Optical Wave Propagation: Reflection & Transmission in Multiple Mediums, Assignments of Electrical and Electronics Engineering

Download Optical Wave Propagation: Reflection & Transmission in Multiple Mediums and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! ECE398SB, Fall 2005 Issued: October 20th, Thursday Problem Set #2 Solutions Posted: November 6 1 Scattering from Homogenous Scatterers (15 points) a) We are given: nwater = 1.33 ntissue = 1.40 nnucleus = 1.50 Figure 1: tissue attenuationLet’s start by computing the coeffi- cients ∣∣∣Γt,r0,1∣∣∣2, ∣∣∣Γt,r1,2∣∣∣2, ∣∣∣Γt,r2,3∣∣∣2, ∣∣∣Γr3,4∣∣∣2. It is useful to point out that these reflec- tion and transmission coefficients have the property that:∣∣∣Γt,ri,j ∣∣∣ = ∣∣∣Γt,rj,i ∣∣∣ (1) In other words, the coefficients are the same whether we are traveling from medium i to medium j or the reverse.∣∣∣Γr0,1∣∣∣2 = ∣∣∣1−1.331+1.33 ∣∣∣2 = 0.020059∣∣∣Γr1,2∣∣∣2 = ∣∣∣1.4−1.331.4+1.33 ∣∣∣2 = 0.00065746∣∣∣Γr2,3∣∣∣2 = ∣∣∣1.4−1.51.4+1.5 ∣∣∣2 = 0.00065746∣∣∣Γr3,4∣∣∣2 = ∣∣∣1−1.51+1.5 ∣∣∣2 = 0.020059 The transmission coefficients may be Figure 2: the important reflectionscomputed using:∣∣∣Γti,j∣∣∣2 = 1− ∣∣∣Γri,j∣∣∣2 (2)∣∣∣Γt0,1∣∣∣2 = 1− ∣∣∣1−1.331+1.33 ∣∣∣2 = 0.97994∣∣∣Γt1,2∣∣∣2 = 1− ∣∣∣1.4−1.331.4+1.33 ∣∣∣2 = 0.99934∣∣∣Γt2,3∣∣∣2 = 1− ∣∣∣1.33−1.41.4+1.5 ∣∣∣2 = 0.99934∣∣∣Γt3,4∣∣∣2 = 1− ∣∣∣1−1.51+1.5 ∣∣∣2 = 0.97994 We now what is needed to compute the amplitudes of the pulses returned from the sample. The equations to compute these amplitudes are given below. Here I use ∣∣∣Γpij ∣∣∣2 to denote the reflection coef- ficient for path i and interface j. |Γp10 | 2 = ∣∣∣Γr0,1∣∣∣2 = 0.02005931 |Γp11 | 2 = [∣∣∣Γt0,1∣∣∣2]2 · ∣∣∣Γr1,2∣∣∣2 = 0.00063135 |Γp12 | 2 = [∣∣∣Γt0,1∣∣∣2]2 · [∣∣∣Γt1,2∣∣∣2]2 · ∣∣∣Γr2,3∣∣∣2 = 0.00063052 |Γp13 | 2 = [∣∣∣Γt0,1∣∣∣2]2 · [∣∣∣Γt1,2∣∣∣2]2 · [∣∣∣Γt2,3∣∣∣2]2 · ∣∣∣Γr3,4∣∣∣2 = 0.01922466 1 Now we may calculate the arrival times for each of these pulses using the fact that t = z·n c in each of the media. t0 = 0 t1 = 2 .01∗1.33 3E8 = 88.67 ps t2 = 2t1 + 22 .01·1.4 3E8 = 182 ps t3 = 2t1 + 2t2 + 2 .01·1.33 3E8 = 270.67 ps We may now take all of this informa- Figure 3: pulse train from path 1tion and write the following expression for the intensity as a function of time. Ij(t) = 3∑ i=0 ∣∣∣Γpji ∣∣∣2 · p(t− ti) (3) where p(t) = Ioe− ct 10−9 2 and the ti’s and the ∣∣∣Γpji ∣∣∣2’s are given above. To make this a little more intuitive, figure 3 shows a plot of this function for the first path I1(t). b) We are told in the problem that we are interested in times where we have perfect overlap of the pulses. Thus, we know that the integral of equation 3 in the homework is 2 √ Isig · Iref . The problem asks us to determine the intensity at the detector due to one of these pulses. I will choose pulse 2 for this solution. For the purposes of this homework, I will ignore the normalization factors for the pulses. If you would like to include them, they are shown just above equation 3 of the homework. Thus we have: Irec(td2) = Io + 0.00063135Io + 2 √ 0.00063135 · Io = 1.0509 · Io (4) Notice that without heterodyning we would get Irec(td2) = 1.0006 · Io. If we evaluate, 2 √ IrefIsig Isig = 79.597 (5) Thus, heterodyning provides an amplification effect. c) Since we know the spatial extent and indices of refraction of our sample, we may compute the path loss to arbitrary depth into the sample. Our image consists of the scattering intensities at different depths. However, as we saw in part a the scattering intensities at deeper depths are weighted by the path loss from the sample above. Fortunately, we know this path loss so we may simply divide through by that loss to recover the reflection coefficients of interfaces at arbitrary depths. The problem arises when we only have noisy estimates of the spatial extent and indices of refraction of the sample available. If our detector has a particular noise floor and the amplitude of the heterodyned reflection that we are interested in is near (or below) the noise floor of this detector our scheme of simply dividing through by the path loss will fail because will will not have a good estimate of the received pulse intensity. Also, we may not have a good estimates of the path losses, which further complicates matters. 2 Resolution and Magnification (5 points) a) Resolution refers to the ability to tell the difference between two closely seperated objects. Mag- nification refers to how large these objects appear in our field of view. Shown below are images 2 the problem since all possible behaviors of the system could be captured with one angle. For this solution, I will choose to set α = 0, which corresponds to fixing the polarization element in place and rotating the polarization of the beam. Setting θ = 0 and changing α would be easier to accomplish experimentally, but the point is that they are actually equivalent. The case of α = 0 is just simpler to analyze. When we make this substitution, we get: p(0, θ) = M(0) [ cos(θ) sin(θ) ] = [ 1 0 0 0 ] [ cos(θ) sin(θ) ] = [ cos(θ) 0 ] (17) According to Table 6.1-1, page 198, of Saleh & Teich this is a linearly polarized wave. So this element took a linearly polarized wave and turned it into a linearly polarized wave at θ = 0. If it looks like a linear polarizer and acts like a linear polarizer, it is a linear polarizer. c) The hint suggested that we cascade two of these elements together where one is at an angle of π 2 with respect to the other. When we do this we get: M(0)M ( π 2 ) = [ 0 0 0 0 ] (18) This gives us the well known result that two crossed polarizers will pass no light. 5 Optical Abberations (10 points) a) The artifact here is axial astigmatism. The orange color in the distorted image does not come from chromatic abberation. Chromatic abberation does not produce new colors; It just means that not all colors come to the same focus. A common sign of chromatic abberation are colored rings around the object under study. In this image, the orange color is present because the objective does not bring all of the colors to the same focus. Since the colors are being blurred the red and the green are being mixed to give orange. b) This image comes from the Hubble space telescope. For further details, see the web. Some of you already have -shall we say. This image has spherical abberation. c) DNA spherulties. The image of a point is focused at sequentially differing heights producing a series of asymmetrical spot shapes of increasing size that result in a comet-like shape. Comatic aberration. d) Crown Wart Fungus. Refractive index of every optical glass formulation varies with wave- length, and blue light is refracted to the greatest extent followed by green and red light. The inability of a lens to bring all the colors into a common focus results in colored fringes surround- ing the image. Chromatic aberration. e) Human Tongue. The intensity ratio of the unit image diminishes with detail and contrast being lost as the angle of the off-axis rays entering the lens increases. Astigmatism. f) Human Astrocytes. This is a mis-shaped image even though each image point is in sharp focus. There is also a difference between the transverse magnification of a lens and the off-axis image distance. Geometrical distortion (barrel) 6 Mie Scattering (10 points) a) Scattering and reduced scattering coefficient vs. Sphere Diameter. 5 b) Scattering and reduced scattering coefficient vs. wavelength c) Scattering and reduced scattering vs index of refraction of the scatterers. 6 d) Scattering and reduced scattering vs density of the scatterers. 7 Quantum Dots and Fluorescent Proteins (15 points) a) The Quantum Dots/nanoparticles will have proteins attached to them that cells recognize. Once the cells recognize these marker proteins, they have a set program for dealing with them. Proteins have specific functions and will be taken to the appropriate sites in the cell. Thus, choos- ing how to target your particles to the correct site in the cell comes down to choosing the right protein and letting the machinery of the cell do the work from there. b) One technique is to incorporate the DNA which codes for the fluorescent protein into the genome of the animal/plant/whatever that you wish to observe. The choice of how to target the proteins is a matter of determining where to insert the code for the protein in the genome of the subject. Certainly, there are other techniques, but I find this one particularly interesting. Speci- ficity is achieved because the fluorescent protein will only be expressed in protein in which we inserted the DNA for the fluorescent protein. c) The problem with using the technique discussed in b is that this may take a long time to grow a transgenic specimen. A more brute force approach to inserting the fluorescent dyes into a speci- men will result in very little specificity. It may also be possible to use the technique of part a with the fluorescent proteins/dyes. The downside to using the technique in part a is that not all of the particles will make it to the appropriate site due to the vagaries of cellular operation. d) Just use the same techniques, but use quantum dots and fluorescent proteins that operate at different wavelengths. e) If the contrast agents have the same excitation wavelengths, it will be necessary to incorporate spectral filters to seperate the signals coming from each of the constrast agents. Otherwise, it would be possible to image one agent at a time by exciting it alone and then creating a composite image. f) Once excited, fluorescent dyes/proteins may decay into toxic non-fluorescent substances. Once enough of these substances build up, the subject will die. Quantum dots have no such restrictions. g) Nanoparticles may constructed to contain a drug which will be released when triggered, per- haps by a pulse of light of the appropriate wavelength. Thus one may imagine that one constructs a molecular marker which targets the nanoparticle to a tumor. The patient is then imaged to make 7 The key to doing this problem in what I think is the ”right” way is to use equation 5.6-15 of Saleh & Teich, page 187. This equation and its important relations are shown below. τ(z) = τo [ 1 + ( z zo )2] 12 (28) zo = π · τ 2o −Dν (29) Dλ = − co λ2o ·Dν (30) If we plug what we know into equation 28, we get: EBF ·∆t = 20× 10−15 = 10−14 1 + (10−3 zo )2 12 (31) If we now solve for zo: zo = 10−3√ 3 = 0.00057735 (32) Using this fact, we may now calculate the dispersion coefficient: Dλ = co λ2o π · τ 2o zo = 3× 10−8 (807× 10−9)2 π · (10−14)2 10−3 = 1.4472× 10−20 (33) We now have the key fact for which we were looking, Dλ. Now we may use this to compute a new zo; then plug this into equation 28 to obtain our final answer. Remember that for this problem, we want to use Dν . So . . . zo = − π · (31.931× 10−15)2 3.1416× 10−25 = 0.010196 [meters] (34) Plug this into equation 28 and do a little computation and we are done. ∆tdispersed = 31.931× 10−15 1 + (2× 10−3 0.010196 = )2 12 = 45.686 [femto− seconds] (35) This answer is satisfactory in the sense that it is longer than the original pulse, which is good. It is also satisfatory since, if you add the ”wrong” answer to the transform limited pulse width you will get approximately this answer. Thus, I think, that equation 20 makes the approximation that the original pulse is so dispersed that its duration depends mostly upon how dispersed it is and not much upon the original pulse width. 10 Photon Counting Detectors (10 points) a) A good diagram of a Photo-Multiplier Tube is shown in Figure 17.0-2b on page 647 of Saleh & Teich. A PMT makes use of the photoelectric effect to detect photons. When a photon strikes the photoelectric material an electron is emitted into a vacuum tube. Inside this tube a series of plates is kept at a series of decreasingly high potentials. When the electron strikes on of these plates (dynodes) it may have enough energy to liberate further electrons from the dynodes. In this way, we get an amplification effect. A PMT is capable of detecting single photons given a high enough potential between the cathode and anode. b) PMT’s are used when no other detector will work; when there are very very few photons to 10 detect. We would NOT want to use a PMT if higher signal levels are present since we could use other schemes such as heterodyning which allows us to determine the phase of the pulse. Also, the PMT has no wavelength selectivity and it is easily damaged or destroyed at higher power levels. c) An APD is simply the semiconductor version of the PMT. Instead of metal plates a high reverse bias is put across a specially doped piece of semiconductor. Since we are using semiconductor, there is no need to have a special photoemissive material. Once an electron is excited to the conduction band it is swept from the junction by the high reverse bias, possibly creating new electron-hole pairs through the same process that occurs in avalanche breakdown, impact ioniza- tion; hence, the name APD. A good diagram of an APD is shown on page 666 (. . . oh oh!) of Saleh & Teich. d) There is a tradeoff in terms of imaging speed versus noise floor. The APD is more efficient, but also much more noisy. The PMT is not as efficient, but has much lower noise. e) Included is a schematic diagram of a multi-photon microscope. Good references are: http://micro.magnet.fsu.edu/primer/techniques/fluorescence/anatomy/fluoromicroanatomy. html http://parkerlab.bio.uci.edu/publication%20attachments/dynamic%20Multiphoton. pdf The schematic is taken from the second reference. Figure 12: Multi-Photon Microscope ‘ 11 Code for Problem #1a %---------------------------------------------------------------------------- % This script will compute the relavent parameters for Problem #1a of HW#2 %---------------------------------------------------------------------------- clear close all %---------------------------------------------------------------------------- % Parameters of the script %---------------------------------------------------------------------------- n_air=1; n_water=1.33; n_tissue=1.4; n_nucleus=1.5; c=3E8; % the speed of light l=.01; % the length of the homogenous medium tn=1E-12; % normalize the times by this factor %------------------------- n0=n_air; n1=n_water; n2=n_tissue; 12