2 Properties of Waves Amplitude – Maximum displacement from equilibrium Period – Time to complete one cycle (wavelength) of motion. Represented by T; Units of sec.Frequency – Number of cycles (wavelengths) per unit time. Represented by f; Units of Hz or sec-1 kHz (AM radio station), MHz (FM radio station), GHz (radar, microwaves), etc..Wavelength – Distance between two adjacent corresponding points on a wave (e.g., crests, troughs, etc.). Represented by lambda, l; Units of length (m, ft, etc.)Note: Period = 1 / Frequency and Frequency = 1 / PeriodEquilibrium PositionWavelengthAmplitude
3 Waves and Wave Motion What is a Wave?? The motion of a disturbance! Example:One person on each end of a long spring (or rope)A pulse is produced in the spring….Wave pulse moves from one end of the spring to the other,BUT no part of the spring is being carried from one person to the other.Direction of WaveDirection of Medium (Spring)
4 Periodic Motion Definition: Back and forth motion over the same path Examples:Mass - Spring SystemBungee JumpingShock Absorbers on VehiclesPendulums:Child on a swing;Trapeze ArtistsPendulum of a grandfather clockWrecking Ball
5 Simple Harmonic Motion Definition:Vibration about an equilibrium position in which a restoring force is proportional to the displacement from equilibriumSine waves describe particles vibrating with SHMExamples:Mass – Spring System (Hooke’s Law)Pendulum (small angles, <15 degrees)Examples: Visible light, radio waves, microwaves, x-rays, etc.
6 Question Is there a direct relationship between the displacement of a mass on a spring and the elastic force of that spring? Is Felastic proportional to x? i.e, Felastic = Constant * x?At equilibrium,Net force is zeroSo,Fg + Felastic = 0Fg = - FelasticFg = - Constant * xx = displacement in metersFgFel
7 Mass-Spring System Worksheet Mass (kgs)xFg0.0000.2500.5000.7501.0001.2501.5001.750x = displacement in metersFgFelUse 250g, 500g, 1000g masses. Eight combinations, including 0 mass. Bring in spring scales from EBL…basis for operation. Know spring constant for selected spring, know displacement for given mass.Do this as a class exercise. Set up 4-5 spring stations. Have each group collect their data, and plot together in class (using one group’s data)..NOTE:Fg = force due to gravity (Fg = m*g)Fel = Elastic Force of the springUse 250g, 500g, 1000g masses. Eight combinations, including 0 mass.
8 Mass-Spring System Plot Fg vs. x Fg, (in Newtons)Displacement, x (in meters)
9 Hooke’s Law For a Spring-Mass System, Robert Hooke established the relationship between Force and Displacement:Felastic = - kxwhere,k is known as the “Spring Constant”, measuring the “stiffness” of the spring. Units for k is N/m.The negative sign in the equation signifies that the direction of the elastic force is opposite the displacement…ie, it’s a restoring force!
10 Hooke’s Law (con’t)Example 1: If a mass of 0.55kg attached to a vertical spring stretches the spring 2 cm from its equilibrium position, what is the spring constant?Given: m = 0.55 kgx = mg = -9.8 m/s2Solution:Fnet = 0 = Felastic + Fg0 = - kx + mg or,kx = mgk = mg/x = (0.55 g)(-9.8 m/s2)/(-0.02 m) = 270 N/mx = mFelFelRecall…”Up” is the convention for positive direction. So, x will be negative values and g will be negative.FgFg
11 Mass-Spring System Period of a Mass-Spring System: T = 2p√ m k Where, k is the spring constant andm is the massNOTE: Changing the amplitude of the vibration (x) does NOT affect the period or frequency of vibration.mk
12 Example: Mass-Spring System A body of a 1275 kg car is supported on a frame by four springs, each of which has a spring constant of 2.0 x 104 N/m. Two people riding in the car have a combined mass of 153 kgs. Find the period of vibration of the car when it is driven over a pothole in the road.Solution: k = 2 x 104 N/mm = 1275 kg kg = 1428 kgBut the mass is evenly distributed over 4 springs, so meff = 1428/4 = 357 kgsT = 2 * p *(357 kgs/2 x 104 N/m)1/2= 2 * p* ( s2)1/2= s
13 Pendulum System Period of a Pendulum System: T = 2p√ Where, L is the length of the pendulum armg is the acceleration due to gravityNOTE: Changing the amplitude of the vibration (q) does NOT affect the period or frequency of vibration.Lg
14 Pendulums and Spring-Mass Systems The period and frequency of motion for each of these systems is INDEPENDENT of:Pendulum: Amplitude (q)Mass on swinging armMass-Spring System:Amplitude (x)
15 Example: PendulumYou need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling almost touches the floor and that its period is 12 s. How tall is the tower?Given: T = 12 s g = 9.8 m/s2Solution: Use the equation for the period of the pendulum and solve for L.T = 2 * p * (L / g )1/2T2 = 4 * p2 * (L / g)(T2 * g) / (4 * p2) = L((12 s2)2 * 9.8 m/s2) / (4 * p2) = m tall
16 Review and Revisit…You are sightseeing in Europe…and curious about the architectural structures….What would be other ways to determine the height of a tower given minimal pieces of data??DataSolution Strategy1. Period of pendulum (suspended from ceiling to floor)Solve for L in equation for period, T.2. Time for an object to fall from towerKinematics: y = ½ gt23. Angle and distance, xTrig. Functions (xTanq)4.
17 Wave Speed, v Speed of wave (v) depends upon: Medium Frequency, f Wavelength, lWave Speed = wavelength x frequencyorEquation: v = l fFreight car analogy-- each car is 5 meters long (wavelength), and 2 cars cross the road each second (frequency). Speed of train = 10 m/s (speed = Distance / time!!)
18 Think About It…. Given the equation for the speed of waves: v = l f Does this mean, for example, that high frequency sounds (high pitches), travel faster than low frequency sounds????NO!!! Wavelength and frequency vary inversely to produce the same speed of all sounds
19 Doppler Effect STATIONARY SOUND-GENERATING OBJECT MOVING Velocity, vABWaves are created at point source and radiate outward creating a wave front with the same frequency as that of the source.Although the frequency of the sound generating object remains constant, wave fronts reach the observer at Point B more frequently than Point A.
20 Doppler Effect Doppler Effect: The frequency shift that is the result of relative motion between the source of waves and an observer.Higher frequency: Object approachingLower frequency: Object recedingSome Applications:Echolocation (e.g., Submarines, Dolphins, Bats, etc.)Police RadarWeather Tracking
21 ResonanceEvery object (all matter!!) vibrates at a characteristic frequency – resonant (“natural”) frequency.Resonance: A condition that exists when the frequency of a force applied to a system matches the natural frequency of the system.Examples:Pushing a swingTuning a radio stationVoice-shattered glass.Tacoma Narrows Bridge Collapse in 1940.—High winds set up standing waves in the bridge, causing the bridge to oscillate at one of its natural frequencies.
22 Wave InteractionsUnlike Matter, more than one wave or vibration can existat the same time andin the same space.This is known as SUPERPOSITION.Superposition Principle:The method of summing the displacements (amplitudes) of 2 or more waves to produce a resultant wave.Applies to all waves types – mechanical and electromagnetic.
23 Interference Patterns +=Interference PatternsThe individual waves can overlap and produce interference patterns.The resultant wave is the sum of the displacements from equilibrium (ie the amplitude) at each point for the individual waves.Constructive Interference = ReinforcementDestructive Interference = Cancellation
25 Constructive and Destructive Interference Definitions Constructive Interference:Interference in which individual displacements on the SAME SIDE of the equilibrium position are added together to form the resultant wave.Destructive Interference:Interference in which individual displacements on OPPOSITE sides of the equilibrium position are added together to form the resultant wave.
26 Wave Superposition-- Demo (1) Using a long coiled spring, generate a transverse pulse wave(s).a. First, from one end while other end fixed.b. Then from both ends of the spring simultaneously and in the same direction.Observe that the amplitudes of traveling waves add as the waves pass one another.
27 Wave Superposition– Activities Using a long coiled spring, generate transverse pulse wave(s) from each end of the spring simultaneously. Observe the pulse that reaches your hand after the pulses have passed through one another.Experiment with the following variables:a. Displacements in opposite directions; same directionsb. Pulses of different amplitudesc. Combinations of a. and b.What did you observe?Which examples were constructive and which were destructive?What can you conclude?Observe that the pulses that pass through from one side to the other are unaffected by the presence of the other pulse!!
28 Beat Frequency equals: BEATSf1f2Beat Frequency equals:fbeat = f1 – f2
29 Standing Waves Standing Waves: Resultant wave created by the interference of two waves traveling at the same frequency, amplitude and wavelength in opposite directions.Standing Waves have Nodes and Antinodes
30 Nodes and Antinodes Nodes: Antinodes Points in the standing wave where the two waves cancel – complete destructive interference– creating a stationary point!AntinodesPoint in the standing wave, halfway between the nodes, at which the largest amplitude occurs.
31 Standing Waves Wavelength, l l1 = 2L l2 = L l3 = 2L/3 l4 = 2L/4 or ½ L Only certain frequencies of vibration produce standing waves for a given string length!!! ….More later when we get to SOUND…The wavelength of each of the standing waves depends on the string length, Lln = 2L/n
32 Standing Waves on a Vibrating String Wavelength, ll1 = 2Ll2 = Ll3 = 2L/3l4 = 2L/4 or ½ LFrequency, ff1 = v / l1f2 = 2 f1f3 = 3 f1f4 = 4f1ANNFundamental Frequency or 1st HarmonicAANNN2nd HarmonicAAANN3rd HarmonicNNAAAANNNN4th HarmonicNln = 2L/nfn = n v/2L n = 1, 2, 3, …
33 Standing Waves on a Vibrating String Fundamental Frequency:The lowest frequency of vibration of a standing wave:f1 = v / l1 = v / 2LWhere,v is the speed of waves on the vibrating string (NOT the speed of the resultant waves in air!!!!)L is the portion of the string that is vibrating
34 Harmonic Series of Standing Waves on a Vibrating String A series of frequencies that includes the fundamental frequency and integral multiples of the fundamental frequency.fn = n v / 2L, n = 1,2,3,….Frequency = harmonic number x (speed of wave on the string) / (2 x length of the vibrating string)
35 Standing Waves in an Air Column OPEN at BOTH ENDS Wavelength, ll1 = 2Ll2 = Ll3 = 2L/3Frequency, ff1 = v / l1f2 = 2 f1f3 = 3 f1ANFundamental Frequency or 1st Harmonic2nd HarmonicAAAA3rd HarmonicNNNNNln = 2L/nfn = n v/2L n = 1, 2, 3, …Example: Organ Pipes; Flute
36 Standing Waves in an Air Column CLOSED at ONE END Wavelength, ll1 = 4Ll3 = 4L/ 3l5 = 4L/5Frequency, ff1 = v / l1f3 = 3 f1f5 = 5 f1ANNANFundamental Frequency or 1st HarmonicNA3rd HarmonicNOTE: The equation for the harmonic series of pipes does not directly apply to these instruments because any deviation from the cylindrical shape affects the harmonic series. As such, for example the bell shape to the open end of the clarinet allows for some even numbered harmonics in the clarinet’s tone at relatively low intensities.5th Harmonicfn = n v/4L n = 1, 3, 5,…ln = 4L/nExample*: Clarinet, Saxophone, Trumpet
37 Waves Types Pulse Waves – A Single non-periodic disturbance Periodic Waves -- A wave whose source is a form of periodic motionTransverse Waves– A wave whose particles vibrate perpendicular to the direction of wave motion.Longitudinal Waves – A wave whose particles vibrate parallel to the direction of the wave motion
38 Wave Motion Mechanical Waves– Propagation requires a medium Examples: Sound waves; ripples in water, etcElectromagnetic Waves – Propagation does NOT require a medium; can travel in a vacuum
39 LIGHT Characteristics of “Light” Electromagnetic Wave: A TRANSVERSE waveConsisting of alternating electric and magnetic fields at right angles to each other,Travels through a vacuumAt the speed of light, c (3 x 108 m/s)Wave-Particle Duality (more later!)– Light can also be described as a “Particle”See Holt T63See Holt transparency 63
40 700 nm (red) > l > 400 nm (violet) Visible LightVisible Light:Small Part of EM SpectrumWavelengths:700 nm (red) > l > 400 nm (violet)Frequencies:4.3 x 1014 Hz < f < x Hz
41 Electromagnetic Spectrum Taken from pg 403 Trefil
42 Speed of LightAll EM radiation travel at the speed of light in a vacuum…but their wavelengths and frequencies will vary!Wave Speed Equation:c = l * fSpeed of light = wavelength x frequency
43 Interactions of EM Radiation with Matter Radiation interacts with matter in 3 principal ways:Scattered ….from the material’s surfaceAbsorbed ….by the materialTransmitted ….through the material, often changing direction in the process.
44 Polarization of Light Unpolarized light: Randomly oscillating charges (electric and magnetic fields)Linear Polarization:The alignment of the electromagnetic waves in such a way that the vibrations of the electric fields in each of the waves are parallel to each other.For example, certain processes can separate waves with electric field oscillations in the vertical direction from those in the horizontal direction.Light can be linearly polarized through:Transmission, and/orReflection and ScatteringSee Holt T 70, 71
45 Polarization of Light via Transmission The transmission axis of the substance is parallel to the plane of polarization of the light–Light passes through freely and “brightly”!Direction of WaveTransmission Axis
46 Polarization of Light via Transmission As the angle between the plane of polarization for the light and the transmission axis of the substance increases from 0 to 90 degrees, amount of light passing through decreases from 100% to 0%,The transmission axis of the substance is perpendicular to the plane of polarization of the light–NO Light passes throughXDirection of WaveTransmission Axis
47 Polarization of Light via Reflection When light is reflected a certain angle from a surface, the reflected light is completely polarized parallel to the reflecting surface.For example, if the reflecting surface is parallel to the ground, then the light is polarized horizontally.Eg, roadways, car hoods, bodies of waterSunglasses application…filter out horizontally polarized “glare” with a “vertical” polarizer.
48 Polarization of Light by Scattering Scattering of light (the absorption and re-radiation of light) by particles in the atmosphere can also cause polarization.Example: SUNLIGHT:When unpolarized beam of sunlight strikes air molecules in the atmosphere, the electrons in the molecules begin to vibrate in the same plane as the electric field of the incoming wave.The re-radiated light is polarized in the direction of the electron oscillations.
49 Physics of ColorAll kinds of interactions of light with matter (scattering, absorption, and transmission) depend on the wavelength of the EM radiation.Rules governing the scattering of EM Waves:If the object causing the scattering is much smaller than the wavelength of radiation, then shorter wavelengths are scattered much more strongly than longer ones.If the object causing the scattering is much larger than the wavelength of incoming radiation, then all wavelengths are scattered equally.
50 Why is the Sky Blue??Sunlight is scattered by air molecules in the atmosphere.Since the size of molecules (tenths of nanometers) is much less than the wavelength of visible light (hundreds of nms), we expect short wavelengths (blue light) to be much more scattered than longer ones (red light).The sky appears blue!
51 Why are Clouds White?When sunlight strikes the clouds, it scatters from droplets of water. These water droplets vary in size, but are typically much larger than the wavelength of visible light.All colors are scattered equally, and the clouds appear white.
52 Why are Sunsets Red?Light from the Sun is white…containing all the colors of the visible spectrum.As the light travels through the atmosphere, blue light is scattered…making the sky appear blue.In the evening, as the sun sets over the horizon, it has to travel a longer distance through the atmosphere, so more scattering occurs. Once the blue light has been removed, yellow and green follow, leaving red.This gradual filtering explains the appearance of the sun at Sunset.
53 “LIGHT RAY” = Direction of Wave Waves as “Rays”Simplified representation of the light wave:Visualize the direction the wave is moving.The line that traces the motion of the wave is called the light “RAY”.“LIGHT RAY” = Direction of WaveMagnetic Field OscillationsElectric Field Oscillations
54 Scattering – Diffuse Reflection Diffuse Scattering (or Diffuse Reflection):Light is reflected from a “rough”, textured surface in all directions.“Rough” must be defined relative to the incidence EM wavelength -- Short wavelengths (eg visible light) require smoother surfaces than long wavelengths (eg radio waves)Examples/Applications:Reading the pages of a bookCan be read from any angleHeating in a microwave ovenEven heating throughout oven
55 Scattering – (Specular) Reflection A beam parallel light rays encountering a smooth, mirrored surface are scattered from that surface in one direction only, leaving the surface as parallel light rays.The angle of incidence = the angle of reflectionq = q’Applies to other forms of EM radiation, but the “smoothness” of the surface is dependent upon wavelength of radiation.Radio waves: Smooth = wire mesh surfaceVisible Light: Smooth = mirrored surfaceqq’Incident RayReflected Ray
56 Reflection with Flat Mirrors A Virtual image is formed by rays that appear to intersect at the image point behind the mirror.The virtual image appears as the same height, h’, as the real image, h.The distance from the virtual image to the mirror (q) is the same as that of the real image to the mirror (p), except that it appears behind the mirror.VIRTUAL IMAGEOBJECTpqh’h
57 Flat Mirrors – Image Location using Rays Pick a point on the object and draw two incident rays to the mirror surface and their reflected rays:1st ray perpendicular to mirror surface2nd ray at an angle from the perpendicular to the mirror surface.Trace both rays back to the point from which they appear to have originated behind the mirror (apparent rays designated with dotted lines)The point at which the apparent rays intersect is the location of the virtual image point.phh’OBJECTVIRTUAL IMAGEq
58 Curved Mirrors Convex Mirrors: Concave Mirrors: Bowed outward Images appear smaller than the objectEx. Used in Stores for surveillance, etc.Concave Mirrors:“Caved” inwardImages appear larger than the objectEx. Used in telescopes, satellite dishes, etc.
59 Concave Parabolic Mirrors A Concave parabolic mirror focuses incoming parallel rays at a focal point.The distance between the focal point and the mirror is the focal length.Focusing the reflected rays at the focal point concentrates the signal, making a weak signal much stronger to detect!!Focal Point
60 Mirror Equation f do di + = + = 1 1 Focal Length Object Distance Image Distance+=Focal Length: -Distance from focal point to mirror;For Concave Mirrors: Positive Value (focal point in front of mirror)For Convex Mirrors: Negative Value (focal point behind mirror)Object distance, do: Distance from mirror to object (Pos. number)Image distance, di: -Distance from mirror to image“Real”: -Able to be projected on a screen (in front of mirror); inverted“Virtual”: -Right-side up behind mirrorSee Hewitt Problem Solving Book, 13-2 pg
61 Example 1: Reflection Plane Mirror Example 1: Plane MirrorSitting in her parlor one night, Gerty sees the reflection of her cat Whiskers, in the living room window. If the image of Whiskers makes an angle of 40 deg with the normal, at what angle does Gerty see him reflected?SOLUTION:Angle of incidence = angle of reflectionSo Gerty must see Whiskers reflected at a 40 deg angle
62 Example 2: Reflection, Convex Mirror EXAMPLE: Convex (Diverging) MirrorWendy the Witch is polishing her crystal ball. It is so shiny that she can see her reflection when she gazes into the ball from a distance of 15 cm.a. What is the focal length of Wendy’s crystal ball if she can see her reflection 4.0 cm behind the surface?b. Is the image real or virtual?
63 Example 2: Reflection EXAMPLE: Convex (Diverging) Mirror Solution: Find fGiven: do = 15 cmdi = -4 cm (behind the mirror)Equation: 1/f = 1/ do + 1/di1/f = 1/ /(-4)1/f = 4/ /60 = -11/60f = -60/11 = cmMinus sign indicates that this is a focal length of a convex (diverging) mirror.It is a virtual image behind the mirror.
64 Example 3: Concave Converging Mirror With his face 6.0 cm from his empty water bowl, Spot sees his refection 12 cm behind the bowl and jumps back.What is the focal length of the bowl?What was surprising about Spot’s reflection that may have caused him to jump?
65 Example 3: Concave Converging Mirror Solution: Find fGiven: do = 6.0 cmdi = -12 cm (behind the mirror!)Equation: 1/f = 1/d0 + 1/di1/f = 1/ /(-12)1/f = 2/ /121/f = 1/12f = 12 cmSince the object’s distance is closer to the mirror than 1 focal length, the image is enlarged!!!...frightening Spot!
66 Concave Mirrors and Focal Length do < f:Object EnlargedObject UprightVirtual Image (Behind mirror)B. do > fObject ReducedObject InvertedReal Image(In front of mirror)Scenario A.do < fScenario B.do > fWhat happens when the object is placed ON the focal point??
67 Refraction Refraction: When light is transmitted through a material substance, its path and speed may change significantly, causing…The bending of light waves as it passes at an angle from one medium to another.Refraction occurs when the velocity of the light changes.AirH2ONote that the angle with which light bends also depends on the wavelength – explains rainbows through prisms (dispersion!!) Blue light bends more than red light.When the barrel rolls onto grass from sidewalk, the grass slows it down, causing the barrel to turn.SidewalkGrass
68 Index of Refraction Index of Refraction, n: The ratio of the speed of light in a vacuum to the speed of light in a particular substance.n = c / vIndex of refraction = (speed of light) / (speed of light in a particular medium)The index of refraction for light in air is nearly that in a vacuum, so we approximate it as n = 1.00.As n increases, more bending from the “normal” occurs.See Holt reference pg 564
69 Refraction: Snell’s Law The angle to which the light will bend on going from one medium to the next depends on:The index of refraction for each medium (n), andThe light’s angle of incidencen1 sin q1 = n2 sin q2Where, q1 is the angle of incidence andq2 is the angle of refractionn1 and n2 are the indices of refraction for medium 1 and medium 2 , respectively
70 Snell’s Law: Special Case-- Critical Angle! Special Case for Refraction:q1 is a critical angle whereby q2 (the refracted beam) is 90 degrees (from the normal)NORMAL REFRACTIONSPECIAL CASEq1q2GlassAirqcq2GlassAirWhat happens when q1 exceeds qc ???
71 Total Internal Reflection When the angle of incidence exceeds the critical angle,Total Internal Reflection occurs…100%!!!...Applications: Optical FibersMachinists, physicians-- to view hard-to-reach areas!Communications – replacing electric circuits and microwave linksMore information can be carried in high frequencies of visible light than lower frequencies of electric current.Critical Angles:Glass ~ 43 degreesDiamond = 24.6 degrees … smallest of all known substances.q1 > qcq2 = q1GlassAir
72 Refraction-- Example 1While fishing out on a lake one summer afternoon, Amy spots a large trout just below the surface of the water at an angle of 60.0 deg to the vertical, and she tries to scoop it out of the water with her net.Draw the fish where Amy sees it.At what angle should Amy aim for the fish nwater = 1.33
73 Refraction-- Example 1 Solution: Solve for q1 Given: n1 = 1.33 (water) n2 = 1.00 (air)q2 = 60 degThe fish will appear to be straight ahead, but since light travel slower in water than air, it is actually closer than she thinks.Sin q1 = (n2sinq2)/n1 = (1.00)(sin 60)/1.33 = 0.651q1 = Sin-1(0.651) = 40.6
74 Example 2: RefractionBinoculars contain prisms inside that reflect light entering at an angle larger than the critical angle. If the index of refraction of a glass prism is 1.58, what is the critical angle for light entering the prism?Given: n1 = 1.58 (glass)n2 = 1.00 (air)q2 = 90 degEquation: n1 sin qc = n2 sin 90Solution: sin qc = (1.00)(sin 90)/1.58 =qc = sin = 39.3 deg
75 Refraction: Lenses Lens: Transparent (translucent) object that refracts light rays, causing them to converge or diverge to create and image.Images can be real or virtual..Applications:Optical instrumentsCameras,TelescopesMicroscopesMagnifying glassBinocularsHuman Eyeball – Lens converges light on retina
76 Types of Lenses Converging Diverging Principal Axis F F f f Object at infinite distance appears as parallel lines entering the lens.These parallel lines (also parallel to the principal axis) will pass through the focal point --- thereby identifying the focal length!**Focal Length is the image distance for an object at infinite distance.**
77 Ray Diagrams Rules for Drawing Reference Rays Ray From Object to Lens From Converging Lens to ImageParallel RayParallel to Principal AxisPasses thru the focal point, FCentral RayTo the center of the lensFrom the center of the lensFocal RayFrom Diverging Lens to ImageDirected away from the focal point, FProceeding toward back focal point , F
78 Characteristics of Lenses Can produce real or virtual imagesSee Handout Holt, pg 571 (or Hewitt Ch 30, pg 469)Some Conditions for Converging LensesObject PositionImage PropertiesSelect Examples: Technology/ApplicationInfinityPoint at FBurning a hole / magnifying glassObject beyond 2FReal, SmallerLens of a camera, human eyeObject at 2FReal, same sizeInverting lens - Field telescopeObject between F and 2FReal MagnifiedSlide Projector, Compound microscope (objective lens)Object a FAt InfinityLighthouse, search lightsObject inside FMagnified, Virtual (same side as object)Magnifying glass, eyepiece lens of binoculars, telescope, and microscopeReal image = can be projected on a screen; virtual image cannot be projected on a screen
79 Example: Magnifying Glass Conditions:Converging LensObject closer to lens than focal point, FFfFocal RayParallel RayCentral RayRESULT: Image is magnified and remains on the same side of the object (VIRTUAL Image!!!)
80 Magnification of a Lens Magnification = image height / object height= (distance from image to lens)(distance from object to lens)M = hi / ho = - di / do-M = Real and inverted+M = Upright and virtualSign Conventions+-dO (or p)Object in front of lensObject in back of lensdi (or q)Image in back of lens (real)Image in front of lens (virtual)fConverging lensDiverging lens“Front” is defined as side where light enters lens
81 Identical to the Mirror Equation!!!! Thin Lens EquationIdentical to the Mirror Equation!!!!Thin Lens Equation:1/(focal length) = 1/(object distance) + 1/(image distance)1/f = 1/do + 1/diAssumption:Lens is “very thin”…thickness is much less than focal length.This allows one to measure the focal length from the center of the lens or the surface of the lens.
83 Example: Human Eyeball Conditions:Converging LensObject outside 2FFfFocal RayParallel RayCentral Ray2FRESULT: Generates a Real, smaller, inverted image between F and 2F
84 Normal Distance Vision Human Eyeball LensFocusing difference between Camera and Eyeball:Camera: Alters distance between lens and filmEyeball Lens Changes Shape and Thickness to focus light on retina through the action of the ciliary muscles.Process is called AccommodationNormal Distance VisionNormal Close Vision
85 Vision Defects Farsightedness Nearsightness Can see clearly at a distanceImage is formed behind retina—eyeball to shortRemedy: Converging LensNearsightnessCan see clearly close upImage is formed in front of retina – eyeball too longRemedy: Diverging Lens
86 Visible Light to split into the Colors of the Rainbow!! DispersionThe average speed of light is less than that of c in a transparent medium.The magnitude of this speed is dictated by the medium and the frequency of the incoming wave.Frequencies closer to the natural frequency of the electron oscillators in the medium travel more slowly through the medium due to more interactions with the mediumThe natural (resonant) frequency of most transparent materials is in the UV part of the EM spectrum….Thus visible light of higher frequencies travel slower than those of lower frequencies….causing….Visible Light to split into the Colors of the Rainbow!!
87 Diffraction and Interference Diffraction is another process whereby light is bent…distinct from refraction and reflection.Diffraction involves the bending of light as it:Passes through a small slit/openingPasses around an objectPasses by sharp edges
88 Terminology Monochromatic – Coherent— Interference– Light waves composed of a single wavelengthWaves do not have to maintain a constant phase relationshipCoherent—Light waves of a identical wavelengthsLight waves maintain a constant phase relationshipInterference–superposition of light waves constructively or destructively
89 Interference Examples Soap BubbleThin layer of Gasoline (or oil, transmission fluid, etc) on paved surfaceCD’sIn all cases above, light waves interfere to form bands of color.This interference pattern depends on the difference in distance traveled between the interfering waves (ie, path difference)
90 Path Difference Conditions for Interference of Light Waves sin q = opposite / hypotenusesin q = Dl / dd sin q = D lqdDldD lThe path length difference, Dl, between the two waves equals a whole-number multiple of the two waves’ wavelengths.Constructive InterferenceSee Holt Transparencies for pictures-- #80-85See Holt Transparencies #80-85
91 Path Difference Conditions for Interference of Light Waves Equation for Constructive Interference:d*sinq = m lwhere, m= 0, +/-1, +/-2, …The path length difference between the two waves equals a whole-number multiple of the two waves’ wavelengths.X-Ray Diffraction Example
92 Path Difference Conditions for Interference of Light Waves Equation for Destructive Interference:d*sinq = (m + ½) lwhere, m= 0, +/-1, +/-2, …The path length difference between the two waves equals an odd number of half wavelengths
93 Examples: Diffraction Grating Monochromatic light shines at the surface of a diffraction grating with 5.0 x 103 lines/cm. The first order maximum is observed at a 15 deg angle. Find the wavelength?Given: d = 1/(5 x 103 lines/cm) = 1/(5 x 105 line/m)q = 15 degm = 1Solution: d sinq = ml(1/(5 x 105 line/m)) * sin (15) = l520 nm = x 10-7 m = lPg 610 Holt
94 Examples: Diffraction Grating Monochromatic light from a He-Ne laser (l = nm) shines at a right angle to the surface of a diffraction grating that contains 150,500 line/m. Find the angles at which one would observe the 1st and 2nd order maxima.Given: l = nm = x10-7d = 1/(150,500 line/m)q1, q2 ???Solution: d sinq = mlsinq1 = 1(150500)(6.328 x10-7 )sinq1 = x 10-2q1 = degreesSo, q2 = degrees
95 Application…X-Ray Crystallography Diffraction of x-rays by the “crystal structure”of a compound…The atoms of the molecules in the crystalline lattice act as a diffraction grating.Result: Identification of molecular structures!!!
96 Application/Devices: Spectrometers Spectrometers: Separate light from a source into its monochromatic componentsLight passes through a grating,Diffracted beams are collected at various anglesWavelengths of light calculatedChemical Composition of Light Source Identified!!!!…………………….HOW??
97 From Classical Physics to Atomic Physics or Quantum Mechanics
98 Some Encouraging Quotes… The very attempt to conjure up a picture of elementary particles and think of them in visual terms is wholly to misinterpret them... Atoms are no things. The electrons, which form an atom’s shell, are no longer things in the sense of classical physics, things which could be unambiguously described by concepts like location, velocity, size. When we get down to the atomic level, the objective world in space and time no longer exists. Werner Heisenberg, Physics and beyond (1971).
99 Some Encouraging Quotes… All of modern physics is governed by that magnificent and thoroughly confusing discipline called quantum mechanics invented more than fifty years ago. It has survived all tests. We suppose that it is exactly correct. Nobody understands it but we all know to use it and to apply it to all problems: thus we have learned to live with the fact nobody can understand it. Murray Gell-Mann.
100 Some Encouraging Quotes… Physics is not about the real world, it is about 'abstractions' from the real world, and this is what makes them so scientific. Anthony Standen, Science is a Sacred Cow (1958).
101 Some Encouraging Quotes… Nothing is more curious than the self-satisfied dogmatism with which mankind at each period of its history cherishes the delusion of the finality of its existing modes of knowledge… At this moment scientists and skeptics are the leading dogmatists. Advance in detail is admitted; fundamental novelty is barred. This dogmatic common sense is the death of philosophic adventure. A.N. Whitehead, Dialogues, recorded by L. Price (1956).
102 Electromagnetic Radiation and Thermodynamics ENERGY is QUANTIZED!!!Max Planck –Classical Physics cannot adequately explain…Electromagnetic Radiation and ThermodynamicsBlackbody Radiation: electromagnetic radiation emitted by a blackbody, which absorbs all incoming radiation and then emits radiation based only on temperature.At low temperatures, radiation is in IR region.As temperature increases, radiation shifts the visible region (higher energy).However, Planck discovered that energy is QUANTIZED!!!
103 Planck’s Quantum Theory Energy is absorbed or emitted in discrete packets of light energy called quanta (photons) by jumping from one energy level (quantum state) to another adjacent level.Energy of a Photon:E = n*h*f or E = n*h*c / lWhere,h = Planck’s constant: x J secf = frequencyc = speed of lightl = wavelengthn = 1, 2, 3, ….(n = 1 for a single quantum of light)Note: 1eV = 1.60 x JE-Mag Radiation at ~4000KWavelengthIntensityPlanck’s TheoryClassical
104 Quantum Energy --Exercises Example 1: How much energy in Joules is carried by 1000 photons of light of the following frequencies:(a) 3.0 x 1014 Hz (infrared)(b) 5.0 x 1014 Hz (orange light)(c) 6.0 x 1014 Hz (UV)Solution: E = nhf(a) = (1000)(6.63 x Js)(3.0 x 1014 Hz) = 2.0 x 10-16J(b) = (1000)(6.63 x Js)(5.0 x 1014 Hz) = 3.3 x 10-16J(c) = (1000)(6.63 x Js)(6.0 x 1014 Hz) = 4.0 x 10-16J
105 Quantum Energy --Exercises Example 2: A photon has 2 eV of energy. What are its frequency and wavelength?Given: E = hf or hc/ lc = 3 x 108 m/sh = 6.63 x Js1eV = 1.60 x J; 2eV = 3.2 x JSolution:(a) 3.2 x J = (6.63 x Js) * f(3.2 x J) / (6.63 x Js) = f = 4.8 x 1014 Hz(b) l = c/ f = (3.0 x 108 m/s)/4.8 x 1014 Hz = 6.2 x 10-7 m = 620 nmHolt pg 833
106 Quantum Energy --Exercises Example 3. A quantum of a certain color of visible light is found to have an energy of 5 x J. What is the wavelength and color of this light?Given: E = hc / lE = 5 x Jc = 3 x 108 m/sh = 6.63 x 10-34JsSolution: l = hc/E= (6.63 x 10-34Js)(3 x 108 m/s) / (5 x J)= 3.98 x 10-7 m = 398 nm ….VIOLET!
107 Positively Charged Protons; Neutrons K ShellL ShellM ShellXXXN ShellEnergy InEnergy OutEnergy Absorbed --Electron Jumps to next Energy LevelEXCITED STATEEnergy Released --Electron Drops down to lower Energy LevelGROUND STATEPg. 833 Holt
108 Bohr Model of the Hydrogen Atom Electron Moves in circular orbits around nucleusElectric force between the positively charged protons in the nucleus and the negatively charged electrons holds the electron in orbit.Only certain orbits are stable…Electrons “never” found in between these orbitsElectrons jump between orbitsEnergy radiated out when jumping from an outer orbit to an inner oneThe frequency of the radiation is related to the change in the atoms energyE = Efinal – Einitial = hf
109 Bohr Model of the Hydrogen Atom Transitions between stable orbits (r1, r2, etc) with different energy levels (E1, E2, etc.) account for discrete spectral lines.Transitions between any two levels are allowed-- resulting in emission or absorption spectra.Example: If an atom has 4 possible energy levels, how many different spectral lines could be emitted?E4-E1, E3-E1, E2-E1, E4-E3, E4-E2, E3-E2Transitions that result in emissions in the visible range are called the Lyman Series.4321
110 Absorption and Emission Spectra Positively Charged Protons; NeutronsK ShellL ShellM ShellN ShellXBohr Model of Hydrogen AtomPhotons of Energy AbsorbedPhotons of Energy EmittedNMLKNote: Transitions to E1 (K) are in the UV regionSample Emission spectral lines
111 Photoelectric EffectEmission of electrons from a surface that occurs when light of certain frequencies shines on that surface.Emitted Electrons: PhotoelectronsResponsive Surfaces: Photosensitive SurfacesThe maximum kinetic energy that the emitted electron can have equals the incoming photon energy minus the energy required to remove the electron from the metal (overcoming the force that binds it to the metal, “work function”).KEmax = hf - hf t or KEmax = hf - Wf = frequency of incoming photonft = threshold frequency specific to the metalh = Planck’s constant
112 Example 1: Photoelectric Effect Example 1: A sodium surface is illuminated with light with a frequency of 1 x 1015 Hz. The work function of sodium is 2.28 eV. Find the maximum kinetic energy of the photoelectrons in electron volts.Given: f = 1 x 1015 Hz hft = 2.28 eV = Wh = 6.63 x Js (1eV = 1.6 x J)Find KEmax ???Solution: KEmax = hf – WKEmax = [(6.63 x Js)(1 x 1015 Hz) / (1.6 x J/eV)] – eV= 4.14 eV – 2.28 eVKEmax = 1.86 eV
113 Example 2: Photoelectric Effect Example 2: Which of the following metals will exhibit the photoelectric effect when light of 7.0 x 1014 Hz frequency is shined on it?Lithium, W = 2.3 eVSilver, W = 4.7 eVCesium, W = 2.14 eVGiven: h = 6.63 x 10-34J*s (or 4.14 x eV*s)f = 7.0 x Hz (428 nm, Violet!)KEmax = hf – W
114 Example 2: Photoelectric Effect…Solution KEmax = (4.14 x eVs)(7 x 1014Hz) – 2.3 eV= 2.89 eV – 2.3 eV = eV for Li YESKEmax = (4.14 x eVs)(7 x 1014Hz) – 4.7 eV= 2.89 – 4.7 eV = eV for Ag NO= 2.89 – 2.14 eV = eV for Cs YESLi and Cs will exhibit the photoelectric effect with this frequency of light, but Ag will not…The input energy is insufficient to overcome silver’s binding energy of its electrons.
115 Example 3: Photoelectric Effect Example 3: Light of wavelength 350 nm (UV) falls on a potassium surface, and the photoelectrons have a maximum KE of 1.3 eV. What is the work function and the threshold frequency for potassium?Given: h = 6.63 x 10-34J*s (or 4.14 x eV*s)l = 3.5 x 10-7 mKEmax = 1.3 eVKEmax = hf – W, where W = hft
116 Example 3: Photoelectric Effect…Solution 1.3 eV = [(4.14 x eV*s)(3.0 x 108 m/s)/(3.5 x 10-7 m)] -W1.3 eV = 3.54 – WW = 3.54 – 1.3 eV = eV Work functionW = hft2.24 eV = (4.14 x eV*s)*ft2.24 ev / (4.14 x eV*s) = ft5.41 x Hz = ft , the threshold frequency(554 nm = Green light!)
117 Arthur H. Compton (1892-1962)… “The Compton Shift” Further Evidence for Quantization of Light– (i.e., Light behaves as Particles (Photons))Theorized–If light behaved like a particle, then a collision between an electron and a photon should be similar to the collision between two billiard balls!Photons should have momentum and energyMomentum and Energy are both conserved in collisions
118 The Compton ShiftStationary electrone-e-p = 8 x kg m/s p=0 p = ?? p = 2.7 x kg m/sEx: If a photon collides with an electron at rest, then the photon should transfer some of its energy to the electron, leaving the scattered photon with lower energy and lower frequency, yet longer wavelength.Compton Shift:The change in wavelength between incoming and scattered electromagnetic waves.Dl = (h / mec )(1-cosq)Compton Wavelength = h / mec where me = x 10-31kgNote: The wavelength change is very small and difficult to detect in visible region; observed with shorter wavelengths (x-rays).Change in wavelength for above example: l = h/mv ; p = mv = 5.3 x kg m/s;lfinal = 1.3 x 10-12m; linitial = 8.3 x m; Change in wavelength = 4.7 x m = ~5 x 10-4 nm
119 Photoelectric Effect Cannot be explained by Classical Physics!!! Classical PredictionsExperimental EvidenceWhether electrons are emitted depends upon…The intensity of the lightThe frequency of the lightThe kinetic energy of ejected electrons depends upon…At low intensity, electron ejection…Takes timeOccurs almost instantaneously above a certain frequencyHolt pg 834
120 Connections… Glowing Objects: As a hot object glows, the color of its glow depends on the object’s temperature. As the temperature increases, the color turns from red to orange to yellow to blue to white.Classical physics cannot explain. What explanation would be given by quantum mechanics??As energy increases, frequency increases.. The color of the glow changes accordingly.
121 Connections… Photoelectric effect: Even though bright red light delivers more total energy per second than dim violet light, the red light cannot eject electrons from a certain metal surface, while the dimmer violet light can.How does Einstein’s photon theory explain this observation?As energy increases, frequency increases.. The color of the glow changes accordingly.
122 Wave-Particle Duality Electromagnetic RadiationWAVEWavePARTICLES (PHOTONS)EVIDENCEInterferenceRefractionDiffractionEVIDENCEPhotoelectric EffectLuminescence: Fluorescence, PhosphorescenceCompton EffectBest Used for L-o-n g Wavelengths(eg, Radiowaves)Best Used for Short Wavelengths(eg, Gamma Rays)VISIBLE
123 Wave Particle DualityConsider L-O-N-G wavelengths...Can one Observe their Particle properties???Ex: Radiowaves at 2.5 Mhz (2.5 x 106 Hz)E = hf=(6.63 x Js)(2.5 x 106 Hz)= 1.7 x JThis energy is too small to be detected as a single photon!!!A sensitive radio receiver might receive 1010 photons per second to produce a detectable signal!!!With such a large number of photons reaching the detector per sec, one would not be able to detect a single photon…it would appear as a continuous wave.
124 Particles as Waves!!! Louis DeBroglie: “If waves can have particle properties, cannot particles have wave properties??....”…question posed as a student…earned him a PhD in Physics…with hesitation!!! (Einstein intervened!)…1st experimental confirmation – electrons show interference patterns – electron diffraction…won the Nobel Prize in physics 5 yrs after PhD
125 Matter Waves…DeBroglie Wavelength “DeBroglie Wavelength”: The wavelength of a particle…!!!All particles of matter – electrons, protons, atoms, marbles, and even YOU and I – have a wavelength that is related to the momentum of the particles by:Wavelength = h / momentuml = h / pl = h / mv,where h is Planck’s constantp = momentum, m = mass, v =velocityAs the particle’s velocity (or mass) increases, its wavelength decreases!!
126 Matter Waves…DeBroglie Frequency “DeBroglie Frequency”: The frequency of matter waves…!!!DeBroglie Frequency = Energy / Planck’s Constantf = E / hDual nature of matter represented:Particle concepts (E and mv), andWave concepts (l and f).
127 Particles as Waves Wavelength vs. Size (mass) of the Particle 1. Macroscopic World:For “larger mass” objects traveling at ordinary speeds:Ex.: mass of ball bearing = 0.02 kgspeed = 330 m/sWavelength = (6.63 x 10-34Js)/((0.02 kg)(330 m/s))= m …1024 times smaller than the diameter of an H atom!!!!!Wavelengths are below detection limits (too SMALL!)Wave properties not “observable”!!Remember…needed long wavelength EM rad to observe wave properties (can’t see as particles); Same is true here, long wavelengths showwaves, short wavelengths show as particles.
128 Particles as Waves Wavelength vs. Size (mass) of the Particle Microscopic World: e.g. ELECTONSFor “tiny particles” traveling at ordinary speeds:Ex.: mass of an electron = x kgspeed = 0.02*cWavelength = (6.63 x 10-34Js) / ((9.109 x kg)(0.02 * 3 x 108 m/s))= 1.2 x m …Roughly equal to the diameter of an H atom!!!!!Wavelengths and wave properties are DECTECTABLE!!!
129 Wave - Particle Duality Matter and Electromagnetic Radiation***Looking at Extremes for Particle and Wave Properties***Waves ObservableLong Wavelengths (Radio waves)Low FrequencyLow EnergySmall MassParticles ObservableShort Wavelengths (Gamma Rays)High FrequencyHigh EnergyLarge MassWave ConceptsParticle Concepts
130 Application of DeBroglie Waves Electron Diffraction:3 years after his theory was proposed, interference patterns were experimentally confirmed using an electron beam!!!....(Scanning) Electron Microscopes1000x shorter wavelengths than visible light, yielding much more detail….i.e., better resolutionRefined the Bohr Planetary Model of the Atom – Early Wave Model….“Electron Clouds”
131 De Broglie’s Early Wave Model of the Atom… Bohr’s Planetary Model– explained...Atomic spectraElements only emit/absorb certain frequencies of lightcorresponds to electron transitions between discrete energy levelsQuestion: Why must the electrons reside in set orbits or energy levels at discrete distances (radii) from the nucleus?DeBroglie’s Response…the electrons travel in orbit as waves rather than particles…
132 De Broglie’s Early Wave Model of the Atom… OK!...6 whole waves yield Standing Wave!!NO GOOD! waves will not yield a Standing Wave (Out of Phase)Wrap the wave around the circumference (length = 2pr) until the ends meet… Standing Waves exhibit Constructive Interference for an integral number of wavelengths at each energy levelWave~~~~Multiple Energy LevelsWave~~~~
133 Review for Test Wave-Particle Duality (of Electromagnetic Radiation) Diffraction, Interference (Evidence for Light as “Waves”)Spectrometersdsinq = ml (for constructive interference)dsinq = (m + ½)l (for destructive interference)Quantization of Light (photons) – Light as “Particles”E = n*hf; E = n*hc/l (Recall, f = c/l)Bohr Model of Atom;Atomic Spectra (Fluorescence; Phosphorescence);e.g., E2 – E1 = hfPhotoelectric EffectKEmax = hf – hft or hf – WCompton ShiftDl = (h/mec)(1-cosq)Matter as Waves –DeBroglie Wavelength and Frequencyl = h/p or l = h/mvf = E/hDeBroglie’s Wave model of atom