Presentation on theme: "Waves (in general) sine waves are nice"— Presentation transcript:
1 Waves (in general) sine waves are nice other types of waves (such as square waves,sawtooth waves, etc.) can be formed by asuperposition of sine waves - this is calledFourier Series . This means that sinewaves can be considered as fundamental.
2 Waves (in general) E = Eo sin() where is a phase angle which describes the location along the wave = 90 degrees is the crest= 270 degrees is the trough
3 Waves (in general) E = Eo sin() where is a phase angle in a moving wave, changes with bothtime (goes 2 radians in time T) anddistance (goes 2 radians in distance )so = (2/)*x +/- (2/T)*twhere 2/T = andwhere 2/ = k and sophase speed: v = distance/time = /T = f = /k
4 Waves (in general) For nice sine waves: E = Eo sin(kx +/- t) For waves in general, can breakinto component sine waves; this iscalled spectral analysis
5 Light and Shadows Consider what we would expect from particle theory: sharp shadowsdarkdarklight
6 Light and Shadows Consider what we would expect from wave theory: shadows NOT sharpcrestcrestcrestdarkdarkdimlightdim
7 Light and Shadows What DOES happen? look at a very bright laser beam going through a vertical slit.(A laser has one frequencyunlike white light.)
8 Double Slit Experiment We will consider this situationbut only after we consider another:the DOUBLE SLIT experiment:
9 Double Slit Experiment Note that along the green linesare places where crests meets crestsand troughs meet troughs.crest on crestfollowed bytrough on trough
10 Double Slit Experiment Note that along the dotted linesare places where crests meets troughsand troughs meet crests.crest on troughfollowed by trough on crestcrest on crestfollowed bytrough on trough
11 Double Slit Experiment Further explanations are in theIntroduction to the Computer HomeworkAssignment on Young’s Double Slit, Vol 5, #3.crest on troughfollowed by trough on crestcrest on crest followed by trough on trough
12 Double Slit Experiment Our question now is: How is the patternof bright and dark areas related to theparameters of the situation: , d, x and L?brightxdbrightLSCREEN
13 Young’s Double Slit Formula λ/d = sin() ≈ tan() = x/LThe two (black) lines from the two slits to the first bright spot are almost parallel, so the two angles are almost 90 degrees, so the two ’s are almost equal.brightxdbrightLλ
14 Double slit: an example n = d sin() = d x / Ld = 0.15 mm = 1.5 x 10-4 mx = ??? measured in classL = ??? measured in classn = 1 (if x measured between adjacent bright spots) = d x / L = (you do the calculation)
15 Interference: Diffraction Grating The same Young’s formula works for multiple slits as it did for 2 slits.lensbrights1s2s3s2 = s1 + s3 = s2 + = s1 + 2s4 = s3 + = s1 + 3s5 = s4 + = s1 + 4dbrights4s5λ
16 Interference: Diffraction Grating With multiple slits, get MORE LIGHT and get sharper bright spots.lensbrights1s2s3s2 = s1 + s3 = s2 + = s1 + 2s4 = s3 + = s1 + 3s5 = s4 + = s1 + 4dbrights4s5
17 Interference: Diffraction Grating With 5 slits, get cancellation when s = 0.8; with two slits, only get complete cancellation when s = 0.5 .lensbrightdarks1s2s3s2 = s1 + .8s3 = s2 + .8 = s s4 = s3 + .8 = s s5 = s4 + .8 = s dbrights4s5
18 Diffraction Grating: demonstrations look at the white light source(incandescent light due to hot filament)look at each of the gas excited sources(one is Helium, one is Mercury)
19 Interference: Thin Films Before, we had several different parts of a wide beam interfering with one another.Can we find other ways of having parts of a beam interfere with other parts?
20 Interference: Thin Films We can also use reflection and refraction to get different parts of a beam to interfere with one another by using a thin film.reflected red interferes withrefracted/reflected/refracted blue.airfilmwater
21 Interference: Thin Films Blue travels an extra distance of 2t in the film.reflected red interferes withrefracted/reflected/refracted blue.airfilmtwater
22 Interference: Thin Films Also, blue undergoes two refractions and reflects off of a different surface.reflected red interferes withrefracted/reflected/refracted blue.airfilmtwater
23 Interference: Thin Films When a wave encounters a new medium:the phase of the refracted wave is NOT affected.the phase of the reflected wave MAY BE affected.
24 Interference: Thin Films When a wave on a string encounters a fixed end, the reflected wave must interfere with the incoming wave so as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.
25 Interference: Thin Films When a wave on a string encounters a fixed end, the reflected wave must interfere with the incoming wave so as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.
26 Interference: Thin Films When a wave on a string encounters a free end, the reflected wave does NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.
27 Interference: Thin Films When a wave on a string encounters a free end, the reflected wave does NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.
28 Interference: Thin Films When light is incident on a SLOWER medium (one of index of refraction higher than the one it is in), the reflected wave is 180 degrees out of phase with the incident wave.When light is incident on a FASTER medium, the reflected wave does NOT undergo a 180 degree phase shift.
29 Interference: Thin Films If na < nf < nw, BOTH red and blue reflected rays will be going from fast to slow, and no difference in phase will be due to reflection.reflected red interferes withrefracted/reflected/refracted blue.airfilmtwater
30 Interference: Thin Films If na < nf > nw, there WILL be a 180 degree phase difference (/2)due to reflection.reflected red interferes withrefracted/reflected/refracted blue.airfilmtwater
31 Interference: Thin Films There will ALWAYS be a phase difference due to the extra distance of 2t/.reflected red interferes withrefracted/reflected/refracted blue.airfilmtwater
32 Interference: Thin Films When t=/2 the phase difference due to path is 360 degrees (equivalent to no difference)reflected red interferes withrefracted/reflected/refracted blue.airfilmtwater
33 Interference: Thin Films When t=/4 the phase difference due to path is 180 degrees.reflected red interferes withrefracted/reflected/refracted blue.airfilmtwater
34 Interference: Thin Films Recall that the light is in the FILM, so the wavelength is not that in AIR: f = a/nf.reflected red interferes withrefracted/reflected/refracted blue.airfilmtwater
35 Interference: Thin Films reflection: no difference if nf < nw;180 degree difference if nf > nw.distance: no difference if t = a/2nf180 degree difference if t = a/4nfTotal phase difference is sum of the above two effects.
36 Interference: Thin Films Total phase difference is sum of the two effects of distance and reflectionFor minimum reflection, need total to be 180 degrees.anti-reflective coating on lensFor maximum reflection, need total to be 0 degrees.colors on oil slick
37 Thin Films: an exampleAn oil slick preferentially reflects green light. The index of refraction of the oil is 1.65, that of water is 1.33, and or course that of air isWhat is the thickness of the oil slick?
38 Thin Films: an exampleSince we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees).
39 Thin Films: an exampleSince we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees).Since we have nf > nw, we have 180 degrees due to reflection.
40 Thin Films: an exampleSince we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees).Since we have nf > nw, we have 180 degrees due to reflection.Therefore, we need 180 degrees due to extra distance, so need t = a/4nf where a = 500 nm, nf = 1.65, and so t = 500 nm / 4(1.65) = 76 nm.
41 Michelson Interferometer Split a beam with a Half Mirror, the use mirrors to recombine the two beams.MirrorHalf MirrorLightsourceMirrorScreen
42 Michelson Interferometer If the red beam goes the same length as the blue beam, then the two beams will constructively interfere and a bright spot will appear on screen.MirrorHalf MirrorLightsourceMirrorScreen
43 Michelson Interferometer If the blue beam goes a little extra distance, s, the the screen will show a different interference pattern.MirrorHalf MirrorLightsourceMirrorsScreen
44 Michelson Interferometer If s = /4, then the interference pattern changes from bright to dark.MirrorHalf MirrorLightsourceMirrorsScreen
45 Michelson Interferometer If s = /2, then the interference pattern changes from bright to dark back to bright (a fringe shift).MirrorHalf MirrorLightsourceMirrorsScreen
46 Michelson Interferometer By counting the number of fringe shifts, we can determine how far s is!MirrorHalf MirrorLightsourceMirrorsScreen
47 Michelson Interferometer If we use the red laser (=632 nm), then each fringe shift corresponds to a distance the mirror moves of 316 nm (about 1/3 of a micron)!MirrorHalf MirrorLightsourceMirrorsScreen
48 Michelson Interferometer We can also use the Michelson interferometer to determine the index of refraction of a gas (such as air).Put a cylinder with transparent ends into one of the beams.
49 Michelson Interferometer Evacuate the cylinder with a vacuum pumpSlowly allow the gas to seep back into the cylinder and count the fringes.MirrorHalf MirrorLightsourceMirrorcylinderScreen
50 Michelson Interferometer In vacuum, #vv = 2L .In the air, #aa = 2L .Since va < c, a < v and #a > #v .Knowing v and L, can calculate #v .
51 Michelson Interferometer Knowing v and L, can calculate #v .By counting the number of fringe shifts, we can determine #.Since # = #a - #v , we can calculate #a .Now knowing L and #a, we can calculate a .
52 Michelson Interferometer We now know v and a, so:with vf = c and af = va , we can usena = c/va = vf / af = v / a .
53 Michelson Interferometer an example If L = 6 cm, and if v = 632 nm, and if 50 fringes are counted when air is let back into the cylinder, then:#v = 2L/v = 2 * .06 m / 632 x 10-9 m = 189,873#a = #v + # = 189, = 189,923a = 2L/#a = 2 * .06 m / 189,923 = nmna = v / a = nm / nm =