3Light Doesn’t Just Bounce It Also Refracts! Reflected: Bounces (Mirrors!)qi = qrqiqrRefracted: Bends (Lenses!)n1 sin(q1) = n2 sin(q2)q1q2n2n1
4Index of Refraction300,000,000 m/second: it’s not just a good idea, it’s the law!Speed of light in vacuumSpeed of light in mediumIndex of refractionsoalways!
5Snell’s Law Preflight n1 sin(q1)= n2 sin(q2) When light travels from one medium to another the speed changes v=c/n, but the frequency is constant. So the light bends:n1 sin(q1)= n2 sin(q2)Preflightn1q11) n1 > n22) n1 = n23) n1 < n2q1 < q2q2n2sinq1 < sinq2n1 > n2Compare n1 to n2.
6Snell’s Law Practice Usually, there is both reflection and refraction! A ray of light traveling through the air (n=1) is incident on water (n=1.33). Part of the beam is reflected at an angle qr = 60. The other part of the beam is refracted. What is q2?q1 =qr =60sin(60) = 1.33 sin(q2)1rq2 = 40.6 degreesn1n2normal
7Total Internal Reflection Recall Snell’s Law: n1 sin(q1)= n2 sin(q2)(n1 > n2 q2 > q1 )q1 = sin-1(n2/n1) then q2 = 90“critical angle”q2Light incident at a larger angle will only have reflection (qi = qr)n2n1qrqiqcThe light ray MUST travel in the direction from larger n to smaller n for a critical angle to be formed.q1normal
8Review Equations (1)n is the refractive index and v is the speed in the mediumSnell’s LawWave equationAlternative versions of Snell’s LawCritical angle for total internal reflection
9UnderstandingLight travels from air into water. If it enters water at 600 to the surface, find the angle at which it is refracted and the speed at which it moves in the water (nwater=1.33)n=1.00n=1.33
10UnderstandingDetermine the critical angle for an air-water boundary (nwater=1.33, nair=1.00)
11UnderstandingA wave moves from medium 1 to medium 2. given that θ2=300, 2=0.50 cm, v1=3.0 cm/s, and the index of refraction for n2 = 1.3 and for n1 = 1.0, find v2, 1, θ1, and f
12Apparent Depth Apparent depth: Apparent Depth n2 n1 d apparent fish d actual fish
13UnderstandingCan the person standing on the edge of the pool be prevented from seeing the light by total internal reflection ?“There are millions of light ’rays’ coming from the light. Some of the rays will be totally reflected back into the water,but most of them will not.”1) Yes 2) No
14Understanding Refraction As we pour more water into bucket, what will happen to the number of people who can see the ball?1) Increase 2) Same 3) Decrease
15Understanding Refraction As we pour more water into bucket, what will happen to the number of people who can see the ball?1) Increase 2) Same 3) Decrease
16Fiber OpticsAt each contact w/ the glass air interface, if the light hits at greater than the critical angle, it undergoes total internal reflection and stays in the fiber.noutsideninsideTelecommunicationsArthroscopyLaser surgeryTotal Internal Reflection only works if noutside < ninside
17Fiber OpticsAt each contact w/ the glass air interface, if the light hits at greater than the critical angle, it undergoes total internal reflection and stays in the fiber.noutsidencladdingFiber optics are used for probing and micro-surgery. Cladding lets optics be used to probe tissues with variable indices of refraction.ninsideAdd “cladding” so outside material doesn’t matter!We can be certain that ncladding < ninside
18Polarization Transverse waves have a polarization (Direction of oscillation of E field for light)Types of PolarizationLinear (Direction of E is constant)Circular (Direction of E rotates with time)Unpolarized (Direction of E changes randomly)xzy
19Linear PolarizersLinear Polarizers absorb all electric fields perpendicular to their transmission axis.Dochroic (Polaroid H) materials have the property of selectively absorbing one of two orthogonal components of ordinary light. The crystal actual align opposite to what you would intuitively expect waves to be blocked. I.e. vertical alignment block vertical electric fields, by absorbing their energy and oscillating the electrons in a vertical manner. Horizontal waves will pass through.
21Unpolarized Light on Linear Polarizer Intensity is proportional to the cross product of E and BMost light comes from electrons accelerating in random directions and is unpolarized.Averaging over all directions: Stransmitted= ½ SincidentAlways true for unpolarized light!
22Linearly Polarized Light on Linear Polarizer (Law of Malus) TAEtranmitted = Eincident cos(q)Stransmitted = Sincident cos2(q)qRecall S is intensityandTransmission axisIncident EEabsorbedq is the angle between the incoming light’s polarization, and the transmission axisqETransmitted=Eincidentcos(q)
24Question zero ½ what it was before ¼ what it was before Unpolarized light (like the light from the sun) passes through a polarizing sunglass (a linear polarizer). The intensity of the light when it emerges iszero½ what it was before¼ what it was before⅓ what it was beforeneed more informationNow, horizontally polarized light passes through the same glasses (which arevertically polarized). The intensity of the light when it emerges iszero½ what it was before¼ what it was before⅓ what it was beforeNeed more information
25Answer for 1Unpolarized light (like the light from the sun) passes through a polarizing sunglass (a linear polarizer). The intensity of the light when it emerges iszero 1/2 what it was before 1/4 what it was before 1/3 what it was before need more information
26Answer for 2Now, horizontally polarized light passes through the same glasses (which are vertically polarized). The intensity of the light when it emerges iszero 1/2 what it was before 1/4 what it was before 1/3 what it was before need more informationSince all light hitting the sunglasses was horizontally polarized and only vertical rays can get thru the glasses, the glasses block everything
27Law of Malus – 2 Polarizers S = S0S1S21) Intensity of unpolarized light incident on linear polarizer is reduced by ½ . S1 = ½ S02) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is q=90º. S2 = S1 cos2(90º) = 0Cool Link
28Law of Malus – 3 Polarizers I1= ½ I0I2= I1cos2(45)1) Light will be vertically polarized with:I1=½I02) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is θ=45º. I2 = I1 cos2 (45º) = ½ I0 cos2 (45º)3) Light transmitted through second polarizer is polarized 45º from vertical. Angle between it and third polarizer is θ=45ºI3 = I2 cos2 (45º)= ½ I0 cos4 (45º)Angle is 45 degrees with respect to last TARemember you can use I or S for intensity.
29ACT: Law of Malus A B S2 S2 S1= S0cos2(60) S1= S0cos2(60) 90TAS1S2S060TAS1S2S060E0E0ABS1= S0cos2(60)S1= S0cos2(60)S2= S1cos2(30)= S0 cos2(60) cos2(30)S2= S1cos2(60)= S0 cos4(60)1) S2A > S2B 2) S2A = S2B 3) S2A < S2B
31LightThe incident light can have the electric field of the light waves lying in the same plane as the that is is traveling. Light with this polarization is said to be p-polarized, because it is parallel to the plane.Light with the perpendicular polarization is said to be s-polarized, from the German senkrecht—perpendicular
32horiz. and vert. polarized Brewster’s angleReflected light is partially polarized (more horizontal than vertical). But…horiz. and vert. polarizedqB90º-qB90ºhoriz. polarized only!n1n2…when angle between reflected beam and refracted beam is exactly 90 degrees, reflected beam is 100% horizontally polarized !n1 sin qB = n2 sin (90-qB)n1 sin qB = n2 cos (qB)
33Polarized so that the oscillation is in same direction as to be flat with reflecting surface. (polarized in the plane of reflecting surface)
35PreflightPolarizing sunglasses are often considered to be better than tinted glasses because they…block more lightblock more glareare safer for your eyesare cheaperWhen glare is around qB, it’s mostly horiz. polarized!Polarizing sunglasses (when worn by someone standing up) work by absorbing light polarized in which direction?horizontalvertical
36Brewster’s Angle from air off glass Air n=1.0Glass n=1.5
37How Sunglasses WorkAny light ray that comes from a reflection will be slightly horizontally polarized, the glasses will only allow vertically (or vertical component of the unpolarized light) to pass through to the eye. Thus reducing the glare.
38ACT: Brewster’s AngleWhen a polarizer is placed between the light source and the surface with transmission axis aligned as shown, the intensity of the reflected light:(1) Increases (2) Unchanged (3) DecreasesT.A.
39Dispersion The index of refraction n depends on color! In glass: nblue = nred = 1.52nblue > nredprismWhite lightBlue light gets deflected more
40Wow look at the variation in index of refraction! Understanding(1)Wow look at the variation in index of refraction!(2)Which is red?Which is blue?(1)Skier sees blue coming up from the bottom (1), and red coming down from the top (2) of the rainbow.(2)Rainbow Power point
42Interference Requirements Need two (or more) wavesMust have same frequencyMust be coherent (i.e. waves must have definite phase relation)
43Interference for Sound … For example, a pair of speakers, driven in phase, producing a tone of a single f and l:hmmm… I’m just far enough away that l2-l1=l/2, and I hear no sound at all!l1l2But this won’t work for light rays--can’t get coherent pair of sources
44Interference for Light … Can’t produce coherent light from separate sources. (f 1014 Hz)Need two waves from single source taking two different pathsTwo slitsReflection (thin films)DiffractionSingle sourceTwo different pathsInterference possible here
45ACT: Young’s Double Slit Light waves from a single source travel through 2 slits before meeting on a screen. The interference will be:ConstructiveDestructiveDepends on LdThe rays start in phase, and travel the same distance, so they will arrive in phase.Single source of monochromatic light L2 slits-separated by dScreen a distance L from slits
46UnderstandingThe experiment is modified so that one of the waves has its phase shifted by ½ l. Now, the interference will be:ConstructiveDestructiveDepends on L½ l shiftdThe rays start out of phase, and travel the same distance, so they will arrive out of phase.60% got this correctSingle source of monochromatic light L2 slits-separated by dScreen a distance L from slits
47Young’s Double Slit Concept At points where the difference in path length is 0, l,2l, …, the screen is bright. (constructive)2 slits-separated by ddAt points where the difference in pathlength isthe screen is dark. (destructive)Single source of monochromatic light LScreen a distance L from slits
48Young’s Double Slit Key Idea Two rays travel almost exactly the same distance. (screen must be very far away: L >> d)Bottom ray travels a little further.Key for interference is this small extra distance.
49Young’s Double Slit Quantitative sin(θ) tan(θ) = y/LL is much larger than d so yellow lines are parallelydNeed l < ddLPath length difference =d sin(q)Constructive interference(Where m = 0, 1, 2, …)Destructive interference(Where m = 0, 1, 2, …)
50If You only know the Distances between Sources and Point (or Large Distances) Constructive InterferencePmDestructive interferencem=0Central Maximumm=0 (first order minimum)m=1 (first order maximum)
51Equations Constructive interference when angle not given Destructive interference when angle not givenConstructive interference, angle givenDestructive interference, angle givenConstructive interference, distance from screen givenDestructive interference, distance from screen given
52UnderstandingLydWhen this Young’s double slit experiment is placed under water. The separation y between minima and maxima1) increases 2) same 3) decreases…wavelength is shorter under water.
53Understanding Need: d sin q = m l => sin q = m l / d In the Young double slit experiment, is it possible to see interference maxima when the distance between slits is smaller than the wavelength of light?1) Yes 2) NoNeed: d sin q = m l=> sin q = m l / dIf l > d then l / d > 1so sin q > 1Not possible!
54Understanding Path length difference = If one source to the screen is 1.2x10-5 m farther than the other source and red light with wavelength 600 nm is used, determine the order number of the bright spot.We don’t know the angle, so we usePath length difference =
55UnderstandingLight of wavelength 450 nm shines through openings 3.0 um apart. At what angle will the first-order maximum occur? What is the angle of the first order minimum?Since we are asked for the angleFirst order max starts at m=1, first order min starts at m=0
56UnderstandingA point P is on the second-order maximum, 65 mm from one source and 45 mm from another source. The sources are 40 mm apart. Find the wavelength of the wave and the angle the point makes in the double slit.
57UnderstandingFor light of wavelength 650 nm, what is the spacing of the bright bands on a screen 1.5 m away if the distance between slits is 6.6x10-6 m?Since we want the spacing on the bands and are given the distance the screen is from the slitsTherefore the distance between the principle bright band and the first order band is 0.15m (this is actually the distance between any two bright bands
58Multiple Slits (Diffraction Grating – N slits with spacing d) 1243ddPath length difference 1-2= d sinq=ldPath length difference 1-3= 2d sinq=2lPath length difference 1-4= 3d sinq=3lConstructive interference for all paths when
59Understanding 1 2 d 3 d these add to zero this one is still there! Each slit contributes amplitude Eo at screen. Etot = 3 Eo. But I a E2. Itot = (3E0)2 = 9 E02 = 9 I0All 3 rays are interfering constructively at the point shown. If the intensity from ray 1 is I0 , what is the combined intensity of all 3 rays? 1) I ) 3 I ) 9 I0When rays 1 and 2 are interfering destructively, is the intensityfrom the three rays a minimum?1) Yes ) No
61Multiple Slit Interference (Diffraction Grating) Peak location depends on wavelength!For many slits, maxima are still atRegion between maxima gets suppressed more and more as no. of slits increases – bright fringes become narrower and brighter.2 slits (N=2)intensityl2l10 slits (N=10)intensityl2lHere do demo using diffraction grating for red and green laser
62Same as for Young’s Double Slit ! Diffraction GratingN slits with spacing dq* screen VERY far awayConstructive Interference Maxima are at:Same as for Young’s Double Slit !
63UnderstandingCompare the spacing produced by a diffraction grating with 5500 lines in 1.1 cm for red light (620 nm) and green light (510 nm). The distance to the screen is 1.1 mFirst we need to determine d the distance between the gratesNow for the spacing, yNote: the larger spacing for red than for green
64UnderstandingA diffraction grating has 1000 slits/cm. When light of wavelength 480 nm is shone through the grating, what is the separation of the bright bands 4.0 m away? How many orders of this light are seen?First we need to determine d the distance between the gratesFor max we will set the angle θ to 900Now for the spacing, y
66Diffraction Rays This is not what is actually seen! Wall shadow bright Can do ripple tank demo here. Not great.This is not what is actually seen!Screen with opening
67Huygens’ Wave ModelIn this theory, every point on a wave front is considered a point source for tiny secondary wavelets, spreading out in front of the wave as the same speed of the wave itself.
68Diffraction/ HuygensEvery point on a wave front acts as a source of tiny wavelets that move forward.•Light waves originating at different points within opening travel different distances to wall, and can interfere!•We will see maxima and minima on the wall.
69Huygens’ PrincipleChristian Huygens( )The Dutch scientist Christian Huygens, a contemporary of Newton, proposed Huygens’ Principle, a geometrical way of understanding the behavior of light waves.Huygens Principle: Consider a wave front of light:Each point on the wave front is a new source of a spherical wavelet that spreads out spherically at wave speed.At some later time, the new wave front is the surface that is tangent to all of the wavelets.
71Single Slit Diffraction As we have seen in the demonstrations with light and water waves, when light goes through a narrow slit, it spreads out to form a diffraction pattern.Now, we want to understand this behavior in more detail.
72Single Slit Diffraction 1122WWhen rays 1 and 1interfere destructively.Rays 2 and 2 also start W/2 apart and have the same path length difference.Under this condition, every ray originating in top half of slit interferes destructively with the corresponding ray originating in bottom half.1st minimum at:
73Single Slit Diffraction 2211wWhen rays 1 and 1will interfere destructively.Rays 2 and 2 also start w/4 apart and have the same path length difference.Under this condition, every ray originating in top quarter of slit interferes destructively with the corresponding ray originating in second quarter.2nd minimum at
74Analyzing Single Slit Diffraction w2wwFor an open slit of width w, subdivide the opening into segments and imagine a Hyugen wavelet originating from the center of each segment. The wavelets going forward (q=0) all travel the same distance to the screen and interfere constructively to produce the central maximum.Now consider the wavelets going at an angle such that l = w sin w q. The wavelet pair (1, 2) has a path length difference Dr12 = l/2, and therefore will cancel. The same is true of wavelet pairs (3,4), (5,6), etc. Moreover, if the aperture is divided into m sub-parts, this procedure can be applied to each sub-part. This procedure locates all of the dark fringes.
76Single Slit Diffraction Summary Condition for halves of slit to destructively interfereCondition for quarters of slit to destructively interfereCondition for sixths of slit to destructively interfere(m=1, 2, 3, …)All together…THIS FORMULA LOCATES MINIMA!!If you do not know the angleNarrower slit => broader patternNote: interference only occurs when the (w)idth > l
77UnderstandingA slit is 2.0x10-5 m wide and is illuminated by light of wavelength 550 nm. Determine the width of the central maximum, in degrees and in centimetres when L=0.30m?We will find the position of the first order minimum, which is one-half the total width of the central maximum.Therefore the width of Central Maximum is 2(.825cm)=1.65 cmThe full width is twice this, 3.20
78UnderstandingLight is beamed through a single slit 1.0 mm wide. A diffraction pattern is observed on a screen 2.0 m away. If the central band has a width of 2.5 mm, determine the wavelength of the light.To use the equations, we need half of the width
79Example: Diffraction of a laser through a slit 1.2 cmLight from a helium-neon laser (l = 633 nm) passes through a narrow slit and is seen on a screen 2.0 m behind the slit. The first minimum of the diffraction pattern is observed to be located 1.2 cm from the central maximum.How wide is the slit?or
80Width of a Single-Slit Diffraction Pattern -y1y1y2y3width
81Understandingl1l2Two single slit diffraction patterns are shown. The distance from the slit to the screen is the same in both cases.Which of the following could be true?(a) The slit width w is the same for both; l1>l2.(b) The slit width w is the same for both; l1<l2.(c) The wavelength is the same for both; width w1<w2.(d) The slit width and wavelength is the same for both; m1<m2.(e) The slit width and wavelength is the same for both; m1>m2.(m=1, 2, 3, …)
82UnderstandingWhat is the wavelength of a ray of light whose first side diffraction maximum is at 15o, given that the width of the single is 2511 nm?We want the first maximum, therefore m is approximately 1.5
83Combined Diffraction and Interference (FYI only) So far, we have treated diffraction and interference independently. However, in a two-slit system both phenomena should be present together.q (degrees)q (degrees)q (degrees)Notice that when d/a is an integer, diffraction minima will fall on top of “missing” interference maxima.
84Diffraction from Circular Aperture Central maximum1st diffraction minimumqHole with diameter DlightMaxima and minima will be a series of bright and dark rings on screenFirst diffraction minimum is at
85UnderstandingA laser is shone at a screen through a very small hole. If you make the hole even smaller, the spot on the screen will get:(1) Larger (2) SmallerWhich drawing correctly depicts the pattern of light on the screen?(1)(2)(3)(4)
86Intensity from Circular Aperture First diffraction minima
87These objects are just resolved Two objects are just resolved when the maximum of one is at the minimum of the other.
88Resolving Power To see two objects distinctly, need qobjects » qmin qobjects is angle between objects and aperture:qmintan qobjects d/yqmin is minimum angular separation that aperture can resolve:Dsin qmin qmin = 1.22 l/DydImprove resolution by increasing qobjects or decreasing qmin
89Understanding Want qobjects > qmin Decrease qmin = 1.22l / D Increase D !Astronaut Joe is standing on a distant planet with binary suns. He wants to see them but knows it’s dangerous to look straight at them. So he decides to build a pinhole camera by poking a hole in a card. Light from both suns shines through the hole onto a second card.But when the camera is built, Astronaut Joe can only see one spot on the second card! To see the two suns clearly, should he make the pinhole larger or smaller?
90ACT: Resolving PowerHow does the maximum resolving power of your eye change when the brightness of the room is decreased.1) Increases 2) Constant 3) DecreasesWhen the light is low, your pupil dilates (D can increase by factor of 10!) But actual limitation is due to density of rods and cones, so you don’t notice an much of an effect!
91Summary Interference: Coherent waves Full wavelength difference = Constructive½ wavelength difference = DestructiveMultiple SlitsConstructive d sin(q) = m l (m=1,2,3…)Destructive d sin(q) = (m + 1/2) l 2 slit onlyMore slits = brighter max, darker minsHuygens’ Principle: Each point on wave front acts as coherent source and can interfere.Single Slit:Destructive: w sin(q) = m l (m=1,2,3…)opposite!
92Thin Film Interference 12n0=1.0 (air)n1 (thin film)tn2Get two waves by reflection off of two different interfaces.Ray 2 travels approximately 2t further than ray 1.
93Reflection + Phase Shifts Incident waveReflected waven1n2Upon reflection from a boundary between two transparent materials, the phase of the reflected light may change.If n1 > n2 - no phase change upon reflection.If n1 < n2 - phase change of 180º upon reflection. (equivalent to the wave shifting by l/2.)
94Note: this is wavelength in film! (lfilm= lair/nfilm) Thin Film Summary (1)Determine d, number of extra wavelengths for each ray.12n = 1.0 (air)n1 (thin film)tn2Note: this is wavelength in film! (lfilm= lair/nfilm)This is important!ReflectionDistanceRay 1: d1 = 0 or ½Ray 2: d2 = 0 or ½+ 2 t/ lfilmIf |(d2 – d1)| = 0, 1, 2, 3 … (m) constructiveIf |(d2 – d1)| = ½ , 1 ½, 2 ½ …. (m + ½) destructive
95Thin Film Summary (2) n1 (thin film) t n2 Phase shift after reflection No phase shift after reflectionNo Phase shiftPhase shift
96Thin Film Practice 1 2 n = 1.0 (air) nglass = 1.5 t nwater= 1.3 d1 = ½ Blue light (lo = 500 nm) incident on a glass (nglass = 1.5) cover slip (t = 167 nm) floating on top of water (nwater = 1.3).Is the interference constructive or destructive or neither?d1 = ½Reflection at air-film interface onlyd2 = 0 + 2t / lglass = 2t nglass/ l0= 1Phase shift = d2 – d1 = ½ wavelength
97ACT: Thin FilmBlue light l = 500 nm is incident on a very thin film (t = 167 nm) of glass on top of plastic. The interference is:(1) constructive(2) destructive(3) neither12n=1 (air)nglass =1.5tnplastic=1.8Brian Dvorkin has had good luck with the following presentation. 1) Look for the ½ wavelength shifts on reflection. Then ask for constructive interference what do I need ray 2 to do? (0 or ½ wavelength more?) If choose 0, increase to 1.d1 = ½Reflection at both interfaces!d2 = ½ + 2t / lglass = ½ + 2t nglass/ l0= ½ + 1Phase shift = d2 – d1 = 1 wavelength
98Understandingt = lnwater=1.3ngas=1.20nair=1.0noil=1.45A thin film of gasoline (ngas=1.20) and a thin film of oil (noil=1.45) are floating on water (nwater=1.33). When the thickness of the two films is exactly one wavelength…The gas looks:brightdarkThe oil looks:brightdarkd1,gas = ½d2,gas = ½ + 2d1,oil = ½d2,oil = 2| d2,gas – d1,gas | = 2| d2,oil – d1,oil | = 3/2constructivedestructive
99Understanding 1 2 n=1.0 (air) d1 = ½ nfilm=1.4 t nwater=1.33 Light travels from air to a film with a refractive index of 1.40 to water (n=1.33). If the film is 1020 nm thick and the wavelength of light is 476 nm in air, will constructive or destructive interference occur?Ray 1: Since nair<nfilm, we have a phase shift of /2 when it reflects of the film.12n=1.0 (air)d1 = ½nfilm=1.4tRay 2: Since nfilm>nwater, we have no phase shift. So we need only determine the number of extra wavelengths travelled by the ray.nwater=1.33Phase shift = |d2 – d1|= 5½ wavelengthDestructive interference
100Understanding 1 2 n=1.0 (air) nfilm=1.25 t nlens=1.50 d1 = ½ A camera lens (n=1.50) is coated with a film of magnesium fluoride (n=1.25). What is the minimum thickness of the film required to minimize reflected light of wavelength 550 nm?Ray 1: Since nair<nfilm, we have a phase shift of /2 when it reflects of the film.12n=1.0 (air)Ray 2: Since nfilm<nlens, we have a phase shift of /2 when it reflects of the lens.nfilm=1.25tnlens=1.50d1 = ½We need d1 and d2 to be out by a minimum ½ if we require destructive interferenced2 = ½ + 2t / lfilmTherefore minimum thickness is 110 nm
101Color Color Addition & Subtraction Spectra LightColorColor Addition & SubtractionSpectra
102What do we see?Our eyes can’t detect intrinsic light from objects (mostly infrared), unless they get “red hot”The light we see is from the sun or from artificial light (bulbs, etc.)When we see objects, we see reflected lightimmediate bouncing of incident light (zero delay)Very occasionally we see light that has been absorbed, then re-emitted at a different wavelengthcalled fluorescence, phosphorescence, luminescence
103ColoursLight is characterized by frequency, or more commonly, by wavelengthVisible light spans from 400 nm to 700 nmor 0.4 m to 0.7 m; mm to mm, etc.
104White lightWhite light is the combination of all wavelengths, with equal representation“red hot” poker has much more red than blue lightexperiment: red, green, and blue light bulbs make whiteRGB monitor combines these colors to display whitecombined, white lightcalled additive colorcombination—workswith light sourceswavelengthblue lightgreen lightred light
105Introduction to Spectra We can make a spectrum out of light, dissecting its constituent colorsA prism is one way to do thisA diffraction grating also does the jobThe spectrum represents the wavelength-by-wavelength content of lightcan represent this in a color graphic like that aboveor can plot intensity vs. wavelength
106Why do things look the way they do? Why are metals shiny?Recall that electromagnetic waves are generated from accelerating charges (i.e., electrons)Electrons are free to roam in conductors (metals)An EM wave incident on metal readily vibrates electrons on the surface, which subsequently generates EM radiation of exactly the same frequency (wavelength)This indiscriminate vibration leads to near perfect reflection, and exact cancellation of the EM field in the interior of the metal—only surface electrons participate
107What about glass? Why is glass clear? The clear piece of glass is transparent to visible light because the available electrons in the material which could absorb the visible photons have no available energy levels above them in the range of the quantum energies of visible photons. The glass atoms do have vibrational energy modes which can absorb infrared photons, so the glass is not transparent in the infrared. This leads to the greenhouse effect.