Chapter 6 Problem Solutions

Nb and substituting into the equations in a and b

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Unformatted text preview: 0+ y HxL y HxL = y HxL A ‰Â k H0L + B ‰- k H0L = F ‰q H0L + G ‰-q H0L F ‰q a + G ‰ - q a = C ‰  k a A  k ‰Â k H0L - B  k ‰- k H0L = F q ‰q H0L - G q ‰-q H0L F q ‰q a - G q ‰- q a = C  k ‰Â k a A+B=F+G F ‰q a + G ‰ - q a = C ‰  k a  k HA - BL = q HF - GL q HF ‰q a - G ‰-q a L = C  k ‰Â k a Solving the equations in F and G A+B=F+G A-B=F ‰q a = G ‰- q a = 1 2 Âq k HF - GL J1 + 1 2 J1 - Âk N C ‰Â k a q Âk N C ‰Â k a q and substituting into the equations in A and B gives A+B= 1 2 CBJ1 + Âq Âk N ‰- q a q A - B = - 2 k CBJ1 + + J1 - Âk N ‰- q a q  k Printed byÂMathematica for Students N ‰q a F ‰ k a q - J1 - Âk N ‰q a F ‰  k a q Ch6ProblemSetKey.nb and substituting into the equations in A and B gives A+B= 1 2 CBJ1 + Âk N ‰- q a q Âq + J1 - Âk N ‰- q a q A - B = - 2 k CBJ1 + Âk N ‰q a F ‰Â k a q Âk N ‰q a F ‰  k a q - J1 - Finally, we eliminate B from these equations via addition to obtain Âk N ‰- q a q A= 1 4 C :BJ1 + = 1 4 CBJ1 - = 1 4 C :B2 -  J k - = 1 4 Âq k N J1 + q C :B2 -  J q2 - k2 kq A = C BcoshHq aL +  J + J1 - Âk N ‰-q a q k NF ‰-q a q NF ‰ -q a q2 - k2 2kq Âk N ‰q a F q - Âq + J1 + k Âq k q q2 - k2 kq Âk N ‰- q a q Âk N ‰q a F > ‰Â k a q - J1 - Âk N ‰q a F ‰Â k a q N J1 - q k NF ‰q a > ‰Â k a ¬༼ k q + B2 +  J k - + B2 +  J BJ1 + qa NF ‰ > ‰ Âka ¬༼ 1 2 1 2 k q = q2 - k2 kq H‰q a + ‰-q a L = coshHq aL H‰q a - ‰-q a L = sinh Hq aL N sinhHq aLF ‰Â k a The transmission coefficient is defined as jx > a T= where jinc j= — 2m Jy* ...
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This note was uploaded on 02/01/2014 for the course PHYC 491/496 taught by Professor Akimasamiyake during the Fall '13 term at New Mexico.

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