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kasap solutions manualChapter 1. Solutions Manual to. Principles of Electronic Materials and Devices. Fourth EditionUniversity of Saskatchewan. Canada. Check author's website for updatesPlease check the above website. Report. Water molecules are polar. A water jet can be bent by bringing a charged comb near the jet. The polarAll rights reserved. No reproduction or distribution without the prior written consent of. McGraw-Hill Education.Chapter 1. Page 31. Oxygen has an atomic mass of 16 whereas it is 14 for nitrogen. The O2 molecule is therefore heavier thanN2 molecules.Page 34, footnote 11. For small extensions, the difference between the engineering and instantaneous strains due to aHistorically, mechanical and civil engineers measured extension by. It is not trivial toHowever, since we know Lo and. Consider a sample of lengthThe engineering definition of strain and hence the thermal expansion coefficient isL Lo T. LoNo reproduction or distribution without the prior written consent of. McGraw-Hill Education.Chapter 1Take the PE and the KE as zero when the electrons and the nucleus are infinitelyNow consider the He-atom shown in Figure 1.75b. There are two electrons. Each electron interacts withUsing the Virial theorem show thatEnergy of He atom. The first ionization energy EI1 is defined as the energy required to remove one electron from the HeNote that the first ionization energy providesThere is one electron in the K-shell orbiting the nucleus that has a. Due to their mutual repulsion, they orbit toSolution. Virial theorem relates the overall energy, the average kinetic energy KE, and average potential energy. PE through the relationsAll rights reserved. McGraw-Hill Education.Chapter 1Now consider Figure 1.75b. Assume that, at all times, the electrons avoid each other by staying inThe PE of this systemFrom the Virial theorem in Equation (1)The eV values were obtained by usingWe can now calculate the radii as follows.http://www.FlashPointIP.com/fckupload/kohler-marine-generator-5ecd-service-manual.xml
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The calculation of r1 involves realizing that Equation (3) is the energy of the whole He atom, with 2All rights reserved. McGraw-Hill Education.Chapter 1The mass of theSimilarly weight fraction of Se isConsider 100 g of CdSe. Then the mass of Cd we need isM Se M Te M P 78. 6 g molM Se M Te M P 78.6 g mol. All rights reserved. McGraw-Hill Education.Chapter 1M Se M Te M P 78. 6 g molThese can be related in a simple way to the atomicSometimes we need to know the number ofConsider a crystal of the material of interest which is a cube of side L as shown in Figure 1.76. To eachSuppose that there are N atoms in the volume L3.If nb is the bulk concentration of atoms, show that the mean separation a between the atoms isA where Mat is the atomic mass.Show that the density of the solid is given by. Calculate the atomic concentration in Si from its density (2.33 g cm )A silicon crystal has been doped with phosphorus. P atoms substitute for Si atoms and are randomly distributed in the crystal. What is the mean separationFigure 1.76 Consider a crystal that has volume L3. This volume is proportioned to each atom, which is a cube ofSolutionFor simplicity, we. M atWe can use Equation (2)Using the Virial Theorem as in. Example 1.1, consider the following:Calculate the total electrostatic potential energy (PE) of all the charges when they are arranged asUsing the Virial theorem, find the change in the totalHow does this compare with the experimental value ofFigure 1.77 A simplified view of the covalent bond in H2. A snapshot at one instant. SolutionIn evaluating the PE of allPE is the sum of the following. Electron 1 interacting with the proton at a distance ro on the left, with the proton at ro on the rightAll rights reserved. McGraw-Hill Education.Chapter 1The negative PE for this particular arrangement indicates that this arrangement of charges is indeedWe can write the last expression above as the change in the total energy.http://www.hotelsinroorkee.com/userfiles/kohler-marine-generator-5e-manual.xmlThe H2 molecule has lower energy than two H-atomsThis is very close to the experimental value of 4.51 eV. (Note. We used a ro value from quantum mechanics - so the calculation was not totally classical!).Energy per ion pair in ionic crystalsSolutionAll rights reserved. McGraw-Hill Education.Chapter 1Richard Christman. Matter (2nd Edition), Clarendon (Oxford University) Press, 1983 (pp. 258-259).Edition, Academic Press, 1984) p. 413, Table 13.5.Further, mostSome authors referCalculate the bonding energy per ion pair in the LiCl crystal. (c) Calculate the atomic cohesive energy ofSolution. Figure 1Q07-1 shows the crystal structure of NaCl. LiCl has the same crystal structure.All rights reserved. McGraw-Hill Education.Chapter 1The corresponding ionic cohesive energy is. Consider the electron affinity of Cl and the ionization energy of Li, then by definition, we have. McGraw-Hill Education.Ionic cohesive. Atomic cohesiveKittel, 7Ed, p73, Table 7Table 20.5, p406T. Moeller et al, Chemistry with Inorganic Qualitative Analysis, Second Edition, Academic Press, 1984) p. 413. Table 13.5. N.W. Ashcroft and N.D. Mermin, Solid State Physics, Saunders College, 1976; Table 20.5 (p406) data, originallyMadelung constant M for NaClSolution. From Coulomb’s law of electrostatic attraction we know that the PE between two charges Q1 and Q2All rights reserved. McGraw-Hill Education.Chapter 1Therefore, the overall PE of theNote: See Appendix A for the explanation on how the PE of ions in the crystal is usually written.General PE curve for bondingThe magnitude of the minimumBulk modulus definition. The bulk modulus K is related to the energy curve. In its simplest form (assuming a simple cubicAll rights reserved. McGraw-Hill Education.Chapter 1A AnAnFrom the definition of Bulk modulus mentioned in the problem statement aboveAnr oAgain, substituting the value of B in the above relation, i.e.All rights reserved. McGraw-Hill Education.https://skazkina.com/ru/boss-me-10-manual-pdf Related documents MAT105(SWE) - Lecture notes 1-7 Strengthofmaterials 4theditionsolutionsmanual singerpytel 2 130904185510 Fluid Mechanics With Engineering Applications, 10th Edition MAT105(SWE)Limit - Note Integer Programming Morris Mano Book- Digital logic Design Preview text Warning: TT: undefined function: 32. No reproduction or distribution without the prior written consent Please check the above website. ReportThis gauge has a maximum strain range ofThe whole gauge is fully encapsulatedThe external solder pads are copper coated.All rights reserved. No reproduction or distribution without the prior written consent The x-component is unchanged. The collisionThe added impurities (Cu) in Al provide an additional scatteringIf each Na atom donates one electron toWhat is the approximateDo you expect the kinetic molecularAll rights reserved. No reproduction or distribution without the prior written consent The Virial TheoremFurther, kinetic theory predictsThe interaction PE is substantial compared with the meanAll rights reserved. No reproduction or distribution without the prior written consent What should be the thickness ofWe have the. We can either. The two given referenceTherefore. No reproduction or distribution without the prior written consent All rights reserved. No reproduction or distribution without the prior written consent How does thisWhat is your conclusion?All rights reserved. No reproduction or distribution without the prior written consent On a log-log. The best power law fit generates All rights reserved. No reproduction or distribution without the prior written consent The best line is a power law fit on a log-logOver the compositionNo reproduction or distribution without the prior written consent In order to keep theThe weight fraction for Cu for this atomic fraction can be calculated from Cu XM X M The weight fraction for Cu for this atomic fraction can be calculatedAll rights reserved.http://www.btrcontrols.com/images/a-manual-of-acupuncture-ebook.pdf No reproduction or distribution without the prior written consent AuTheir properties arePlot these properties all in graph (using a log-scale for theCompositions are wt. . Assume the Cu-Sn is a solid solution over this composition range. Sn No reproduction or distribution without the prior written consent All rights reserved. No reproduction or distribution without the prior written consent. From the slope of the best-fit line find the mean (effective) NordheimNordheim equation. An intuitive correction is then done by increasing the conduction electronThere is, however, a modern physics explanation that involvesAll rights reserved. No reproduction or distribution without the prior written consent ResistivityNo reproduction or distribution without the prior written consent It is observed that the valuesIn general, the resistivity due to the introduction of soluteWe can now plot Principles of Electronic Materials and Devices. Connect with us to Instant Access ISBNs are for individuals purchasing with credit cards or PayPal. Savvas Learning Company is a trademark of Savvas Learning Company LLC.Please try again. Chapter 2 If each Na atom donates one electron What is the approximate Do you expect the kinetic molecular Why should the mean kinetic energy be comparable to the mean If d is the mean separation be. To browse Academia.edu and the wider internet faster and more securely, please take a few seconds to upgrade your browser. You can download the paper by clicking the button above. And by having access to our ebooks online or by storing it on your computer, you have convenient answers with Optoelectronics And Photonics Solutions Manual Kasap. To get started finding Optoelectronics And Photonics Solutions Manual Kasap, you are right to find our website which has a comprehensive collection of manuals listed. Our library is the biggest of these that have literally hundreds of thousands of different products represented. I get my most wanted eBook Many thanks If there is a survey it only takes 5 minutes, try any survey which works for you. And by having access to our ebooks online or by storing it on your computer, you have convenient answers with Optoelectronics And Photonics Kasap Solution Manual. To get started finding Optoelectronics And Photonics Kasap Solution Manual, you are right to find our website which has a comprehensive collection of manuals listed. And by having access to our ebooks online or by storing it on your computer, you have convenient answers with Kasap Optoelectronics And Photonics Solution Manual Pdf. To get started finding Kasap Optoelectronics And Photonics Solution Manual Pdf, you are right to find our website which has a comprehensive collection of manuals listed. Our system will send you a confirmation message that contains the download-able links. Download links do not expire and you can re-download your files anytime. We strongly recommend keeping a copy of the email message you receive. And it is not an issue of what device or operating system you have got. Read all your test banks and solutions manuals on any computer, tablet, or smartphone. It holds plenty of questions with their appropriate answers related to an academic textbook. Test banks normally contain true-false questions, multiple-choice questions, and writing questions. 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See our User Agreement and Privacy Policy.If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.If you wish to opt out, please close your SlideShare account. Learn more. You can change your ad preferences anytime. Second Edition. Safa Kasap. Revised: 11 December 2012. Check author's website for updatesPlease check the above website. ReportChapter 2 2.2Chapter 2. Note: Printing errors and corrections are indicated in dark red. Currently none reported.E(y, z,t) 2Eo cos(m y 2The planar waveguide is symmetric, which means that the intensity, E2Choose suitable values and plot the relativeSolution. E(y) Eo cos(t m z m z ) Eo cos(t m z m z). E(y, z,t) 2Eo cos(m y ? 1The first cosine term represents the field distribution along y and the second term is the propagation ofThus, the amplitude is. The intensity is maximum or minimum at the center.Chapter 2 2.3. Figure 2Q1-1 Amplitude of the electric field across the planar dielectric waveguide.Allowed upward and downward traveling waves inside the core of the planarPut differently, a wave startingFigure 2.51 Upward and downward traveling waves along y set-up a standing wave. The condition for setting-up aSolution. From Figure 2.51 it can be seen that the optical path is. AB BA 4a. With the ray under going a phase change with each reflection the total phase change isChapter 2 2.4. The wave will replicate itself, is the phase is same after the one round-trip, thusHence find the amplitude of the field variation along y, across the guide. What is your conclusion? Figure 2.52 Rays 1 and 2 are initially in phase as they belong to the same wavefront. Ray 1 experiences total internalSolutionThe phase difference between the waves meeting at C isChapter 2 2.5Then, m 2k1(a y)cosm m 2k1(a y)Thus, the amplitude Eo is. E 2AcosSee Figure 2Q1-1.Plot the fieldChapter 2 2.6Ray 1 experiences total internalSolution. The two waves interfering at C are out phase by,E 2AcosThus, the amplitude Eo is. E 2AcosExample 2.1.1 in the textbook, the waveguide condition isChapter 2 2.7The above two equations can be solved graphically as in Example 2.1.1 to find m for each choice of m. Alternatively one can use a computer program for finding the roots of a function. The above equationsThere is no significant difference between the TE and TM modes (the reason is that n1 and n2 are veryFigure 2Q4-1 Field distribution across the core of a planar dielectric waveguide. E 2cosThis is shown in Figure 2Q4-1.Chapter 2 2.8Solution. The waveguide condition isThe above two equations can be solved graphically as in Example 2.1.1 to find m for each choice of m. The above equationsChapter 2 2.9Suppose that is the delay time between the TE and TM waves over a length L. Then,It is assumed that the math-software package can carryBoth and are now a function of in Eqs (1) and (2). Then the group velocity is found byChapter 2 2.10. Solution. The results shown in Figure 2.11, and Figure 2Q6-1 were generated by the author using LiveMath basedThe important conclusionThus, it is only approximate. Figure 2Q6-1 Group velocity vs.The refractive index of GaAs is 3.66 and that of the. AlGaAs layers is 3.40. What is the cut-off wavelength beyond which only a single mode can propagateChapter 2 2.11What is the mode field widthSolutionChapter 2 2.12Propagation constants () at other wavelengths and hence frequencies () can be similarly calculated. The results are listed in Table 2Q8-1 and plotted in Figure 2Q8-1. This is the dispersion diagram. ForThey are drawn so that the. Thus, the solutions of the waveguide condition as in Example 2.1.1 generates the data in Table 2Q8-1Chapter 2 2.13Author's Note: Remember that the slope at a particular frequency is the group velocity at thatCompare your MFW calculation. SolutionChapter 2 2.14Solve the waveguide conditionThen useWe can also calculate MFW fromAuthor's Note: Consider a more extreme caseChapter 2 2.15Calculate the maximum acceptance angle. (e) Calculate the modal dispersion and hence the bit rateSolutionNumber of modes M,The numerical aperture NA is. If max is the maximum acceptance angle, then,Modal dispersion is given byL c 3108Given that 0.29, maximum bit-rate isWe neglected material dispersion at this wavelength which would further decrease BL. MaterialChapter 2 2.16. For example, assuming an LED with a spectral rms deviation of about 20 nm, and a Dm 200 ps km-1The refractive index of water is 1.330. Consider a water jet of diameter 3 mm that is illuminated by green light of wavelength 560 nm. What isHow many modes are there? WhatThe diameter of the jet increases (slowly) as the jet flows away from theLight guided along a thin water jet. A small hole is made in a plastic soda drinkWhen the hole is illuminated with aWater with air bubbles (produced by shakingAir bubbles scatter light andFirst such demonstration has been attributed to Jean-. Daniel Colladon, a Swiss scientist, who demonstrated a water jet guiding light inV-numberNumerical apertureThe large difference in refractive indices between the water and the air ensures that total internalChapter 2 2.17Is this a single mode fiber? (b) Calculate the wavelength below which theSolutionChapter 2 2.18Is this a single mode fiber? (b) Calculate theCalculate the maximum total acceptance angle. (e) Calculate the material, waveguide and chromaticSolutionV-number is given byChapter 2 2.19The first equation can beSolutionTaylor expansion around x 0 and truncating the expression, keeping only the linear term yields,Chapter 2 2.20Change the operating wavelength to by a small amount, sayThen determine the group velocity vg ofHow do your resultsSolution. From example 2.3.4, we haveGroup VelocityComparing to Example 2.3.4Chapter 2 2.21Solution. The Sellmeier equation isSellmeier A1 A2 A3 1SiO2 (fused silica) 0.696749 0.408218 0.890815 0.0690660 0.115662 9.900559Apply NA n2The refractive index n(x) of SiO2-x mol.GeO2, assuming a linear relationship, can be written asSince Ng will depend on the wavelength, show that the material dispersion coefficient Dm is givenDmLd? c d2. Using the Sellmeier equation and the constants in Table 1.2 in Ch. 1, evaluate the material dispersion. From Ch. 1 we know thatChapter 2 2.22Differentiate with respect to wavelength using the above relationship between Ng and n. L LNg1Thus, Dm. Ld?The 1, 2, 3 are in m.Figure 2Q17-1 Materials dispersion Dm vs.Matlab can also be used.)Chapter 2 2.23It is present even when theLet us suppose that n1 and n2 are wavelength (or k)The group velocity is defined and given byShow that the propagation time, or the group delay time, of the mode is. L Ln2 Ln2 d(kb)Given the definition of V,Show thatV (6)w. Ld? c dV 2Chapter 2 2.24Waveguide dispersion coefficient (9)Suppose that a 1.3 m laser diode with a spectralEstimate the waveguide dispersion perFigure 2.53 d2Electronics Engineer, 51, 313, 1981.). Waveguide dispersion arises as a result of the dependence of the propagation constant on the V-numberIt is present even when the refractive index is constant; no materialSuppose that is the. Then the normalizedThe group velocity is defined and given byThus, the propagation time of the mode isChapter 2 2.25Given the definition of V,From Eq. (5),This means that depends on V as. Ln2 Ln2 d(Vb)Dispersion, that is, spread in due to a spread can be found by differentiating Eq. (6) to obtain,The waveguide dispersion coefficient is defined asV (8)w. Ld? c dV 2. In the range 2 Chapter 2 2.26Consider a fiber with a core of diameter of 8 m and refractive index of 1.468 and a claddingSuppose that a1.3 m laser diode with aEstimate the waveguide dispersionFrom the graph, Vd2Dw 4.610-6Using Eq. (10). Dw ?Dw 4.610-6However, there is an additional dispersion mechanismConsider a light source with a range of wavelengthsWe can view this as a change in the input wavelength. Suppose that n1, nThe propagation time, or the group delay time, g per unit length is. Since depends on n1, and V,Chapter 2 2.27The mathematics turns out to be complicated but the statement in Eq. (2) is equivalent toEven the statement in Eq.The total intramodeSolution. Total dispersion in a single mode step index fiber is primarily due to material dispersion and waveguideSuppose that n1, n 2, hence depends on the wavelength. The propagation time, or the group delay time, g per unit length isSince depends on n1, and V, let us consider g as a function of n1, (thus n2) and V. A change?Using the partial differential chain rule,Chapter 2 2.28The mathematics turns out to be complicated but the statement in Eq. (2) is equivalent toEven the above statement in Eq.The total intramodeVb d. D ?g1Waveguide dispersion. Dw ?Dw 5.6 ps km-1Profile dispersion is more than 10 times smaller than waveguide dispersion.Chapter 2 2.29Consider theSolution. Consider the Taylor expansion for, a function of wavelength, about its center around, say at 0, when weThen, Taylor's expansion gives,L 2 1 km ? -2 -1 2What would be theSolution. DispersionBit rateOptical bandwidthChapter 2 2.30What should be the linewidth of the laser source so that over 100 km,SolutionThe rms dispersion isAssume that the input light spectral width is 1 nm. SolutionSolution. For the single mode fiber, the small core diameter is to ensure that the V-number is below the cutoffChapter 2 2.31For multimode fibers, the larger core size allows multiple modes to propagate in the fiber and thereforeFurther, the larger diameter results in a greater acceptance angle. Thus. LEDs, which are cheaper and easier to use than lasers, are highly suitable. The total diameter of the coreWhat is its NA at the fiber axis, and its effective NA? SupposeThe material dispersion coefficient at this wavelength is about 5 ps km-1How does this compare with theWhat would the total dispersion. Modal dispersion for 1 km of graded index fiber is. Ln1 2 (1000)(1.474) 2Assuming a Gaussian output light pulse shape, rms material dispersion is. Total dispersion isL c. Maximum bit-rate isChapter 2 2.32The corresponding B for 1 km would be around 13 Mb s-1. With LED excitation, again assuming a Gaussian output light pulse shape, rms material dispersion isTotal dispersion isCalculate, the total dispersion and estimate the bit-rate distance product of theSolution. Ln1 2 (1000)(1.474) 2Assuming a Gaussian output light pulse shaper. Total dispersion isL c 3108Chapter 2 2.33Maximum bit-rate would beThe corresponding B for 1 km would be around 13 Mb s-1Assume that a laser is to be used with this fiber and the laser. What type of dispersion do you thinkSolutionWe can now calculate intermodal dispersionThe total dispersion for 850 nm isSo the intramodal dispersion isChapter 2 2.34If is the profile index then the rms intermodal dispersion is given by2. LnThe optimum profileSolutionChapter 2 2.35. From the graph in Figure 2Q28-1From the graph in Figure 2Q28-1, a 3.4 changeFigure 2Q28?1What are O and O? (b) How wouldWhat is O? (c) Sketch ray paths forWhere is O? (d). What use is 0.23P GRIN rod lens in Figure 2.32(c)? Figure 2.32 Graded index (GRIN) rod lenses of different pitches. (a) Point O is on the rod face center and the lens focusesSolutionChapter 2 2.36O and O are the focal points of the lenses (approximately). (Schematic only). (b) TheFigure 2Q29-2: Ray paths in a GRIN rod that has a pitch between 0.25P to 0.5P. (Schematic only.)Possible AnswersReduce scattering by reducing density and hence refractive index nUse a glass material with a lower glass transition temperatureAssume that all the light was launched into the fiber. The fiber is quoted asChapter 2 2.37Solution. Pout Pin exp(L)The first part of the experiment involvesThe output power Pnear is measured at the near end from the short cut fiber. The attenuation is then givenUsually a mode scrambler (mode stripper) is used forThe power output from a particular fiber is measured to be 13 nW. Then, 10 km of fiber is cut-out and the power output is measured again and found to be 43 nW. What isFigure 2.54 Illustration of the cut-back method for measuring the fiber attenuation. S is an optical source and D is aChapter 2 2.38Actual values would be less. SolutionWhen the wavelength is 1490 nmAR 0.918dB km ?mAt 850 nm and 1300 nm the FIR. At 850 nm,Chapter 2 2.39T B fAR 1.111dB km ?mAt 1300 nm,AR 1.111dB km ?mRayleigh scattering process decreases with wavelength, and as mentioned in Ch. 1, it is inverselyRayleigh scattering is approximately given by two sets of different equations in the literature3Fiber is drawn at high temperatures and asHowever, note that the reportedChapter 2 2.40It is found that for a single mode fiber withSolution. We expect the bending loss vs. R on a semilogarithmic plot to be as in Figure 2Q35-1 (schematic). Figure 2Q35-1 Microbending loss decreases sharply with the bend radius R. (Schematic only.). Expected R with V (1). Expected R with (3)? Eqs. (3) and (4) correspond to the general statement that microbending loss gets worse whenObservation R with (5). Consider the penetration depth into a second medium (Example 2.1.3),Chapter 2 2.41Thus, increases with decreasing. Thus, from Eqs. (3) and (6), we expect. Expected R with with (7) Thus Eq, (7)Thus, takingR constant (9). We are interested in the R behavior at a constant. We can lump the constant into ln and obtain,Figure 2Q35-2 The relationship between and the radius of curvature R for a given amount of bending loss.Chapter 2 2.42The trench fibers have a trench placed in the cladding where the refractivePlot the bending lossSuppose that we set our maximum acceptable bending loss toData over 1.55 - 1.65 m range. (Note, data used from a number of sources: (a). M.-J. Li et al. J. Light Wave Technol., 27, 376, 2009; (b) K. Himeno et al, J. Light Wave Technol., 23, 3494, 2005; (c) L.- A. Standard SMFTrench Fiber 1bTrench Fiber 2cNanoengineered FiberaChapter 2 2.43Trench 1. Trench 2. Nanoengineered. Expon. (Standard SMF). Expon. (Trench 1). Expon. (Trench 2). Expon. (Nanoengineered)Figure 2Q36-1 Attenuation per turn as a function of bend radius. Standard SMF, 13 mm; trench 1, 10 mm; trench 2, 6 mm; nanoengineered, 3 mm.We will calculate approximately given various fiber parameters using the single mode fiberWhat is your conclusion? (You might wish to compareChapter 2 2.44Chapter 2 2.45Figure 2Q37-1 Bending loss B vs.Results compare reasonably with the experiments in Figure 2.56 given the approximate nature of theWhat should beIf the amplitude of the index variation n is 2?10-5FBG is 5 mm, what are the maximum reflectance at the Bragg wavelength and the bandwidth of the. FBG? Assume that the effective refractive index n is 1.460. What are the reflectance and the bandwidthChapter 2 2.46R tanh2The parameterTwo neighboring sensors haveWhat would beWhat is the strain at fracture if the fiber fractures roughly at anWhat is your conclusion? SolutionThe shift due to strain is (only B is strained). Maximum strain occurs whenChapter 2 2.47The separation between the Bragg. Shift due to strain is now. Which must be B2 B1 so thatThe main problem is precise compensation of temperature. Optoelectronics and Photonics Principles and Practices 2nd Edition KasapSolution Manual Of Kasap. KasapNow customize the name of a clipboard to store your clips.