnetic relates spectrum, 1 Section Chapter Overview CHAPTER 13 Chapter 13 identifies thecomponents of shows how image location applies thelaws ofreflection investigates additive and Lab Preview Untitled-280 440 Polarization of Sunlight Curved Mirrors QuickL Polarization ofLight(Probeware) Curved Mirrors (STEM) Brightness ofLight(Core Skill) Light andMirrors L concepts presented inthischapter. The following investigations the support abs 440 poised abovethedish. and focusestheraysatreceiver telescope reflects theradiowaves spectrum. Thedishofaradio radio andmicrowave regions ofthe electromagnetic radiationinthe in diameter. Theseantennasdetect 27 radioantennas,each25meters Socorro, NewMexico,consistsof The Very Large Array, locatednear abs Polarizing by Transmission Beams Reflectedfrom aConcave Mirror Focal Point ofaConcave Mirror Image Formed by aConcave Mirror Flat Mirror Images Specular Reflection Diffuse Reflection Radio Waves Infrared Light Demons tra tions 5/20/2011 6:21:20AM

©Laurence Parent CHAPTER 13 SECTION 1 Characteristics of Light SECTION 2 Focus and Motivate Light and Flat Mirrors SECTION 3 Activate Prior Curved Mirrors Knowledge SECTION 4 Knowledge to Review Re ection Color and • Waves transport energy. They can be Polarization transverse or longitudinal. Waves have amplitude, frequency, wavelength, and velocity.

Why It Matters Items to probe Mirrors have many • The ability to describe spatial applications both for relationships in geometric scientists and in everyday life. For example, a terms: Make sure students understand reflector telescope uses terms such as perpendicular and two mirrors to gather, focus, and reflect light parallel and can solve equations of onto the eyepiece. the form The reflector telescope 1 1 1 _ + _ = _ . remains one of the ( A) ( B) ( C) most popular designs used by amateur • Preconceptions about light: Ask astronomers, even students to describe the path of sound though it was invented over 300 years ago. waves when they hear something and the path of light rays when they see an object.

ONLINE Physics HMDScience.com

ONLINE LABS Light and Mirrors PREMIUM Brightness of Light CONTENT Designing a Device to Trace Physics Drawings HMDScience.com

Polarization of Light Curved Mirrors ©Laurence Parent Why It Matters 441 Connecting to History piece. This design concept is still very popular; however, today’s mirrors are much larger and are Untitled-280 441 Dark pools of water or other natural reflective 5/20/2011 6:21:28 AM surfaces such as polished stone gave rise to the often manufactured in special labs at large invention of mirrors. What began as a tool for research universities. reflecting peoples’ appearance now has many After briefly explaining this history to practical and even scientific uses. As early as the students, ask them if they are aware of other seventeenth century, scientists such as Isaac uses of mirrors. Discuss with them whether the Newton and Laurent Cassegrain began experi- mirrors are necessary or not. For instance, a car menting with mirrors to develop reflector can be driven without rear or side-view mirrors, telescopes, which work by using two mirrors to but not as safely. gather, focus, and reflect light onto the eye-

Light and Reflection 441

Plan andPrepare Demonstration 442 sensitive film. photographically detected with infrared- from thesubstance asheat canbe temperature. Energy transferred away netic radiation, increases asubstance’s one example ofnonvisibleelectromag- Explain that infrared radiation, whichis the leftofred ofthespectrum. part thermometer bulbinthedarkregion to minutes. 5 after temperature the and temperature initial the record student a Have bulb. thermometer the on shines spectrum the of light red the that so paper the on sensor) temperature probeware(or thermometer Tape the paper. the on cast is beam’sspectrum light the that so prism the of side other the on paper white the place and lights, classroom the Lower slit. the from the path ofthebeam oflightemerging in prism the Set box. the inside source light the place and box, black the Procedure with probeware system thermometer ortemperature sensor black box, prism,white paper, Materials invisible electromagnetic radiation. Purpose Infrared Light definition ofelectromagnetic waves. in waves, andthenrelate thisto the netic energy islightenergy that travels Emphasizethatparticles. electromag- interaction between electrically charged referred to astheforce that causes heard thedescriptor electromagnetic Scientific Meanings Preview Vocabulary SECTION Teach Repeat thisprocedure, placingthe Chapter 13 Demonstrate oneform of incandescent lightsource, Make a slit in one side of of side one in slit a Make 1 Students may have Differentiated Instruction Untitled-281 442 different frequencies. as differences in color, explain it as waves having spectrum. Rather than explaining the spectrum seeing white light separated into the color have limited or no vision and are incapable of above will not make sense to students who the humaneye. Theprismexample discussed waves interms ofthose visibleornotto difficulty comprehending adescription oflight Students with visual impairment may have Inclusion component colors. Prism electromagnetic wave from thesourceatspeedoflight magnetic fields,whichradiateoutward consists ofoscillatingelectricand 442 FIGURE 1.1 by distance. of a light source is affected Describe how the brightness a finite speed. Recognize that light has radiation. wavelength of electromagnetic Calculate the frequency or electromagnetic spectrum. Identify the components of the SECTION 1 A prism separates light into its Chapter 13 Objectives awavethat Electromagnetic Waves electromagnetic wave Key Term Light Characteristics of wave theories, but the wave model will be used in this section. this in used be will wave model the but wavetheories, of the two. The current model incorporates aspects of both particle and will see sunlight see will pass through theglass andemerge asaband ofcolors. The spectrum includes more than visible light. as theonein orange, yellow, green, blue, andviolet. Ifyou examineaglass prism, such : red, separatedbe colors ofthevisible spectrum into six elementary colors. What your eyesrecognize as “white” light isactually light that can you green see light pass through. forother Thistrue phenomenonisalso youWhen ofgreen holdapiece plastic infront ofasource ofwhite light, bulb orthesun. However, there ismore to light than examples. these examples include thebright,Some white light that isproduced by alight thinkoflight, mostWhen people thinkofthelight they that can they see. light. The reason isthat are they allexamples of waves, andradio waves—have many ofthesame properties asvisible one you ofradiation—including see. offorms Avariety Xrays, micro- you wouldfindthat thefilmrecords than the amuch spectrum wider of photographic filmto examinethelight through dispersed aprism, Not alllight isvisible to thehuman eye. Ifyou were types certain to use Light has been described as a particle, a wave, and even a combination combination a wave, a even and particle, a as described been has Light Figure 1.1, orany thick, triangular-shaped ofglass, piece you electromagnetic waves. 5/20/2011 6:22:04AM

©Photo Researchers, Inc. Electromagnetic waves vary depending on FIGURE 1.2 frequency and wavelength. Electromagnetic Wave An electromagnetic wave consists In classical electromagnetic wave theory, light is of electric and magnetic field waves at right angles to each other. considered to be a wave composed of oscillating Demonstration electric and magnetic fields. These fields are perpendicular to the direction in which the wave moves, as Oscillating magnetic ﬁeld Radio Waves shown in Figure 1.2. Therefore, electromagnetic Purpose Demonstrate another example waves are transverse waves. The electric and magnetic fields are also at right angles to each other. of radiation in the electromagnetic Electromagnetic waves are distinguished by their spectrum. different frequencies and wavelengths. In visible Materials pocket-size transistor radio, light, these differences in frequency and wavelength account for different colors. The difference in metal can with metal lid, glass jar,

frequencies and wavelengths also distinguishes Oscillating electric ﬁeld aluminum foil, paper, plastic wrap visible light from invisible electromagnetic radiation, such as X rays. Procedure Tell students that radio Direction of the electromagnetic wave Types of electromagnetic waves are listed in waves have a longer wavelength than Figure 1.3. Note the wide range of wavelengths and infrared waves and that these waves can frequencies. Although specific ranges are indicated be detected with a radio. Turn on the in the table, the electromagnetic spectrum is, in reality, continuous. There is no sharp division PHYSICS radio, and tune it to a station. Have Spec. Number PH 99 PE C14-001-002-A between one kind of wave and the next. Some types Boston Graphics, Inc.students notice changes in reception as of waves even have overlapping ranges. 617.523.1333 you place the radio in the metal can and then in the glass jar. Wrap the radio in FIGURE 1.3 the paper, in the aluminum foil, and THE ELECTROMAGNETIC SPECTRUM in the plastic wrap. Have students Classification Range Applications notice that paper is “transparent” to radio waves. Point out that different radio waves λ > 30 cm AM and FM radio; television f < 1.0 × 109 Hz types of radiation can penetrate different materials. microwaves 30 cm > λ > 1 mm radar; atomic and molecular research; 1.0 × 109 Hz < f < 3.0 × 1011 Hz aircraft navigation; microwave ovens

infrared (IR) waves 1 mm > λ > 700 nm molecular vibrational spectra; infrared 3.0 × 1011 Hz < f < 4.3 × 1014 Hz photography; physical therapy Teaching Tip Point out that all of the frequencies in visible light 700 nm (red) > λ > 400 nm (violet) visible-light photography; optical 4.3 × 1014 Hz < f < 7.5 × 1014 Hz microscopy; optical astronomy Figure 1.3 are expressed in hertz but that the wavelengths are in centimeters, ultraviolet (UV) light 400 nm > λ > 60 nm sterilization of medical instruments; millimeters, and nanometers, which can 7.5 × 1014 Hz < f < 5.0 × 1015 Hz identification of fluorescent minerals be expressed as 10−2, 10−3, and 10−9 m, X rays 60 nm > λ > 10−4 nm medical examination of bones, teeth, and respectively. Ask students if the wave- 5.0 × 1015 Hz < f < 3.0 × 1021 Hz vital organs; treatment for types of cancer lengths increase or decrease as the gamma rays 0.1 nm > λ > 10−5 nm examination of thick materials for frequencies decrease. increase 3.0 × 1018 Hz < f < 3.0 × 1022 Hz structural flaws; treatment for types of cancer; food irradiation ©Photo Researchers, ©Photo Inc. Researchers, Problem Solving Light and Reflection 443 Alternative Approaches Students will see the wave speed equation on the next page. However, manipulating the Untitled-281 443 Expand the teaching tip by explaining that 5/20/2011 6:22:06 AM frequency and wavelength are inversely speed equation so that students are asked to proportional. An increase in the distance solve for frequency demonstrates the relation- between troughs and peaks of a wave means ship discussed in the teaching tip. that the frequency with which the wave passes a fixed point will decrease. Frequency and wavelength can be related with the following equation: _λ f = v

Light and Reflection 443 All electromagnetic waves move at the speed of light. All forms of electromagnetic radiation travel at a single high speed in continued a vacuum. Early experimental attempts to determine the speed of light Teach failed because this speed is so great. As experimental techniques improved, especially during the nineteenth and early twentieth centuries, the speed of light was determined with increasing accuracy and precision. By the mid-twentieth century, the experimental error was less Demonstration than 0.001 percent. The currently accepted value for light traveling in a vacuum is 2.997 924 58 × 108 m/s. Light travels slightly slower in air, How Light Travels with a speed of 2.997 09 × 108 m/s. For calculations in this book, the value used for both situations will be 3.00 × 108 m/s. Purpose Demonstrate that light waves The relationship between frequency, wavelength, and speed described can be approximated as rays. in the chapter on vibrations and waves also holds true for light waves. Materials laser, index card, two dusty chalkboard erasers, plane mirror Wave Speed Equation Caution Avoid pointing the laser beam c = f λ near students’ eyes; retinal damage may occur. speed of light = frequency × wavelength Procedure Direct the laser beam across the room. Point out that in order for PREMIUM CONTENT students to see the beam, it is necessary Electromagnetic Waves Interactive Demo HMDScience.com to have an object in the path of the Sample Problem A The AM radio band extends from beam that will reflect some light. 5.4 × 105 Hz to 1.7 × 106 Hz. What are the longest and shortest Place the index card in the path of wavelengths in this frequency range? the beam, and slowly walk across the 5 6 ANALYZE Given: f1 = 5.4 × 10 Hz f2 = 1.7 × 10 Hz room while keeping the beam centered c = 3.00 × 108 m/s on the card. Students should see that Unknown: λ = ? λ = ? the beam travels in a straight line. 1 2 Stand beside the beam, and tap SOLVE Use the wave speed equation on this page to find the wavelengths: the erasers together above the beam. c = f λ λ = _ c As the chalk dust falls from the erasers, f 8 Calculator Solution the beam will become visible. Quickly λ = __3.00 × 10 m/s 1 5 Although the calculator solutions walk along the length of the beam, 5.4 × 10 Hz are 555.5555556 m and λ = × 2 176.470588 m, both answers tapping the erasers until the entire beam 1 5.6 10 m must be rounded to two digits becomes visible. 8 because the frequencies have λ = __3.00 × 10 m/s only two significant figures. 2 1.7 × 106 Hz 2 λ2 = 1.8 × 10 m

Continued

Problem444 Chapter Solving 13 Reality Check Untitled-281Because 444 the relationship between frequency 5/20/2011 6:22:06 AM and wavelength is inversely proportional, students should double-check their answers to problems of this type to assure that wavelength increases as frequency decreases.

444 Chapter 13 Electromagnetic Waves (continued) PROBLEM guide A 1. Gamma-ray bursters are objects in the universe that emit pulses of gamma rays Use this guide to assign problems. with high energies. The frequency of the most energetic bursts has been measured at around 3.0 × 1021 Hz. What is the wavelength of these gamma rays? SE = Student Edition Textbook 2. What is the wavelength range for the FM radio band (88 MHz–108 MHz)? PW = Sample Problem Set I (online) 3. Shortwave radio is broadcast between 3.50 and 29.7 MHz. To what range of PB = Sample Problem Set II (online) wavelengths does this correspond? Why do you suppose this part of the spectrum is called shortwave radio? Solving for: 4. What is the frequency of an electromagnetic wave if it has a wavelength of 1.0 km? λ SE Sample, 1–3; 5. The portion of the visible spectrum that appears brightest to the human eye is Ch. Rvw. 10–13 around 560 nm in wavelength, which corresponds to yellow-green. What is the frequency of 560 nm light? PW 5–7 6. What is the frequency of highly energetic ultraviolet radiation that has PB 7–10 a wavelength of 125 nm? f SE 4–6 PW Sample, 1–4 Waves can be approximated as rays. PB 3–6 Consider an ocean wave coming toward the shore. The broad crest of the wave that is perpendicular to the wave’s motion consists of a line of water c PW 8 particles. Similarly, another line of water particles forms a low-lying trough PB Sample, 1, 2 in the wave, and still another line of particles forms the crest of a second wave. In any type of wave, these lines of particles are called wave fronts. *Challenging Problem All the points on the wave front of a plane wave can be treated as point sources, that is, coming from a source of negligible size. A few of these points are shown on the initial wave front in Figure 1.4. Each of these point sources produces a circular or spherical secondary wave, or wavelet. Answers: The radii of these wavelets are indicated by the blue arrows in Figure 1.4. The line that is tangent to each of these wavelets at some later time Practice A determines the new position of the initial wave front (the new wave 1. 1.0 × 10−13 m front in Figure 1.4). This approach to analyzing waves is called Huygens’s principle, named for the physicist Christian Huygens, who developed it. 2. 3.4 m –2.78 m Huygens’s principle can be used to derive the properties of any wave 3. 85.7 m –10.1 m; The wavelengths (including light) that interacts with matter, but the same results can be are shorter than those of the obtained by treating the propagating wave as a straight line perpendicular AM radio band. to the wave front. This line is called a ray, and this simplification is called the ray approximation. 4. 3.0 × 105 Hz 5. 5.4 × 1014 Hz 15 FIGURE 1.4 Tangent line 6. 2.40 × 10 Hz (new wave front) Huygens’s Principle According to Huygens’s principle, a wave front can be divided into point sources. The line tangent to the wavelets from these sources marks the wave front’s new position.

Initial wave front

Light and Reflection 445 HRW • Holt Physics PH99PE-C14-001-004-A Deconstructing Problems Equation: c = f λ 8 Untitled-281 445 5/20/2011 6:22:07 AM Practice Problem 3 asks students to synthesize 3.00 × 10 m/s = 3.50 MHz × λ1 8 their answer into an educated guess as to why 3.00 × 10 m/s = 29.7 MHz × λ2 shortwave radio is referred to as such. Students Solve: should follow the same steps for solving for the 3.00 × 108 m/s __ = λ = 1.01 × 10 1 m unknown variable in this problem as the steps 3.50 MHz 1 8 demonstrated in Sample Problem A. 3.00__ × 10 m/s 1 = λ2 = 8.57 × 10 m Unknown: λ = ? 29.7 MHz Students should compare these wavelengths, Given: c = 3.00 × 108 m/s 1.01 × 10 1 to 8.57 × 10 1 m, to the wavelengths in f = 3.50 MHz - 29.7 MHz Sample Problem A, 1.8 × 10 2 to 5.6 × 10 2 m.

Light and Reflection 445 Illuminance decreases as the square of the distance from the source. You have probably noticed that it is easier to read a book beside a lamp Teach continued using a 100 W bulb rather than a 25 W bulb. It is also easier to read nearer to a lamp than farther from a lamp. These experiences suggest that the Teaching Tip intensity of light depends on both the amount of light energy emitted from a source and the distance from the light source. Point out that the illuminance at any Light bulbs are rated by their power input (measured in watts) and location is analogous to the power per their light output. The rate at which light is emitted from a source is called unit of area at that location. Ask the luminous flux and is measured in lumens (lm). Luminous flux is students to compare the surface areas a measure of power output but is weighted to take into account the response of the human eye to light. Luminous flux helps us understand of two spherical lampshades and the why the illumination on a book page is reduced as you move away from a brightness of the light that hits their FIGURE 1.5 light. Imagine spherical surfaces surface when identical light bulbs are of different sizes with a point light Luminous Flux Less light falls on each unit source at the center of the sphere, placed at the center of each. Use values square as the distance from the source increases. shown in Figure 1.5. A point source for their radii, such as 10 cm and 50 cm provides light equally in all (surface area of a sphere = 4πr2). directions. The principle of conservation of energy requires that the luminous flux is the same Teaching Tip on each sphere. However, the luminous flux divided by the area As compact fluorescent light bulbs of the surface, which is called the become more prevalent, the use of illuminance (measured in lm/m2, luminous flux (lumens) rather than or lux), decreases as the radius 1m 2m 3m squared when you move away power rating (watts) is becoming more from a light source. common in rating a bulb’s light output. Incandescent bulbs generally produce SECTION 1 FORMATIVE ASSESSMENTPHYSICS about 15 lumens/watt, whereas fluores- Spec. Number PH 99 PE C14-001-005-A Boston Graphics, Inc. cent bulbs produce 50–100 lumens/watt. Reviewing Main Ideas 617.523.1333 Ask students which type of light bulb is 1. Identify which portions of the electromagnetic spectrum are used in each more efficient. Have them compare the of the devices listed. average power usage over a week, a. a microwave oven b. a television set month, or year of normal usage for both c. a single-lens reflex camera types of light bulbs. 2. If an electromagnetic wave has a frequency of 7.57 × 1014 Hz, what is its wavelength? To what part of the spectrum does this wave belong? 3. Galileo performed an experiment to measure the speed of light by timing how long it took light to travel from a lamp he was holding to an assistant Assess and Reteach about 1.5 km away and back again. Why was Galileo unable to conclude that light had a finite speed? Assess Use the Formative Assessment on this page to evaluate student Critical Thinking mastery of the section. 4. How bright would the sun appear to an observer on Earth if the sun were four times farther from Earth than it actually is? Express your answer as a fraction of the sun’s brightness on Earth’s surface. Reteach For students who need additional instruction, download the Answers446 Chapter to Section 13 Assessment Section Study Guide. 1. a. microwave Response to Intervention To reassess Untitled-281 b. 446 radio waves, visible light 5/20/2011 6:22:08 AM students’ mastery, use the Section Quiz, c. visible light available to print or to take directly −7 online at HMDScience.com. 2. 3.96 × 10 m; near ultraviolet 3. The speed of light is too great to be measured over such a short distance. The time of travel for the light in Galileo’s experiment was about 1.0 × 10−5 s. __1

4. 16 of the sun’s brightness on Earth’s surface

446 Chapter 13 SECTION 2 SECTION 2

Objectives Flat Mirrors Distinguish between specular Plan and Prepare and diffuse reflection of light. Key Terms reflection angle of reflection Apply the law of reflection for Preview Vocabulary angle of incidence virtual image flat mirrors. Latin Word Origins The word Describe the nature of images reflection comes from the Latin verb Reflection of Light formed by flat mirrors. reflectere, which means “to bend back.” Suppose you have just had your hair cut and you want to know what the back of your head looks like. You can do this seemingly impossible task Relate the literal meaning of the root to by using two mirrors to direct light from behind your head to your eyes. the scientific definition of eflection (e.g., Redirecting light with mirrors reveals a basic property of light’s interaction the wave moves or bends away from the with matter. surface). Light traveling through a uniform substance, whether it is air, water, reflection the change in direction of an or a vacuum, always travels in a straight line. However, when the light electromagnetic wave at a surface that causes it to move away from the surface encounters a different substance, its path will change. If a material is opaque to the light, such as the dark, highly polished surface of a wooden table, the light will not pass into the table more than a few wavelengths. Part Teach FIGURE 2.1 of the light is absorbed, and the rest of it is deflected at the surface. This change in the direction of the light is called reflection. All substances absorb Reflection Mirrors reflect nearly at least some incoming light and reflect the rest. A good mirror can reflect all incoming light, so multiple images about 90 percent of the incident light, but no surface is a perfect reflector. of an object between two mirrors are Demonstration Notice in Figure 2.1 that the images of the golf ball get successively darker. easily formed. Diffuse Reflection The texture of a surface affects how it reflects light. Purpose Demonstrate that light The manner in which light is reflected from a surface depends on the reflected from a rough surface reflects surface’s smoothness. Light that is reflected from a rough, textured surface, in many directions. such as paper, cloth, or unpolished wood, is reflected in many different directions, as shown in Figure 2.2(a). This type of reflection is called diffuse Materials laser, index card reflection and is covered later in the chapter. Caution Avoid directing the laser beam Light reflected from smooth, shiny surfaces, such as a mirror or water near students’ eyes. in a pond, is reflected in one direction only, as shown in Figure 2.2(b). This type of reflection is called specular reflection. A surface is considered Procedure Tape the index card to a wall smooth if its surface variations are small compared with the wavelength in the classroom. Direct the laser beam of the incoming light. For our discussion, reflection will be used to mean only specular reflection. onto the card from across the room. Make sure the beam is not perpendicular

FIGURE 2.2 to the card’s surface. Explain to students Diffuse and Specular that the light strikes the index card as a Reflection Diffusely reflected single beam, but because the card is a light is reflected in many directions diffuse reflector, the beam undergoes (a), whereas specularly reflected light is reflected in the same forward (a) (b) reflection in all directions. Only a small direction only (b). part of the reflected light will go to each part of the room, enabling everyone to (br) ©Richard Megna/Fundamental NYC Photographs, see the spot where the laser beam HRW • Holt Physics HRW • Holt Physics Differentiated InstructionPH99PE-C14-002-003-A PH99PE-C14-002-004-ALight and Reflection 447 strikes the card. Inclusion TEACH FROM VISUALS Untitled-282 447 Accommodate auditory learners with 5/20/2011 6:22:39 AM a sound-driven mnemonic device for remem- FIGURE 2.2 Point out that the incident bering the distinction between diffuse and light rays are parallel. specular. “Diffuse” and “different” both begin Ask Why are the reflected rays from with diff, which can serve as a hint for the surface in (a) not parallel? students to remember that diffuse reflection means that light is reflected in many Answer: Because the surface in (a) has different directions. irregularities, it forms a different angle with the incident ray at each point.

Light and Reflection 447 FIGURE 2.3 Symmetry of Reflected Teach continued Light The symmetry of reflected light (a) is described by the law Incoming Re ected of reflection, which states that the light Normal light TEACH FROM VISUALS angles of the incoming and reflected rays are equal (b). FIGURE 2.3 In the image, the light from the flashlight strikes the surface. Re ecting surface Ask Why is there a bright spot where

the light strikes the surface? (a) TSI Graphics(b) HRW • Holt Physics Answer: Light is scattered off the PH99PE-C14-002-005-A imperfect or dirty surface. Incoming and reflected angles are equal. Ask Below the bright spot, there seem You probably have noticed that when incoming rays of light strike a to be rays at the lower left and lower smooth reflecting surface, such as a polished table or mirror, at an angle close to the surface, the reflected rays are also close to the surface. When right. What causes these rays? the incoming rays are high above the reflecting surface, the reflected rays Answer: We are seeing reflections off are also high above the surface. An example of this similarity between incoming and reflected rays is shown in Figure 2.3(a). the mirror surface of the incident and If a straight line is drawn perpendicular to the reflecting surface at the reflected beams. angle of incidence the angle point where the incoming ray strikes the surface, the angle of incidence between a ray that strikes a surface and and the angle of reflection can be defined with respect to the line. Careful the line perpendicular to that surface at meas urements of the incident and reflected angles θ an θ', respectively, the point of contact reveal that the angles are equal, as illustrated in Figure 2.3(b). angle of reflection the angle formed Demonstration by the line perpendicular to a surface θ = θ' and the direction in which a reflected angle of incoming light ray = angle of reflected light ray Specular Reflection ray moves Purpose Demonstrate that all parallel The line perpendicular to the reflecting surface is referred to as the normal to the surface. It therefore follows that the angle between rays of light reflected from a smooth the incoming ray and the surface equals 90° − θ, and the angle surface are reflected in the same between the reflected ray and the surface equals 90° − θ'. direction. FIGURE 2.4 Materials laser, flat mirror, dusty Flat Mirror Light reflecting off of a flat mirror Flat Mirrors creates an image that appears to be behind the mirror. chalkboard erasers The simplest mirror is the flat mirror. If an object, such as a Caution Avoid directing the Flat mirror pencil, is placed at a distance in front of a flat mirror and light is bounced off the pencil, light rays will spread out from the primary beam of the laser and the pencil and reflect from the mirror’s surface. To an observer reflected beam from the mirror toward Object Image looking at the mirror, these rays appear to come from a location the students. on the other side of the mirror. As a convention, an object’s image is said to be at this location behind the mirror because Procedure Tape the flat mirror to a wall the light appears to come from that point. The relationship in the classroom. Direct the beam of the between the object distance from the mirror, represented as p, p q and the image distance, represented as q, is such that the object laser onto the mirror from across the Object distance = Image distance and image distances are equal, as shown in Figure 2.4. Simi larly,

room. Make sure the beam is not the image of the object is the same size as the object. ©Richard Megna/Fundamental York New Photographs, ©Richard Megna/Fundamental York New Photographs, perpendicular to the mirror’s surface. 448 Chapter 13 Explain to students that the beam is Differentiated Instruction reflected at the mirror’s surface and that Below Level the angle of incidence is equal to the angle of reflection. This can be shown Untitled-282Give 448below-level students a concrete example 5/20/2011 6:22:40 AM qualitatively by gently tapping the of the flat (or plane) mirrors. For example, a erasers in front of the mirror so that two-foot-tall person standing two feet from a both the incoming and reflected beams flat mirror will be reflected as two feet tall and become visible. two feet from the mirror. If, in the instance of a car’s rearview mirror, the object is reflected smaller or farther away, the mirror is not flat.

448 Chapter 13 PHY_CNLAESE586694_811A.ai Sam Valentino 2.8.11 1st pass The image formed by rays that appear to come from the image point behind the mirror—but never really do—is called a virtual image. As virtual image an image from which shown in Figure 2.5(a), a flat mirror always forms a virtual image, which light rays appear to diverge, even though they are not actually focused always appears as if it is behind the surface of the mirror. For this reason, there; a virtual image cannot be Demonstration a virtual image can never be displayed on a physical surface. projected on a screen Flat Mirror Images Image location can be predicted with ray diagrams. Purpose Demonstrate that the image Ray diagrams, such as the one shown in Figure 2.5(b), are drawings that behind a mirror is virtual. use simple geometry to locate an image formed by a mirror. Suppose you want to make a ray diagram for a pencil placed in front of a flat mirror. Materials sheet of high-quality First, sketch the situation. Draw the location and arrangement of the plate glass (0.5 m × 0.5 m), two identical mirror and the position of the pencil with respect to the mirror. Construct candles, black chalkboard or black the drawing so that the object and the image distances (p and q, respectively) are proportional to their actual sizes. To simplify matters, we surface will consider only the tip of the pencil. Procedure Place the sheet of glass To pinpoint the location of the pencil tip’s image, draw two rays on vertically in front of the chalkboard. your diagram. Draw the first ray from the pencil tip perpendicular to the mirror’s surface. Because this ray makes an angle of 0° with a line Place one candle about 30 cm in front perpendicular (or normal) to the mirror, the angle of reflection also of the glass, and place the other candle equals 0°, causing the ray to reflect back on itself. In Figure 2.5(b), this ray is 30 cm behind it. denoted by the number 1 and is shown with arrows pointing in both directions because the incident ray reflects back on itself. Have students who might be able to Draw the second ray from the tip of the pencil to the mirror, but this see the candle behind the glass without time place the ray at an angle that is not perpendicular to the surface of looking through the glass move to the the mirror. The second ray is denoted in Figure 2.5(b) by the number 2. Then, draw the reflected ray, keeping in mind that it will reflect away from rear of the classroom. Have students the surface of the mirror at an angle, θ', equal to the angle of incidence, θ. close their eyes while you light the front Next, trace both reflected rays candle. Discreetly adjust the position back to the point from which they FIGURE 2.5 of the candle behind the glass so that appear to have originated, that is, Ray Diagram The position and size of the virtual image that forms in behind the mirror. Use dotted lines the image of the front candle as seen a flat mirror (a) can be predicted by constructing a ray diagram (b). when drawing these rays that in the glass coincides with the position appear to emerge from behind the of the back candle. When properly mirror to distinguish them from the rays of light in front of the positioned, the back candle will appear mirror. The point at which these lit when it is viewed through the glass dotted lines meet is the image sheet. point, which in this case is where Eye p q the image of the pencil’s tip forms. 1 Hold a match near the back candle

By continuing this process for 2 and tell students to open their eyes. all of the other parts of the pencil, h h' Raise the match and blow it out, giving you can locate the complete virtual ' the impression that you have just image of the pencil. Note that the Object Image pencil’s image appears as far finished lighting the back candle. Ask behind the mirror as the pencil is (b) Mirror how many candles are lit. Then lift the in front of the mirror (p = q). glass. Explain that the image of the flame Likewise, the object height, h, (a) (b) equals the image height, h'. appeared to be at the same distance

©Richard Megna/Fundamental York New Photographs, ©Richard Megna/Fundamental York New Photographs, HRW • Holt Physics behind the glass as the rear candle. PH99PE-C14-002-007A Light and Reflection 449

English Learners Teaching Tip Untitled-282 449 English learners may have heard the word virtual 5/20/2011 6:22:41 AM Point out that the object and its image being used in a figurative sense are equidistant from the mirror’s surface (e.g., virtual reality, virtual chat, etc.). (p = q). Students can demonstrate this Call attention to this context, and apply the conclusion using simple geometry to figurative meaning of virtual to the literal show that the ray-tracing procedure meaning of the word in the definition for virtual produces congruent triangles. image. Reiterate that the term virtual doesn’t mean “pretend” or “practice” or “simulated,” as it does in other instances in which students may have heard it.

Light and Reflection 449 FIGURE 2.6 This ray-tracing procedure will work for any object placed in front of a flat mirror. By selecting a single point on the object Mirror Reversal The front of an object (usually its uppermost tip or edge), you can use ray tracing to Assess and Reteach becomes the back of its image. locate the same point on the image. The rest of the image can be added once the image point and image distance have been Assess Use the Formative Assessment determined. on this page to evaluate student The image formed by a flat mirror appears reversed to an mastery of the section. observer in front of the mirror. You can easily observe this effect by placing a piece of writing in front of a mirror, as Reteach For students who need shown in Figure 2.6. In the mirror, each of the letters is reversed. additional instruction, download the You may also notice that the angle the word and its reflection make with respect to the mirror is the same. Section Study Guide.

Response to Intervention To reassess students’ mastery, use the Section Quiz, available to print or to take directly online at HMDScience.com. SECTION 2 FORMATIVE ASSESSMENT

Reviewing Main Ideas 1. Which of the following are examples of specular reflection, and which are examples of diffuse reflection? a. reflection of light from the surface of a lake on a calm day b. reflection of light from a plastic trash bag c. reflection of light from the lens of eyeglasses d. reflection of light from a carpet 2. Suppose you are holding a flat mirror and standing at the center of a gi- ant clock face built into the floor. Someone standing at 12 o’clock shines a beam of light toward you, and you want to use the mirror to reflect the beam toward an observer standing at 5 o’clock. What should the angle of incidence be to achieve this? What should the angle of reflection be? 3. Some department-store windows are slanted inward at the bottom. This is to decrease the glare from brightly illuminated buildings across the street, which would make it difficult for shoppers to see the display inside and near the bottom of the window. Sketch a light ray reflecting from such a window to show how this technique works.

Interpreting Graphics 4. The photograph in Figure 2.1 shows multiple images that were created by multiple reflections between two flat mirrors. What conclusion can you make about the relative orientation of the mirrors? Explain your answer.

Critical Thinking 5. If one wall of a room consists of a large flat mirror, how much larger will the room appear to be? Explain your answer. 6. Why does a flat mirror appear to reverse the person looking into a mirror left to right, but not up and down? Answers to Section Assessment 450 Chapter 13

1. a. specular 4. The light must be reflected from one 6. An object facing the mirror produces an

b. diffuse Untitled-282 mirror 450 to the other in order to form image that faces the object, and the front 5/20/2011 6:22:42 AM c. specular images of images. Therefore, the mirrors of the object corresponds to the back of must be exactly or very nearly parallel to the image. It appears as if the image is an d. diffuse each other, with their mirrored surfaces object located behind the mirror that is 2. 75°; 75° facing each other. left-right reversed. Diagrams will vary. Verify that the rays’ 3. 5. It will seem twice as large, because the angles of incidence equal the angles of object distance (the distance between the reflection in the students’ drawings. mirror and any point in the room, including the back wall) equals the image distance.

450 Chapter 13 SECTION 3 SECTION 3

Objectives Curved Mirrors Calculate distances and focal Plan and Prepare lengths using the mirror equation for concave and Key Terms convex spherical mirrors. Preview Vocabulary concave spherical mirror real image convex spherical mirror Draw ray diagrams to find the Latin Word Origins The word convex Concave Spherical Mirrors image distance and comes from the Latin convexus, which magnification for concave and means “arched.” The word concave Small, circular mirrors, such as those used on dressing tables, may appear convex spherical mirrors. at first glance to be the same as flat mirrors. However, the images they comes from the Latin concavus, which form differ from those formed by flat mirrors. The images for objects close Distinguish between real and means “hollow.” Students can use these to the mirror are larger than the object, as shown in Figure 3.1(a), whereas virtual images. word origins to remember the differ- the images of objects far from the mirror are smaller and upside down, as Describe how parabolic mirrors ence between the two terms. Relate the shown in Figure 3.1(b). Images such as these are characteristic of curved differ from spherical mirrors. mirrors. The image in Figure 3.1(a) is a virtual image like those created by Latin origins of the words to the flat mirrors. In contrast, the image in Figure 3.1(b) is a real image. reflecting surfaces in convex and concave mirrors. Concave mirrors can be used to form real images. One basic type of curved mirror is the spherical mirror. A spherical mirror, as its name implies, has the shape of part of a sphere’s surface. A spherical mirror with light reflecting from its silvered, concave surface Teach (that is, the inner surface of a sphere) is called a concave spherical mirror. concave spherical mirror a mirror Concave mirrors are used whenever a magnified image of an object is whose reflecting surface is a segment of the inside of a sphere needed, as in the case of the dressing-table mirror. One factor that determines where the image will appear in a concave Demonstration spherical mirror and how large that image will be is the radius of curvature, R, of the mirror. The radius of curvature is the same as the radius of the IMage Formed by a Concave spherical shell of which the mirror is a small part; R is therefore the Mirror distance from the mirror’s surface to the center of curvature, C. Purpose Demonstrate that changing FIGURE 3.1 the curvature of a concave mirror Concave Spherical Mirror Curved mirrors can be used produces different images. to form images that are larger (a) or smaller (b) than the object. Materials flexible polyester film with reflective coating (about 1 m × 0.5 m), poster board, bottle Procedure Tape the reflective film to the poster board and place a bottle with a label about 30 cm away. Let students observe that the image in the film is like one they would see in a flat mirror. Slowly bend the sides of the film on the (a) (b) poster board to turn it into a concave

(r) ©Richard Megna/Fundamental Photographs, New York; (r) (l) ©Richard ©Richard Megna/Fundamental Megna/Fundamental York; York New Photographs, New Photographs, mirror (until it has a cylindrical shape).

Light and Reflection 451 Have students describe changes in the Differentiated Instruction image as you roll the film. The virtual Inclusion image grows wider and moves farther away, eventually disappearing when the Untitled-499 451 5/25/2011 5:01:52 AM Employ tactile techniques so that alternative radius of curvature becomes too small. learners can distinguish between convex and Explain that by bending the mirror, concave curves. Allow visually-impaired or you changed its radius of curvature. tactile learners to touch concave objects Have a student hold the mirror at one (for example, a spoon) and convex objects (for curvature as you move the object closer example, a ball) to feel the difference in curves. to the mirror and then farther away. Ask students to describe how changes in the object’s distance affect the position and size of the image.

Light and Reflection 451 FIGURE 3.2 Images and Concave Mirrors Teach continued (a) The rays from a light bulb converge to form a real image in front of a concave mirror.

TEACH FROM VISUALS Mirror

Object Figure 3.2 Image (b) shows a real Front of mirror image of a light bulb on a glass plate. h Image The bulb itself is off to the left, too far Principal C away to fit in the image frame. There is axis h

actually an image of the entire bulb on (b) In this lab setup, the real image of a light-bulb filament the plate, but only the image of the Back of appears on a glass plate in front of a concave mirror. mirror filament is bright enough to be seen in R the photograph. f p In diagram , what represents the Ask (a) (a) q Imagine a light bulb placed upright at a distance p from a concave size of the object? What represents its spherical mirror, as shown in Figure 3.2(a). The base of the bulb is along the distance from the mirror? What repre- mirror’s principal axis, which is the line that extends infinitely from the center of the mirror’s surface through the center of curvature, C. Light rays sents the size of the image? What diverge from the light bulb, reflect from the mirror’s surface, and converge represents the image’s distance from C ? at some distance (q) in front of the mirror. Because the light rays reflected What does f refer to? by the mirror actually pass through the image point—which in this case is below the principal axis—the image forms in front of the mirror. Answer: h; p; h′; R − q; distance from If you place a piece of paper at the image point, you will see on the focal point to mirror paper a sharp and clear image of the light bulb. As you move the paper in either direction away from the image point, the rays diverge and the real image an image that is formed by image becomes unfocused. An image of this type is called a real image. the intersection of light rays; a real Unlike the virtual images that appear behind a flat mirror, real images can image can be projected on a screen be displayed on a surface, like the images on a movie screen. Figure 3.2(b) Key Models and shows a real image of a light-bulb filament on a glass plate in front of a Analogies concave mirror. This light bulb itself is outside the photograph, to the left. To help students understand the reflections in Figure 3.2, point out that Images created by spherical mirrors suffer from spherical aberration. As you draw ray diagrams, you may notice that certain rays do not exactly light striking the mirror is reflected intersect at the image point. This phenomenon is particularly noticeable according to the law of reflection, as for rays that are far from the principal axis and for mirrors with a small if the curved mirror were made of many radius of curvature. This situation, called spherical aberration, also occurs small plane mirrors positioned to form with real light rays and real spherical mirrors, and will be discussed further at the end of this section when we introduce parabolic mirrors. In a circle. The ray through C would be the next pages of this section, you will learn about the mirror equation normal to such a mirror, so it is reflected and ray diagrams. Both of these concepts are valid only for paraxial rays, back in the same direction from which but they do provide quite useful approximations. Paraxial rays are those light rays that are very near the principal axis of the mirror. We will it came. assume that all of the rays used in our drawings and calculations with spherical mirrors are paraxial, even though they may not appear to be so

in all of the diagrams accompanying the text. (bl) ©Richard Megna/Fundamental York New Photographs, Differentiated452 Chapter 13 Instruction Pre-AP For plane mirrors, the distance of the image is equal to the distance of the object: Untitled-499Ask students 452 to research the disclaimer 5/25/2011 5:01:53 AM

printed on the rearview mirrors of vehicles. p = d image and q = d object = - d image Ask them to determine, using the information Equation for Magnification:

they find, if the images reflected in rearview h i q M = __ = – __ . For plane mirrors, M = 1. mirrors are virtual or real and if a rearview h (p) mirror is a plane mirror. (Provide information Although a rearview mirror produces a virtual on plane mirrors and the formula for image, it is not a plane mirror because images magnification.) are smaller than the object, and magnification is less than 1.

452 Chapter 13 Image location can be predicted with the mirror equation. Looking at Figure 3.2(a), you can see that object distance, image distance, and radius of curvature are interdependent. If the object distance and radius of curvature of the mirror are known, you can predict where the MATERIALS QuickLab image will appear. Alternatively, the radius of curvature of a mirror can be • stainless-steel or silver spoon calculated if you know where the image is for a given object distance. The • short pencil Teacher’s Notes following equation relates object distance, p, image distance, q, and the This activity is intended to explore how radius of curvature, R, and is called the mirror equation. CURVED MIRRORS an object’s distance affects the object’s _ 1 + _ 1 = _2 Observe the pencil’s reflection p q R in the inner portion of the image in concave and convex mirrors. If the light bulb is placed very far from the mirror, the object distance, spoon. Slowly move the spoon This experiment works best with a very p, is great enough compared with R that 1/p is almost 0. In this case, q is closer to the pencil. Note any shiny spoon that has a large radius of almost R/2, so the image forms about halfway between the center of changes in the appearance of curvature and the center of the mirror’s surface. The image point, as the pencil’s reflection. Repeat curvature. shown in Figure 3.3, is in this special case called the focal point of the these steps using the other Homework Options This QuickLab can side of the spoon as the mirror. mirror and is denoted by the capital letter F. Because the light rays are easily be performed outside of the reversible, the reflected rays from a light source at the focal point will emerge parallel to each other and will not form an image. physics lab room. For light emerging from a source very far away from a mirror, the light rays are essentially parallel to one another. In this case, an image forms at the focal point, F, and the image distance is called the focal length, denoted by the lowercase letter f. For a spherical mirror, the focal length Demonstration is equal to half the radius of curvature of the mirror. The mirror equation can therefore be expressed in terms of the focal length. Focal Point of a Concave Mirror Purpose Demonstrate that rays parallel Mirror Equation to the principal axis are reflected _1 + _ 1 = _ 1 p q f through the focal point, and show that __1 __1 __1 __R + = object distance image distance focal length f = 2 . Materials light source, ray filter, concave mirror, white paper FIGURE 3.3 Procedure Use the ray filter to produce Parallel Light Rays Light rays that are parallel converge at a single point (a), which can be represented in a diagram (b), when the rays are assumed to be from a distant object (p ≈ ∞). five beams. Dim the lights, and hold the sheet of paper in front of the beams to let students observe that the incident Mirror rays are parallel. Place the concave mirror

C F 20 to 30 cm from the light source, and Principal axis let students observe the beams con- verging. Tell them that the point of convergence is called the focal point. f Explain that past that point, the beams (b) R diverge. Draw the mirror’s curve and (a) (b) HRW • Holt Physics principal axis, and mark the focal point

(bl) ©Richard Megna/Fundamental Photographs, New York (bl) ©Richard Megna/Fundamental York New Photographs, PH99PE-C14-003-005-A on the chalkboard. Ask students if that Problem Solving Light and Reflection 453 point could be the center of the circle from which the mirror was cut. no Take It Further object on film so that variables can be Have students mark where the approxi- adjusted for a larger or smaller image. mate center of the circle is. Measure R, Untitled-499 453 Explain that there are practical applications for 5/25/2011 5:01:54 AM and compare it with f. the magnification equation. If you know where an object’s image will form for a given object, you can determine the magnification of the image. However, you can also use the magnification equation to solve for other variables, such as height of a magnified object or distance of a magnified object. This can be used in photography and astronomy to determine the size of an image of an

Light and Reflection 453 A set of sign conventions for the three variables must be established for use with the mirror equation. The region in which light rays reflect and form real images is called the front side of the mirror. The other side, Teach continued where light rays do not exist—and where virtual images are formed—is called the back side of the mirror. Misconception Alert! Object and image distances have a positive sign when measured from the center of the mirror to any point on the mirror’s front side. Distances The term magnification sometimes for images that form on the back side of the mirror always have a negative leads students to think that the image is sign. Because the mirrored surface is on the front side of a concave larger than the object. Use a numerical mirror, its focal length always has a positive sign. The object and image heights are positive when both are above the principal axis and negative example to show that this is not always when either is below. the case. Ask students to calculate the image height of an 8 cm object when Magnification relates image and object sizes. p = 12 cm and q = 3 cm. h’ = −2 cm Unlike flat mirrors, curved mirrors form images that are not the same size as the object. The measure of how large or small the image is with respect to the original object’s size is called the magnification of the image. If you know where an object’s image will form for a given object Demonstration distance, you can determine the magnification of the image. Magnification, M, is defined as the ratio of the height of the bulb’s image to the bulb’s Beams Reflected from a actual height. M also equals the negative of the ratio of the image distance to the object distance. If an image is smaller than the object, the magnitude Concave Mirror of its magnification is less than 1. If the image is larger than the object, the Purpose Demonstrate the formation magnitude of its magnification is greater than 1. Magnification is a unitless of a virtual image. quantity. Materials light source, ray filter, Equation for Magnification concave mirror q M = _ h′ = -_ Procedure Ask students if reflected h p image height image distance beams always converge from concave magnification = __ = - __ mirrors. Lower the lights in the room object height object distance so that the light can be easily seen. Place the mirror as far away as possible For an image in front of the mirror, M is negative and the image is from the front of the light source, and upside down, or inverted, with respect to the object. When the image is behind the mirror, M is positive and the image is upright with respect to move the mirror closer and closer to the the object. The conventions for magnification are listed in Figure 3.4. light source until the reflected beams diverge. Tell students that the point Figure 3.4 behind the mirror from which the beam seems to come is a virtual image of that Sign ConventionS for MagnifiCation Orientation of image Type of image point on the source. Sign of M with respect to object this applies to

upright + virtual

inverted - real

Problem454 Chapter Solving 13

Deconstructing Problems M = 2 Untitled-499Provide 454 this example to help students use the h = 25 cm 5/25/2011 5:01:54 AM

equation for magnification. hi = 50 cm (magnified x2) You are working with an object 25 cm tall q = 10 cm and want it to have a magnification magnitude p = ? of 2. How far away should you place the Determine the equation you wish to use: object? Assume the image distance is 10 cm. __h i __q M = = - p Break down the problem, by determining the h ( ) Solve for the unknown. givens and unknowns. ____50 cm ____10 cm 2 = 25 cm = - ( p ) p = 5 cm

454 Chapter 13 Ray diagrams can be used for concave spherical mirrors. Ray diagrams are useful for checking values calculated from the mirror and magnification equations. The techniques for ray diagrams that were Teaching Tip used to locate the image for an object in front of a flat mirror can also be Offer a shortcut for drawing rays. If the used for concave spherical mirrors. When drawing ray diagrams for concave mirrors, follow the basic procedure for a flat mirror, but also object is a vertical line, then the image is measure all distances along the principal axis and mark the center of also a vertical line. For such simplified curvature, C, and the focal point, F. As with a flat mirror, draw the diagram situations, the image is a vertical line to scale. For instance, if the object distance is 50 cm, you can draw the object distance as 5 cm. with the lower extremity located upon For spherical mirrors, three reference rays are used to find the image the principal axis. point. The intersection of any two rays locates the image. The third ray should intersect at the same point and can be used to check the diagram. These reference rays are described in Figure 3.5.

Figure 3.5 Rules foR DRawing RefeRence Rays Line drawn from Line drawn from mirror to Ray object to mirror image after reflection

1 parallel to principal axis through focal point F

2 through focal point F parallel to principal axis

3 through center of curvature C back along itself through C

The image distance in the diagram should agree with the value for q calculated from the mirror equation. However, the image distance may differ because of inaccuracies that arise from drawing the ray diagrams at a reduced scale and far from the principal axis. Ray diagrams should therefore be used to obtain approximate values only; they should not be relied on for the best quantitative results.

Concave mirrors can produce both real and virtual images. When an object is moved toward a concave spherical mirror, its image changes, as shown in Figure 3.6. If the object is very far from the mirror, the light rays converge very near the focal point, F, of the mirror and form an image there. For objects at a finite distance greater than the radius of curvature, C, the image is real, smaller than the object, inverted, and located between C and F. When the object is at C, the image is real, located at C, and inverted. For an object at C, the image is the same size as the object. If the object is located between C and F, the image will be real, inverted, larger than the object, and located outside of C. When the object is at the focal point, no image is formed. When the object lies between F and the mirror surface, the image forms again, but now it becomes virtual, upright, and larger.

Differentiated Instruction Light and Reflection 455 Below Level Untitled-499 455 Students can use the acronym SALT to 5/25/2011 5:01:55 AM remember the uses of ray diagrams. Ray diagrams can be drawn to predict the Size, Alignment, Location, and Type of image produced by a convex mirror.

Light and Reflection 455 Figure 3.6 Images created by concave mIrrors

Teach continued Ray Diagrams

1. 2. Mirror 1. 3 TEACH FROM VISUALS Mirror 1 2 FIGURE 3.6 Point out that from all the Principal 3 possible rays coming from the tip of the C F axis Image pencil, these three rays are selected Principal Object C F axis 2 because they are reflected according to the simple rules listed in Figure 3.5.

Make sure that students understand Front of Back of Front of Back of the application of these rules in the six mirror mirror 1 mirror mirror diagrams. Configuration: object outside C Configuration: object at infinity Image: real image between C and F, Image: real image at F inverted with magnification < 1

3. 4. Mirror Mirror

Object 1 2 Object 2 Principal Principal axis axis C F C F 3 Image 1 1 2

Image 2 Front of Back of 1 Front of Back of mirror mirror mirror mirror

C13003020c C13003020dA Configuration: object at C Configuration: object between C and F Image: real image at C, inverted with magnification = 1 Image: real image at C, inverted with magnification > 1

5. 6. Mirror Mirror 2 2 3 3 1 1

Principal 3 Object C F axis 3 Principal Object Image axis C F 1 Front of Back of 1 Front of Back of mirror mirror mirror mirror

Configuration: object at F Configuration: object inside F Image: image at infinity (no image) Image: virtual, upright image at C with magnification >1

Problem456 Chapter Solving 13 Take It Further Untitled-499All ray 456 lines drawn must be drawn according to 5/25/2011 5:01:55 AM the rules described in Figure 3.5. Rules 1 and 2 combined and Rule 3 alone illustrate the Principle of Reversibility: if a ray travels a particular path through a mirror in one direction, it must travel the same path in the opposite direction.

456 Chapter 13 PREMIUM CONTENT

Imaging with Concave Mirrors Interactive Demo HMDScience.com Sample Problem B A concave spherical mirror has a focal Classroom Practice length of 10.0 cm. Locate the image of a pencil that is placed Imaging with Concave Mirrors upright 30.0 cm from the mirror. Find the magnification of the image. Draw a ray diagram to confirm your answer. When an object is placed 30.0 cm in front of a concave mirror, a real image is ANALYZE Determine the sign and magnitude of the focal length and formed 60.0 cm from the mirror’s object size. f = +10.0 cm p = +30.0 cm surface. Find the focal length. The mirror is concave, so f is positive. The object is in front of the Answer: 20.0 cm mirror, so p is positive. A square object is placed 15 cm in front Unknown: = q ? of a concave mirror with a focal length M = ? of 25 cm. A round object is placed 45 cm Diagram: Draw a ray diagram using the rules given in Figure 3.5. in front of the same mirror. Find the image distance, magnification, and type

1 of image formed for each object. Draw 2 ray diagrams for each object to confirm 3 your answers. C F Answer: q = −38 cm, M = 2.5, 2 square square virtual and upright; qround = 56 cm,

Mround = −1.2, real and inverted 1 3

f = 10.0 cm p = 30.0 cm q = ?

HRW • Holt Physics PLAN Use the mirror equation PH99PE-C14-003-008-Ato relate the object and image distances to the focal length. _1 + _ 1 = _ 1 p q f Use the magnification equation in terms of object and image distances. _q M = - p Rearrange the equation to isolate the image distance, and calculate. Subtract the reciprocal of the object distance from the reciprocal of the focal length to obtain an expression for the unknown image distance.

_1 = _ 1 – _ 1 q f p

Continued Light and Reflection 457

Take It Further Untitled-499 457 Ask the students to solve Sample Problem B 5/25/2011 5:01:56 AM again from a new perspective. What happens if the pencil is inverted rather than upright? Which variables are affected by this change? Is it necessary to redo the ray diagram? Answer: If the pencil is inverted, it affects the variable M. The sign of M determines whether an object is inverted (negative) or upright (positive). It is necessary to redo the diagram.

Light and Reflection 457

458 Practice B Answers Teach continued 4. 2. 3. *Challenging Problem h, h’ M f R, p q Solving for: ProblemPB =Sample SetII(online) PW SE =StudentEditionTextbook Use thisguide to assign problems. PR 1. M =−0.286;real image f =6.00cm;M−1.20;q7.71 cm; image virtual R =1.00×10 inverted image q =53 cm;M=−0.57, real, M =2.00; p =5.00cm:q−10.0cm, p =10.0cm:noimage (infinite q); OBLEM guide Sample Problem =Sample SetI(online) Chapter 13 PB PW PB PW SE PB PW SE PB PW SE PB PW SE virtual, uprightimage virtual, Sample, 1–3,7 Sample, 5, 9 8–10 6–8 3–4, 34–36, 49* Ch.Rvw. 1–4; Sample 4–6 1,2 Sample, Ch.Rvw.3–4; 46–47 1–3 Sample, 6–9 Ch. Rvw. 36,46,48 7–10 3–5 35*, 49* 1–2;Ch.Rvw.Sample, 2 2.00; cm;M=2.00; Problem Solving Untitled-499 458 images have positive magnification). distance results image, andupright invirtual distance, andlocation (e.g., negative image tween positive ornegative measurements, Ask studentsto make theconnection be- meanings ofreal, upright, andinverted virtual, . Encourage students to interpret the physical lity Check Rea 458 Chapter 13

4. 3. 2. 1. Imaging withConcaveMirrors new image real or virtual? Draw ray diagrams to yourconfirm results. the new position of the image? What is the magnification of the new image? Is the magnification of the image? If the pen is placed 27.0 cm from the mirror, what is 13.2 cm from the mirror. What is the focal length of the mirror? What is the or virtual? curvature of the mirror? What is the magnification of the image? Is the image real upright image at a distance of 50.0 cm behind the mirror. What is the radius of respectwith to the object. or upright? Draw a ray diagram to show where the image and forms how large it is the magnification of the image. Is the image real or virtual? Is the image inverted of a cologne bottle placed in front of the mirror at a distance of 93 cm. Calculate your results. Are the images inverted or upright? Draw a ray diagram for each case to confirm when the object distances are 10.0 cm and 5.00 cm. Are the images real or virtual? A pen placed 11.0 cm from a concave spherical mirror produces a real image A concave makeup mirror is designed so that a 25.0 person cm in front of it an sees A concave shaving mirror has a focal length of 33 cm. Calculate the image position Find the image distance and magnification of the mirror in the sample problem

work check your solve confirmed byconfirmed theray diagram. The image istherefore real. smaller than andinverted theobject (−1

(continued) M =- q =15cm _ q 1

= _ 10.0 cm _ p q 1

- _ 30.0 cm 15 cm _ 30.0 cm 1

=-0.05

= _ 0.100 1 cm

- _ 0.003 1 cm

= _ 0.067 1 cm

5/25/2011 5:01:57AM

(br) ©Richard Megna/Fundamental Photographs, New York Convex Spherical Mirrors On recent models of automobiles, there is a side-view mirror on the passenger’s side of the car. Unlike the flat mirror on the driver’s side, Demonstration which produces unmagnified images, the passenger’s mirror bulges outward at the center. Images in this mirror are distorted near the mirror’s edges, and the image is smaller than the object. This type of Convex Mirror mirror is called a convex spherical mirror. convex spherical mirror a mirror Purpose Demonstrate that parallel whose reflecting surface is an A convex spherical mirror is a segment of a sphere that is silvered so that outward-curved segment of a sphere beams reflected by convex mirrors are light is reflected from the sphere’s outer, convex surface. This type of mirror diverging. is also called a diverging mirror because the incoming rays diverge after reflection as though they were coming from some point behind the mirror. Materials light source, ray filter, convex The resulting image is therefore always virtual, and the image distance is mirror, white paper always negative. Because the mirrored surface is on the side opposite the radius of curvature, a convex spherical mirror also has a negative focal Procedure Use the ray filter to produce length. The sign conventions for all mirrors are summarized in Figure 3.8. five beams. Dim the lights. Place the The technique for drawing ray diagrams for a convex mirror differs paper in front of the beams, and let slightly from that for concave mirrors. The focal point and center of curvature are situated behind the mirror’s surface. Dotted lines are students observe that the incident rays extended along the reflected reference rays to points behind the mirror, are parallel. Place the convex mirror as as shown in Figure 3.7(a). A virtual, upright image forms where the three far as possible from the front of the rays apparently intersect. Magnification for convex mirrors is always less light source, and let students observe than 1, as shown in Figure 3.7(b). Convex spherical mirrors take the objects in a large field of view and the beams diverging. Move the mirror produce a small image, so they are well suited for providing a fixed closer to the light source. Students will observer with a complete view of a large area. Convex mirrors are often notice that the beam is always diverging. placed in stores to help employees monitor customers and at the inter- Remind students that concave mirrors sections of busy hallways so that people in both hallways can tell when others are approaching. produce both real and virtual images The side-view mirror on the passenger’s side of a car is another applica- from real objects. In contrast, convex tion of the convex mirror. This mirror usually carries the warning, “objects mirrors produce only virtual images from are closer than they appear.” Without this warning, a driver might think that real objects. he or she is looking into a flat mirror, which does not alter the size of the image. The driver could therefore be fooled into believing that a vehicle is farther away than it is because the image is smaller than the actual object.

FIGURE 3.7

Reflection from a Convex Mirror Light rays diverge upon reflection from a convex mirror (a), forming a virtual image that is always smaller than the object (b).

1 Eye Mirror 3 1 3 1 2 2 2

F 3 C Principal Object Image axis Front of Back of (a)(a) mirror mirror (b) (br) ©Richard Megna/Fundamental York New Photographs,

Light and Reflection 459 Differentiated InstructionHRW • Holt Physics PH99PE-C14-003-011-A Below Level Use the analogy of a ball bouncing off the floor. If a ball is thrown straight down at the Untitled-499 459 Students with a below-level understanding of 5/25/2011 5:01:57 AM reflection may believe that lenses work the floor, it will bounce up. If you throw the ball at same way as mirrors (e.g., concave mirrors an angle, it will bounce away at the same spread out light and convex mirrors focus the angle. Throw a bouncy ball at a concave shape light). They may be thrown off by seeing that (e.g., a bowl) and it will tend to bounce toward the opposite is true when they are introduced the middle. Throw a bouncy ball at a convex to descriptions of how concave and convex shape (a dome), and it will bounce all over the mirrors work. Explain that when light hits a place. This shows why convex mirrors spread mirror, it bounces off the surface rather than out light rather than focusing it. going through the mirror.

Light and Reflection 459 Did YOU Know? Figure 3.8 There are certain circumstances in Sign ConventionS for MirrorS Teach continued which the object for one mirror is the image that appears behind another Symbol Situation Sign mirror. In these cases, the object TEACH FROM VISUALS is virtual and has a negative object p object is in front of the + distance. Because of the rar ity of these mirror (real object) FIGURE 3.8 Make sure that students situations, virtual object distance ( p < 0) has not been listed in Figure 3.8. p > 0 properly interpret information related to all the cases listed in Figure 3.8. Point q image is in front of the + out that as a general rule, distances in mirror (real image) HRW • Holt Physics PH99PE-C14-003-010a-A front of the mirror are assigned a positive sign and distances behind the q > 0 mirror are assigned a negative sign. q image is behind the - Ask The image formed by a concave mirror (virtual image) HRW • Holt Physics mirror is upright and virtual. What would PH99PE-C14-003-010b-A be the signs of R, f, q, and h’? q < 0 Answer: +, +, -, + R, f center of curvature is in + R > 0 Ask The image formed by a convex front of the mirror HRW • Holt Physics mirror is also upright and virtual. What (concave spherical mirror) PH99PE-C14-003-010c-A

would be the signs of R, f, q, and h’ ? f > 0 Answer: -, -, -, + R, f center of curvature is - R < 0 behind the mirror (convex HRW • Holt Physics spherical mirror) PH99PE-C14-003-010d-A

f < 0

R, f mirror has no curvature ∞ R, f • (flat mirror) HRW • Holt Physics PH99PE-C14-003-010e-A

h ′ image is above the + principal axis HRW • Holt Physics h PH99PE-C14-003-010f-Ah'

h, h' > 0

h ′ image is below the - HRW • Holt Physics principal axis h PH99PE-C14-003-010g-A

h' h > 0, h' < 0

HRW • Holt Physics 460 Chapter 13 Problem Solving PH99PE-C14-003-010h-A Take It Further concentrate or focus them). Examples are below. Untitled-499Now 460that students have studied the images 5/25/2011 5:01:58 AM produced by convex and concave mirrors, 1. Dentist uses a mirror close to the tooth to provide an activity that allows them to make it appear much larger. concave synthesize this information. Provide a list of 2. Streetlamps use mirrors as reflectors to common mirrors and functions and ask diverge light over a large area. convex students to determine whether the mirror is 3. Solar-powered products use mirrors to convex or concave based on how it is being reflect the sun’s rays so that the rays are used. Tell students to look for keywords that concentrated in one place. concave explain the mirror’s function (e.g., convex mirrors diverge rays; concave mirrors

460 Chapter 13 PREMIUM CONTENT

Convex Mirrors Interactive Demo HMDScience.com Sample Problem C An upright pencil is placed in front of a Classroom Practice convex spherical mirror with a focal length of 8.00 cm. An erect Convex Mirrors image 2.50 cm tall is formed 4.44 cm behind the mirror. Find the position of the object, the magnification of the image, and the The radius of curvature of a convex height of the pencil. mirror is 12.0 cm. Where is the focal point located? ANALYZE Given: f = −8.00 cm q = −4.44 cm h� = 2.50 cm Answer: 6.00 cm behind the mirror Because the mirror is convex, the focal length is negative. The image is behind the mirror, so q is also negative. ( f = −6.00 cm) Unknown: p = ? h = ? M = ? Find the position of the image for an Diagram: Construct a ray diagram. object placed at the following distances from the mirror in the previous question: p = 1.00 cm, 2.00 cm, 3.00 cm, 6.00 cm,

1 12.0 cm, 30.0 cm, 50.0 cm. 3 1 2 3 1 Answer: −0.855 cm, −1.50 cm, 2 2 −2.00 cm, −2.99 cm, −4.00 cm, 3 −5.00 cm, −5.35 cm F C p = ? How does the position of the image f = -8.00 cm vary as the object in the previous

q = -4.44 cm question moves farther away from the mirror? Answer: The image is always behind PLAN Choose an equation or situation: Use the mirror equation. HRW • Holt Physics the mirror and between the mirror and _1 + _ 1 = PH99PE-C14-003-012-A_ 1 the focal point. It moves from = 0 to p q f q Use the magnification formula. q = f as the object moves away from q the mirror (from p = 0 to infinity). M = _ h� = -_ h p Rearrange the equation to isolate the unknown: p _1 = _ 1 - _ 1 and h = -_ h � p f q q

SOLVE Substitute the values into the equation and solve: _1 = _ 1 − _ 1 p −8.00 cm −4.44 cm

_1 = _ -0.125 - _ -0.225 = _ 0.100 p 1 cm 1 cm 1 cm p = 10.0 cm

Continued Light and Reflection 461

Reality Check by using the magnification equation to solve for p when M = 2 and q = -4.44 cm. Untitled-499 461 Students should interpret the physical effects 5/25/2011 5:01:59 AM of changing the variables in the equation for magnitude. In this case, the magnitude of the image is .44; does this mean that the image is smaller or larger than the object? Smaller, because the magnitude is less than 1. What if the desired magnitude was 2.14? What would need to happen to the position of the object? The object would need to be positioned at +8.88 cm. Students can determine this figure

Light and Reflection 461 Convex Mirrors (continued)

Teach continued Substitute the values for p and q to find the magnification of the image. q M = -_ = -_ -4.44 cm PROBLEM guide C p 10.0 cm Use this guide to assign problems. M = 0.444 SE = Student Edition Textbook Substitute the values for p, q, and h′ to find the height of the object. p PW = Sample Problem Set I (online) h = -_ h′ = -_ 10.0 cm (2.50 cm) q -4.44 cm PB = Sample Problem Set II (online) h = 5.63 cm Solving for: p SE Sample, 1–3; Ch. Rvw. 36, 51* PW 7–9 1. The image of a crayon appears to be 23.0 cm behind the surface of a convex mirror and is 1.70 cm tall. If the mirror’s focal length is 46.0 cm, how far in front PB 3–6 of the mirror is the crayon positioned? What is the magnification of the image? Is the image virtual or real? Is the image inverted or upright? How tall is the q SE 4–6 actual crayon? PW Sample, 1–3 2. A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an PB 7–10 automobile at a distance of 0.24 m behind the mirror. What is the magnification R, f SE Ch. Rvw. 36, 48, 50, 55 of the image? Where is the car located, and what is its height? Is the image real or virtual? Is the image upright or inverted? PW 4–6 3. A convex mirror of focal length 33 cm forms an image of a soda bottle at a PB Sample, 1, 2 distance of 19 cm behind the mirror. If the height of the image is 7.0 cm, where is M SE 1–6; Ch. Rvw. 50 the object located, and how tall is it? What is the magnification of the image? Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram PW Sample, 1–2, 4–5, 7–8 to confirm your results. PB Sample, 1, 3–5, 7–8 4. A convex mirror with a radius of curvature of 0.550 m is placed above the aisles in a store. Determine the image distance and magnification of a h, h’ PW 3, 6, 9 customer lying on the floor 3.1 m below the mirror. Is the image virtual or PB 2, 6, 9 real? Is the image inverted or upright?

*Challenging Problem 5. A spherical glass ornament is 6.00 cm in diameter. If an object is placed 10.5 cm away from the ornament, where will its image form? What is the magnification? Is the image virtual or real? Is the image inverted or upright?

6. A candle is 49 cm in front of a convex spherical mirror that has a focal length Answers of 35 cm. What are the image distance and magnification? Is the image Practice B virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results. 1. p = 46.0 cm; M = 0.500; virtual, upright image; h = 3.40 cm 2. M = 0.04; p = 6 m; h = 2 m; virtual, upright image 3. p = 45 cm; h = 17 cm; 462 Chapter 13 M = 0.41; virtual, upright image Problem Solving 4. q = −0.25 m; M = 0.081; Reality Check virtual, upright image Untitled-499Students 462 should pay attention to key words 5/25/2011 5:01:59 AM 5. q = −1.31 cm; M = 0.125; virtual, upright image within the question. If a problem states that an object is behind the mirror, the mirror is 6. q = −2.0 × 101 cm; M = 0.41; virtual, virtual. When asked to determine the location upright image of the object, they are being asked to determine whether the image is virtual or real.

462 Chapter 13 Parabolic Mirrors FIGURE 3.9 You have probably noticed that certain rays in ray diagrams do not Spherical Aberration TEACH FROM VISUALS intersect exactly at the image point. This occurs especially with rays that Spherical aberration occurs when reflect at the mirror’s surface far from the principal axis. The situation parallel rays far from the principal Figure 3.9 Remind students that for also occurs with real light rays and real spherical mirrors. axis converge away from the mirror’s focal point. a spherical mirror, the normal to the If light rays from an object are near the principal axis, all of the reflected rays pass through the image point. Rays that reflect at points on surface at any point lies along the radius the mirror far from the principal axis converge at slightly different points at that point. Point out that each pair of on the principal axis, as shown in Figure 3.9. This produces a blurred reflected rays at equal distances on image. This effect, called spherical aberration, is present to some extent in any spherical mirror. opposite sides of the principal axis cross C F one another on the principal axis. Parabolic mirrors eliminate spherical aberration. Ask How does the angle of incidence A simple way to reduce the effect of spherical aberration is to use a mirror vary when the incoming ray is farther with a small diameter; that way, the rays are never far from the principal away from the principal axis of a axis. If the mirror is large to begin with, shielding its outer portion will HRW • Holt Physics limit how much of the mirror is used and thus will accomplish the same PH99PE-C14-003-013-A spherical mirror? Mark the points where effect. However, many concave mirrors, such as those used in astronomi- each pair of reflected rays intersects. cal telescopes, are made large so that they will collect a large amount of Where will the next pair intersect if the light. An alternative approach is to use a mirror that is not a segment of a sphere but still focuses light rays in a manner similar to a small spherical mirror’s surface is extended? concave mirror. This is accomplished with a parabolic mirror. Answer: The angle of incidence increases; Parabolic mirrors are segments of a paraboloid (a three-dimensional FIGURE 3.10 The points should be marked on the parabola) whose inner surface is reflecting. All rays parallel to the principal axis converge at the focal point regardless of where on the mirror’s Parabolic Mirror All parallel principal axis, to the right of F; The points surface the rays reflect. Thus, a real image forms without spherical aberra- rays converge at a parabolic mirror’s of intersection come closer to the focal point. The curvature in this tion, as illustrated in Figure 3.10. Similarly, light rays from an object at the figure is much greater than it is in mirror’s surface, and the focal point focal point of a parabolic mirror will be reflected from the mirror in real parabolic mirrors. parallel rays. Parabolic reflectors are ideal for flashlights and automobile becomes more ill-defined. headlights. (Spherical mirrors are extensively used because they are easier to manufacture than parabolic mirrors, and thus are less expensive.)

Reflecting telescopes use parabolic mirrors. F A telescope permits you to view distant objects, whether they are buildings a few kilometers away or galaxies that are millions of light-years from Earth. Not all telescopes are intended for visible light. Because all electromagnetic radiation obeys the law of reflection, parabolic surfaces can be constructed to reflect and focus electromagnetic radiation of different wavelengths. For instance, a radio telescope consists of a large metal parabolic surface HRW • Holt Physics that reflects radio waves in order to receive radio signals from objects PH99PE-C14-003-014-A in space. There are two types of telescopes that use visible light. One type, called a refracting telescope, uses a combination of lenses to form an image. The other kind uses a curved mirror and small lenses to form an image. This type of telescope is called a reflecting telescope.

Differentiated Instruction Light and Reflection 463 Below Level Untitled-499 463 Students may be unclear about the difference 5/25/2011 5:02:00 AM between segments of a circle and of a parabolic curve. Ask them to draw the continuation of the circular and parabolic curves of the mirrors to see that the curves are similar for a short segment. Have students compare the curves of a circle and parabola by using their graphing calculators to compare the curves whose equations are y = √1 - x2 and y = x2.

Light and Reflection 463 FIGURE 3.11 Reflecting telescopes employ a parabolic mirror (called an objective mirror) to focus light. One type of reflecting Reflecting Telescope The parabolic objective telescope, called a Cassegrain reflector, is shown in Figure 3.11. C14-003-015-A Assess and Reteach mirror in a Cassegrain reflector focuses incoming light. Parallel light rays pass down the barrel of the telescope and are reflected by the parabolic objective mirror at the tele- Assess Use the Formative Assessment scope’s base. These rays converge toward the objective mirror’s focal point, F , where a real image would normally on this page to evaluate student F form. However, a small curved mirror that lies in the path of Small mirror mastery of the section. the light rays reflects the light back toward the center of the Reteach For students who need objective mirror. The light then passes through a small hole in the center of the objective mirror and comes to a focus at additional instruction, download the point A. An eyepiece near point A magnifies the image. Section Study Guide. You may wonder how a hole can be placed in the objective mirror without affecting the final image formed by the Response to Intervention To reassess Parabolic telescope. Each part of the mirror’s surface reflects light from students’ mastery, use the Section Quiz, objective mirror distant objects, so a complete image is always formed. The available to print or to take directly presence of the hole merely reduces the amount of light that online at HMDScience.com. is reflected. Even that is not severely affected by the hole because the light-gathering capacity of an objective mirror is dependent on the mirror’s area. For instance, a 1 m diameter A hole in a mirror that is 4 m in diameter reduces the mirror’s Eyepiece 1 __ reflecting surface by only 16 , or 6.25 percent.

SECTION 3 FORMATIVE ASSESSMENT

Reviewing Main Ideas 1. A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s surface. Determine the image distance and magnification. Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results. 2. A spherical mirror is to be used in a motion-picture projector to form an inverted, real image 95 times as tall as the picture in a single frame of film. The image is projected onto a screen 13 m from the mirror. What type of mirror is required, and how far should it be from the film? 3. Which of the following images are real and which are virtual? a. the image of a distant illuminated building projected onto a piece of heavy, white cardboard by a small reflecting telescope b. the image of an automobile in a flat rearview mirror c. the image of shop aisles in a convex observation mirror

Critical Thinking 4. Why is an image formed by a parabolic mirror sharper than the image of the same object formed by a concave spherical mirror? 5. The reflector of the radio telescope at Arecibo Observatory has a radius of curvature of 265.0 m. How far above the reflector must the radio- detecting equipment be placed in order to obtain clear radio images?

Answers464 Chapter to S ection13 Assessment

1. q = −0.45 cm; M = 0.41; virtual, 5. 132.5 m, which is the telescope’s focal

Untitled-499upright 464 image length 5/25/2011 5:02:01 AM 2. concave mirror; p = 0.14 m 3. a. real b. virtual c. virtual 4. The rays reflected by a parabolic mirror all focus at one point, whereas the rays reflected by a concave spherical mirror reflect along a line that includes the mirror’s focal point.

464 Chapter 13 SECTION 4 SECTION 4

Objectives Color and Recognize how additive colors Plan and Prepare affect the color of light. Polarization Recognize how pigments affect Preview Vocabulary the color of reflected light. Visual Vocabulary In the introductory Key Term Explain how linearly polarized linear polarization paragraph, students will be indirectly light is formed and detected. asked to compare reflection, diffusion, Color and absorption. Provide students with You have probably noticed that the color of an object can appear a concrete visual analogy of each. different under different lighting conditions. These differences are due Reflection of light is like a ball bouncing to differences in the reflecting and light-absorbing properties of the off of a surface; the way it bounces object being illuminated. depends on the surface at which it is So far, we have assumed that objects are either like mirrors, which reflect almost all light uniformly, or like rough objects, which reflect light thrown. Diffusion of light is like placing a diffusely in several directions. However, objects absorb certain wave- cube of sugar in a cup of tea. The sugar lengths from the light striking them and reflect the rest. The color of an dissolves and disperses throughout the object depends on which wavelengths of light shine on the object and which wavelengths are reflected (see Figure 4.1). tea. Absorption of light is like a sponge If all wavelengths of incoming light are completely reflected by an being dipped in water. The water object, that object appears to have the same color as the light illuminating permeates the sponge. it. This gives the object the same appearance as a white object illuminated by the light. An object of a particular color, such as the green leaf in Figure 4.1, absorbs light of all colors except the light whose color is the same as the object’s color. By contrast, an object that reflects no light Teach appears black. In truth, leaves appear green only when their primary pigment, chlorophyll, is present. In the autumn, when the green pigment is destroyed, other colors are reflected by the leaves. Demonstration Additive primary colors produce FIGURE 4.1 white light when combined. Reflection and Absorption Color and Reflection A leafC14-004-001-A appears green under white light Because white light can be dis- because the primary pigment in the leaf reflects only green light. of Color persed into its elementary colors, it is reasonable to suppose that Purpose Help students realize that the elementary colors can be combined colors we see depend on the properties R to form white light. One way of G O Y of the light incident on an object and on doing this is to use a lens to recom- G B the colors the object reflects. bine light that has been dispersed V by a prism. Another way is to Materials white paper; crayons of combine light that has been passed different colors; projector with red, blue, through red, green, and blue filters. These colors are called the additive and green filters or cellophane primary colors because when they Procedure Write a sentence on the are added in varying proportions, they can form all of the colors of paper, alternating crayons for each word. the spectrum. Turn off the lights and challenge

Light and Reflection 465 students to read the sentence while you Differentiated Instruction cover the projector with one or two Below Level filters at a time. Students should have their eyes closed while the filters are Untitled-284 465 5/20/2011 6:24:54 AM Below-level learners may benefit from an being changed. Ask students to write overview of the differences between lenses/ their guesses for the complete text that prisms and mirrors. This will help to clarify the has been partially revealed under explanation of additive primary colors as well different lights. Finally, have students as the introductory paragraph’s explanation of explain what combination of light color how light waves behave with different objects. and crayon color made a particular word Remind students that mirrors reflect light visible or invisible. (e.g., light bounces off them) whereas lenses and prisms refract (bend) light. A lens is simply varying amounts of prism.

Light and Reflection 465

466 percentages. blue) orCMY (cyan-magenta-yellow) choose thecolor withRGB (red-green- rectangle. Manyprograms allow you to drawing program to create afilled you mightsuggest usingasimple mixing. Different computers vary, but a personal computer to explore color Students canuse color controls on T Teach continued eaching Tip Chapter 13 Differentiated Instruction Untitled-284 466 tion withFigure 4.3. a basiccolor wheelto beviewed in conjunc- complementary colors. For example, supply learners understandtheconcepts of Incorporate elementsthat willhelp visual Inclusion 466 FIGURE 4.2 additive primary color.additive primary circles color produces of the the complementary third The combination of colors the in additive any primary two Colors Combining Additive Primary Chapter 13 yellow (red magenta blue) cyan (blue green) blue green red Colors FIGURE 4.3 be perceived.be a combination receptors ofthese that so range awide ofcolors can either red, green, orbluelight. Light ofdifferent wavelengths stimulates receptors intheeye. Each receptor, called cell,issensitiveto acone appears to have many colors present at thesame time. different pixels indifferent ofthepicture parts produces apicture that pixels television screen. Atelevision screen consists ofsmall, luminous dots, or Humans incolorbecause can there see are three kinds ofcolor Another example ofadditive colors istheimage produced onacolor ADDITIVE ANDSUBTRACTIVEPRIMARY COLORS , that glow eitherred, green, orblue. Varying thebrightness of colorless. green andmagenta light, andtheresulting glass appears compoundsof these produces combination anequal of magenta are colors, complementary theright so proportion give glass amagenta, orreddish-blue, color. Green and compounds give glass agreen color. Manganese compounds chemical compoundscertain to give colorto glass. Iron colorless, be will or “white,” asshownin yellow light bluelight, iscombined with theresulting light green light are produce combined, they yellow light. Ifthis For colorisformed. complementary example, whenred and light colors are appears. combined, a primary two When green light produced agreen with filter, apatch ofyellow cyan, magenta,cyan, andyellow, respectively, asindicated in colors red, green,primary andblueproduce complements of Figure 4.3. One application of additive primary colors is the use of colors istheuse applicationOne ofadditive primary lightWhen through passed ared filter iscombined with complementary to blue complementary to green complementary to red complementary primary primary primary (mixing light) Additive primary primary primary to yellow complementary to magenta complementary to cyancomplementary (mixing pigments) Subtractive Figure 4.2. The 5/20/2011 6:24:56AM

©Leonard Lessin/Peter Arnold, Inc. Subtractive primary colors filter out all light when FIGURE 4.4 combined. Combining Subtractive Primary Colors When blue light and yellow light are mixed, white light The combination of the subtractive primary colors by any Key Models and results. However, if you mix a blue pigment (such as paint two filters produces the complementary color of the third Analogies or the colored wax of a crayon) with a yellow pigment, the subtractive primary color. resulting color is green, not white. This difference is due to The names of the primary additive and the fact that pigments rely on colors of light that are subtractive colors can be arranged on a absorbed, or subtracted, from the incoming light. triangle, as shown below. For example, yellow pigment subtracts blue and violet colors from white light and reflects red, orange, yellow, and green light. Blue pigment subtracts red, orange, and Red yellow from the light and reflects green, blue, and violet. When yellow and blue pigments are combined, only green light is reflected. When pigments are mixed, each one subtracts certain Yellow Magenta colors from white light, and the resulting color depends on the frequencies that are not absorbed. The primary pigments (or primary subtractive colors, as they are sometimes called) are cyan, magenta, and yellow. These Cyan are the same colors that are complementary to the addi- Green Blue tive primary colors (see Figure 4.3). When any two primary subtractive colors are combined, they produce either red, green, or blue pigments. When the three primary pigments are mixed together in the Each side of the triangle is the comple- proper proportions, all of the colors are subtracted from white light, and mentary color of the color at the the mixture is black, as shown in Figure 4.4. opposite HRWvertex. • This Holt diagram Physics also shows Combining yellow pigment and its complementary color, blue, should the results of combining pigments: the produce a black pigment. Yet earlier, blue and yellow were combined to PH99PE-014-004-001A produce green. The difference between these two situations is explained colors at any two sides of the triangle by the broad use of color names. The “blue” pigment that is added to a combine to form the color at the vertex “yellow” pigment to produce green is not a pure blue. If it were, only blue between them. For example, pure light would be reflected from it. Similarly, a pure yellow pigment will reflect only yellow light. Because most pigments found in paints and dyes magenta pigment mixed with pure cyan are combinations of different substances, they reflect light from nearby pigment will only reflect blue light. The parts of the visible spectrum. Without knowledge of the light-absorption pigment whose name is on one side characteristics of these pigments, it is hard to predict exactly what colors will result from different combinations. reflects the light composed of the colors that are indicated at the vertices on either end of that side. This same Conceptual Challenge pigment absorbs the light with the color

Colors in a Blanket Brown is Blueprints If a blueprint (a blue whose name is on the opposite vertex. a mixture of yellow with small drawing on a white background) is amounts of red and green. If you viewed under blue light, will you still shine red light on a brown woolen be able to perceive the drawing? blanket, what color will the blanket What will the blueprint look like Answers appear? Will it appear lighter or under yellow light? darker than it would under white Conceptual Challenge light? Explain your answers. 1. red; darker; Some of the red light is

Light and Reflection 467 (assume that yellow pigment consists of green and red pigments). Below level 2. No, the entire page will look blue (the blueprint may appear darker, Untitled-284 467 Below-level students may need a refresher 5/20/2011 6:24:57 AM depending on the light source); The about the difference between primary color blue drawing will become black, and and primary pigment. Primary colors are specific the page will be yellow. wavelengths of visible or white light as the prism splits it. Primary pigments are molecules that reflect light at wavelengths closest to the spectral definition of the color.

Light and Reflection 467 FIGURE 4.5 Beam of unpolarized light Unpolarized Light Teach continued Randomly oscillating electric fields Oscillating electric ﬁelds produce unpolarized light.

Oscillating Demonstration magnetic ﬁelds

Polarizing Light by Electromagnetic waves Transmission Purpose Demonstrate linear polarization of light. FIGURE 4.6 Beam of linearly polarizedPHYSICS light Materials two sheets of polarizing film Linearly Polarized Light Spec. Number PH 99 PE C14-004-005-A Light waves with aligned electric fields Oscillating electric ﬁelds Boston Graphics, Inc. (approximately 20 cm × 25 cm), over- are linearly polarized. 617.523.1333 head projector, projection screen Oscillating Procedure Turn on the overhead magnetic ﬁelds projector, and have students observe the intensity of the light on the screen. Electromagnetic waves Place one of the sheets of polarizing film on the overhead projector. Have students note the decreased intensity

of the light. Explain to the class that the Polarization of Light Waves PHYSICS light from the projector is randomly Spec. Number PH 99 PE C14-004-006-A You have probably seen sunglasses with polarized lenses thatBoston reduce Graphics, Inc. oriented in all directions. Only the glare without blocking the light entirely. There is a property617.523.1333 of light that components of light parallel to the allows some of the light to be filtered by certain materials in the lenses. polarizer’s transmission axis are transmit- In an electromagnetic wave, the electric field is at right angles to both the magnetic field and the direction of propagation. Light from a typical ted, and therefore the intensity of light source consists of waves that have electric fields oscillating in random on the screen is reduced. directions, as shown in Figure 4.5. Light of this sort is said to be unpolarized. Hold the second polarizer in front of Electric-field oscillations of unpolarized light waves can be treated as combinations of vertical and horizontal electric-field oscillations. There the top lens of the overhead projector. are certain processes that separate waves with electric-field oscillations Have students note that the intensity in the vertical direction from those in the horizontal direction, producing of the light remains constant when the a beam of light with electric field waves oriented in the same direction, linear polarization the alignment of as shown in Figure 4.6. These waves are said to have linear polarization. second polarizer’s transmission axis is electromagnetic waves in such a way parallel to the transmission axis of the that the vibrations of the electric fields in each of the waves are parallel to Light can be linearly polarized through transmission. original polarizer. Rotate the second each other Certain transparent crystals cause unpolarized light that passes through polarizer 90° so that its transmission axis them to become linearly polarized. The direction in which the electric is perpendicular to the transmission axis fields are polarized is determined by the arrangement of the atoms of the original polarizer. Have students or molecules in the crystal. For substances that polarize light by transmission, the line along which light is polarized is called the transmission axis note that the intensity of the light is almost zero. Differentiated468 Chapter 13 Instruction Below level vibrations pass to nearby atoms and re-emit on the opposite side of the object. This is Untitled-284Below-level 468 learners may need a refresher on 5/20/2011 6:24:58 AM transmission. Explain that, like reflection, known as transmission. transmission of light waves occurs because the frequencies of light waves don’t match the natural frequencies of the vibration of objects. When these waves strike an object, the electrons of the object’s atoms vibrate. Because they do not vibrate in resonance at a large amplitude, the energy is re-emitted as a light wave. If the object is transparent, the

468 Chapter 13 of the substance. Only light waves that are linearly polarized with respect FIGURE 4.7 to the transmission axis of the polarizing substance can pass freely through the substance. All light that is polarized at an angle of 90° to the Polarizing Films transmission axis does not pass through. (a) Light will pass through a pair Demonstration of polarizing films when their When two polarizing films are held with the transmission axes parallel, polarization axes are aligned in the light will pass through the films, as shown in Figure 4.7(a). If they are held same direction. (b) When the axes Transmitting Light with with the transmission axes perpendicular to each other, as in Figure 4.7(b), are at right angles to one another, Crossed Polarizers no light will pass through the films. light will not get through. Purpose Demonstrate the vector A polarizing substance can be used not only to linearly polarize light (a) but also to determine if and how light is linearly polarized. By rotating nature of light. a polarizing substance as a beam of polarized light passes through it, Materials three sheets of polarizing a change in the intensity of the light can be seen (see Figure 4.8). The light is brightest when its plane of polarization is parallel to the transmission film, overhead projector, projection axis. The larger the angle is between the electric-field waves and the screen transmission axis, the smaller the component of light that passes through the polarizer will be and the less bright the light will be. When the trans- Procedure This demonstration is best mission axis is perpendicular to the plane of polarization for the light, (b) done following the previous one, no light passes through. “Polarizing Light by Transmission.” Set up two of the films with their axes FIGURE 4.8 of transmission crossed at 90° to each Polarization and Brightness The brightness of the polarized other. Ask students to predict what will light decreases as the angle, θ, increases between the transmission axis happen if a third film is inserted of the second polarizer and the plane of polarization of the light. between the two films. Insert the third Unpolarized light sheet of polarizing film between the first two at an angle of 45°. Have

Polarizer students note that some light is now passed through the three films. What is happening? Light, which Polarized light consists of oscillating electric and

Transmission axis Polarized light transmitted magnetic fields, has a vector nature. by second polarizer The light passing through the first film is polarized in the direction of that film’s HRW • Holt Physics axis of polarization. When this polarized PH99PE-C14-004-008-A POLARIZATION OF SUNLIGHT light reaches the second film, the second film then passes the component During mid-morning or mid-after- polarizer make the sky darker and noon, when the sun is well above thus best reduce the amount of MATERIALS of the polarized light parallel to its axis the horizon but not directly transmitted light. • a sheet of polarizing filter or of polarization. The light is now polar- sunglasses with polarizing overhead, look directly up at the Repeat the test with light from lenses ized at 45° when it reaches the third film. sky through the polarizing filter. other parts of the sky. Test light Note how the light’s intensity is reflected off a table near a window. SAFETY Again, the third film passes the compo- Never look directly at the reduced. Compare the results of these nent of the light parallel to its axis sun. Rotate the polarizer. Take note various experiments. of polarization. of which orientations of the

Problem Solving Light and Reflection 469 Take It Further Untitled-284 469 How can linearly polarized light be used 5/20/2011 6:25:00 AM to make a three-dimensional image? In order for this to work, linearly polarized light must be converted to circularly polarized light by slowing one component of the electrical field. When the vertical and horizontal parts of an image are projected onto a movie screen, a filter slows down the vertical component. This makes the light appear to rotate.

Light and Reflection 469 FIGURE 4.9 Light can be polarized by reflection and scattering. Polarized Sunglasses At a particular angle, When light is reflected at a certain Teach continued reflected light is polarized horizontally. This light can be blocked by aligning the transmission axes of the Polarizer with transmission angle from a surface, the reflected sunglasses vertically. axis in vertical orientation light is completely polarized parallel to the reflecting surface. If the surface Unpolarized light is parallel to the ground, the light is Demonstration polarized horizontally. This is the case with glaring light that reflects at a low angle from bodies of water. Polarizing Light by Reﬂected light Reflection (polarized perpendicular to page) Because the light that causes glare is in most cases horizontally polar- Purpose Demonstrate that a reflected ized, it can be filtered out by a beam is polarized. polarizing substance whose trans- Reﬂecting surface mission axis is oriented vertically. Materials focusable flashlight or This is the case with polarizing projector, pane of glass (or the glass FIGURE 4.10 sunglasses. As shown in Figure 4.9, the angle between the surface of an overhead projector), sheet Polarized Sunlight The sunlight scattered by air polarized reflected light and the transmission axis of the of polarizing film, projection screen molecules is polarized for an observerHRW •on Holt Earth’s Physics surface. polarizer is 90°. Thus, none of the polarized light passes PH99PE-C14-004-010-A through. Procedure Dim the classroom lights. Unpolarized In addition to reflection and absorption, scattering can sunlight also polarize light. Scattering, or the absorption and reradia- Shine a flashlight beam directly on the Molecule of tion of light by particles in the atmosphere, causes sunlight screen. Place the sheet of polarizing film atmospheric gas to be polarized, as shown in Figure 4.10. When an unpolar- across the beam, and rotate the sheet. ized beam of sunlight strikes air molecules, the electrons in Have students observe that the initial the molecules begin vibrating with the electric field of the incoming wave. A horizontally polarized wave is emitted by intensity is reduced and that rotating the Scattered polarized light the electrons as a result of their horizontal motion, and a polarizing sheet does not make a vertically polarized wave is emitted parallel to Earth as a difference. Explain that this is because result of their vertical motion. Thus, an observer with his or her back to the sun will see polarized light when looking up the light is randomly polarized. Now, Observer toward the sky. shine the light beam onto the glass pane. Place the polarizing film across the reflected beam, and rotate the film. The SECTIONHRW •4 Holt FORMATIVE Physics ASSESSMENT PH99PE-C14-004-011A brightness of the light on the screen will Reviewing Main Ideas vary with the rotation. With a particular 1. A lens for a spotlight is coated so that it does not transmit yellow light. angle (about 34° between the beam and If the light source is white, what color is the spotlight? the glass plate), complete polarization 2. A house is painted with pigments that reflect red and blue light but absorb may be obtained, and the image on the all other colors. What color does the house appear to be when it is illuminated by white light? What color does it appear to be under red light? screen will completely fade. 3. What primary pigments would an artist need to mix to obtain a pale yellow green color? What primary additive colors would a theater-lighting designer need to mix in order to produce the same color with light?

Assess and Reteach Critical Thinking 4. The light reflected from the surface of a pool of water is observed through Assess Use the Formative Assessment a polarizer. How can you tell if the reflected light is polarized? on this page to evaluate student 470 Chapter 13 mastery of the section. Answers to Section Assessment

Reteach For students who need 1. blue

additional instruction, download the Untitled-284 2. magenta; 470 red 5/20/2011 6:25:01 AM Section Study Guide. 3. yellow and cyan (more yellow than cyan); green and red (more green than red) Response to Intervention To reassess 4. If you turn the polarizer and very little students’ mastery, use the Section Quiz, light is transmitted, then the reflected light available to print or to take directly is polarized. online at HMDScience.com.

470 Chapter 13 Chapter summary CHAPTER 13 Summary Teaching Tip SECTION 1 Characteristics of Light KEY TERM Writing down difficult concepts • Light is electromagnetic radiation that consists of oscillating electric and electromagnetic wave can help students better understand magnetic fields with different wavelengths. them and can enhance students’ • The frequency times the wavelength of electromagnetic radiation is equal to c, the speed of light. communication skills. Have students • The brightness of light is inversely proportional to the square of the summarize the differences between distance from the light source. images formed by convex mirrors and images formed by concave mirrors. Their KEY TERMS SECTION 2 Flat Mirrors writings should include a thorough • Light obeys the law of reflection, which states that the incident and reflection explanation of the mirror equation, sign reflected angles of light are equal. angle of incidence conventions, and ray diagrams for each • Flat mirrors form virtual images that are the same distance from the angle of reflection mirror’s surface as the object is. case. Be sure students explain concepts virtual image clearly and correctly, and use good SECTION 3 Curved Mirrors KEY TERMS sentence structure.

• The mirror equation relates object distance, image distance, and focal concave spherical mirror length of a spherical mirror. real image • The magnification equation relates image height or distance to object convex spherical mirror height or distance, respectively.

SECTION 4 Color and Polarization KEY TERM

• Light of different colors can be produced by adding light consisting of the linear polarization primary additive colors (red, green, and blue). • Pigments can be produced by combining subtractive colors (magenta, yellow, and cyan). • Light can be linearly polarized by transmission, reflection, or scattering.

VARIABLE SYMBOLS DIAGRAM SYMBOLS Light rays Quantities Units (real) Light rays object distance m meters p (apparent)

q image distance m meters Normal HRW • Holt Physics lines PH99PE-C14-CHS-001A R radius of m meters HRW • Holt Physics curvature PH99PE-C14-CHS-002A Problem Solving Flat mirror See Appendix D: Equations for a summary f focal length m meters HRW • Holt Physics Concave PH99PE-C14-CHS-003AConvex of the equations introduced in this chapter. If mirror mirror you need more problem-solving practice, M magnification (unitless) see Appendix I: Additional Problems. HRW • Holt Physics PH99PE-C14-CHS-005A

HRW • Holt Physics Chapter Summary 471 PH99PE-C14-CHS-010A

Untitled-279 471 5/20/2011 6:20:50 AM

Light and Reflection 471 CHAPTER REVIEW CHAPTER 13 Review Answers 7. How fast do X rays travel in a vacuum? 1. a. radio waves Characteristics of Light 8. Why do astronomers observing distant galaxies talk REVIEWING MAIN IDEAS b. gamma rays about looking backward in time? 1. Which band of the electromagnetic spectrum has 2. b 9. Do the brightest stars that you see in the night sky a. the lowest frequency? necessarily give off more light than dimmer stars? 3. Its speed is accurately known. b. the shortest wavelength? Measuring the time it takes light to Explain your answer. 2. Which of the following electromagnetic waves has the travel a distance allows the distance highest frequency? PRACTICE PROBLEMS to be determined. (Alternatively, if a. radio For problems 10–13, see Sample Problem A. the source’s brightness is known, its b. ultraviolet radiation c. blue light 10. The compound eyes of bees and other insects are apparent brightness can be meas- d. infrared radiation highly sensitive to light in the ultraviolet portion of ured and its distance calculated.) 3. Why can light be used to measure distances the spectrum, particularly light with frequencies 14 15 The wave front at B would be an arc accurately? What must be known in order to between 7.5 × 10 Hz and 1.0 × 10 Hz. To what 4. wavelengths do these frequencies correspond? of a large circle. The rays would point make distance measurements? 4. For the diagram below, use Huygens’s principle to 11. The brightest light detected from the star Antares radially outward from A to B. 14 show what the wave front at point A will look like has a frequency of about 3 × 10 Hz. What is the 5. Apparent brightness equals the at point B. How would you represent this wave front wavelength of this light? actual brightness divided by the in the ray approximation? 12. What is the wavelength for an FM radio signal if the number on the dial reads 99.5 MHz? square of the distance between Source New wavefront position observer and source. A B 13. What is the wavelength of a radar signal that has a frequency of 33 GHz? 6. 1999 + 2(95) = 2189 8 7. 3.00 × 10 m/s Flat Mirrors 8. The light from galaxies was emitted 5. What is the relationship between the actual REVIEWING MAIN IDEAS millions of years ago. HRW • Holt Physics brightness of a light source and its apparent PH99PE-C14-CHR-001A 14. For each of the objects listed below, identify whether brightness from where you see it? 9. no; Those stars may be closer so light is reflected diffusely or specularly. they appear brighter. a. a concrete driveway CONCEPTUAL QUESTIONS 10. 4.0 × 10−7 m, 3.0 × 10−7 m b. an undisturbed pond 6. Suppose an intelligent society capable of receiving c. a polished silver tray −6 11. 1 × 10 m and transmitting radio signals lives on a planet d. a sheet of paper e. a mercury column in a thermometer 12. 3.02 m orbiting Procyon, a star 95 light-years away from Earth. If a signal were sent toward Procyon in 1999, 3 13. 9.1 × 10− m (9.1 mm) what is the earliest year that Earth could expect to 14. a. diffusely receive a return message? (Hint: A light-year is the distance a ray of light travels in one year.) b. specularly c. specularly d. diffusely e. specularly

472 Chapter 13

Untitled-279 472 5/20/2011 6:20:51 AM

472 Chapter 13 CHAPTER REVIEW

15. If you are stranded on an island, where would you 24. If an object is placed outside the focal length of a align a mirror to use sunlight to signal a searching concave mirror, which type of image will be formed? 15. point normal halfway between sun aircraft? Will it appear in front of or behind the mirror? and aircraft 16. If you are standing 2 m in front of a flat mirror, how 25. Can you use a convex mirror to burn a hole in paper 16. 2 m behind; M = 1 far behind the mirror is your image? What is the by focusing light rays from the sun at the mirror’s magnification of the image? focal point? 17. The gas molecules in air do not

26. A convex mirror forms an image from a real object. reflect the light. CONCEPTUAL QUESTIONS Can the image ever be larger than the object? 18. Reflection is diffuse if λ is 17. When you shine a flashlight across a room, you see 27. Why are parabolic mirrors preferred over spherical smaller than surface irregularities the beam of light on the wall. Why do you not see the concave mirrors for use in reflecting telescopes? light in the air? of reflector.

18. How can an object be a specular reflector for some CONCEPTUAL QUESTIONS 19. no; Diagram should show that the ray electromagnetic waves yet be diffuse for others? from feet reflected at bottom of 28. Where does a ray of light that is parallel to the 19. A flat mirror that is 0.85 m tall is attached to a wall principal axis of a concave mirror go after it is mirror goes to the top of the head so that its upper edge is 1.7 m above the floor. Use the reflected at the mirror’s surface? above the eyes. law of reflection and a ray diagram to determine if 29. What happens to the real image produced by a ´ this mirror will show a person who is 1.7 m tall his 20. θ 2 = 55°; Ray reflected from the or her complete reflection. concave mirror if you move the original object to the second mirror is always parallel to location of the image? 20. Two flat mirrors make an the incoming ray. 30. Consider a concave spherical mirror and a real angle of 90.0° with each 21. 1.2 m/s; The image moves toward the other, as diagrammed at object. Is the image always inverted? Is the image = 35 always real? Give conditions for your answers. mirror’s surface.

right. An incoming ray B Mirror makes an angle of 35° with 31. Explain why enlarged images seem dimmer than the 22. Images serve as objects for more the normal of mirror A. Use Mirror A original objects. the law of reflection to images. Each reflection doubles the determine the angle of reflection from mirror B. What 32. What test could you perform to determine if an image apparent distance from “object” is real or virtual? is unusual about the incoming andHRW reflected • Holt Physics rays of to mirror. light for this arrangement of mirrors?PH99PE-C14-CHR-002A 33. You’ve been given a concave mirror that may or may 23. concave 21. If you walk 1.2 m/s toward a flat mirror, how fast does not be parabolic. What test could you perform to your image move with respect to the mirror? In what determine whether it is parabolic? 24. real, inverted image; in front direction does your image move with respect to you? PRACTICE PROBLEMS 25. No, rays always diverge from 22. Why do the images produced by two opposing flat a convex mirror. mirrors appear to be progressively smaller? For problems 34–35, see Sample Problem B. 26. no, h' < h for convex mirrors 34. A concave shaving mirror has a radius of curvature of Curved Mirrors 25.0 cm. For each of the following cases, find the 27. no spherical aberration magnification, and determine whether the image 28. through the focal point REVIEWING MAIN IDEAS formed is real or virtual and upright or inverted. a. an upright pencil placed 45.0 cm from the mirror 29. A real image appears at the former 23. Which type of mirror should be used to project movie b. an upright pencil placed 25.0 cm from the mirror images on a large screen? object position. c. an upright pencil placed 5.00 cm from the mirror 30. no; no; Image is upright and virtual when p < f. 31. The light is spread out more in the larger image. 32. try to project image on paper

Chapter Review 473 33. Produce rays parallel to and far from the principal axis. All rays focus at F for a parabolic mirror. 34. a. M = −0.384; real, inverted Untitled-279 473 5/20/2011 6:20:52 AM b. M = −1.00; real, inverted c. M = 1.67; virtual, upright

Light and Reflection 473 CHAPTER REVIEW CHAPTER REVIEW

35. A concave spherical mirror can be used to project an 42. A substance is known to reflect green and blue light. 35. q = 26 cm; real, inverted; M = −2.0 image onto a sheet of paper, allowing the magnified What color would it appear to be when it is 36. p = 52.9 cm; h = 5.69 cm; image of an illuminated real object to be accurately illuminated by white light? By blue light? traced. If you have a concave mirror with a focal 43. How can you tell if a pair of sunglasses has polarizing M = 0.299; virtual, upright length of 8.5 cm, where would you place a sheet of lenses? 37. red, green, blue; They make paper so that the image projected onto it is twice as far from the mirror as the object is? Is the image 44. Why would sunglasses with polarizing lenses remove white light. upright or inverted, real or virtual? What would the the glare from your view of the hood of your car or a 38. cyan, magenta, yellow; They make magnification of the image be? distant body of water but not from a tall metal tank black pigment. For problem 36, see Sample Problem C. used for storing liquids? 45. 39. The polarized light from the first 36. A convex mirror with a radius of curvature of 45.0 cm Is light from the sky polarized? Why do clouds seen through polarizing glasses stand out in bold contrast polarizer is blocked by the second forms a 1.70 cm tall image of a pencil at a distance of 15.8 cm behind the mirror. Calculate the object to the sky? polarizer when the component distance for the pencil and its height. Is the image real of the light that is parallel to the or virtual? What is the magnification? Is the image inverted or upright? Mixed Review second polarizer’s transmission axis REVIEWING MAIN IDEAS equals zero; The light must be perpendicular (90°) to the second Color and Polarization 46. The real image of a tree is magnified −0.085 times by a telescope’s primary mirror. If the tree’s image forms polarizer’s transmission axis. REVIEWING MAIN IDEAS 35 cm in front of the mirror, what is the distance between the mirror and the tree? What is the focal 40. a. green pigment 37. What are the three primary additive colors? What length of the mirror? What is the value for the mirror’s happens when you mix them? b. white light radius of curvature? Is the image virtual or real? Is the 38. What are the three primary subtractive colors image inverted or upright? c. black pigment (or primary pigments)? What happens when you 47. A candlestick holder has a concave reflector behind mix them? d. yellow light the candle, as shown below. The reflector magnifies a e. cyan light 39. Explain why a polarizing disk used to analyze light candle −0.75 times and forms an image 4.6 cm away can block light from a beam that has been passed from the reflector’s surface. Is the image inverted or 41. a. magenta through another polarizer. What is the relative upright? What are the object distance and the b. red orientation of the two polarizing disks? reflector’s focal length? Is the image virtual or real?

c. blue CONCEPTUAL QUESTIONS

d. black 40. Explain what could happen when you mix the e. red following: a. cyan and yellow pigment 42. cyan; blue b. blue and yellow light 43. Rotate the sunglasses while c. pure blue and pure yellow pigment d. green and red light looking at the sky or sunlight e. green and blue light reflecting off a horizontal surface. 41. What color would an opaque magenta shirt appear to If brightness changes, the glasses be under the following colors of light? 48. A child holds a candy bar 15.5 cm in front of the have polarizing lenses. a. white d. green convex side-view mirror of an automobile. The image b. red e. yellow height is reduced by one-half. What is the radius of PHYSICS 44. Light reflected from a horizontal c. c y a n curvature of the mirror? Spec. Number PH 99 PE C14-CHR-003-A surface like an auto hood is polarized Boston Graphics, Inc. horizontally and is blocked by the 617.523.1333 lenses. Light reflected from tall narrow surfaces like the tank will be 474 Chapter 13 vertically polarized, and almost all of it will pass through the lenses. 45. yes; Light from the sky is polarized, Untitled-279 474 5/20/2011 6:20:53 AM but light from the clouds is not polarized. 46. p = 4.1 × 102 cm; f = 32 cm; R = 64 cm; real, inverted image 47. inverted; p = 6.1 cm; f = 2.6 cm; real 48. R = −31.0 cm

474 Chapter 13 CHAPTER REVIEW CHAPTER REVIEW

49. A glowing electric light bulb placed 15 cm from a 55. A real object is placed at the zero end of a meterstick.

concave spherical mirror produces a real image A large concave mirror at the 100.0 cm end of the 49. q2 = 6.7 cm; real; M1 = −0.57, 8.5 cm from the mirror. If the light bulb is moved to meterstick forms an image of the object at the M = −0.27; inverted a position 25 cm from the mirror, what is the position 70.0 cm position. A small convex mirror placed at the 2 of the image? Is the final image real or virtual? What 20.0 cm position forms a final image at the 10.0 cm 50. f = −13.7 cm; M = 0.0656; are the magnifications of the first and final images? point. What is the radius of curvature of the convex virtual, upright Are the two images inverted or upright? mirror? (Hint: The first image created by the concave mirror acts as an object for the convex mirror.) 51. p = 11.3 cm 50. A convex mirror is placed on the ceiling at the intersection of two hallways. If a person stands 56. A dedicated sports-car enthusiast polishes the inside 52. concave; M = −20.0; real; inverted directly underneath the mirror, the person’s shoe is a and outside surfaces of a hubcap that is a section of a 53. (Go online to see the full solution.) distance of 195 cm from the mirror. The mirror forms sphere. When he looks into one side of the hubcap, an image of the shoe that appears 12.8 cm behind the he sees an image of his face 30.0 cm behind the 54. (Go online to see the full solution.) mirror’s surface. What is the mirror’s focal length? hubcap. He then turns the hubcap over and sees What is the magnification of the image? Is the image another image of his face 10.0 cm behind the hubcap. 55. R = −25.0 cm real or virtual? Is the image upright or inverted? a. How far is his face from the hubcap? 56. a. 15.0 cm b. What is the radius of curvature of the hubcap? 51. The side-view mirror of an automobile has a radius of c. What is the magnification for each image? b. 59.9 cm curvature of 11.3 cm. The mirror produces a virtual d. Are the images real or virtual? c. M = 2.00, M = 0.667 image one-third the size of the object. How far is the e. Are the images upright or inverted? convex concave object from the mirror? d. virtual 57. An object 2.70 cm tall is placed 12.0 cm in front of a 52. An object is placed 10.0 cm in front of a mirror. mirror. What type of mirror and what radius of e. upright What type must the mirror be to form an image of curvature are needed to create an upright image that 57. concave, R = 48.1 cm; the object on a wall 2.00 m away from the mirror? is 5.40 cm in height? What is the magnification of the What is the magnification of the image? Is the image image? Is the image real or virtual? M = 2.00; virtual real or virtual? Is the image inverted or upright? 58. A “floating coin” illusion consists of two parabolic 58. (Go online to see the full solution.) 53. The reflecting surfaces of two intersecting flat mirrors mirrors, each with a focal length of 7.5 cm, facing θ < θ < are at an angle of (0° 90°), as shown in the each other so that their centers are 7.5 cm apart figure below. A light ray strikes the horizontal mirror. (see the figure below). If a few coins are placed on the Use the law of reflection to show that the emerging lower mirror, an image of the coins forms in the small ray will intersect the incident ray at an angle of opening at the center of the top mirror. Use the mirror ϕ = − θ 180° 2 . equation, and draw a ray diagram to show that the final image forms at that location. Show that the magnification is 1 and that the image is real and ø upright. (Note: A flashlight beam shined on these images has a very startling effect. Even at a glancing angle, the incoming light beam is seemingly reflected off the images of the coins. Do you understand why?) 54. Show that if a flat mirror is assumed to have an “infinite” radius of curvature, the mirror equation Small hole reduces to q = −p. HRW • Holt Physics Parabolic PH99PE-C14-CHR-004A mirrors

Coins

HRW • Holt Physics PH99PE-C14-CHR-006A

Chapter Review 475

Untitled-279 475 5/20/2011 6:20:53 AM

Light and Reflection 475 CHAPTER REVIEW CHAPTER REVIEW

59. Use the mirror equation and the equation for 60. Use trigonometry to derive the mirror and 59. (Go online to see the full solution.) magnification to prove that the image of a real object magnification equations. (Hint: Note that the 60. (Go online to see the full solution.) formed by a convex mirror is always upright, virtual, incoming ray between the object and the mirror and smaller than the object. Use the same equations forms the hypo t enuse of a right triangle. The to prove that the image of a real object placed in front reflected ray between the image point and the of any spherical mirror is always virtual and upright mirror is also the hypo tenuse of a right triangle.) Alternative when p < | f |. Assessment Answers ALTERNATIVE ASSESSMENT 4. Research the characteristics, effects, and applications 1. Students’ answers will vary. Be sure of a specific type of electromagnetic wave in the plans are safe and experimental 1. Suntan lotions include compounds that absorb the spectrum. Find information about the range of wavelengths, frequencies, and energies; natural and designs reflect a controlled ultraviolet radiation in sunlight and therefore prevent the ultraviolet radiation from damaging skin cells. artificial sources of the waves; and the methods used experiment. Design experiments to test the properties of varying to detect them. Find out how they were discovered and how they affect matter. Learn about any dangers 2. Alhazen (965–1038) studied light, grades (SPFs) of suntan lotions. Plan to use blueprint paper, film, plants, or other light-sensitive items. associated with them and about their uses in pinhole cameras, and parabolic mirrors. Write down the questions that will guide your technology. Work together with others in the class When he failed to invent a machine to inquiry, the materials you will need, the pro cedures who are researching other parts of the spectrum to build a group presentation, brochure, chart, or control the Nile’s floods, the Caliph you plan to follow, and the measurements you will take. If your teacher approves your plan, perform the webpage that covers the entire spectrum. sentenced him to death. Alhazen experiments and report or demonstrate your findings 5. The Chinese astronomer Chang Heng (78–139 ) escaped by pretending to be insane. in class. recognized that moonlight was a reflection of 3. Students’ discussions will vary. 2. The Egyptian scholar Alhazen studied lenses, mirrors, sunlight. He applied this theory to explain lunar eclipses. Make diagrams showing how Heng might Students should recognize that rainbows, and other light phenomena early in the Middle Ages. Research his scholarly work, his life, have represented the moon’s illumination and the convex mirrors will work best. and his relationship with the Caliph al-Hakim. path of light when the Earth, moon, and sun were in Check that plans take into account How advanced were Alhazen’s inventions and various positions on ordinary nights and on nights theories? Summarize your findings and report them when there were lunar eclipses. Find out more about the law of reflection. to the class. Heng’s other scientific work, and report your findings to the class. 4. Students’ answers will vary. Be sure the 3. Work in cooperative groups to explore the use of class covers the entire spectrum. corner and ceiling mirrors as low-tech surveillance 6. Explore how many images are produced when you stand between two flat mirrors whose reflecting Students should provide source devices. Make a floor plan of an existing store, or devise a floor plan for an imaginary one. Determine surfaces face each other. What are the locations of references. how much of the store could be monitored by a clerk the images? Are they identical? Investigate these questions with diagrams and calculations. Then 5. Students’ answers will vary but should if flat mirrors were placed in the corners. If you could use curved mirrors in such a system, would you use test your calculated results with parallel mirrors, indicate that when the sun is on one concave or convex mirrors? Where would you place perpendicular mirrors, and mirrors at angles in side of Earth and the moon is on the them? Identify which parts of the store could be between. Which angles produce one, two, three, five, and seven images? Summarize your results with a other side, the moon will be bright observed with the curved mirrors in place. Note any disadvantages that your choice of mirrors may have. chart, diagram, or computer presentation. at night. 6. Students’ summaries will vary. Parallel mirrors produce an infinite number of images that are smaller and smaller. The angles for one, two, three, five, and seven images are 180°, 120°, 90°, 60°, and 45°, respectively. 476 Chapter 13

Untitled-279 476 5/20/2011 6:20:54 AM

476 Chapter 13 CHAPTER REVIEW CHAPTER REVIEW

Mirrors

Mirrors produce many types of images: virtual or real, Magnification values that are greater than 1 or less than –1 enlarged or reduced, and upright or inverted. The mirror indicate that the image of an object is larger than the object equation and the magnification equation can help sort things itself. Negative magnification values indicate that an image out. The mirror equation relates the object distance (p ), is real and inverted, while positive magnification values image distance (q ), and focal length (f ) to one another. indicate that an image is virtual and upright. 1 1 1 In this graphing calculator activity, the calculator will _ + _ = _ p q f produce a table of image distance and magnification for Image size can be determined from the magnification various object distances for a mirror with a known focal equation. length. You will use this table to determine the characteris- q tics of the images produced by a variety of mirrors and M = -_ p object distances. Go online to HMDScience.com to find this graphing calculator activity.

Chapter Review 477

Untitled-279 477 5/20/2011 6:20:55 AM

Light and Reflection 477 STANDARDS-BASED ASSESSMENT Standards-Based Assessment Answers MULTIPLE CHOICE Use the ray diagram below to answer questions 5–7. 1. B 2. H 1. Which equation is correct for calculating the focal point of a spherical mirror? 3. C A. 1/f = 1/p − 1/q B. 1/f = 1/p + 1/q 4. H C. 1/p = 1/f + 1/q 5. B D. 1/q = 1/f + 1/p 6. F 2. Which of the following statements is true about the speeds of gamma rays and radio waves in a p = 15.0 cm q = –6.00 cm 7. A vacuum? F. Gamma rays travel faster than radio waves. 8. G C13TEP001A G. Radio rays travel faster than gamma rays. 5. Which kind of mirror is shown in the ray diagram? H. Gamma rays and radio waves travel at the same A. flat speed in a vacuum. B. convex J. The speed of gamma rays and radio waves in C. concave a vacuum depends on their frequencies. D. Not enough information is available to draw a conclusion. 3. Which of the following correctly states the law of reflection? 6. What is true of the image formed by the mirror? A. The angle between an incident ray of light F. virtual, upright, and diminished and the normal to the mirror’s surface equals G. real, inverted, and diminished the angle between the mirror’s surface and H. virtual, upright, and enlarged the reflected light ray. J. real, inverted, and enlarged B. The angle between an incident ray of light and 7. What is the focal length of the mirror? the mirror’s surface equals the angle between the A. −10.0 cm normal to the mirror’s surface and the reflected B. −4.30 cm light ray. C. 4.30 cm C. The angle between an incident ray of light and D. 10.0 cm the normal to the mirror’s surface equals the angle between the normal and the reflected 8. Which combination of primary additive colors will light ray. produce magenta-colored light? D. The angle between an incident ray of light F. green and blue and the normal to the mirror’s surface is G. red and blue complementary to the angle between the H. green and red normal and the reflected light ray. J. cyan and yellow

4. Which of the following processes does not linearly polarize light? F. scattering G. transmission H. refraction J. reflection

478 Chapter 13

Untitled-285 478 5/20/2011 6:25:14 AM

478 Chapter 13 TEST PREP

9. C 9. What is the frequency of an infrared wave that has 13. X rays emitted from material around compact 10. J a vacuum wavelength of 5.5 µm? massive stars, such as neutron stars or black holes, 11. The blue fabric appears blue. The red A. 165 Hz serve to help locate and identify such objects. What B. 5.5 × 1010 Hz would be the wavelength of the X rays emitted from fabric appears black. 13 C. 5.5 × 10 Hz material around such an object if the X rays have 12. 65° D. 5.5 × 1016 Hz a frequency of 5.0 × 1019 Hz? 13. 6.0 × 10−12 m = 6.0 pm 10. If the distance from a light source is increased by a factor of 5, by how many times brighter does the EXTENDED RESPONSE 14. Polarized light will pass through the light appear? 14. Explain how you can use a piece of polarizing plastic when the transmission axis of F. 25 plastic to determine if light is linearly polarized. G. 5 the plastic is parallel with the light’s H. 1/5 plane of polarization. Rotating the J. 1/25 Use the ray diagram below to answer questions 15–19. plastic 90° will prevent the polarized

SHORT RESPONSE light from passing through the p = 30.0 cm plastic, making the plastic appear R = 20.0 cm 11. White light is passed through a filter that allows only dark. If light is not linearly polarized, yellow, green, and blue light to pass through it. This light is then shone on a piece of blue fabric and on a rotating the plastic 90° will have no piece of red fabric. Which colors do the two pieces effect on the light’s intensity. of fabric appear to have under this light? 15. 15.0 cm 12. The clothing department of a store has a mirror that consists of three flat mirrors, each arranged so that a 16. 10.0 cm person standing before the mirrors can see how an 17. −0.500 article of clothing looks from the side and back. A candle is placed 30.0 cm from the reflecting surface Suppose a ray from a flashlight is shined on the of a concave mirror. The radius of curvature of the 18. −6.0 cm mirror on the left. If the incident ray makes an angle mirror is 20.0 cm. of 65° with respect to the normal to the mirror’s 19. real; inverted 15. surface, what will be the angle θ of the ray reflected What is the distance between the surface of the from the mirror on the right? mirror and the image? 16. What is the focal length of the mirror?

17. What is the magnification of the image?

18. If the candle is 12 cm tall, what is the image height?

19. Is the image real or virtual? Is it upright or inverted? 65˚

11 12 1 Test Tip 10 2 9 3 Double-check the signs of all C13TEP002A 8 4 values to be used in the mirror and 7 6 5 magnification equations.

Standards-Based Assessment 479

Untitled-285 479 5/20/2011 6:25:14 AM

Light and Reflection 479