2.710 Optics
Problem Set 8 Solutions
1. (a) For a diffraction limited system the slopes of the OTF are constant.
mu=25mm−1 = 68.75% = 0.6875
x i
1h
Iin =
1 + cos 2π
2
Λ 1
1
1
1
1
ˆ
+ δ u+
⇒ Iin = δ(u) + δ u −
2
4
Λ
4
Λ
1
a
1
1
a
ˆ
ˆ
Iout = Iin · OTF = δ(u) + δ u −
+ δ u+
2
4
Λ
4
Λ
0
1
x
1 + a cos 2π
Iout (x0 ) =
2
Λ
1
a
1
a
( + ) − (2 − 2)
=a
mu= 1 = 21 a2
Λ
( 2 + 2 ) + ( 12 − a2 )
∴ the contrastis the normalized value of the OTF at that frequency. Using similar
triangles, if mu=25mm−1 = 0.6875 = (1 − 0.3125), then
mu=50mm−1 = (1 − 0.6250) = 0.3750 = 37.5 %
(b) The cut-off frequency for incoherent imaging is u0 = 80mm−1 . The cut-off frequency of the coherently illuminated system is 40mm−1 . Hence 50mm−1 frequencies do NOT go through if it is coherently illuminated.
2.
1
1
x
1
3x
I(x) =
1 + cos 2π
+ cos 2π
2
2
40µm
2
40µm
(a) The contrast m is given by:
m=
At the input, m =
1−0
1+0
= 1.
1
Imax − Imin
Imax + Imin
(b) The Fourier transform of I(x) is:
1
1
1
3
3
1
˜ =
I(u)
δ u−
+δ u+
+δ u−
+δ u+
+ δ(u)
8
40
40
40
40
2
The Fourier transform of the output intensity is:
˜
I˜0 (u) = (MTF) · I(u)
1
1
1
1
+δ u+
= δ(u) + (0.25) δ u −
2
8
40
40
1
1 1
1
1
1
= δ(u) +
δ u−
+ δ u+
2
16 2
40
2
40
0
1
x
1
cos 2π
I0 (x0 ) = +
2 16
40
1
1
1
1
( 2 + 16 ) − ( 2 − 16
)
1
mout = 1
= = 0.125
1
1
1
8
( 2 + 16 ) + ( 2 − 16 )
(c) The incoherent transfer function is an autocorrelation of the coherent transfer
function. The coherent transfer function in this case is probably a triangle function
with half the cut-off frequency.
H(u) = F{h(x)}
2
3.
h(x) = sinc2
x
b
(a) Incoherent iPSF
2
4
h̃(x) = |h(x)| = sinc
x
b
(b) MTF = H̃(u)
n
o
2 x
2 x
H̃(u) = F{h̃(x)} = F sinc
· sinc
n
x o
n b x o b
⊗ F sinc2
= F sinc2
b
b
= bΛ(bu) ⊗ bΛ(bu)
bΛ(bu)
MTF
3
2.710 Optics
Solutions to Problem Set #8
Spring ’09
Due Wednesday, May 13, 2009
Problem 4:
a) Consider the system shown in Figure 1. At the input we place a sinusoidal
amplitude grating of perfect contrast,
x i
1h
1 + cos 2
;
(1)
gt (x) =
2
where = 0:5 m, f1 = f2 = f = 20cm, = 10 m, and the two apertures at the pupil
plane have a diameter of 1cm and are separated by 1cm from the optical axis.
We begin by computing the input intensity (assuming uniform intensity illumination),
Iin =
=
=
=
x i2
1h
1 + cos 2
jgt (x)j =
4
h
x
x i
1
2
1 + cos 2
+ 2 cos 2
4
1
x
x
1 1
+ 2 cos 2
1 + + cos 4
4
2 2
x
1
x
3 1
+ cos 4
+ cos 2
;
8 8
2
2
(2)
where we used the following trigonometric identity,
cos2 ( ) =
1 1
+ cos (2 ) :
2 2
Now we compute the frequency spectra of the input signal,
Figure 1: Optical system for problem 4.
1
(3)
Figure 2: Input signal spectrum.
Figure 3: Graphical calculation of OTF.
Gin = zfIin g
1
3
(u) +
=
8
4
(4)
(u
1
) + (u +
1
) +
1
16
2
(u
) + (u +
2
) ;
which can be represented graphically as shown in Figure 2.
The ATF of the system is,
H = rect
x
00
00
xo
x + xo
+ rect
d
d
;
(5)
where xo is the lateral shift (1cm) and d is the aperture diamter (also 1cm). We compute the OTF of the optical system graphically as shown in Figure 3. The OTF is the
normalized cross-correlation of the ATF of the system.
To compute the spectrum of the output signal we multiply the OTF by the input spectra,
2
^ in
Gout = HG
3
1
=
(u) +
8
32
(6)
2
(u
) + (u +
2
)
00
:
u= x f
The output intensity is,
Iout = zfGout g
2
3
1 4e
=
+
8 16
i2
00
2x
+ ei2
2
1
2x
3
+
cos 2
=
8 16
00
2x
(7)
3
5
00
:
(b) The constrast or visibility is given by,
V =
Imax Imin
:
Imax + Imin
(8)
For the output intensity of equation 7, the contrast is V = 0:167:
(c) Finally we compare our results with the coherent case where the spectrum of the
input …eld is,
Gin = Ffgt (x)g =
1
1
(u) +
2
4
1
(u
) + (u +
1
) :
(9)
The ATF of the system acts as a pass-band …lter as shown in Figure 4 and only allows
the …rst di¤raction order to pass,
Gout =
1
4
(u
1
) + (u +
1
)
00
:
(10)
u= x f
The output …eld is,
gout = zfGout g
"
x0
1 e i2 + ei2
=
2
2
=
1
x0
cos 2
2
3
;
x0
#
(11)
Figure 4: Band-pass …lter - coherent system.
and the output intensity is,
Iout = jgout j2
1
x0
2
=
cos 2
4
2x0
1 1 1
+ cos 2
=
4 2 2
1 1
2x0
=
+ cos 2
:
8 8
For the coherent case we note the the contrast is V = 1.
4
(12)
2.710 Optics
Solutions to Problem Set #8
Spring ’09
Due Wednesday, May 13, 2009
Problem 5:
a) Consider the system shown in Figure 1. The input transparency is a binary amplitude grating with the following parameters: m = 1; duty cycle = = 1=3; = 10 m.
The operating wavelength is = 0:5 m, and the focal lengths are f1 = f2 = f = 20cm.
At the Fourier plane, the pupil mask has two apertures with a diameter of 1cm shifted
2cm from the optical axis.
We begin by expressing the binary amplitude grating in a Fourier Series,
1
X
gt (x) =
sinc ( q) ei2
qx
(1)
:
q= 1
Since equation 1 has a binary amplitude dependence that goes from 0 to 1, the intensity
is also binary, Iin = jg(t)j2 . The spectrum of the input signal is given by,
Gin (u) =
"
1
X
sinc ( q)
u
q
q= 1
00
! Gin (x ) =
1
X
sinc ( q)
#
(2)
00
u= x f
x
fq
00
;
q= 1
and is shown in Figure 2.
^ of the optical system is computed in the same way as in problem 4 and
The OTF, H,
it is shown in Figure 3.
The output …eld spectrum is given by,
^ in
Gout = HG
1
00
(x )
=
3
p
3
16
(3)
x
00
The output intensity is,
1
f4
+
00
x +
f4
:
Figure 1: Optical system for problem 5.
Figure 2: Input signal spectrum.
Figure 3: System’s OTF.
2
Iout = zfGout g
2
0
p
x 4
1
3 4 e i2
+ ei2
=
3
8
2
p
0
3
x4
1
cos 2
:
=
3
8
0
x 4
3
(4)
5
(b) The resulting contrast is,
V =
Imax Imin
= 0:2067:
Imax + Imin
(5)
(c) Comparying with the results from Lecture 19, p. 24 , for the coherent case, the output
intensity is,
Iout =
3
8
2
+
0
3
8
2
and the contrast V = 1.
3
cos 2
x4
;
2.710 Optics
Solutions to Problem Set #8
Spring ’09
Due Wednesday, May 13, 2009
Problem 6:
Consider the optical system shown in Figure 1. The input transparency is a binary
amplitude grating with the same parameters as in problem 5. The operating wavelength is = 0:5 m, and the focal lengths are f1 = f2 = f = 20cm. The pupil mask
is given by,
00
00
00
gP M (x ) =
x
rect
a
+ (i
x
1) rect
b
! H(u) = gP M (u)
u
= rect
+ (i 1) rect
u
(1)
x00 =u f
;
where = a= f , = b= f , a = 3cm and b = 1cm. The magnitude and phase of the
pupil mask are shown in Figure 2.
We now compute the coherent PSF,
h(x) = F 1 fH(u)g
= sinc ( x) + (i 1) sinc ( x)
= sinc ( x)
sinc ( x) + i sinc ( x) :
(2)
The incoherent PSF is given by,
^
h(x)
= jh(x)j2 = h h
= [ sinc ( x)
sinc ( x)]2 +
2
sinc2 ( x) + 2 2 sinc2 ( x)
(3)
2
2
sinc ( x)
2 sinc ( x) sinc ( x) :
The OTF of the system is found by computing the Fourier transform of equation 3,
u
^
H(u)
= triag
+ 2 triag
u
and is shown in Figure 3.
1
2 rect
u
rect
u
;
(4)
Figure 1: Optical system for problem 6.
Figure 2: Magnitud and phase of the pupil mask.
2
Figure 3: OTF and spectrum of the input signal.
From problem 5, the spectrum of the input signal is,
Gin (u) =
1
X
sinc ( q)
u
q
:
(5)
q= 1
To compute the spectrum of the output signal we multiply equations 5 and 6,
p
1
3
2
2
Gout (u) =
(u) +
u
+
u+
:
3
12
(6)
The output intensity is given by,
Iout = FfGout g
2
0
p
x 2
1
3 4 e i2
+ ei2
=
+
3
6
2
p
0
1
3
x2
=
+
cos 2
:
3
6
0
x 2
3
(7)
5
(b) The resulting contrast is,
V =
Imax Imin
= 0:2757:
Imax + Imin
3
(8)
(c) Finally, we compare our results to those from the coherent case as discussed in
Lecture 19, p. 27, where the output intensity is,
0
Iout (x ) =
1
3
2
+
3
2
2
and the contrast is V = 0:2548:
4
+
0
3
2
2
cos 2
x2
;
(9)
2.710 Optics
Solutions to Problem Set #8
Spring ’09
Due Wednesday, May 13, 2009
Problem 7:
In this problem we are required to show that the intensity at the Fourier plane of
the system shown in Figure 1 is proportional to the autocorrelation of the input …eld,
gin = gillum (x) gt (x). From the supplement to lecture 18 , we can see that the optical
…eld at the Fourier plane for the special case of z = f is,
2f
ei2
G
gF (x ; y ) =
i f
00
00
00
00
x y
;
f f
:
(1)
The intensity at that plane is,
IF = gF gF _ G G ;
(2)
which is proportional to the autocorrelation of the input …eld as they are related by a
Fourier transform,
=fgin
gin g = G G :
Figure 1: Fourier transforming lens.
1
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