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Waves Ensure Sound is on: Light Doesn’t Just Bounce It Also Refracts! Reflected: Bounces (Mirrors!) Refracted: Bends (Lenses!) ii rr 11 22.

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Presentation on theme: "Waves Ensure Sound is on: Light Doesn’t Just Bounce It Also Refracts! Reflected: Bounces (Mirrors!) Refracted: Bends (Lenses!) ii rr 11 22."— Presentation transcript:

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3 Waves Ensure Sound is on:

4 Light Doesn’t Just Bounce It Also Refracts! Reflected: Bounces (Mirrors!) Refracted: Bends (Lenses!) ii rr 11 22 n2n2 n1n1  i =  r n 1 sin(  1 ) = n 2 sin(  2 )

5 Speed of light in medium Index of refraction Speed of light in vacuum so Index of Refraction 300,000,000 m/second: it’s not just a good idea, it’s the law! always!

6 n1n1 n2n2 When light travels from one medium to another the speed changes v=c/n, but the frequency is constant. So the light bends: n 1 sin(  1 )= n 2 sin(  2 ) 11 22 1) n 1 > n 2 2) n 1 = n 2 3) n 1 < n 2 Preflight Compare n 1 to n 2. Snell’s Law  1 <  2 sin  1 < sin  2 n 1 > n 2

7 n1n1 n2n2 Snell’s Law Practice normal 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  r = 60. The other part of the beam is refracted. What is  2 ? 11 rr Usually, there is both reflection and refraction! sin(60) = 1.33 sin(  2 )  2 = 40.6 degrees  1 =  r = 

8 Total Internal Reflection normal 22 11 n2n2 n1n1 Recall Snell’s Law: n 1 sin(  1 )= n 2 sin(  2 ) (n 1 > n 2   2 >  1 )  1 = sin -1 (n 2 /n 1 ) then  2 = 90 cc Light incident at a larger angle will only have reflection (  i =  r ) ii rr “critical angle” The light ray MUST travel in the direction from larger n to smaller n for a critical angle to be formed.

9 Review Equations (1) n is the refractive index and v is the speed in the medium Snell’s Law Wave equation Alternative versions of Snell’s Law Critical angle for total internal reflection

10 Light travels from air into water. If it enters water at 60 0 to the surface, find the angle at which it is refracted and the speed at which it moves in the water (n water =1.33) Understanding n=1.00 n=1.33

11 Determine the critical angle for an air-water boundary (n water =1.33, n air =1.00) Understanding

12 A wave moves from medium 1 to medium 2. given that θ 2 =30 0, 2 =0.50 cm, v 1 =3.0 cm/s, and the index of refraction for n 2 = 1.3 and for n 1 = 1.0, find v 2, 1, θ 1, and f Understanding

13 n2 n1 d d Apparent depth: Apparent Depth actual fish apparent fish Apparent Depth

14 Can the person standing on the edge of the pool be prevented from seeing the light by total internal reflection ? 1) Yes2) No Understanding “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.”

15 As we pour more water into bucket, what will happen to the number of people who can see the ball? 1) Increase2) Same3) Decrease Understanding Refraction

16 As we pour more water into bucket, what will happen to the number of people who can see the ball? 1) Increase2) Same3) Decrease Understanding Refraction

17 Fiber Optics Telecommunications Arthroscopy Laser surgery Total Internal Reflection only works if n outside < n inside At 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. n inside n outside

18 Fiber Optics At 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. We can be certain that n cladding < n inside n inside Add “cladding” so outside material doesn’t matter! n cladding n outside

19 Polarization Transverse waves have a polarization Transverse waves have a polarization (Direction of oscillation of E field for light) (Direction of oscillation of E field for light) Types of Polarization Types of Polarization Linear (Direction of E is constant) Linear (Direction of E is constant) Circular (Direction of E rotates with time) Circular (Direction of E rotates with time) Unpolarized (Direction of E changes randomly) Unpolarized (Direction of E changes randomly) x z y

20 Linear Polarizers Linear 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.

21 Reflected Light is Polarized

22 Unpolarized Light on Linear Polarizer Most light comes from electrons accelerating in random directions and is unpolarized. S transmitted = ½ S incident Averaging over all directions: S transmitted = ½ S incident Always true for unpolarized light! Intensity is proportional to the cross product of E and B

23 Linearly Polarized Light on Linear Polarizer (Law of Malus) E tranmitted = E incident cos(  ) S transmitted = S incident cos 2 (  ) TA   is the angle between the incoming light’s polarization, and the transmission axis  Transmission axis Incident E E Transmitted E absorbed =E incident cos(  ) Recall S is intensity and Recall S is intensity and

24 Intensity The average of

25 Question Unpolarized light (like the light from the sun) passes through a polarizing sunglass (a linear polarizer). The intensity of the light when it emerges is 1.zero 2.½ what it was before 3.¼ what it was before 4.⅓ what it was before 5.need more information Now, horizontally polarized light passes through the same glasses (which are vertically polarized). The intensity of the light when it emerges is 1.zero 2.½ what it was before 3.¼ what it was before 4.⅓ what it was before 5.Need more information

26 Answer for 1 Unpolarized light (like the light from the sun) passes through a polarizing sunglass (a linear polarizer). The intensity of the light when it emerges is 1. zero 2. 1/2 what it was before 3. 1/4 what it was before 4. 1/3 what it was before 5. need more information

27 Answer for 2 Now, horizontally polarized light passes through the same glasses (which are vertically polarized). The intensity of the light when it emerges is zero 1/2 what it was before 1/4 what it was before 1/3 what it was before need more information Since all light hitting the sunglasses was horizontally polarized and only vertical rays can get thru the glasses, the glasses block everything

28 Law of Malus – 2 Polarizers Cool Link 1) Intensity of unpolarized light incident on linear polarizer is reduced by ½. S 1 = ½ S 0 S = S 0 S1S1 S2S2 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is  =90º. S 2 = S 1 cos 2 (90º) = 0

29 Law of Malus – 3 Polarizers 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is θ=45º 3) Light transmitted through second polarizer is polarized 45º from vertical. Angle between it and third polarizer is θ=45º I 2 = I 1 cos 2 (45) I 3 = I 2 cos 2 (45º) = ½ I 0 cos 4 (45º) I 1 = ½ I 0. I 2 = I 1 cos 2 (45º) = ½ I 0 cos 2 (45º) Angle is 45 degrees with respect to last TA I 1 =½I 0 1) Light will be vertically polarized with: Remember you can use I or S for intensity.

30  TA S1S1 S2S2 S0S0  TA S1S1 S2S2 S0S0  ACT: Law of Malus AB 1) S 2 A > S 2 B 2) S 2 A = S 2 B 3) S 2 A < S 2 B S 1 = S 0 cos 2 (60) S 2 = S 1 cos 2 (30)= S 0 cos 2 (60) cos 2 (30) S 1 = S 0 cos 2 (60) S 2 = S 1 cos 2 (60) = S 0 cos 4 (60) E0E0 E0E0

31 Brewster’s Angle

32 Light The 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.electric fieldplane Light with the perpendicular polarization is said to be s-polarized, from the German senkrecht— perpendicularGerman

33 Brewster’s angle …when angle between reflected beam and refracted beam is exactly 90 degrees, reflected beam is 100% horizontally polarized ! Reflected light is partially polarized (more horizontal than vertical). But… n 1 sin  B = n 2 sin (90-  B ) n 1 sin  B = n 2 cos (  B ) horiz. and vert. polarized BB BB 90º-  B 90º horiz. polarized only! n1n1 n2n2

34 Polarized so that the oscillation is in same direction as to be flat with reflecting surface. (polarized in the plane of reflecting surface)

35 Brewster’s Angle Since then

36 Polarizing sunglasses are often considered to be better than tinted glasses because they… Preflight block more light block more light block more glare block more glare are safer for your eyes are safer for your eyes are cheaper are cheaper When glare is around  B, it’s mostly horiz. polarized! Polarizing sunglasses (when worn by someone standing up) work by absorbing light polarized in which direction? horizontal vertical

37 Brewster’s Angle from air off glass Air n=1.0 Glass n=1.5

38 How Sunglasses Work Any 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.

39 ACT: Brewster’s Angle When 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) Decreases T.A.

40 Dispersion prism White light Blue light gets deflected more n blue > n red The index of refraction n depends on color! In glass: n blue = 1.53n red = 1.52

41 Skier sees blue coming up from the bottom (1), and red coming down from the top (2) of the rainbow. Understanding Wow look at the variation in index of refraction! Which is red? Which is blue? Rainbow Power point (1) (2)

42 LIKE SO!In second rainbow pattern is reversed

43 Interference Requirements Need two (or more) waves Must have same frequency Must be coherent (i.e. waves must have definite phase relation)

44 Interference for Sound … For example, a pair of speakers, driven in phase, producing a tone of a single f and : l1l1 l2l2 But this won’t work for light rays--can’t get coherent pair of sources hmmm… I’m just far enough away that l 2 - l 1 = /2, and I hear no sound at all!

45 Interference for Light … Can’t produce coherent light from separate sources. (f  Hz) Single source Two different paths Interference possible here Need two waves from single source taking two different paths –Two slits –Reflection (thin films) –Diffraction

46 ACT: Young’s Double Slit Screen a distance L from slits Single source of monochromatic light d 2 slits- separated by d 1)Constructive 2)Destructive 3)Depends on L L Light waves from a single source travel through 2 slits before meeting on a screen. The interference will be: The rays start in phase, and travel the same distance, so they will arrive in phase.

47 Understanding Screen a distance L from slits Single source of monochromatic light d 2 slits- separated by d 1)Constructive 2)Destructive 3)Depends on L The rays start out of phase, and travel the same distance, so they will arrive out of phase. L ½ shift The experiment is modified so that one of the waves has its phase shifted by ½. Now, the interference will be:

48 Young’s Double Slit Concept Screen a distance L from slits Single source of monochromatic light d 2 slits- separated by d L At points where the difference in path length is 0,,2, …, the screen is bright. (constructive) At points where the difference in path length is the screen is dark. (destructive)

49 Young’s Double Slit Key Idea L 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. d

50 d Path length difference = Young’s Double Slit Quantitative d sin(  d Constructive interference Destructive interference (Where m = 0, 1, 2, …) sin( θ )  tan( θ ) = y/L Need  < d y L (Where m = 0, 1, 2, …) L is much larger than d so yellow lines are parallel

51 If You only know the Distances between Sources and Point (or Large Distances) PmPm Constructive Interference Destructive interference m=0 Central Maximum m=1 (first order maximum) m=0 (first order minimum)

52 Equations Constructive interference when angle not given Destructive interference when angle not given Constructive interference, angle given Destructive interference, angle given Constructive interference, distance from screen given Destructive interference, distance from screen given

53 d L Understanding y When this Young’s double slit experiment is placed under water. The separation y between minima and maxima 1) increases 2) same3) decreases …wavelength is shorter under water.

54 Understanding 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) Yes2) No Need: d sin  = m => sin  = m   d If   d then   d > 1 so sin  > 1 Not possible!

55 Understanding 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. Path length difference = We don’t know the angle, so we use

56 Understanding Light 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 angle First order max starts at m=1, first order min starts at m=0

57 Understanding A 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.

58 Understanding For 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 slits Therefore 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

59 d Path length difference 1-2 Multiple Slits (Diffraction Grating – N slits with spacing d) L = d sin  Constructive interference for all paths when d Path length difference 1-3 = 2d sin  d Path length difference 1-4 = 3d sin     4

60 d Understanding L d All 3 rays are interfering constructively at the point shown. If the intensity from ray 1 is I 0, what is the combined intensity of all 3 rays? 1) I 0 2) 3 I 0 3) 9 I 0 When rays 1 and 2 are interfering destructively, is the intensity from the three rays a minimum? 1) Yes 2) No Each slit contributes amplitude E o at screen. E tot = 3 E o. But I  E 2. I tot = (3E 0 ) 2 = 9 E 0 2 = 9 I 0 this one is still there! these add to zero

61 Three slit interference I0I0 9I 0

62 For many slits, maxima are still at Region between maxima gets suppressed more and more as no. of slits increases – bright fringes become narrower and brighter. 10 slits (N=10) intensity  0 2 slits (N=2) intensity  0 Multiple Slit Interference (Diffraction Grating) Peak location depends on wavelength!

63 N slits with spacing d Constructive Interference Maxima are at: * screen VERY far away  Diffraction Grating Same as for Young’s Double Slit !

64 Understanding Compare 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 m First we need to determine d the distance between the grates Now for the spacing, y Note: the larger spacing for red than for green

65 Understanding A 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 grates Now for the spacing, y For max we will set the angle θ to 90 0

66 Single Slit Interference?!

67 Wall Screen with opening shadow bright This is not what is actually seen! Diffraction Rays

68 Huygens’ Wave Model In 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.

69 Diffraction/ Huygens Every point on a wave front acts as a source of tiny wavelets that move forward. We will see maxima and minima on the wall. Light waves originating at different points within opening travel different distances to wall, and can interfere!

70 Huygens’ Principle 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: 1.Each point on the wave front is a new source of a spherical wavelet that spreads out spherically at wave speed. 2.At some later time, the new wave front is the surface that is tangent to all of the wavelets. Christian Huygens ( )

71 1 st minima Central maximum

72 Single 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.

73 W 1 1 Rays 2 and 2 also start W/2 apart and have the same path length difference st minimum at: When rays 1 and 1 interfere destructively. Under this condition, every ray originating in top half of slit interferes destructively with the corresponding ray originating in bottom half. Single Slit Diffraction

74 w Rays 2 and 2 also start w/4 apart and have the same path length difference. 2 nd minimum at Under this condition, every ray originating in top quarter of slit interferes destructively with the corresponding ray originating in second quarter. Single Slit Diffraction When rays 1 and 1 will interfere destructively.

75 Analyzing Single Slit Diffraction For 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 (  =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  w sin  w . The wavelet pair (1, 2) has a path length difference  r 12 =, 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. w w w2w2

76 Conditions for Diffraction Minima w w wwwwww w2w w

77 Condition for quarters of slit to destructively interfere (m=1, 2, 3, …) Single Slit Diffraction Summary Condition for halves of slit to destructively interfere Condition for sixths of slit to destructively interfere THIS FORMULA LOCATES MINIMA!! Narrower slit => broader pattern All together… Note: interference only occurs when the (w)idth > If you do not know the angle

78 Understanding A 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. The full width is twice this, Therefore the width of Central Maximum is 2(.825cm)=1.65 cm

79 Understanding Light 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

80 Example: Diffraction of a laser through a slit Light from a helium-neon laser ( = 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? 1.2 cm or

81 Width of a Single-Slit Diffraction Pattern width -y 1 y1y1 y2y2 y3y3 0

82 Understanding Two 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;   . (b) The slit width w is the same for both;   . (c) The wavelength is the same for both; width w 1 m 2. (m=1, 2, 3, …)

83 Understanding What is the wavelength of a ray of light whose first side diffraction maximum is at 15 o, given that the width of the single is 2511 nm? We want the first maximum, therefore m is approximately 1.5

84  (degrees) Combined 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. d a a Notice that when d/a is an integer, diffraction minima will fall on top of “missing” interference maxima.

85 Maxima and minima will be a series of bright and dark rings on screen Central maximum First diffraction minimum is at  Hole with diameter D light Diffraction from Circular Aperture 1 st diffraction minimum

86 Understanding A 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) Smaller Which drawing correctly depicts the pattern of light on the screen? (1)(2)(3)(4)

87 I Intensity from Circular Aperture First diffraction minima

88 These objects are just resolved Two objects are just resolved when the maximum of one is at the minimum of the other.

89 Resolving Power To see two objects distinctly, need  objects »  min  min  objects Improve resolution by increasing  objects or decreasing  min  objects is angle between objects and aperture: tan  objects  d/y sin  min   min = 1.22 /D  min is minimum angular separation that aperture can resolve: D d y

90 Understanding 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? Decrease  min = 1.22  / D Want  objects >  min Increase D !

91 ACT: Resolving Power How does the maximum resolving power of your eye change when the brightness of the room is decreased. 1) Increases2) Constant3) Decreases When 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!

92 Summary Interference: Coherent waves Interference: Coherent waves Full wavelength difference = Constructive Full wavelength difference = Constructive ½ wavelength difference = Destructive ½ wavelength difference = Destructive Multiple Slits Multiple Slits Constructive d sin(  ) = m m=1,2,3…) Constructive d sin(  ) = m m=1,2,3…) Destructive d sin(  ) = (m + 1/2)  2 slit only Destructive d sin(  ) = (m + 1/2)  2 slit only More slits = brighter max, darker mins More slits = brighter max, darker mins Huygens’ Principle: Each point on wave front acts as coherent source and can interfere. Huygens’ Principle: Each point on wave front acts as coherent source and can interfere. Single Slit: Single Slit: Destructive: w sin(  ) = m m=1,2,3…) Destructive: w sin(  ) = m m=1,2,3…) opposite!

93 Thin Film Interference n 1 (thin film) n2n2 n 0 =1.0 (air) t 1 2 Get two waves by reflection off of two different interfaces. Ray 2 travels approximately 2t further than ray 1.

94 Reflection + Phase Shifts n1n1 n2n2 Upon reflection from a boundary between two transparent materials, the phase of the reflected light may change. If n 1 > n 2 - no phase change upon reflection. If n 1 < n 2 - phase change of 180º upon reflection. (equivalent to the wave shifting by /2.) Incident wave Reflected wave

95 Thin Film Summary (1) n 1 (thin film) n2n2 n = 1.0 (air) t 1 2 Ray 1:  1 = 0 or ½ Determine  number of extra wavelengths for each ray. If |(  2 –  1 )| = ½, 1 ½, 2 ½ …. (m + ½) destructive If |(  2 –  1 )| = 0, 1, 2, 3 …. (m) constructive Note: this is wavelength in film! ( film = air /n film ) + 2 t/ film ReflectionDistance Ray 2:  2 = 0 or ½ This is important!

96 Thin Film Summary (2) n 1 (thin film) n2n2 t Phase shift after reflection No phase shift after reflection No Phase shift Phase shift

97 Thin Film Practice n glass = 1.5 n water = 1.3 n = 1.0 (air) t 1 2  1 = ½  2 = 0 + 2t / glass = 2t n glass / 0 = 1 Blue light (  = 500 nm) incident on a glass (n glass = 1.5) cover slip (t = 167 nm) floating on top of water (n water = 1.3). Is the interference constructive or destructive or neither? Phase shift =  2 –  1 = ½ wavelength Reflection at air-film interface only

98 ACT: Thin Film n glass =1.5 n plastic =1.8 n=1 (air) t 1 2  1 = ½  2 = ½ + 2t / glass = ½ + 2t n glass / 0 = ½ + 1 Blue light = 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) neither Phase shift =  2 –  1 = 1 wavelength Reflection at both interfaces!

99 Understanding The gas looks: bright dark A thin film of gasoline (n gas =1.20) and a thin film of oil (n oil =1.45) are floating on water (n water =1.33). When the thickness of the two films is exactly one wavelength… t = n water =1.3 n gas =1.20 n air =1.0 n oil =1.45  1,gas = ½ The oil looks: bright dark  2,gas = ½ + 2  1,oil = ½  2,oil = 2 |  2,gas –  1,gas | = 2 |  2,oil –  1,oil | = 3/2 constructivedestructive

100 Understanding 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? n film =1.4 n water =1.33 n=1.0 (air) t 1 2 Ray 1: Since n air n water, we have no phase shift. So we need only determine the number of extra wavelengths travelled by the ray.  1 = ½ Phase shift = |  2 –  1 |= 5½ wavelength Destructive interference

101 Understanding 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? n film =1.25 n lens =1.50 n=1.0 (air) t 1 2 Ray 1: Since n air { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/2511684/9/slides/slide_100.jpg", "name": "Understanding A camera lens (n=1.50) is coated with a film of magnesium fluoride (n=1.25).", "description": "What is the minimum thickness of the film required to minimize reflected light of wavelength 550 nm. n film =1.25 n lens =1.50 n=1.0 (air) t 1 2 Ray 1: Since n air

102 Light Color Color Addition & Subtraction Spectra

103 What do we see? Our eyes can’t detect intrinsic light from objects (mostly infrared), unless they get “red hot” 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.) The light we see is from the sun or from artificial light (bulbs, etc.) When we see objects, we see reflected light When we see objects, we see reflected light immediate bouncing of incident light (zero delay) immediate bouncing of incident light (zero delay) Very occasionally we see light that has been absorbed, then re-emitted at a different wavelength Very occasionally we see light that has been absorbed, then re-emitted at a different wavelength called fluorescence, phosphorescence, luminescence called fluorescence, phosphorescence, luminescence

104 Colours Light is characterized by frequency, or more commonly, by wavelength Light is characterized by frequency, or more commonly, by wavelength Visible light spans from 400 nm to 700 nm Visible light spans from 400 nm to 700 nm or 0.4  m to 0.7  m; mm to mm, etc. or 0.4  m to 0.7  m; mm to mm, etc.

105 White light White light is the combination of all wavelengths, with equal representation White light is the combination of all wavelengths, with equal representation “red hot” poker has much more red than blue light “red hot” poker has much more red than blue light experiment: red, green, and blue light bulbs make white experiment: red, green, and blue light bulbs make white RGB monitor combines these colors to display white RGB monitor combines these colors to display white blue light green light red light wavelength combined, white light called additive color combination—works with light sources

106 Introduction to Spectra We can make a spectrum out of light, dissecting its constituent colors We can make a spectrum out of light, dissecting its constituent colors A prism is one way to do this A prism is one way to do this A diffraction grating also does the job A diffraction grating also does the job The spectrum represents the wavelength-by- wavelength content of light The spectrum represents the wavelength-by- wavelength content of light can represent this in a color graphic like that above can represent this in a color graphic like that above or can plot intensity vs. wavelength or can plot intensity vs. wavelength

107 Why do things look the way they do? Why are metals shiny? Why are metals shiny? Recall that electromagnetic waves are generated from accelerating charges (i.e., electrons) Recall that electromagnetic waves are generated from accelerating charges (i.e., electrons) Electrons are free to roam in conductors (metals) 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) 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 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

108 What about glass? Why is glass clear? 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. 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.


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