Optique de Fourier, suite: Application à la di raction par des

T(x, y) = exp iφ(x, y)
T(x, y)

x y
Σ
d
T=X
p
δ(y−pd)×1x= ΠΠd(y)×1x
Ox 1x
~
k
ψ(~ k) = ψ0ZT(x, y) exp[−i(kxx+kyy)]dxdy =ψ0b T(kx, ky)

ψ(~ k) = ψ0.b 1(kx).d ΠΠd(ky)
2πδ
ψ(~ k) = ψ0.2π.δ(kx).2π
dΠΠ2π
d(ky)
kx
ky=q2π
d
q= 0,±1,±2...
θ~ k
ky=ksin θ=2π
λsin θ
sin θ=qλ
d
~
kikix = 0
kiy =ksin θiθi
ψΣ(x, y) = ψ0exp(+ikiyy)T(x, y) = ψ0exp(+ik sin θiy)T(x, y)
+ exp(i~ ki.~r)
~
ki