2.71: Solutions to Home work 1Problem 1: Optics BuzzwordsYou will get full credit for this problem if your comments about the topicindicate a certain depth of understanding about what you have written andthat you have done some research.Problem 2: Why is the sky blue?This problem seeks to help you recollect some of the linear systems theorythat will be useful in this class. It is also instructive to understand thedynamics of why the sky is blue. To solve this problem, you need to formulatea simple 1 Degree of freedom (DOF) model for the Nitrogen atom. We willfind the response of the Nitrogen atom subjected to a periodic excitation bya photon and then apply Rayleigh’s scattering assumption to see why thesky is blue.Part 2aThe model for a Nitrogen atom is shown in Figure 1. The mass of theelectron is ‘m’, ‘k’ and ‘b’ model the stiffness and damping between themassive nucleus (M m) and the light electron. The equation of motionfor the electron subjected to a force ‘F’ ism¨x + b ˙x + kx = F.Taking the Laplace transform of this equation we get the transfer function ,XF=1ms2+ bs + k.1mkMbFigure 1: 1 DOF model of Nitrogen atom.Applying Rayleigh’s approximation,|A(ω)|2=s2XF2.The extra s2in the numerator is on account of the fact that the scattering isproportional to the acceleration of the particle. The driving force is assumedto be sinusoidal. Recall that the Fourier transform of a linear system tells usabout its response to sinusoidal inputs. Hence, we need to find the transferfunction in the Fourier domain. This is obtained by replacing s by iω,|A(ω)|2=−ω2(k − mω2)+iωb2.⇒|A(ω)|2=ω4(k − mω2)2+(ωb)2.Part 2bFor a simple second order system, recall thatNatural frequency, ωn=kmDamping ratio, ζ =b2√mk210010510101015102010−7010−6010−5010−4010−3010−2010−101001010ω|A(ω)|2ωrFigure 2: Transfer function |A(ω)|2.Resonance frequency, ωr= ωn 1 − ζ2This is not a simple second order system on account of the −ω2term in thenumerator. You can find that the maximum of the transfer occurs forωr=2k22mk − b2.However, for b =0,ωr=km.So the resonance frequency is equal to the natural frequency if ζ ≈ 0. Thetransfer function is shown in Figure 2Part 2cHere, we assume that the damping ratio is negligible in the Nitrogen atom(This is justified because typically dissipation within atoms is usually very3small). In this case,ωr≈ ωn=1409.31 × 10−31=1.24 × 1016rad/secIn the visible spectrum, violet has the highest frequency ωviolet=4.71 ×1015rad/sec. Thus, the resonance frequency of the Nitrogen atom is higher thanall optical frequencies within the visible spectrum of light.Part 2dAssuming zero dissipation, the transfer function of the system is|A(ω)|2=ω4(k − mω2)2,⇒|A(ω)|2= c ×ω4(1 − (ω/ωr)2)2.where c is a constant. If ω ωr, the denominator reduces to 1 and theresulting expression is|A(ω)|2= c × ω4.Thus we see that in the visible range, higher frequencies are scattered morestrongly than the lower frequencies. As a result of greater scattering, theblue end of the spectrum is what is visible most when we look at the
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