Announcements 3/23/11 Prayer Two labs this week (telescope, interferometer) Review: phaseshift 2 (PL ) Approx.1: PL d sin E stuff e e E stuff 1 ei i 2 i 2 E stuff cos( 2) I I0 cos2 ( 2) Approx.2: sin y L Fourier Transforms? From last time: what did our two-slit analysis have to do with Fourier transforms? E stuff 1 e e E~ e E~ i ieach slit i dY (this is the y-coordinate on the slits, not the y-coordinate on the screen) open areas E~ compare to: 2 " aperture function " e dY an L i L2 L 2 2 nx f ( x) cos dx L Adding up phases … In short, we need to add up a bunch of vectors that have the same magnitude (1), but angles (phases) that go like 0, 20, 40, 60, etc. For a different position on the screen (measured by y slits or , we need to add up a different set of phases… irel.to ref. islit1 islit 2 screen ifinal slit Etot perhaps E0e e42, 63, ... etc. e likee0, 21, 2 y PL I E two-slit PL d sin d 2 for each slit L For an equally-spaced pattern of slits, how do the PLs compare? Each is a multiple of 1! (Could have an overall reference phase…not too important.) Adding up phases, cont. … slits Etot E0e irel.to ref. PL 2 for each slit e islit1 e islit 2 ... e ifinal slit two-slit PL d sin d y L screen I Quick writing: graphically add these three vectors: 10 + 120 + 140 What about 10 + 190 + 1180 E 2 Three Slit Problem: Scanning Theta Credit: this animation and the next one are from Dr. Durfee Note: for some reason he picked the overall reference phase to be about 20 Thought question How many “sub” peaks are there between the “main” peaks in a 5-slit interference pattern? a. 1 b. 2 c. 3 d. 4 e. 5 Five Slit Problem: Scanning Theta Note: for some reason he picked the overall reference phase to be about 20-30 Reading Quiz When a wave on a string moves from a fast velocity section to a slow velocity section, the reflected wave is phase-shifted by 180. When a wave moves from a slow section to a fast section, however, the reflected wave has no phase shift. How does the phase shift of a light wave moving from one medium to another compare to that? (close to normal incidence) a. The phase shift obeys the same rules b. The phase shift is the same for fast-to-slow, but reversed for slow-to-fast c. The phase shift is the same for slow-to-fast, but reversed for fast-to-slow d. The phase shift is reversed for both cases Remember these? “Fresnel Equations” Just the same as strings If near perpendicular (1-D problem) v2 v1 n1 n2 r v1 v2 n1 n2 R r t 2 The Truth (overlooked by textbook): Fordon’t arbitrary angle you always get a phase shift, even if going n1 cosfast 1 tonslow. 2 cos 2 rs(Brewster polar . marks boundary) n1 cos1 n2 cos2 n1 cos2 n2 cos1 rp polar. n1 cos2 n2 cos1 2v2 2n1 v1 v2 n1 n2 T 1 r 2 2n1 cos1 ts polar . More Truth: sometimes phase n1 cos 1 n180: 2 have 2 coscan shifts not just complex n, complex , etc. 2n1 cos1 t p polar. n1 cos2 n2 cos1 Back to 1D case From low to high index: 180 phase shift From high to low index: no phase shift What does the thickness of this slab need to be to get constructive interference between the two rays? air thin glass air What changes if rays really are at an angle? Rays drawn at an angle to make viewing easier. They’re really perpendicular to surface. thickness t Optical path length OPL = Path Length n since wavelength inside the material is reduced by a factor of n, the distance “looks” bigger than it actually is Constructive interference: OPL ( any phase shifts) = m Destructive interference: OPL ( any phase shifts) = (m+1/2) New situation What does the thickness of the COATING need to be to get constructive interference between the two rays? air thin coating, n = 1.3 thick glass, n = 1.5 Rays drawn at an angle to make viewing easier. They’re really perpendicular to surface. thickness t Pretty pictures What’s going on here? http://superphysics.netfirms. com/pp_optics.html http://twilit.wordpress.com/2008/ 03/15/bubbles-and-interference/ Demo Demo: Soap film Interferometer From lab 9: changing optical path length, yields ngas Interference! How does this disprove the ether?