7.3 Problem 7.3. ~B(~x) = ~ k ~ E(~x)=! but we also have a reected wave. ~E(~x) = ~ E 2 e i~ k 2 ~x i!t. ~B R (~x) = ~ k R ~ E R (~x)=!

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1 7. Problem 7. We hve two semi-innite slbs of dielectric mteril with nd equl indices of refrction n >, with n ir g (n ) of thickness d between them. Let the surfces be in the x; y lne, with the g being z [; d] nd the incident wve coming from z <. In the rst mteril we hve n incident wve ~E(~x) ~ E e i~ k ~x i!t ~B(~x) ~ k ~ E(~x)! but we lso hve reected wve ~E R (~x) ~ E R e i~ k R ~x i!t ~B R (~x) ~ k R ~ E R (~x)! In the ir g, we my hve two wves with oscilltory behvior or exonentil behvior in z. We my write either cse s ~E(~x) ~ E g e i~ k g ~x i!t + ~ E g e i~ k g ~x i!t ~B(~x) ~ k g ~ E g (~x)! + ~ k g ~ E g (~x)! but remembering tht the z comonents of the wvenumbers my be imginry. The second slb hs only n outgoing wve ~E(~x) ~ E e i~ k ~x i!t ~B(~x) ~ k ~ E (~x)! The squres of the wvenumbers re determined by the indices of refrction nd!: but we must kee in mind tht jk g z j. ~kg k k R k n!c k g k g! c z my be imginry, in which cse k g k g x + k g y s for the single interfce discussion, we my chose x so tht the incident wve is in the x; z lne. s the wve equtions nd the boundry conditions re invrint under trnsltions in the x nd y directions, we cn Fourier trnsform in those directions nd see tht the equtions involve only the sme vlues for the k x 's nd for the k y 's, so k x k R x k g x k g x k x k y k R y k g y k g y k y

2 From the equlity of the k x 's nd q the reltions mong the k ' we hve k z k R z k z k cos i, nd k g z k g z kg k x k n sin i k cos rn, with the ngle of reection for ~ k R nd the ngle of ~ k equl to the ngle of incidence i, nd the ngle of reection, r given by Snell's lw n sin i sin r. Note k gi x will be imginry if n sin i >, nd r will then be comlex. Finlly, we cn divide the roblem into rt (E? ) for which ~ E is erendiculr to the lne of incidence ( ~ E k ^e y ) nd rt (E ) in which it lies in the lne of incidence (E y ), in which cse ~ B k ^e y in the rst mteril. s the roblem is invrint under reection in the y lne, in the (E? ) cse ll elds re reversed, so ll of the ~ E's re in the y direction, nd ll the ~ Bs re in the x; z lne. In the (E k ) cse reection in the y lne chnges none of the incident elds, nd therefore none of the others, so ll the ~ E y 's vnish, nd ll the ~ B's re erendiculr to the lne of incidence. The boundry conditions re continuity of D z, B z, E x, E y, H x, nd H y t ech of the two boundries. The ~ E k cse: We will now consider the cse where ll the ~ E elds lie in the lne of incidence. See below for icture with ll of the E x 's ositive if the corresonding mlitudes E re ositive. This is dierent from wht Jckson did. From the continuity of E x, (E + E R ) cos i (E g + E g ) cos r E g e ikd(cos r)n + E g e r)n ikd(cos cos r E e ikd cos i cos i nd continuity of H y gives n(e E R ) E g E g ne e ikd cos i ikd(cos r)n E g e E g e ikd(cos r)n

3 ikd(cos r)n To simlify our lgebr, let e nd equtions become E B cos i E g + E g cos r E Bn E g E g B e ikd cos i, so the second nd fourth giving E g cos i E g cos i E g n cos r E g n cos r with: ) E g E g n cos r cos i n cos r + cos i E g + Plugging bck into the rst interfce forms, (E R + E ) n (E E R ) E g E g cos i n cos r + + Eg n ( ) + ( + ) n( + ) + + E g E g so E + E R E E R + ( + ) + + ( ) Note s e i' with ' kd(cos r)n, + +ei' e i' i cot ', we get: E R E ( )( ) + ( + ) + ( ) + i cos ' + Provided the ngle of incidence is less thn the ngle of totl reection, nd cot ' re rel, nd the reection coecient is E R E ( ) ( + ) + r cot ' This will hve mxim nd minim ccording to the hse, mxim when it is multile of nd minim hlfwy between those. On the other hnd, if n sin i >, in n sin i nd cot is imginry nd goes to i, without oscilltions, Of course in either cse the trnsmission coecient is the reection coecient. Below is lot of the trnsmission coecient with n :5 nd \i rd, for kd [; ].

4 7.6 Problem Prt Strting with equtions 7. nd ssuming tht ~ B nd ~ E hve solutions with hrmonic time deendence: r E ~ + B ~ t ) r E ~ i! B ~ r ~ E B " ~ t ) r B ~ + i! D ~ Fourier trnsforming the bove two equtions, noting tht r, i ~ k: i ~ k ~ E i! ~ B i ~ k ~ B + i! ~ D Solving the rst eqution for ~ B nd lugging it into the second eqution yields: i ~ k ~ k ~ E!! + i! ~ D ~ k ~k ~ E +! ~ D 4

5 7.6. Prt b We strt by exnding the result from the revious roblem using the BC-CB rule: k ^n ^n E ~ +! D ~ k ^n ^n E ~ E ~ (^n ^n) +! D ~ n i (n j E j ) E i +! k {z} {z} D i " v i E i n i n j E j E i + v " i E i {z} v i n i n j E j E i + v E i v i We now nd the three comonents of both sides of the bove eqution, remembering tht we're imlicitly summing over j in the bove eqution. B 4 n (n E + n E + n E ) E n (n E + n E + n E ) E 5 + v 4 E v E v 5 n (n E + n E + n E ) E E v 4 (n n ) E + n n E + n n E n n E + (n n ) E + n n E 5 + v 4 E v E v 5 n n E + n n E + (n n ) E E v 4 n n n n n n n n n n n n n n n 5 +v 4 v v v B 5 C 4 E 5 Hence, solutions to v re eigenvlues of. We cn solve for v using the chrcteristic eqution, det( + v B). Tking the determinnt of + v B in Mle yields the following (where we hve rerrnged some terms for resons which will become immeditely rent), nd using the fct tht n + n + n : v v v v v 4 + v v n v v n v v + n v v v v n v v + vv n v v n vv v v (n +n ) n v v E E + n v v v v + n v v v v + vv n v v {z n vv } v v (n +n ) v v (n +n ) + n + n + n 5 n v v

6 v v v v n (v 4 + v v v v v v ) +n (v 4 + v v + v v v v ) +n (v 4 + v v v v + v v ) v v v v n (v v )(v v ) + n (v v )(v v ) + n (v v )(v v ) Dividing both sides of this eqution by (v v)(v v n v v v v v + n v v " v X v v v v )(v v ) yields: + n i v v n i v v i There re three solutions for v:, v +, nd v (where the lst two solutions cn be found using the qudrtic eqution, which yields two solutions). The two nontrivil solutions stisfy the Fresnel eqution, which is when the term in brckets in the bove eqution is zero. Tht is: X i n i v v i # 7.6. Prt c From our solution to rt, we obtin, for wve in mode : ~n ~ E ~n ~ E! k " ~ E v ~ D where ~n ~ kk is unit vector in the ~ k direction, nd v! k, which my be dierent for the dierent modes, s the hse velocity my deend on the olriztion. Dot this into D ~ b for nother mode with the sme ~n, giving: ~D b D ~ v D ~ b ~E ~n E ~ ~n D ~ b E ~ becuse ~n ~ E. Of course, the sme lies with $ b, so: But: ~D b ~ D v v b ~ Db ~ E ~ D ~ Eb 6

7 ~D b ~ E X i " i E i E b i ~ D ~ E b so, ~D b ~ D v v b nd if the two modes hve dierent hse velocities (v 6 v b ), then: ~D b ~ D 8.5 Problem Prt For the simly-connected tringulr region the modes will be either TE or TM, with the longitudinl B z or E z given by solution of the Helmholtz eqution (r t + ) with boundry conditions nj S or j S resectively. ny such solution on the tringle x, y x cn be extended to solution on the squre x, y by dening (x; y)j y>x (y; x), with the lus sign for the Neumnn (TE) cse nd the minus sign for the Dirichlet (TM) cse. The vnishing on x y in the TM cse insures continuity, which is utomtic with the lus sign in the TE cse. The norml derivtive is is zero nd continuous due to the lus sign in the TE cse, but is utomtic with the minus sign for the TM cse. So the solutions for the tringle must be combintions of solutions for the squre, but with symmetry or ntisymmetry under x $ y for TE nd TM modes resectively. In terms of the functions in 8.5 nd 8.6, this mens mx ny my nx i sin sin sin T M : E zmn E hsin mx T E : H zmn H hcos cos ny my cos nx cos s mn m + n. For our uroses we do not need to clculte the normliztion constnts E nd H. Thus ll the modes re the sme s for the squre of side, excet tht m n is forbidden for the TM mode, nd for ech ir m n, there is only one mode rther thn two. s for the rectngle, m n is forbidden s it leds to zero trnsverse elds. The cuto frequencies re s! mn m + n " 7 i

8 8.5. Prt b The lowest modes re TM ; nd TE ;. The ttenution coecients deend on nd, which involve the rtio of integrls over the boundry to those over the re. For the TE ; mode,. Tke (x; y) cos(x) cos(x) + cos(y) ( + ) dx du j (u; )j + j (; u)j + j (u; u)j du ( + cos(u)) + (cos(u) ) + [ cos(u)] dy (x; y) dx C + dy[cos(x) + cos(y)] TE ; C j^n r t j R lso, of course, r ~ t ( + ) ( + ) (x; )dx + x " + dx h sin (u) du R,. So ^n ~ r + + # y x y + (; y)dy (x; x)! + + i ( + ) ~ r C ( + ) ( + ) 8

9 . Thus, r " TE +!! +!!!! For the full squre of side, the mode is the sme, the re R j j, both twice s big, C 4, R j j 6, nd so j^n r t j + x x (x; )dx + (x; )dx + y y TE C ( ) 4 TE 6 4 (; y)dy (; y)dy r " TE TE!!!! +!! The rtio of ttenution of the tringle to the squre is not frequency indeendent, but t ll frequencies it is greter thn. TM ; mode: We cn tke sin(x) sin(y) sin(x) sin(y), ; 5, sin (x) sin (y) + sin (x) sin (xy) sin(x) sin(y) sin(x) sin(y)

10 n du " (u; ) + y + x y x (u; u)# (; u) du ( sin(u) sin(u)) + ( sin(u) sin(u)) + ( cos(u) sin(u) 4 sin(u) cos(u)) 5 ( + ) where it is useful to write the exression to be squred in the lst term s sin(u) sin(u). Thus TM "!! r q n j j 4 + " q!!!! For the squre R nd so n 4 du ( sin(u) sin(u)) TM r " TM 4 4 q!! 5 r " q!!!! Thus the tringle ttenution is + :7 times tht of the squre.!!!!

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