This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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
12.3f
12.3g
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. ...
View
Full
Document
This note was uploaded on 12/12/2010 for the course ECE 604 taught by Professor Staff during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 Staff
 Electromagnet

Click to edit the document details