1. Optical path length opl = path length x η Δopl = Δ path length x Δη δ (phase shift) = 2π x Δopl / λ 0o0o 90 o 270 o 360 o 180 o How far out of phase have these 2 waves become? 0o0o 90 o 270 o 360 o 180 o 0π.5 π 1π 1.5π 2π degs. rads.

2. http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

3. One full wavelength λ, one full cycle, 360 degrees or 2 π Relative phase difference of 1/4 λ or 1/2 π. resultant energy in space Combine black ray or wave with grey one and derive resultant (orange). All are same wavelength, only relative phase and amplitude changes. Remember, this energy is not destroyed, only redistributed to another place in space (where is not always obvious). Relative phase difference of 1/2 λ of π. In a phase contrast microsccope, a ring at the back plane of the objective advances undiffracted rays (those that do not interact with the specimen) by one quarter wavelength. This shift combined with a ~one quarter λ retardation at the sample leads to 1/2 λ difference or destructive interference at the image plane; differences in phase shifting properties of sample can become amplitude differences.

4. One full wavelength λ, one full cycle, 360 degrees or 2 π resultant energy in space Relative phase difference of 1 λ or 2π.

5. polymer, η =2, thickness = 312.5 nm glass, η =1.5 1 2t 3t 3r 2r 4 resultant reflection resultant transmitted Solve for following wavelengths: Wavelengths of light maximally reflected Wavelengths of light maximally transmitted Wavelengths of light minimally reflected Wavelengths of light minimally transmitted What thickness of polymer would give maximum reflection at 580nm? See text on next slide.

6. t-transmitted, r-reflected 1 is a ray of white light When 1 hits polymer surface, it is split into 2t and 2r, 2r is phase shifted relative to 2t by 1/2 λ or π (low to high η interface). 2t continues on and is split into 3t and 3r at the glass surface, there is no phase shift with this reflection because it is a high to low η interface. 3r (in phase with 2t) will combine and interfere with 2r ( 3r + 2r = resultant, constructive + or destructive - depends on the λ). Since 2r has already shifted relative to 2t by 1/2 λ, for + interference to occur, 3r must be shifted in phase by 1/2 λ, 1 1/2 λ, 2 1/2 λ, etc. For - interference to occur, 3r must be shifted in phase by 0 λ, 1 λ, 2 λ, etc. So to solve this for maximum constructive reflective interference, we take: OPL (2t + 3r) must equal 1/2 λ or 1 1/2 λ or 2 1/2 λ OPL (2t + 3r) = 2(thickness of polymer x η polymer) so.... 1/2 λ or 1 1/2 λ or 2 1/2 λ = 2(thickness of polymer x η polymer) So to solve for this in maximum destructive reflective interference, we take: OPL (2t + 3r) must equal 0 λ or 1 λ or 2 λ OPL (2t + 3r) = 2(thickness of polymer x η polymer) so.... 0 λ or 1 λ or 2 λ or 3λ = 2(thickness of polymer x η polymer) Now try one in transmission: 2t is split into 3t and 3r; at this point, all these rays are in phase (no shift high to low η). 3r reflects back as 4 within polymer, its phase then changes relative to 2t. 3t + 4 = resultant, constructive + or destructive - depends on the λ. So to solve this for max. constructive transmitted interference, we take: OPL (3r + 4) must equal 0 λ or 1 λ or 2 λ OPL (3r + 4) = 2(thickness of polymer x η polymer) so.... 0 λ or 1 λ or 2 λ or 3 λ = 2(thickness of polymer x η polymer) These integer values (1, 2, 3, etc.) for wavelengths are called 'orders'. As we can see, we lose energy with each reflection so that the most important orders (with regard to amplitude) are usually the lowest order effects.

7. η - hi for phase shifts on reflection: low to high, shift in π high to low, no shift in π air = white glass = blue light rays = black semi-silvered surfaces = grey resultant waves/rays = orange η - hi η - air single semi-silvered surfaces reflect 50% and transmit 50% of light (all λ ) This diagram may help you solve the extra credit question on the midterm exam.

8. They won’t let you use their calculator for normal (90 degree) incidence so to try their calculator use a 30 degree angle (answers are close to those for 90). http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html Murphy 2001 WOW!

9. www.jasondoucette.com/ graphics/interfrs.gif resources.yesican-science.ca Note that energy is neither destroyed nor created during interference. out of phase by π,3π,5π, etc.

10. Phase change of 180 deg (pi) when reflection is rare (low refractive index) to dense (high refractive index), no phase change when reflection is dense to rare as in TIR or total internal reflection. Laser design inc.

11. http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html Georgia State U. AR coating Oil on water