Announcements 11/12/12 Prayer Labs 8 & 9 due Saturday Progress Reports – replies sent In the Bleachers Review

From warmup Extra time on? a. a.How do beam splitters work? Other comments? a. a.Is there a demo on Newton's rings? b. 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? (this is the y-coordinate on the slits, not the y-coordinate on the screen) compare to:

Adding up phases … 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.) slits screen 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 or , we need to add up a different set of phases… perhaps like 0 , 21 , 42 , 63 , etc.

Adding up phases, cont. … Quick writing: graphically add these three vectors: 1  0  + 1  20  + 1  40  What about 1  0  + 1  90  + 1  180  slits screen

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. a.1 b. b.2 c. c.3 d. d.4 e. e.5

Five Slit Problem: Scanning Theta Note: for some reason he picked the overall reference phase to be about 

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. a.0  b. b.45  c. c.90  d. d.180  e. e.depends on whether it is s- or p-polarization

From warmup In Fig (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. a.I don't have the 8th edition b. 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. 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) For arbitrary angle Same as strings The Truth (overlooked by textbook): you don’t always get a phase shift, even if going fast to slow. (Brewster marks boundary) More Truth: sometimes phase shifts not just 180  : can have complex n, complex , etc. You don’t need this much truth!

Air to glass (n=1 to n=1.5) p-polarization r t field amplitudes vs  Brewster 180  phase shift (close to perpendicular) 0  phase shift (close to glancing)

Back to “near normal incidence” Rays drawn at an angle to make viewing easier. They’re really perpendicular to surface. 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? air thin glass thickness t If rays at an angle… determine if above/ below Brewster angle (if p-polarization).

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, 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. 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 Rays drawn at an angle to make viewing easier. They’re really perpendicular to surface. What does the thickness of the COATING need to be to get constructive interference between the two rays? air thick glass, n = 1.5 thin coating, n = 1.3 thickness t

Pretty pictures What’s going on here? 03/15/bubbles-and-interference/ com/pp_optics.html

Demo Demo: Soap film

Interferometer From lab 9: Interference! How does this disprove the ether? changing optical path length, yields n gas