HWsol10_v2 - Physics 2214 Fall 2008 - Solutions to Problem...

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Physics 2214 Fall 2008 - Solutions to Problem Set #10 Author: Itay Nachshon 1. YF 35.35: The two re ected beams have to interfere such that they cancel each other. Any part of the path which is similar won't change the relative phase between the two beams. This includes re ecting from the plastic, traveling through the re ective coating above the pit, and transmitting out of the coating. For example, the beams may have a phase change of - 1 from the re ection of the plastic, but since both have this change, this will not e ect the end result. The part of the beam traveling into the pit travels 2 times the depth d , so the di erence between the beam's is a phase e ik 2 d , where k is the wave number in the re ected coating. (One can also keep k 0 , the wave number in vacuum and use the OPL instead of d . Both choices will give the right answer) Using the principle of superposition: ± E out = ± E out of pit + ± E in pit (1) The two parts of the eld don't necessarily have the same magnitude, but lets assume they do for simplicity (otherwise we would not have complete destructive interference). We get ± E out = ± E out of pit 1 + e ik 2 d · = ± E out of pit 1 + e ink 0 2 d · (2) Here we are also assuming that all other phase changes occur for both beams. For example, re ection would add a phase of - 1 to both beam and in this equation it is hidden in ± E out of pit . If one wants complete destructive interference then ± E out =0 e ink 0 2 d = - 1 (3) which means that 2 nk 0 d = π (1 + 2 n ) (4) The minimal d will be for n = 0 , 2 nk 0 d = π d = 1 4 2 π nk 0 = 1 4 λ 0 n =109 nm (5) 2. YF 35.56 (a) Lets think through it carefully before you plunge into the icy waters of algebra: Notice that the optical path length is THE SAME in all of the layers (since n G d G = n C d C ). There's a π phase shift on re ection at only every second layer boundary. The re ection coe cients that re ects R 2 , R 4 , . . . are the same and positive (as we are going from n = 1 . 80 to n = 1 . 33 , or from a medium of smaller light speed to one with greater speed). The re ection coe cients that re ects R 1 , R 3 , R 5 , . . . are the same and negative (as they are for a medium change from n = 1 . 33 to n = 1 . 8 ). 1
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As the light travels through the Guanine it picks up a phase e ik 0 n G d G and as it travels through the Cytoplasm it picks up a phase e ik 0 n C d C . As the transmission coe cient is always positive it does not change the phase of the re ected light. So we would like R 1 and R 2 to have the same phase. If R 1 picks a phase of - 1 then R 2 should also pick a phase like that. Then because the OPL's are the same R 3 will also pick an additional phase of - 1 from traveling in the the Cytoplasm × - 1 from re ection, so it will be in the same phase as
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This note was uploaded on 12/24/2008 for the course PHYS 2214 taught by Professor Giambattista,a during the Fall '07 term at Cornell University (Engineering School).

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HWsol10_v2 - Physics 2214 Fall 2008 - Solutions to Problem...

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