Question

1. This problem is intended to take you through an example of how the optical dispersion formula obtained in class is often used in combination with experimental data to produce a useful expression for the refractive-index of a dielectric medium. We start from the expression for the frequency dependent refractive-index n() in which absorption is neglected This expression is valid for optical frequencies w far below the material resonant frequencywo >>w (a) Starting from the above, and using the relation between refractive index and susceptibility, show that the inverse of susceptibility, 1/x(can be written in the form This is all one question. I need all parts clearly x(u) labeled please including the MATLAB code. Thank A, and slope B. (2pts) Write the expressions f (b) Here is a table of experimentally measured values for the susceptibility of water you Please do not copy the answer already given to this question. it's wrong. Original answer needed w10's 3.0 4.5 6.0 0.77647 08024 0.84020 Write a Matlab (or other) code to plot y x) for the water data above, and attach a copy of your code with your solution. You should find approximately linear relation between y and r as predicted by the theory. Using this plot extract the intercept A from the data, and estimate the slope B working from the plot, or data in the table. (Reasonable estimates of these parameters from your plots will be accepted. For more precision you may try the Matlab or Mathematica fitting routines, but this is not essential.) It may be convenient, but not necessary, to work with units of u given in the table. (4pts) (c) Using your numerical values of the slope B and intercept A B determined from part (b) calculate the value of the resonant wavelength λο in nm. (You should find λο--100 nm. which is in the UV, and if this does not agree with your answer you need to revisit part (b)) (2pts) (d) Given your value for λο explain whether the conditions underlying Eq. (1) are satisfied for wavelengths in the visible range? (2pts) (e) Go to http://refractiveindex.info, look up oxides and among them water, and find out what is the refractive index of water at λ-550 nm. Use the fit (2) with A and B you obtained to evaluate the refractive index for the same wavelength. You should obtain agreement to two digits (three if you used a fit from Matlab/Mathematica). Give some possible reasons for the difference. (3pts) (f) Explain whether water exhibits normal or anomalous dispersion in the visible region. (2pts)

Answer 1

n^2(w) = 1 + (N_q^2 e/m_e epsilon_0) 1/W_c^2 - w^2 1/x(w) = ? h^2(w) = 1 + x(w) 1 + x(w) = 1 + (N_q^2 e/m_e epsilon_0) 1/W_0^2 - w^2 x(w) = N_q^2 e/m_e epsilon_0 (1/W_0^2 - w^2) 1/x(w) = m_e epsilon_0/N_q^2 e (w_0^2 - w^2) Therefore A = m_e epsilon_0/N_q_e^2 w_0^2 B = m_e epsilon_0/N_q_e^2 W_0 = squareroot K/m m = 18 g/mol q_e = 1.6 times 10^-19 m_e = 9.11 times 10^-31 epsilon_0 = 8.85 times 10^-12 Therefore W_0 = squareroot K/m = 8.758 times 10^-13 x(w) = 0.77647, 0.80245, 0.84020 \r\n Enter the following code in MATLAB: - W_0 = 0.8758e - 13: N = 3: me = 9.11e - 31: e = 1.6e - 19: eo = 8.85e - 12: W = [34.56]: Psi = (me * eo/(N * e^2)) * (wo^2 - w^2): Psil = [0.77647 0.80245 0.84020]: plot(w^2, Psil) P = polyfit (w^2, Psil, l): Y intercept = P(2) The plot is as follows A = Y intercept = 0.7550

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