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Unformatted text preview: Notes: ECE604 Homework 13
Out: Tuesday, April 19, 2005
Due: Tuesday, April 26, 2005 Final: Monday, May 2, 3:20 — 5:20 Room to be announced
Review Session: Friday, April 29, 4:30 — 5:30 in EE226 . See attached 3.190
12.6a 12.6b Problem #1 Consider a symmetrical optical resonantor (Fig. 1a) formed by two parallel mirrors with
reflectivities R = r2, where R is the power reﬂectivity and r is the field reﬂectivity. For this
problem you will model the resonator as a series of dielectrics (Fig. lb) with impedances 711,712, and 111, respectively. Regions 1 and 3 are assumed to be made of the same material. The ratio 111/712 is chosen to be consistent with the field reﬂectivity r, i.e., we take r = m _ n2 . T11 +712 a) For part (a) We'Will assume- therresonance condition is satisfied, i.e., k2! = 1111:, where k2 is the propagation constant in region 2. As discussed in class, when k2! = nm, there is
no reflected wave in region 1. On resonance we may write the E—field in the three regions as a function of 2 as follows: E1 = El+e‘jklZ , z < 0
E2 = E2+C—jkzz + E2_C+jk2z , 0 < Z < 3
E3 = E3+e_jk1(z_z) , I! < 2 Similar expressions may be written for H. i) Use the boundary conditions at z = 0 to obtain expressions for E2+ and E2- in terms of
E“. Then use the boundary conditions at z = E to obtain an expression for E3+. ii) Use these results to obtain expressions for the output time—averaged intensity P3+, and the
intensities of the forward travelling and backward travelling waves in region 2 (PZJr and P2-
, respectively), all in terms of the input time—averaged intensity P1+ and the power
reﬂectivity R. Also express P2- in terms of P2+. Then evaluate your expressions numerically for
R = 99%. iii) Comment on the physical meaning of all the expressions from (ii). (b) Now we no longer assume the resonant condition, i.e., we do not assume kzé = mat.
Instead we wish to calculate the fraction of the power transmitted as a function of
frequency. Plot the fraction of the power transmitted assuming R = 99%, E = 150 um, and refractive index equal to one for optical frequencies in the range 2.975 x 1014 Hz to 3.025
x 1014 Hz. (Hint: there is no need to repeat the calculations of part (a). Rather, this
problem can be solved easily by using the impedance transformation formula and writing
a simple numerical code). Compare the width of your resonant peaks with the simple expression discussed in class. ...
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This note was uploaded on 12/12/2010 for the course ECE 604 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.
- Spring '08