Fourier transforming property of lenses
MIT 2.71/2.710
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Fourier transform by far field propagation or lens
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Spherical-plane wave duality
The two pictures above are interpretations of the same physical phenomenon.
On the left, the transparency is interpreted in the Huygens sense as a superposition of “spherical wavelets.”
Each spherical wavelet is collimated by the lens and contributes to the output a plane wave, propagating at the
appropriate angle (scaled by f.)
On the right, the transparency is interpreted in the Fourier sense as a superposition of plane waves (“angular” or
“spatial frequencies.”) Each plane wave is transformed to a converging spherical wave by the lens and contributes
to the output, f to the right of the lens, a point image that carries all the energy that departed from the input at the
corresponding spatial frequency.
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Fourier transforming by lenses
lens
f
lens
f
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f
f
lens
f
f
lens
f
f
Imaging: the 4F system
The 4F system (telescope with finite conjugates
one focal distance to the left of the objective and
one focal distance to the right of the collector,
respectively) consists of a cascade of two Fourier
transforms
collector lens
f
objective lens
f
plane wave
illumination
Fourier
plane
(pupil plane)
thin
transparency
MIT 2.71/2.710
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f
f
image
plane
Spatial filtering: the 4F system
Spatial frequencies which have the misfortune of
hitting the opaque portions of the pupil plane
transparency vanish from the output. Of course the
transparency may be gray scale (partial block) or a
phase mask; the latter would introduce
relative phase delay between
collector lens
spatial frequencies.
f
block
pass
objective lens
f
plane wave
illumination
thin
transparency
MIT 2.71/2.710
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f
Fourier
plane
(pupil plane)
transparency
f
image
plane
Imaging and spatial filtering: physical justification
input transparency:
decomposed into
Huygens wavelets
image
plane
g
divergin
l
a
spheric
wave
con
v
sph ergin
g
er
wa ical
(foc ve
usi
ng)
collimate
d
plane
wave
illumination
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conv
erg
sphe ing
rica
wave l
(focu
sing)
Fourier (pupil)
plane
diverging
spherical
wave
diffraction order
comes to focus
collector
input transparency:
decomposed into
spatial frequencies
wave
plane order)
ction
(diffra
objective
plane
wave
illumination
image
plane
collimate
d
Today
•
•
Spatial filtering in the 4F system
The Point-Spread Function (PSF)
and Amplitude Transfer Function (ATF)
next Wednesday
•
•
•
•
Lateral and angular magnification
The Numerical Aperture (NA) revisited
Sampling the space and frequency domains, and
the Space-Bandwidth Product (SBP)
Pupil engineering
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t
+1s
0th
or
ct
lle
Fourier (pupil)
plane
tion o
r
d
e
rs
focus
ing
diffrac
co
je
ob
input transparency:
sinusoidal amplitude
grating
ct
iv
e
Spatial filtering by a telescope (4F system)
image
replicates the object
collimate
d
−1 st
plane
wave
illumination
input transparency:
sinusoidal amplitude
grating
diffraction orders
focused
t
+1s
0th
diffrac
tion o
focus rders
ing
pupil
mask
−1 st
plane
wave
illumination
MIT 2.71/2.710
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diffraction orders diffraction order
blocked
passing
blocked spatial frequencies
are missing from the image
t
+1s
0th
tion o
focus rders
ing
ct
pupil
mask
lle
diffrac
co
je
ob
input transparency:
sinusoidal amplitude
grating
ct
iv
or
e
Low-pass filtering: analysis
blocked spatial frequencies
are missing from the image
−1 st
plane
wave
illumination
diffraction orders diffraction order
blocked
passing
field after input transparency
field before pupil mask
blocked by the pupil filter
field after pupil mask
field at output
(image plane)
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tion o
focus rders
ing
ct
lle
diffrac
co
je
nd
+2
t
+1s
0th
−1 st
−2 nd
ob
input transparency
binary amplitude
grating
ct
iv
e
pupil
mask
or
Example: low-pass filtering a binary amplitude grating
blurred (smoothened)
image
plane
wave
illumination
diffraction orders diffraction orders
blocked
passing
Consider a binary amplitude grating, with perfect contrast m=1, period Λ=10µm, duty cycle 1/3 (33.3%),
illuminated by an on-axis plane wave at wavelength λ=0.5µm.
The 4F system consists of two identical lenses of focal length f=20cm.
A pupil mask of diameter (aperture) 3cm is placed at the Fourier plane, symmetrically about the optical axis.
What is the intensity observed at the output (image) plane?
The sequence to solve this kind of problem is:
➡ calculate the Fourier transform of the input transparency and scale to the pupil plane coordinates x”=uλf1
➡ multiply by the complex amplitude transmittance of the pupil mask
➡ Fourier transform the product and scale to the output plane coordinates x’=uλf2
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Example: low-pass filtering a binary amplitude grating
pupil mask
1
1
0.75
0.75
|gP(x’’)|2 [a.u.]
|gt(x)|2 [a.u.]
binary amplitude grating
0.5
diffraction
orders
passing
diffraction
orders
blocked
0.5
0.25
0.25
0
diffraction
orders
blocked
−30 −25 −20 −15 −10 −5
0 5
x [µm]
10 15 20 25 30
0
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
1
1.5
A binary amplitude grating of duty cycle α is expressed in a Fourier series harmonics expansion as
The field at the pupil plane to the left of the pupil mask is
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2
2.5
3
Example: low-pass filtering a binary amplitude grating
pupil mask
1
1
0.75
0.75
|gP(x’’)|2 [a.u.]
|gt(x)|2 [a.u.]
binary amplitude grating
0.5
diffraction
orders
passing
diffraction
orders
blocked
0.5
0.25
0.25
0
diffraction
orders
blocked
−30 −25 −20 −15 −10 −5
0 5
x [µm]
10 15 20 25 30
The pupil mask itself is
Its Fourier transform is
from which we may obtain the output field as
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0
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
1
1.5
2
2.5
3
so the field at the pupil plane to the right of the pupil mask is
Example: low-pass filtering a binary amplitude grating
pupil mask
1
1
0.75
0.75
|gP(x’’)|2 [a.u.]
|gt(x)|2 [a.u.]
binary amplitude grating
0.5
diffraction
orders
passing
diffraction
orders
blocked
0.5
0.25
0.25
0
diffraction
orders
blocked
−30 −25 −20 −15 −10 −5
0 5
x [µm]
10 15 20 25 30
0
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
1
1.5
2
2.5
3
Note the 2nd harmonic term
in the intensity, due to the
magnitude-square operation!
This term explains the
“ringing” in coherent low-pass
filtering systems
The output intensity is
MIT 2.71/2.710
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Example: low-pass filtering a binary amplitude grating
pupil mask
1
1
0.75
0.75
|gP(x’’)|2 [a.u.]
|gt(x)|2 [a.u.]
binary amplitude grating
0.5
diffraction
orders
blocked
diffraction
orders
passing
diffraction
orders
blocked
0.5
0.25
0.25
1
−30 −25 −20 −15 −10 −5
0 0.9
5
x [µm]
0.8
|gout(x’)|2 [a.u.]
0
0
10 15 20 25 30
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
low-pass filtered
binary amplitude grating
0.7
0.6
0.5
0.4
0.3
0.2
0.1
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0
−30 −25 −20 −15 −10 −5
0 5
x’ [µm]
10 15 20 25 30
1
1.5
2
2.5
3
tion o
focus rders
ing
ct
lle
diffrac
co
je
nd
+2
t
+1s
0th
−1 st
−2 nd
ob
input transparency
binary amplitude
grating
ct
iv
e
pupil
mask
or
Example: band-pass filtering a binary amplitude grating
amplitude: 1st harmonic
intensity: 2nd harmonic
plane
wave
illumination
diffraction orders diffraction orders
blocked
passing
Now consider the same optical system, but with a new pupil mask consisting of two holes, each of diameter
(aperture) 1cm and centered at ±1cm from the optical axis, respectively.
What is the intensity observed at the output (image) plane?
MIT 2.71/2.710
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Example: band-pass filtering a binary amplitude grating
pupil mask
1
1
0.75
0.75
diffraction
orders
passing
blocked
2
|gP(x’’)| [a.u.]
|gt(x)|2 [a.u.]
binary amplitude grating
0.5
0.25
0.25
0
0.5
−30 −25 −20 −15 −10 −5
0 5
x [µm]
10 15 20 25 30
The new pupil mask is
Its Fourier transform is
so the output field and intensity are
MIT 2.71/2.710
04/15/09 wk10-b- 10
0
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
1
1.5
2
2.5
so the field at the pupil plane to
the right of the pupil mask is
3
Example: band-pass filtering a binary amplitude grating
pupil mask
1
1
0.75
0.75
diffraction
orders
passing
blocked
2
|gP(x’’)| [a.u.]
|gt(x)|2 [a.u.]
binary amplitude grating
0.5
0.5
0.25
0.25
0.1
−30 −25 −20 −15 −10 −5
0.09
0 5 10 15 20 25 30
x [µm]
0.08
|gout(x’)|2 [a.u.]
0
0
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
1.5
2
2.5
3
band-pass filtered
binary amplitude grating
0.07
0.06
0.04
field: 1st harmonic
intensity: 2nd harmonic
(because of squaring)
0.03
contrast = 1
0.05
0.02
0.01
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1
0
−30 −25 −20 −15 −10 −5
0 5
x’ [µm]
10 15 20 25 30
tion o
focus rders
ing
ct
lle
diffrac
co
je
st
+1
0th
−1st
−2 nd
−3 rd
ob
input transparency
binary amplitude
grating
ct
iv
e
pupil
mask
or
Example: band-pass filtering a binary amplitude grating
with tilted illumination
amplitude: 1st harmonic
intensity: 2nd harmonic
plane
wave
illumination
diffraction orders diffraction orders
blocked
passing
Now consider the same optical system, again with the pupil mask consisting of two holes, each of diameter
(aperture) 1cm and centered at ±1cm from the optical axis, respectively. We illuminate this grating with an
off-axis plane wave at angle θ0=2.865o.
What is the intensity observed at the output (image) plane?
As you saw in a homework problem, the effect of rotating the input illumination is that the entire diffraction
pattern from the grating rotates by the same amount; so in this case the 0th order is propagating at angle θ0
off-axis, the +1st order at angle θ0+λ/Λ, etc.
Analytically, we find this by expressing the illuminating plane wave as
and the field after the input transparency gt(x) as
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Example: band-pass filtering a binary amplitude grating
with tilted illumination
pupil mask
1
1
0.75
0.75
diffraction
orders
passing
blocked
2
|gP(x’’)| [a.u.]
|gt(x)|2 [a.u.]
binary amplitude grating
0.5
0.25
0.25
0
0.5
−30 −25 −20 −15 −10 −5
0 5
x [µm]
10 15 20 25 30
0
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
1
1.5
The field to the left of the pupil mask is the Fourier transform of gin(x). Using the shift theorem,
all diffracted orders are
displaced by 1cm
in the positive x” direction
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2
2.5
3
Example: band-pass filtering a binary amplitude grating
with tilted illumination
pupil mask
1
1
0.75
0.75
diffraction
orders
passing
blocked
2
|gP(x’’)| [a.u.]
|gt(x)|2 [a.u.]
binary amplitude grating
0.5
0.25
0.25
0
0.5
−30 −25 −20 −15 −10 −5
0 5
x [µm]
10 15 20 25 30
After passing through the pupil mask
so the output field and intensity are
MIT 2.71/2.710
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0
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
1
1.5
, the field is
2
2.5
3
Example: band-pass filtering a binary amplitude grating
with tilted illumination
pupil mask
1
1
0.75
0.75
diffraction
orders
passing
blocked
2
|gP(x’’)| [a.u.]
|gt(x)|2 [a.u.]
binary amplitude grating
0.5
0.5
0.25
0.25
0.3
0
−30 −25 −20 −15 −10 −5
0 0.25
5 10 15 20 25 30
x [µm]
0
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
1.5
2
2.5
0.15
field: 1st harmonic
intensity: 2nd harmonic
(because of squaring)
0.1
contrast ≈ 0.7
0.05
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0
3
band-pass filtered
binary amplitude grating
with tilted illumination
0.2
|gout(x’)|2 [a.u.]
1
−30 −25 −20 −15 −10 −5
0 5
x’ [µm]
10 15 20 25 30
tion o
focus rders
ing
ct
lle
diffrac
co
je
nd
+2
t
+1s
0th
−1 st
−2 nd
ob
input transparency
binary amplitude
grating
ct
iv
e
pupil
mask
or
Example: band-pass filtering a binary amplitude grating
amplitude: 2nd harmonic
intensity: 4th harmonic
plane
wave
illumination
diffraction orders diffraction orders
blocked
passing
Consider the same optical system yet again, with a new pupil mask consisting of two holes, each of
diameter (aperture) 1cm and centered further away from the axis at ±2cm from the optical axis, respectively.
What is the intensity observed at the output (image) plane?
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Example: band-pass filtering a binary amplitude grating
pupil mask
1
1
0.75
0.75
diffraction
orders
passing
blocked
2
|gP(x’’)| [a.u.]
|gt(x)|2 [a.u.]
binary amplitude grating
0.5
0.25
0.25
0
0.5
−30 −25 −20 −15 −10 −5
0 5
x [µm]
10 15 20 25 30
The new pupil mask is
Its Fourier transform is
so the output field and intensity are
MIT 2.71/2.710
04/15/09 wk10-b- 17
0
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
1
1.5
2
2.5
so the field at the pupil plane to
the right of the pupil mask is
3
Example: band-pass filtering a binary amplitude grating
pupil mask
1
1
0.75
0.75
diffraction
orders
passing
blocked
2
|gP(x’’)| [a.u.]
|gt(x)|2 [a.u.]
binary amplitude grating
0.5
0.5
0.25
0.25
0.025
−30 −25 −20 −15 −10 −5
0 0.02
5 10 15 20 25 30
x [µm]
|gout(x’)|2 [a.u.]
0
0
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
1
1.5
2
2.5
band-pass filtered
binary amplitude grating
0.015
field: 2nd harmonic
intensity: 4th harmonic
(because of squaring)
0.01
contrast = 1
0.005
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0
3
−30 −25 −20 −15 −10 −5
0 5
x’ [µm]
10 15 20 25 30
Example: binary amplitude grating through phase pupil mask
plane
wave
illumination
tion o
focus rders
ing
or
ct
lle
diffrac
co
nd
+2
t
+1s
0th
−1 st
−2 nd
ob
input transparency
binary amplitude
grating
je
ct
iv
e
pupil
mask
phase
delayed
arbitrary
1cm
3cm
diffraction orders diffraction orders
blocked
passing
s=0.25µm
We finally consider a pupil mask consisting of 3cm aperture placed
symmetrically with respect to the optical axis, and filled with a glass
transparency that is thicker by 0.25µm in its central 1cm-wide portion.
This is known as a “phase pupil mask” or “pupil phase mask.”
What is the intensity observed at the output (image) plane?
The phase mask imparts a phase delay in the portion of the optical field
that strikes the region where the glass is thicker. The phase delay is
in this case.
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Example: binary amplitude grating through phase pupil mask
binary amplitude grating
pupil
mask
1
|gPM| [a.u.]
1
diffraction
orders
blocked
diffraction
orders
blocked
0.5
0.25
0
−4
−3
−2
−1
0
x’’ [cm]
1
2
3
4
1
2
3
4
0.5
pi
phase(gPM) [rad]
|gt(x)|2 [a.u.]
0.75
0.75
passing
0.25
0
−30 −25 −20 −15 −10 −5
0 5
x [µm]
10 15 20 25 30
delayed
pi/2
0
−pi/2
−pi
−4
−3
−2
−1
This pupil mask is
so the field at the pupil plane
to the right of the pupil mask is
and the output field and
intensity are
MIT 2.71/2.710
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compare with slide 14 (low-pass filter without phase mask)
0
x’’ [cm]
Example: binary amplitude grating through phase pupil mask
binary amplitude grating
pupil
mask
1
|gPM| [a.u.]
1
diffraction
orders
blocked
diffraction
orders
blocked
0.5
0.25
0
−4
−3
−2
−1
0
x’’ [cm]
1
2
3
4
1
2
3
4
0.5
pi
phase(gPM) [rad]
|gt(x)|2 [a.u.]
0.75
0.75
passing
0.25
0.3
0
−30 −25 −20 −15 −10 −5 0 0.25
5 10 15 20 25 30
x [µm]
delayed
pi/2
0
−pi/2
−pi
−4
−3
−2
−1
binary amplitude grating
filtered by
pupil phase mask
2
|gout(x’)| [a.u.]
0.2
0.15
field: 1st harmonic
intensity: 2nd harmonic
(because of squaring)
0.1
0.05 with slide 14 (low-pass filter without phase mask)
compare
MIT 2.71/2.710
04/15/09 wk10-b-21
0
0
x’’ [cm]
−30 −25 −20 −15 −10 −5
0 5
x’ [µm]
10 15 20 25 30
or
ct
lle
co
je
pupil
mask
ob
input transparency
ideal point source
ct
iv
e
The Point-Spread Function (PSF) of a low-pass filter
PSF
plane
wave
illumination
ideal
(infinitely wide)
spherical wave
ideal
(infinitely wide)
plane wave
truncated
plane wave
wave converging
to form the PSF
at the output plane
Now consider the same 4F system but replace the input transparency with an ideal point source,
implemented as an opaque sheet with an infinitesimally small transparent hole and illuminated with a plane
wave on axis (actually, any illumination will result in a point source in this case, according to Huygens.)
In Systems terminology, we are exciting this linear system with an impulse (delta-function);
therefore, the response is known as Impulse Response.
In Optics terminology, we use instead the term Point-Spread Function (PSF) and we denote it as h(x’,y’).
The sequence to compute the PSF of a 4F system is:
➡ observe that the Fourier transform of the input transparency δ(x) is simply 1 everywhere at the pupil plane
➡ multiply 1 by the complex amplitude transmittance of the pupil mask
➡ Fourier transform the product and scale to the output plane coordinates x’=uλf2.
➡Therefore, the PSF is simply the Fourier transform of the pupil mask, scaled to the output coordinates x’=uλf2
MIT 2.71/2.710
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Example: PSF of a low-pass filter
pupil mask
PSF
1
3
h(x’) [a.u.]
2
2
|gP(x’’)| [a.u.]
0.75
0.5
0.25
0
1
0
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5
x’’ [cm]
The pupil mask is
1
1.5
2
2.5
3
−1
−20
−15
−10
−5
0
x’ [µm]
5
10
15
. If the input transparency is δ(x), the field at the pupil plane to the
right of the pupil mask is
The output field, i.e. the PSF is
The scaling factor (3×) in the PSF ensures that the integral ∫|h(x’)|2dx equals the portion of the input energy
transmitted through the system
MIT 2.71/2.710
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20
Example: PSF of a phase pupil filter
PSF
5
|h(x’)|2 [a.u.]
|gPM| [a.u.]
pupil
mask
1
0.75
0.5
0.25
0
−4
−3
−2
−1
0
x’’ [cm]
1
2
3
2
0
−20
4
−15
−10
−5
0
x’ [µm]
5
10
15
20
−15
−10
−5
0
x’ [µm]
5
10
15
20
pi
phase[h(x’)] [rad]
phase(gPM) [rad]
3
1
pi
pi/2
0
−pi/2
−pi
−4
4
−3
−2
The pupil mask is
The PSF is
MIT 2.71/2.710
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−1
0
x’’ [cm]
1
2
3
4
pi/2
0
−pi/2
−pi
−20
Comparison: low-pass filter vs phase pupil mask filter
PSF: phase pupil mask filter
5
8
|h(x’)|2 [a.u.]
|h(x’)|2 [a.u.]
PSF:
low-pass filter
10
6
4
2
0
−20
−15
−10
−5
0
x’ [µm]
5
10
15
2
0
−20
20
−15
−10
−5
0
x’ [µm]
5
10
15
20
−15
−10
−5
0
x’ [µm]
5
10
15
20
pi
phase[h(x’)] [rad]
phase[h(x’)] [rad]
3
1
pi
pi/2
0
−pi/2
−pi
−20
4
−15
−10
MIT 2.71/2.710
04/15/09 wk10-b-25
−5
0
x’ [µm]
5
10
15
20
pi/2
0
−pi/2
−pi
−20
MIT OpenCourseWare
http://ocw.mit.edu
2.71 / 2.710 Optics
Spring 2009
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