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1 This is a note

2 Waves Ensure Sound is on:

3 Light Doesn’t Just Bounce It Also Refracts!
Reflected: Bounces (Mirrors!) qi = qr qi qr Refracted: Bends (Lenses!) n1 sin(q1) = n2 sin(q2) q1 q2 n2 n1

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

5 Snell’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) Preflight n1 q1 1) n1 > n2 2) n1 = n2 3) n1 < n2 q1 < q2 q2 n2 sinq1 < sinq2 n1 > n2 Compare n1 to n2.

6 Snell’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 =60 sin(60) = 1.33 sin(q2) 1 r q2 = 40.6 degrees n1 n2 normal

7 Total 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” q2 Light incident at a larger angle will only have reflection (qi = qr) n2 n1 qr qi qc The light ray MUST travel in the direction from larger n to smaller n for a critical angle to be formed. q1 normal

8 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

9 Understanding Light 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=1.00 n=1.33

10 Understanding Determine the critical angle for an air-water boundary (nwater=1.33, nair=1.00)

11 Understanding A 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

12 Apparent Depth Apparent depth: Apparent Depth n2 n1 d apparent fish d
actual fish

13 Understanding Can 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

14 Understanding 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

15 Understanding 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

16 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. noutside ninside Telecommunications Arthroscopy Laser surgery Total Internal Reflection only works if noutside < ninside

17 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. noutside ncladding Fiber optics are used for probing and micro-surgery. Cladding lets optics be used to probe tissues with variable indices of refraction. ninside Add “cladding” so outside material doesn’t matter! We can be certain that ncladding < ninside

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

19 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.

20 Reflected Light is Polarized

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

22 Linearly Polarized Light on Linear Polarizer (Law of Malus)
TA Etranmitted = Eincident cos(q) Stransmitted = Sincident cos2(q) q Recall S is intensity and Transmission axis Incident E Eabsorbed q is the angle between the incoming light’s polarization, and the transmission axis q ETransmitted =Eincidentcos(q)

23 Intensity The average of

24 Question 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 is zero ½ what it was before ¼ what it was before ⅓ what it was before 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 zero ½ what it was before ¼ what it was before ⅓ what it was before Need more information

25 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 zero      1/2 what it was before      1/4 what it was before      1/3 what it was before      need more information

26 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

27 Law of Malus – 2 Polarizers
S = S0 S1 S2 1) Intensity of unpolarized light incident on linear polarizer is reduced by ½ . S1 = ½ S0 2) Light transmitted through first polarizer is vertically polarized. Angle between it and second polarizer is q=90º. S2 = S1 cos2(90º) = 0 Cool Link

28 Law of Malus – 3 Polarizers
I1= ½ I0 I2= I1cos2(45) 1) Light will be vertically polarized with: I1=½I0 2) 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=45º I3 = I2 cos2 (45º) = ½ I0 cos4 (45º) Angle is 45 degrees with respect to last TA Remember you can use I or S for intensity.

29 ACT: Law of Malus A B S2 S2 S1= S0cos2(60) S1= S0cos2(60)
90 TA S1 S2 S0 60 TA S1 S2 S0 60 E0 E0 A B S1= 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

30 Brewster’s Angle 1

31 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. Light with the perpendicular polarization is said to be s-polarized, from the German senkrecht—perpendicular

32 horiz. and vert. polarized
Brewster’s angle Reflected light is partially polarized (more horizontal than vertical). But… horiz. and vert. polarized qB 90º-qB 90º horiz. polarized only! n1 n2 …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)

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

34 Brewster’s Angle Since then

35 Preflight Polarizing sunglasses are often considered to be better than tinted glasses because they… block more light block more glare are safer for your eyes are cheaper When 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? horizontal vertical

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

37 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.

38 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.

39 Dispersion The index of refraction n depends on color!
In glass: nblue = nred = 1.52 nblue > nred prism White light Blue light gets deflected more

40 Wow 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

41 LIKE SO! In second rainbow pattern is reversed

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

43 Interference 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! l1 l2 But this won’t work for light rays--can’t get coherent pair of sources

44 Interference for Light …
Can’t produce coherent light from separate sources. (f  1014 Hz) Need two waves from single source taking two different paths Two slits Reflection (thin films) Diffraction Single source Two different paths Interference possible here

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

46 Understanding The experiment is modified so that one of the waves has its phase shifted by ½ l. Now, the interference will be: Constructive Destructive Depends on L ½ l shift d The rays start out of phase, and travel the same distance, so they will arrive out of phase. 60% got this correct Single source of monochromatic light  L 2 slits-separated by d Screen a distance L from slits

47 Young’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 d d At points where the difference in path length is the screen is dark. (destructive) Single source of monochromatic light  L Screen a distance L from slits

48 Young’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.

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

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

51 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

52 Understanding L y d When this Young’s double slit experiment is placed under water. The separation y between minima and maxima 1) increases 2) same 3) decreases …wavelength is shorter under water.

53 Understanding 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) No Need: d sin q = m l => sin q = m l / d If l > d then l / d > 1 so sin q > 1 Not possible!

54 Understanding 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 use Path length difference =

55 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

56 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.

57 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

58 Multiple Slits (Diffraction Grating – N slits with spacing d)
1 2 4 3 d d Path length difference 1-2 = d sinq =l d Path length difference 1-3 = 2d sinq =2l Path length difference 1-4 = 3d sinq =3l Constructive interference for all paths when

59 Understanding 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 I0 All 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 I0 When rays 1 and 2 are interfering destructively, is the intensity from the three rays a minimum? 1) Yes ) No

60 Three slit interference

61 Multiple Slit Interference (Diffraction Grating)
Peak location depends on wavelength! 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. 2 slits (N=2) intensity l 2l 10 slits (N=10) intensity l 2l Here do demo using diffraction grating for red and green laser

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

63 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

64 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 For max we will set the angle θ to 900 Now for the spacing, y

65 Single Slit Interference?!

66 Diffraction 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

67 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.

68 Diffraction/ Huygens Every 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.

69 Huygens’ Principle Christian 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.

70 Central maximum 1st minima

71 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.

72 Single Slit Diffraction
1 1 2 2 W When rays 1 and 1 interfere 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:

73 Single Slit Diffraction
2 2 1 1 w When rays 1 and 1 will 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

74 Analyzing Single Slit Diffraction
w2 w w 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 (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.

75 Conditions for Diffraction Minima
w w w 2w w w w w w w w

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

77 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. Therefore the width of Central Maximum is 2(.825cm)=1.65 cm The full width is twice this, 3.20

78 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

79 Example: Diffraction of a laser through a slit
1.2 cm Light 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

80 Width of a Single-Slit Diffraction Pattern
-y1 y1 y2 y3 width

81 Understanding l1 l2 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; 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, …)

82 Understanding What 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

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

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

85 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)

86 Intensity from Circular Aperture
First diffraction minima

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

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

89 Understanding 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?

90 ACT: Resolving Power How does the maximum resolving power of your eye change when the brightness of the room is decreased. 1) Increases 2) Constant 3) 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!

91 Summary Interference: Coherent waves
Full wavelength difference = Constructive ½ wavelength difference = Destructive Multiple Slits Constructive d sin(q) = m l (m=1,2,3…) Destructive d sin(q) = (m + 1/2) l 2 slit only More slits = brighter max, darker mins Huygens’ 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!

92 Thin Film Interference
1 2 n0=1.0 (air) n1 (thin film) t n2 Get two waves by reflection off of two different interfaces. Ray 2 travels approximately 2t further than ray 1.

93 Reflection + Phase Shifts
Incident wave Reflected wave n1 n2 Upon 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.)

94 Note: this is wavelength in film! (lfilm= lair/nfilm)
Thin Film Summary (1) Determine d, number of extra wavelengths for each ray. 1 2 n = 1.0 (air) n1 (thin film) t n2 Note: this is wavelength in film! (lfilm= lair/nfilm) This is important! Reflection Distance Ray 1: d1 = 0 or ½ Ray 2: d2 = 0 or ½ + 2 t/ lfilm If |(d2 – d1)| = 0, 1, 2, 3 … (m) constructive If |(d2 – d1)| = ½ , 1 ½, 2 ½ …. (m + ½) destructive

95 Thin Film Summary (2) n1 (thin film) t n2 Phase shift after reflection
No phase shift after reflection No Phase shift Phase shift

96 Thin 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 = ½ Reflection at air-film interface only d2 = 0 + 2t / lglass = 2t nglass/ l0= 1 Phase shift = d2 – d1 = ½ wavelength

97 ACT: Thin Film Blue 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) neither 1 2 n=1 (air) nglass =1.5 t nplastic=1.8 Brian 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 = ½ Reflection at both interfaces! d2 = ½ + 2t / lglass = ½ + 2t nglass/ l0= ½ + 1 Phase shift = d2 – d1 = 1 wavelength

98 Understanding t = l nwater=1.3 ngas=1.20 nair=1.0 noil=1.45 A 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: bright dark The oil looks: bright dark d1,gas = ½ d2,gas = ½ + 2 d1,oil = ½ d2,oil = 2 | d2,gas – d1,gas | = 2 | d2,oil – d1,oil | = 3/2 constructive destructive

99 Understanding 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. 1 2 n=1.0 (air) d1 = ½ nfilm=1.4 t Ray 2: Since nfilm>nwater, we have no phase shift. So we need only determine the number of extra wavelengths travelled by the ray. nwater=1.33 Phase shift = |d2 – d1|= 5½ wavelength Destructive interference

100 Understanding 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. 1 2 n=1.0 (air) Ray 2: Since nfilm<nlens, we have a phase shift of /2 when it reflects of the lens. nfilm=1.25 t nlens=1.50 d1 = ½ We need d1 and d2 to be out by a minimum ½ if we require destructive interference d2 = ½ + 2t / lfilm Therefore minimum thickness is 110 nm

101 Color Color Addition & Subtraction Spectra
Light Color Color Addition & Subtraction Spectra

102 What 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 light immediate bouncing of incident light (zero delay) Very occasionally we see light that has been absorbed, then re-emitted at a different wavelength called fluorescence, phosphorescence, luminescence

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

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

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

106 Why 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

107 What 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.


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