Announcements 11/12/12 Prayer Labs 8 & 9 due Saturday Progress Reports – replies sent In the Bleachers Review phaseshift 2 (PL ) Approx.1: PL d sin E stuff ei 2 ei E stuff cos( 2) I I0 cos2 ( 2) Approx.2: sin y L 2 From warmup Extra time on? a. How do beam splitters work? Other comments? a. Is there a demo on Newton's rings? b. I worked on a research group at my previous university. The leading professor in our group is/was a researcher at VIRGO and and LIGO. For LIGO (in the US), he said they could detect vibrations in the earth's crust that varied due to the rush hour(s) in nearby large cities. I thought that was pretty cool. Fourier Transforms How can our two-slit analysis possibly have anything to do with Fourier transforms? E stuff ei 2 ei e E~ e E~ 2 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 2! (Could have an overall reference phase for 1…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 Clicker 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 Clicker question: When a light wave travels from a low index to a high index material at normal incidence (perpendicular to surface), what is the phase shift of the reflected wave? a. 0 b. 45 c. 90 d. 180 e. depends on whether it is s- or p-polarization From warmup In Fig. 37.9 (8th edition) the textbook compares the reflection of light from a surface to the reflection of a wave on a rope. How does the analogy work? a. I don't have the 8th edition b. Light reflecting off a higher n substance is like a wave reflecting off a rigid support, and reflecting off a lower n substance is like reflecting off a free support. The n value correlates to how easily the light can move through the substance, which is similar to a rope being free vs rigid. c. Colton: CAUTION! this analogy only works for close to normal incidence, i.e. 0 deg. (this fact not mentioned in book). Remember these? “Fresnel Equations” 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 Same as strings 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 Air to glass (n=1 to n=1.5) p-polarization field amplitudes vs 1.0 t 0.8 0.6 0.4 r 0.2 20 40 60 80 0.2 Brewster 180 phase shift (close to perpendicular) 0 phase shift (close to glancing) Back to “near normal incidence” From low to high index: 180 phase shift From high to low index: no phase shift Quick writing: What does the thickness of this slab need to be to get constructive interference between the two rays? If rays at an angle… air thin glass air determine if above/ Rays drawn at an angle to make below Brewster angle viewing easier. They’re really perpendicular to surface. (if p-polarization). thickness t From warmup The concept of "optical path length" (OPL) is used in many places to analyze optical problems. (Unfortunately your textbook doesn't use it.) See for example Wikipedia, http://en.wikipedia.org/wiki/Optical_path_length. In materials with a constant index of refraction n, the OPL is just the path length times n. In the context of today's reading, why might that be a helpful concept? a. I think it would help when we analyze reflections off of thin films because the destructive interference depends on the index of refraction of the film times the thickness times two. That would be the optical path length. 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?