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Fourier transforming property of lenses
MIT 2.71/2.710 04/13/09 wk10-a- 6
Fourier transform by far field propagation or lens
MIT 2.71/2.710 04/13/09 wk10-a- 7
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.
MIT 2.71/2.710 04/13/09 wk10-a- 8
Fourier transforming by lenses
MIT 2.71/2.710 04/13/09 wk10-a- 9
f flens
f flens
f flens
f flens
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
image objective lens plane
plane wave Fourierillumination plane
(pupil plane)
f f
f f
thin transparency
MIT 2.71/2.710 04/13/09 wk10-a-10
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 thetransparency may be gray scale (partial block) or aphase mask; the latter would introduce relative phase delay between spatial frequencies. collector lens
image
objective lens plane
plane wave Fourierillumination plane
(pupil plane) thin transparency
transparency MIT 2.71/2.710 04/13/09 wk10-a- 11
f f
f f
pass block
Imaging and spatial filtering: physical justificationinput transparency: image decomposed into
plane wave
illumination
plane wave
illumination
Huygens wavelets
input transparency: decomposed into
spatial frequencies
diverging
sphericalwave
plane wave
(diffraction order)
collimated
convergingsphericalwave(focusing)
convergingsphericalwave(focusing)
divergingspherical
wave collimated
plane
image plane
Fourier (pupil) plane
obje
ctiv
e
colle
ctor
diffraction order comes to focus
MIT 2.71/2.710 04/13/09 wk10-a-12
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
MIT 2.71/2.710 04/15/09 wk10-b- 1
objec
tive
colle
ctor
Spatial filtering by a telescope (4F system)
input transparency: image
plane wave
illumination
sinusoidal amplitude grating +1st
diffraction ordersfocusing collimated
Fourier (pupil)
diffraction orders focused
−1st
0th
plane replicates the object
pupil blocked spatial frequenciesinput transparency:
plane wave
illumination
sinusoidal amplitude grating +1st
diffraction ordersfocusing
diffraction orders blocked
−1st
0th
diffraction order passing
mask are missing from the image
MIT 2.71/2.710 04/15/09 wk10-b- 2
Low-pass filtering: analysis
plane wave
illumination
input transparency: sinusoidal amplitude
grating +1st diffraction ordersfocusing
blocked spatial frequencies are missing from the image
pupil mask
diffraction orders blocked
−1st
0th
diffraction order passing
objec
tive
colle
ctor
field after input transparency
blocked by the pupil filter
field before pupil mask
field after pupil mask
field at output(image plane)
MIT 2.71/2.710 04/15/09 wk10-b- 3
Example: low-pass filtering a binary amplitude grating
plane wave
illumination
+1st diffraction ordersfocusing
pupil mask
−1st 0th
colle
ctor
+2nd
−2nd
objec
tive
blurred (smoothened) image
input transparency binary amplitude
grating
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
MIT 2.71/2.710 04/15/09 wk10-b- 4
Example: low-pass filtering a binary amplitude grating
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
MIT 2.71/2.710 04/15/09 wk10-b- 5
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
binary amplitude grating pupil mask
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
diffraction orders
passing
diffraction orders
blocked
diffraction orders
blocked
Example: low-pass filtering a binary amplitude grating
The pupil mask itself is so the field at the pupil plane to the right of the pupil mask is
Its Fourier transform is
from which we may obtain the output field as
6
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
MIT 2.71/2.710 04/15/09 wk10-b-
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
binary amplitude grating pupil mask
diffraction diffraction diffraction orders orders
blocked passing blocked orders
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
MIT 2.71/2.710 04/15/09 wk10-b-
Example: low-pass filtering a binary amplitude grating
7
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
binary amplitude grating pupil mask
diffraction orders
passing
diffraction orders
blocked
diffraction orders
blocked
The output intensity is
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
Example: low-pass filtering a binary amplitude grating
MIT 2.71/2.710 04/15/09 wk10-b- 8
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x’ [µm]
|gou
t(x’)|
2 [a.u
.]
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
binary amplitude grating pupil mask
diffraction diffraction diffraction orders orders
blocked passing blocked orders
low-pass filtered binary amplitude grating
Example: band-pass filtering a binary amplitude grating
plane wave
illumination
+1st diffraction ordersfocusing
pupil mask
−1st 0th
colle
ctor
+2nd
−2nd
objec
tive
amplitude: 1st harmonic intensity: 2nd harmonic
input transparency binary amplitude
grating
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 04/15/09 wk10-b- 9
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
MIT 2.71/2.710 04/15/09 wk10-b-
Example: band-pass filtering a binary amplitude grating
10
binary amplitude grating pupil mask
Its Fourier transform is
The new pupil mask is so the field at the pupil plane to the right of the pupil mask is
so the output field and intensity are
diffraction orders
passing
blocked
Example: band-pass filtering a binary amplitude grating
MIT 2.71/2.710 04/15/09 wk10-b- 11
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
x’ [µm]
|gou
t(x’)|
2 [a.u
.]
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
diffraction orders
passing
blocked
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
binary amplitude grating pupil mask
band-pass filtered binary amplitude grating
field: 1st harmonic intensity: 2nd harmonic (because of squaring)
contrast = 1
Example: band-pass filtering a binary amplitude gratingwith tilted illumination
plane wave
illumination
0th diffraction ordersfocusing
pupil mask
−2nd −1st
colle
ctor
+1st
−3rd
objec
tive
amplitude: 1st harmonic intensity: 2nd harmonic
input transparency binary amplitude
grating
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.
MIT 2.71/2.710 04/15/09 wk10-b-12
Analytically, we find this by expressing the illuminating plane wave as and the field after the input transparency gt(x) as
Example: band-pass filtering a binary amplitude grating
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
binary amplitude grating pupil mask
diffraction orders
passing
blocked
with tilted illumination
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
MIT 2.71/2.710 04/15/09 wk10-b-13
Example: band-pass filtering a binary amplitude gratingwith tilted illumination
After passing through the pupil mask , the field is
so the output field and intensity are
MIT 2.71/2.710 04/15/09 wk10-b-14
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
binary amplitude grating pupil mask
diffraction orders
passing
blocked
Example: band-pass filtering a binary amplitude gratingwith tilted illumination
MIT 2.71/2.710 04/15/09 wk10-b-15
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.05
0.1
0.15
0.2
0.25
0.3
x’ [µm]
|gou
t(x’)|
2 [a.u
.]
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
binary amplitude grating pupil mask
diffraction orders
passing
blocked
band-pass filtered binary amplitude grating
with tilted illumination field: 1st harmonic
intensity: 2nd harmonic (because of squaring)
contrast ≈ 0.7
Example: band-pass filtering a binary amplitude grating
plane wave
illumination
+1st diffraction ordersfocusing
pupil mask
−1st 0th
colle
ctor
+2nd
−2nd
objec
tive
input transparency binary amplitude
grating
amplitude: 2nd harmonic intensity: 4th harmonic
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?
MIT 2.71/2.710 04/15/09 wk10-b-16
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
MIT 2.71/2.710 04/15/09 wk10-b-
Example: band-pass filtering a binary amplitude grating
17
binary amplitude grating pupil mask
Its Fourier transform is
diffraction orders
passing
blocked
The new pupil mask is so the field at the pupil plane to the right of the pupil mask is
so the output field and intensity are
Example: band-pass filtering a binary amplitude grating
MIT 2.71/2.710 04/15/09 wk10-b-18
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.005
0.01
0.015
0.02
0.025
x’ [µm]
|gou
t(x’)|
2 [a.u
.]
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
diffraction orders
passing
blocked
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
binary amplitude grating pupil mask
band-pass filtered binary amplitude grating
field: 2nd harmonic intensity: 4th harmonic (because of squaring)
contrast = 1
Example: binary amplitude grating through phase pupil mask
MIT 2.71/2.710 04/15/09 wk10-b-19
plane wave
illumination
+1st diffraction ordersfocusing
pupil mask
diffraction orders blocked
−1st 0th
diffraction orders passing
colle
ctor
+2nd
−2nd
objec
tive
phase delayed
arbitrary
3cm
1cm
s=0.25µm 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.
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?
input transparency binary amplitude
grating
−4 −3 −2 −1 0 1 2 3 40
0.25
0.5
0.75
1
x’’ [cm]
|gPM
| [a.
u.]
−4 −3 −2 −1 0 1 2 3 4 −pi
−pi/2
0
pi/2
pi
x’’ [cm]
phas
e(g PM
) [ra
d]
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
MIT 2.71/2.710 04/15/09 wk10-b-
Example: binary amplitude grating through phase pupil mask
20
binary amplitude grating pupil mask diffraction
orders blocked
diffraction orders
blocked
so the field at the pupil plane to the right of the pupil mask is
and the output field and intensity are
This pupil mask is
compare with slide 14 (low-pass filter without phase mask)
delayed
passing
Example: binary amplitude grating through phase pupil mask
MIT 2.71/2.710 04/15/09 wk10-b-21
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.05
0.1
0.15
0.2
0.25
0.3
x’ [µm]
|gou
t(x’)|
2 [a.u
.]
−4 −3 −2 −1 0 1 2 3 40
0.25
0.5
0.75
1
x’’ [cm]
|gPM
| [a.
u.]
−4 −3 −2 −1 0 1 2 3 4 −pi
−pi/2
0
pi/2
pi
x’’ [cm]
phas
e(g PM
) [ra
d]
−30 −25 −20 −15 −10 −5 0 5 10 15 20 25 300
0.25
0.5
0.75
1
x [µm]
|gt(x
)|2 [a.u
.]
binary amplitude grating pupil mask diffraction diffraction
passing
orders blockedblocked
orders
compare with slide 14 (low-pass filter without phase mask)
delayed
binary amplitude grating filtered by
pupil phase mask field: 1st harmonic
intensity: 2nd harmonic (because of squaring)
The Point-Spread Function (PSF) of a low-pass filter
plane wave
illumination
pupil mask
ideal truncated
colle
ctor
objec
tive
PSFinput transparency ideal point source
ideal wave converging (infinitely wide) (infinitely wide) plane wave to form the PSF spherical wave plane wave 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
MIT 2.71/2.710 04/15/09 wk10-b-22
➡ 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
Example: PSF of a low-pass filterPSF
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 30
0.25
0.5
0.75
1
x’’ [cm]
|gP(x
’’)|2 [a
.u.]
pupil mask
The output field, i.e. the PSF is
The pupil mask is . If the input transparency is δ(x), the field at the pupil plane to the
right of the pupil mask is
−20 −15 −10 −5 0 5 10 15 20−1
0
1
2
3
x’ [µm]
h(x’
) [a.
u.]
The scaling factor (3×) in the PSF ensures that the integral ∫|h(x’)|2dx equals the portion of the input energy MIT 2.71/2.710 transmitted through the system 04/15/09 wk10-b-23
Example: PSF of a phase pupil filter
The pupil mask is
The PSF is
MIT 2.71/2.710 04/15/09 wk10-b-24
−4 −3 −2 −1 0 1 2 3 40
0.25
0.5
0.75
1
x’’ [cm]
|gPM
| [a.
u.]
−4 −3 −2 −1 0 1 2 3 4 −pi
−pi/2
0
pi/2
pi
x’’ [cm]
phas
e(g PM
) [ra
d]
pupil mask
−20 −15 −10 −5 0 5 10 15 200
1
2
3
4
5
x’ [µm]
|h(x
’)|2 [a
.u.]
−20 −15 −10 −5 0 5 10 15 20 −pi
−pi/2
0
pi/2
pi
x’ [µm]
phas
e[h(
x’)]
[rad]
PSF
Comparison: low-pass filter vs phase pupil mask filter
−20 −15 −10 −5 0 5 10 15 200
1
2
3
4
5
x’ [µm]
|h(x
’)|2 [a
.u.]
−20 −15 −10 −5 0 5 10 15 20 −pi
−pi/2
0
pi/2
pi
x’ [µm]
phas
e[h(
x’)]
[rad]
PSF: phase pupil mask filter
−20 −15 −10 −5 0 5 10 15 200
2
4
6
8
10
x’ [µm]
|h(x
’)|2 [a
.u.]
−20 −15 −10 −5 0 5 10 15 20 −pi
−pi/2
0
pi/2
pi
x’ [µm]
phas
e[h(
x’)]
[rad]
PSF: low-pass filter
MIT 2.71/2.710 04/15/09 wk10-b-25
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2.71 / 2.710 Optics Spring 2009
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