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# Electromagnetism PH 317

RHIT

GPA 3.69

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This 5 page Study Guide was uploaded by Jalon Willms on Monday October 19, 2015. The Study Guide belongs to PH 317 at Rose-Hulman Institute of Technology taught by Staff in Fall. Since its upload, it has received 31 views. For similar materials see /class/225119/ph-317-rose-hulman-institute-of-technology in Physics 2 at Rose-Hulman Institute of Technology.

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Date Created: 10/19/15

Conducting waveguides and slab optical waveguides PH 317 February 9 2002 rev 1 The zdirection is parallel to the aXis of the waveguide Energy in the form of electromagnetic waves is con ned within the waveguide Assume waves with zdependence of eXpikZz iioat The X and y dependence is that of superposed plane waves These combine to form standing waves between the boundaries in X and y Solutions will be things like sin kxX coskyy eXpikZz iioat kx ky will be determined by boundary conditions kZ will be determined from kZ Woazc2 7 kx2 7 ky2 or v2 instead of c2 in glass First write down all 6 components of both curl equations in free space Then solve for X and y components of E and B in terms of EZ and BZ or their derivatives Separate two distinct kinds of waves 0 TE in which EZ is zero so E components are transverse to z 0 TM in which BZ is zero so B components are transverse to 2 For TE waves nd the form of BZ which will satisfy the boundary conditions For TM waves nd the form of EZ which will satisfy the boundary conditions This will X the values of kx and ky In a conducting waveguide with perfect conductors for walls walls at X0 Xa y0 yb O Eparanel 0 Bpemendicular 0 at the walls 0 at X0 Xa Ey 0 EZ 0 Bx 0 0 at y0 yb Ex0 EZ 0 By 0 In a slab waveguide waves travel in the Xz plane and boundaries are at X a2 a 2 For waves to be con ned in the slab nomside ltninside and 9 gt9 critical E and B vectors outside the slab must be eXponentially decreasing for the wave to be con ned For both types of waveguides solve separately for even and odd symmetries within TE and TM modes For curl E 6B6t we get BExay BEyBX imBZ 1 Keeping in mind that 662 on any component results in ikz we have further BEyBz BEZay ime ikZ Ey 6EZ6y ime 2 BEZBX BExBz imBy BEZBX ikZ Ex imBy 3 For curl B us 6E6t we have anay aByax inc2 EZ 4 ikz By 6BZ6y inc2 Ex 5 aBZax ikz Bx inc2 Ey 6 Waveguide Notes PH 3 17 MJM February 9 2002 page 2 To solve for Ex in terms of B2 EZ and their derivatives we choose 3 and 5 In 3 Ex appears by itself no derivatives and then for By in 3 we will need 5 where By appears without derivatives BEZBX ikZ Ex imBy 3 ikz By 6BZ6y inc2 Ex 5 From this comes Ex likz BEZBX nkZ nkZc2 Ex likz6BZ6y which simpli es to Ex Fikz aEZax oakz aBZay where F 11k2kzz with k oac 7 This expresses Ex in terms of derivatives of BZ and E2 Because of 5 we can obtain By likewise By likz BBZay nkzc2 Fikz 6EZ6x nkz BBZay or By Fikz nkzcz aEZax aBZay 8 Using 2 and 6 we obtain Ey Fikz 6EZ6y nkz BBZax 9 Ex Fikz BBZax Jokch BEZay 10 In a conducting waveguide at X0 and Xa we require EZ0 Ey0 and BK 0 In a TE mode EZ0 everywhere For Ey0 and for 13K 0 we need BBZBX 0 The function to satisfy this will be BZ A cos kxX provided that kxa integer 11 n 11 Thus we require BZ A cos nnXa TE mode In a conducting waveguide at y0 and yb we require EZ0 Ex0 and By 0 With EZ and its derivatives all zero in the TE mode we need BBZay 0 at y0 and yb This is satisfied by BZ B cos mnyb So the overall expression for BZ is BZ A cosn11xa cosm11yb eXpikZz iioat From this we derive the four transverse components of E and B according to 7 7 10 PH 317 Waveguide Notes Feb 9 2002 page 3 For a TM mode BZ and all its derivatives are zero and we still have to satisfy the same boundary conditions Now the requirement at x0 and xa for Ey0 translates into EZ 0 and BEZay 0 To satisfy EZ 0 at x0 and xa we may use EZ A sinn11xa where n 12 and for EZ 0 at y0 and yb we may use EZ A sinn11xa sinm11yb But we must also satisfy BEZBy 0 at x0 BEZBy mnb A sinn11xa cosm11yb and this is indeed zero at x0 so setting EZ 0 satisfied both conditions Slab optical waveguide A slab of width a and index n1 is the waveguide surrounded on both sides by material of index nzltn1 For modes to propagate in the slab waveguide we need to launch waves at an angle greater than the critical angle This causes the fields in n to exponentially decay We will separately solve for EVEN waveguide modes where the function of x is cos kxx and ODD waveguide modes where the function of x is sinkxx The sketch shows x a TE ray EZ0 n lt n1 in index n1 91 Ht re ecting from z a boundary and xa2 showing a decaying n1 Hi 9 Hr field in n2 We All E vectors are in will want the the y direction parallel component into the paper of E and the parallel component of H The waves travel in the xz plane to be continuous across the boundary x a2 The net E field in n1 is Enemlyy Ei expikxx ikzz imt Er expikxx ikzz int or Enemlyy A coskxx B sin kxx exp ikzz int We will tackle these one at a time The EVEN TE mode in the slab waveguide and the ODD TE mode in the slab waveguide For the TM modes HZ 0 we must also work out both even and odd solutions EVEN TE mode conditions PH 317 Waveguide Notes Feb 9 2002 page 4 For the parallel component of E at the boundary we have Ey A cos kx a2 Et eXpi km a2 1 Since 9 in n1 is greater than the critical angle cos 91 is pure imaginary And km kt cos 91 so km is pure imaginary or km i k y where y l cos 9 l So the elds in n are exponential in X eXpkt y X We also recall that k1 oavl n1 oac and likewise kt n2 oac For matching the H eld parallel components we want HZ To obtain HZ we appeal to the curl of E curl EZ u 6HZ6t BEyBX BExBy imu HZ We apply this both inside and outside In n1 then evaluated at Xa2 kx A sinkxa2 imu HZ In n2 then evaluated at Xa2 E i km eXplkX a2 imu th Setting parallel components of H namely HZ equal at the boundary gives kx A sinkxa2 E i km eXplkX a2 Dividing one equation by the other for EZ and HZ we nd kxtankx a2lkm l kty This is ofthe form utanuqwhereukx a2 and q l km l a2 kmaZ Squaring both sides and adding u2 to both sides gives u2 tanu2 u2 ucosu2 q2 u2 Next we write out k2 on both sides of the boundary ktz ktzz kmz n22 DzCZ and k2 kzz kxz n12 DZc2 Because kZ is the same on both sides for constant phase along the boundary and k 2 km2 n12 DzCZ n22 DzCZ kx2 k y2 Multiplying both sides by a22 we get 112 q2 R2 where R2 a22 n12 DZCZ n22 Dzoz PH 317 Waveguide Notes Feb 9 2002 page 5 Now we want to solve ucosu2 q2 u2 R2 where R2 depends only on the slab width and indices In particular we want solutions of cos u i39uR But because u tan u q we only want solutions where tan u is positive namely the rst and third quadrants indeed all odd quadrants One can plot simultaneously cos u and uR and uR then accept solutions in the odd quadrants Homework Due Friday February 15 2002 No late work accepted Turn in at the start of class 1 Show that when uR l we are at the critical angle from n1 to n2 2 Work out the condition for odd TE modes analogous to cos u iu R odd quadrants 3 For a slab whose thickness is 10 microns at an omega of l X 1015 rads and n1 15 and n2 146 determine the valid even TE solutions Determine the launch angle for each of these theta in n1 This is readily done in Maple Could also be done on a spreadsheet

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