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Reading Quiz In the text’s treatment of the double boundary problem, in the middle region all of the reflections that end up traveling to the right are.

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Presentation on theme: "Reading Quiz In the text’s treatment of the double boundary problem, in the middle region all of the reflections that end up traveling to the right are."— Presentation transcript:

1 Reading Quiz In the text’s treatment of the double boundary problem, in the middle region all of the reflections that end up traveling to the right are ___________. lumped into one field handled with an infinite series

2 Reading Quiz In the text’s treatment of the double boundary problem, the electric field at the right side of the middle region is related to the electric field at the left side of the middle region via: the Fabry-Perot theorem the Fresnel coefficients a phase factor a polarization transformation

3 Total internal reflection
Derivation from Snell’s law: Above this angle, A

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5 Simple way to think about complex functions
(probably doesn’t help much in using them) A In TIR the wave decays not from ik, but from ik Photon picture of TIR

6

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8 Thin film constr.,destr. interference
Phys 123 Thin film constr.,destr. interference A thin film floats between two liquids. In this problem how many 180o reflection phase shifts are there? (Normal incidence) A. from top B. from bottom C. from both D. from neither n=2.2 n=1.8 n=1.5

9 Thin film constr.,destr. interference
Phys 123 Thin film constr.,destr. interference Thickness for greatest reflection if l = 300 nm? n=2.2 n=1.8 n=1.5 Principles Constructive int. Reflection phase shifts When I say “l = 300 nm” it means l in vacuum! Pathlength phase shifts “Optical thickness” is n times thickness

10 Find the smallest thickness for greatest reflection if l = 300 nm (vacuum), at normal incidence.

11 Why does top of soap film become black in R?
So thin at top that only reflection phase shift matters

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13 A common cheap anti-reflection coating on glass is to put a (lower index) layer thickness of
l /(4n1 ) l /2 l /(2n1 ) n=1 n2 =1.8 n1 =1.5

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15 An anti-reflection coated window, shown at a 45° and a 0° angle of incidence.

16 More complicated Effect of significant incident angle:
To get high reflection, will I need a thicker or thinner film (vs normal incidence)? Non-intuitive principle: only the part of k perpendicular to the interface causes phase shifts. Why: Snell’s law already enforces phase matching in parallel direction. Need thicker d at higher angles. Comprehensive theory will give us all this.

17 Double boundary r,t

18 Double boundary r,t Key ideas:
How do we handle the multiple reflections in the thin film? How do we join the fields across each interface? How do we introduce phase shifts due to thickness of thin film? lumped into one left and one right-going waves A Fresnel coefficients Use only the phase shifts in the z direction:

19 A

20 Where do n’s enter into the above?
Where do phase shifts due to reflections enter?

21 http://www.youtube.com/watch?v=tu3T8f-8Fgw Start at 3:26 .
Soap film vs time

22 above: http://media.efluids.com/galleries/interfacial?medium=518
video of turbulence


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