Question

Problem 1: This problem is just a reminder of material covered earlier in class. It is meant as a preparation for a problem on quantum mechanics, Problem 2 below. Consider the Helmholtz equation (5.3-16) SU )+ (2x U4)=0 for the case of an ideal (lossless) planar mirror resonator with vacuum between two metallic mirrors. Assume the mirrors to be located at -d/2 and +d/2, and the wave amplitude to be zero at the boundaries. Determine all possible solutions , wave functions U,(z) and frequencies V, . Note that the functions U(z) are either cos functions (even as function of z) or sin functions (odd as function of z). Sketch U,(2) for the lowest four frequency levels. Assuming the width to be 1.188 micron, which level (mode number q) is closest to frequency of ħo= 5.22 eV? (10 points)

Answer 1

U (zl=0 Ige U (2) 127 pifc to u(-d/2) = 0 Uld/2}=0 I let 2kuak C DU(z) + K² U (2)= 0 - ý. Səz²4 let us assume the solution of the form Ulz) = ê d²Ucz) - 2 piz az² Substituting these in equation o x² e 2 + k? et2 = 0 - 02+K? 20 3) da tek hence the general solution of Tequation o is U(z)=G DER G ē ekz a and q are abitrary et constants in we can convert U(z) in terms of I sine and cosine using enler's I identity tu (Z) = C, eikztheek? T a coska ti sinkz7+ Cel cos (K2) 7 i sin(ke)]

= (Cite) coska & (64-ic) sinka Is a Coskz + b sinkz te (a=uter , b = 12 27 = a costred) +bsint- Red ) = 0 acosko) - bsin (ko) = 0 tu (6/2) = 0 => acos(kd.) + bsin(k)=0 I adding ② and ③. 20 cos(kd) = 0 = a cos(kd) = 0 subtracting 2 from o absinl kd] =0 =) bisin (d = 0

Ос?-) | O, (2) -d/ Кај, 7 . d- 1. 188 тіс, ой - ), )ЅУ x16 T Since.ke С. . 2 и k: - 2 по 9 с C - di. since año a w 5 q, и .cd wa Qис d ГК : Б. 22 ev

to eac . 5.22 ev = 5.22x1-68105 = 3:22 e 23:2270ne 9. = 5.22 X1.6X10X74188 X106 1 x 3X 108 . 33 :53:307392 x 10 ñ X.105 457.71034 q sa 10 hence leved frequency of to in closest to tw = 5.22ev