中国科学技术大学：光学：5

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 μ?ù? bW2;D2￥ àb V[ μQ i 2 V[??]￥i Q  qs1?a;Ea??;E
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 θsin)1( 1 dnLL n ?+= M # ?íW￥;?μ θδ sind= M # ?íW￥êMμ
 θ λ π θ? sin 2 sin dkd ==? 1 a?^ ;D ??c  ;E ?β ?=2
 θ λ π β sind=
 N? O  náM¤
GQV9
' 2&? b β β β θ θ N a NROBA N sin sin 2/ 2sin2 === β β θ sin sin N a= β β sin sinsin~ 0 N u u U= 2a¨ Fresnel-Kirchhoff sp3 ü?;? 
 ?@íàHqb ∫∫∫∫ ΣΣ Σ=Σ= de r UKFd r e FQUKPU ikr ikr 0 000 1 )0( ~ ),()( ~ )( ~ θθ ? ài;?sps
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 ';E￥?ùb ò.????￥;o?rà ￥;?1 L j
5 μ δδ 2, 1312 +=+= LLLL
 δ)1( 1 ?+= nLL n
 θδ sind= 1M # .????￥ ;?r P?￥;?μb ? j?.? ?
ê? x j ￥?;÷? ￥;D. ???? ￥;?r P?￥;?μ1 θsin jj xr ?=?
' θsin jjj xLr ?=
  ?s?1 ]][ sin)0( ~ [ 1 )0( ~ 1 )0( ~1 )0( ~ )( ~ 1 0 0 1 2/ 2/ sin 0 0 11 2/ 1/ sin 0 0 2/ 2/ 0 0 ∑∑ ∫ ∑∑ ∫∫ == ? ? == ? ? ? == == N j ikL N j a a j ikxikL N j N j a a j ikxikL a a j ikr jjj jjj e u u a r U KFdxee r UKF dxe r UKFdxe r UKFPU θ θ Vn- ? []=1?? ￥2T
ò?.?? ^B"￥
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1?í
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 y0
?\f ? %?b )( sin )sin( 1 1 )( ~ )1()1( sin2 sin2 1 sin)1( θ β β θ ββ ββ ββ β β θ λ π θ λ π θ Ne N e ee ee e e e e eN NiNi ii iNiN i iN d i d iN N j djik ?? ? ? = ? == ? ? = ? ? == ∑ θ λ π θβ sinsin d kd == )(θN
 Ní? y0b )()( ~ )( ~ )( 0 θθ θ? NeUUPU i =
 22 0 ) sin sin () sin ()( β βN u u IPI = 3 a?^ ;D ??c  ;E -3-2-101234567 π bsinθ /λ U(θ ) 2 N(θ ) 2 β =π dsinθ /λ u=π bsinθ /λ (sinNβ /sinβ ) 2 (sinu/u) 2 =

?? 
 N=2 2 2 2 0 sin cos4)( u u IPI β=
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 u=0
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 n=1
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 =0
 sinN& =0
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 1(/ /N)
 2(/ /N)ll(N-1)(/ /N)
 N(/ /N)
 (N+1)(/ /N)
 (N+2)(/ /N)
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 λθθ jd =+ )sin(sin 0 ? a ;Eü ?EL]§
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 jjj θθθ ?+=′ μ Nd jjjj λ θθθθ =?=?′ cossinsin
 jj j LNd θ λ θ λ θ coscos =? L=Nd
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 δλθδθ jd =cos
 θ δλ δθ cosd j= b ¨ Rayleigh ? 
 δθθ =? H
 VsO
μ θ δλ θ λ coscos d j Nd =
 VsO￥Klo éW?1 jN λ δλ =
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 j v
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 θ?l H
 djdd // =λθ
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 oéS?1 λλλλ ?+= mMm ~
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 j=1 H
 K<￥oé1 BB d θλ sin2 1 =
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 ?[
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B )ú 2oé 1 BB d θλ 2sin 1 = b λθθθθ jd BB =?′?+′ )]sin()[sin( 0 )sin( B d θθδ ?′?=′ )sin( 00 B d θθδ +′= λθθ jd =? ]sin[sin 0
 B θθθ ?′= B θθθ +′= 00 8 a?^ ;D ??c  ;E ?a??;E D?a?; E?]
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 iV q 1 x d t π2 cos1+∝
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?í y01 ∫ ? ? += 2/ 2/ sin 0 0 ) 2 cos1()( ~ 0 d d ikx ikr dxex dr e KFUu θ π θ ∫ ? ? ? ++= 2/ 2/ sin 22 0 0 ) 2 1 2 1 1( 0 d d ikx x d ix d i ikr dxeee r e KFU θ ππ ∫ ? ??? ? ++= 2/ 2/ )sin 2 ()sin 2 ( sin 0 0 ] 2 1 2 1 [ 0 d d xk d ixik d i ikx ikr dxeee r e KFU θ π θ π θ ∫ ? ??? ? ++= 2/ 2/ )sin 2 ()sin 2 ( sin 0 0 ] 2 1 2 1 [ 0 d d xk d ixk d i ikx ikr dxeee r e KFU θ π θ π θ ] )sin( 2 1)sin( 2 1sin [ 0 0 0 πβ πβ πβ πβ β β + + + ? ? += r e dKFU ikr
∵ ∫∫ ? ? ? ? ? ? = 2/ 2/ )sin 2 (2/ 2/ )sin 2 ( ])sin 2 ([ )sin 2 ( 1 2 1 2 1 d d xk d id d xk d i xk d ide k d i dxe θ π θ π θ π θ π ][ )sin 2 ( 1 2 1 2 )sin 2 ( 2 )sin 2 ( d k d i d k d i ee k d i θ π θ π θ π ??? ? ? = )sin 2 ( )]sin 2 sin[(2 2 1 θ π θπ k d i d ki ? ? = θπ θπ sin 2 )]sin 2 sin[( 2 d k d k d ? ? = βπ βπ ? ? = )sin( 2 d
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