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Fourier optics
Fourier relations in Optics
Near field Far field
Frequency Pulse duration
Frequency Coherence length
Beam waist Beam divergence
Focal plane of lens The other focal plane
Spatial dimension Angular dimension
Review of Fourier theorem
A complex function f(t) may be decomposed as a superposition integral of harmonic function of all frequencies and complex amplitude
∞ −∞
f t = F e dv
j ω t
( ) ( ω) (inverse Fourier transform) (1)
The component with frequency ν has a complex amplitude F( ν ), given by
∞ −∞
−
F = f t e dt
j ω t
( ω ) () (Fourier transform) (2)
Selected function and their Fourier transforms, between t and ν, and between x and θ.
f ( t )= 1 F ( ω )= δ( ω) (3)
f ( t )= δ( t ) F ( ω)= 1 (4)
elsewhere
f t t 0
ωτ
ωτ ω τ sin( / 2 ) F ( )= 2 (5)
(sinc function)
elsewhere
f t t t
0
( ) cos( 0 )
]
[( )
]
sin[( ) ( ) 0
0 τ ω ω
τ ω ω ω τ −
F = (6)
2
2
( )^ τ
t
f t e
−
2 2 ( )
τ ω ω
F = e (7)
elsewhere
n N
f t nT t nT
τ τ
sin(
sin(
sin( ( ) T
NT
F
ω
ω
τ ω
τ ω ω = (8)
Huygens’ principle-
Huygens’ Principle
E(r) E(R)
d r R r
E re ER
j kR r
−
( ) ⋅^ (^ −)
Applications of Eq. (5)
Single- slit diffraction
When a single slit of width a illuminated by radiation of wavelength λ , the angular distribution of electric field observed at infinity is
λ
π
θ^ θ
a
a
a
a
k
a
k
E eikx dx a
sin( )
sin
sin )
sin(
( )= (^) sin^ = ≈
∞ −∞ (^) a
λ
j (^) xx j yy f x y Ae
2 πν 2 πν ( , )
− − = (^) (11)
where the ν ’s are the spatial frequencies in the x and y directions. The spatial frequencies are the inverse of the periods.
When this pattern is emitting at wavelength λ, a plane wave of the following form is generated:
j (^) xx j yy jkz z U x y z Ae
=
2 πν 2 πν ( , , ) (12)
where
2 2 1 / 2 2 )
kz 2 ( ν x ν y
λ
This can be proven by considering the diffraction of a sinusoidal grating illuminated by a plane wave of wavelength λ.
Thus by decomposing a spatial distribution of electric field into spatial harmonics, each component can be treated separately.
Define a transfer function in free space for the spatial harmonics of spatial frequency νx and νy to travel from z=0 to z =d as
j d x y
x y H e
2 21 / 2 2 ) 2 (^1 ( , )
ν ν λ
π ν ν
− − − = (^) (14)
This is the multiplication factor to a sinusoidal spatial pattern of spatial frequency νx and νy. The redirecting of an incident plane wave into another direction by a periodic structure can be illustrated in the following figure. If the spatial periodic structure is stationary, the outgoing wave number is the same as the incident wave number.
A simple rule to remember:
When a plane wave of wave vector k 1 is incident on a spatial harmonics, such as a grating with spacing d extending in the x-direction, the wave vector, k 2 of the outgoing wave can be related to k 1 by the following relation.
( k 2 ± k 1 )• x ˆ= 2 πν x N
Where N is an integer.
k=2 ππππ / λλλλ
1 / νννν x
k
kz
θθθθ^2^ ππππ /^ νννν x x
k 1
1/ ν x k 2
k2z
2 πν x ± k1x
The H factor can be simplified in the limit of small spatial period compared to the wavelength. (1/ν <<λ). ( What is the meaning ?)
Using the Fresnel approximation (small angle
approximation) 1 /^2
2 2 2 1 / 2 21 / 2 2 )^2 (^1 )^2 (^12 )
d d x y d
Thus (14) becomes
( 2 2 ) ( , )
jkd j d x y H x y e e
πλ ν ν ν ν
− − + = (^) (15)
Thus the phase change as a function of propagation distance d in free space result in a phase
change that is quadratic function of spatial frequency ν ’s.
First the function f(x,y) at z=0 can be expanded using the spatial harmonics:
− + = (^) x y j x y f x y F x y e d d ( , ) (ν ,ν )^2 π^ (ν x^ ν y ) ν ν (17)
The integrand is then propagate in free space to plane A, using (15). Then after the lens to plane B using (16)
f
x x y y j x y
j x y x y
f j d
x y j jkd B B
A e
U x y z e e e F e x y x y
λ
π
π λ πλ ν ν πν ν
ν ν
ν ν 2 0 2 0
2 2
2 2
( ) ( )
( ) 2 ( )
( , )
( , , ) ( , )
− + −
− + − +
−
=
=
where (^ , ) ( , )
( )(^22 ) x y
jkd j d f A (^) x y e e F ν ν = − πλ^ − ν x +^ ν y ν ν
, x 0 = λν xf , and y 0 = λν yf.
The waveform further propagates from plane B to the focal plane.
Results:
( , ) ( , ) f
y f
x g x y hfF
where (^) l e jkf f
j h =( ) −^2
A lens has the effect of bring the far-field angular distribution into the focal plane.
Z=
f f Focal plane
f(x,y) g(x,y)
Holography
Interference fringes caused by two plane waves
Two plane waves whose wave vector subtends an angle θ create interference fringe lines in
the direction bisecting the angle and with a separation
2 sin(
θ
d =. (19)
Example: Counter-propagating beams create standing waves of spacing λ /.
Figure 1
Thin planar holography Show that a grating with the spacing of the fringes, Eq. (19) is illuminated by wave k 1 , one of the diffracted waves is in the direction of k 2.
Figure 2
Actually, there may be two diffractive waves, corresponding to different grating orders.
Bragg condition
Path difference of waves from successive layers is
∆ = AB + BC − CD = 2 d cos θ (20)
Maximum reflection occurs when ∆ equals the wavelength.
2 sin(^ θ 2 )
d = λ
k 1 k^2
2 sin(^ θ 2 )
d = λ
θθθθ
Consider am arbitrary object waves at ( x,y,z ) to be E ( x , z ) e −^ j (^ kxx + kzz ) making an local angle θ
at ( x,y,z ). Assuming that the reference wave is a plane wave Er eikz in the z-direction.
The superposition of the two result in interference fringes pattern. At z=0, the intensity is
( , ) ( , ) ( , ) ( , )
(^222) E E x y e E E E x y e E E x y e E x y
jkx r
jkx r r
jkx r
If the fringe pattern is recorded on a transparency and illuminated by plane waves propagating in the z-direction, the diffractive waves are
jkx jkz jkz r
jkx jkz r
jkz Ediffracted Ere E E x y e E E x y e E x y e ∝ 2 −^ + ( , ) − x^ − z + ( , ) x − z +^2 ( , ) −
The first the last terms are the transmitted reference waves, and the filtered intensity. The second term is the reconstructed the object waves and the third term is the conjugation of the object waves. Often the reconstructed and the conjugate waves form the image and virtual image. . Problems:
- Prove Eqs. (5)-(8). You don’t need to turn in the answer, but must prove it once.
- Prove (19).
- The beam divergence at infinity of a plane wave passing through a single slit of width
is known to be D
θ = where the angle is measured from the maximum intensity at
the center to the first intensity minimum on either side, and D is the slit width for D>>
Use the transfer function approach to obtain the same relation. (a) First, express F(x), the x-dependency of the wave, after the slit using Fourier transform. (b) Apply the transfer function (15).
(c) Express the resultant function as a function of z
x θ = for z approaching
infinity,
D
F(x) H( ν x)F(x)
z=0 z →∞
- Find all possible directions of the diffracted beams when the planar grating, shown in Figure 2, is illuminated at an angle θ/2 from the normal. The grating is created by the interference of two waves shown in Figure 1.
- Prove the Bragg condition Eq(20).
- Design a holographic converging lens of focal length f for wavelength λ. Specify the fringe spacing as a function of position in the plane of the hologram.
