1 Physics 4150 Optics, Fall 2017 M. Goldman, Instructor HW-1 (total value = 40 pts) due Tues, Sept. 12, in class 1) Spherical waves (20 pts) a) Find the general waveform of a spherically symmetric scalar wave, A(r,t) with radial wavenumber, kr, which is a sol'n to the wave eqn, v 2 ∇ 2 −∂ t 2 ( ) A r, t ( ) = 0 . (8 pts) Your expression for A(r,t) should be real and depend on v, kr , initial phase φ (at r = 0 and t = 0) and real amplitude. In spherical coordinates the Laplacian is ∇ 2 A r, t ( ) = 1 r 2 ∂ r r 2 ∂ r A ( ) + 1 r 2 sinθ ∂ θ sinθ ∂ θ A ( ) + 1 r 2 sin 2 θ ∂ φ A . Since the wave is spherically symmetric, A(r,t) = A(r,t) is independent of θ and φ, so only the first term is non-zero. Hence we must solve v 2 r 2 ∂ r r 2 ∂ r Ar, t ( ) ( ) −∂ t 2 Ar, t ( ) = 0 . To solve, first let A(r,t) = a(r,t)/r, so that r 2 ∂rA = - a(r,t) + r[∂r a(r,t)] and ∂r(r 2 ∂rA) = r∂r 2 a. This gives the following eqn for a: v 2 ∂ r 2 −∂ t 2 ( ) ar, t ( ) = 0 . Try a solution of form a(r,t) = a0(t)Exp[ikrr], by analogy with how we solved a similar wave eqn for plane waves. Now the wave eqn. becomes, k r 2 v 2 + ∂ t 2 ( ) a 0 t () = 0 This is a harmonic oscillator eqn with solution, a0(t) = sExp[-iωt] where s is a complex number of form s0·Exp[iφ] and ω =krc is the angular frequency of the wave. Hence the full solution (waveform) is Re A r, t ( ) ⎡ ⎣ ⎤ ⎦ = s 0 r Re e ik r r−k r vt +φ ( ) ⎡ ⎣ ⎤ ⎦ = s 0 cos k r r − k r vt + φ [ ] r
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Physics 4150 Optics, Fall 201 Goldman, Instructor HW-1 …€¦ · · 2017-09-13Physics 4150 Optics, Fall 2017 M. Goldman, Instructor HW-1 (total value = 40 pts) due Tues, Sept.
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Physics4150Optics,Fall2017M. Goldman,Instructor
HW-1(totalvalue=40pts)dueTues,Sept.12,inclass
1) Sphericalwaves(20pts)
a) Findthegeneralwaveformofasphericallysymmetricscalarwave,A(r,t)withradialwavenumber,kr,whichisasol'ntothewaveeqn,v2∇2 −∂t
ik ·r( )whichisevaluatedbythefollowingdeterminant:
x y z∂x ∂y ∂z
Axeik ·r Aye
ik ·r Azeik ·r
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
= x ∂y Azeik ·r −∂z Aye
ik ·r( )− y ∂x Azeik ·r −∂z Axeik ·r( )+ z ∂x Aye
ik ·r −∂y Axeik ·r( )
= x ikyAzeik ·r − ikzAye
ik ·r( )− y ikxAzeik ·r − ikzAxeik ·r( )+ z ikxAye
ik ·r − ikyAxeik ·r( ) = ik× Aeik ·r( )
Hence, ik×E = −∂tB / c ,soB r,t( ) = ckωk
×Akei k ·r−ωkt+ϕk( ) = k×Ak( )ei k ·r−ωkt+ϕk( )
b) AfterexplaininghowtofindthedirectionofBintermsofkingeneral
answerthefollowing.(3pts)ThegeneralresultisthatthedirectionofBisgivenbykcrossedintoE.Ifkisinthex-directionandAkisiny-direction,whatisthedirectionofB?x× y = z soBisinthez-direction,orthogonaltobothxandy.AreBandEinphase?YesDoBandEhavethesameamplitude?Yes(incgsunits),butonlyinavacuum