2.710 Optics Problem Set 8 Solutions

1. (a) For a diffraction limited system the slopes of the OTF are constant.

m∣∣u=25mm−1 = 68.75% = 0.6875

Iin =1

2

[1 + cos

(2π

x

Λ

)]⇒ Iin =

1

2δ(u) +

1

4δ

(u− 1

Λ

)+

1

4δ

(u +

1

Λ

)Iout = Iin ·OTF =

1

2δ(u) +

a

4δ

(u− 1

Λ

)+

a

4δ

(u +

1

Λ

)Iout(x

′) =1

2

[1 + a cos

(2π

x′

Λ

)]m

∣∣u= 1

Λ

=(1

2+ a

2)− (1

2− a

2)

(12

+ a2) + (1

2− a

2)

= a

∴ the contrast is the normalized value of the OTF at that frequency. Using similartriangles, if m

∣∣u=25mm−1 = 0.6875 = (1− 0.3125), then

m∣∣u=50mm−1 = (1− 0.6250) = 0.3750 = 37.5 %

(b) The cut-off frequency for incoherent imaging is u0 = 80mm−1. The cut-off fre-quency of the coherently illuminated system is 40mm−1. Hence 50mm−1 frequen-cies do NOT go through if it is coherently illuminated.

2.

I(x) =1

2

[1 +

1

2cos

(2π

x

40µm

)+

1

2cos

(2π

3x

40µm

)](a) The contrast m is given by:

m =Imax − Imin

Imax + Imin

At the input, m = 1−01+0

= 1.

1

(b) The Fourier transform of I(x) is:

I(u) =1

8

[δ

(u− 1

40

)+ δ

(u +

1

40

)+ δ

(u− 3

40

)+ δ

(u +

3

40

)]+

1

2δ(u)

The Fourier transform of the output intensity is:

I0(u) = (MTF) · I(u)

=1

2δ(u) + (0.25)

1

8

[δ

(u− 1

40

)+ δ

(u +

1

40

)]=

1

2δ(u) +

1

16

[1

2δ

(u− 1

40

)+

1

2δ

(u +

1

40

)]I0(x

′) =1

2+

1

16cos

(2π

x′

40

)mout =

(12

+ 116

)− (12− 1

16)

(12

+ 116

) + (12− 1

16)

=1

8= 0.125

(c) The incoherent transfer function is an autocorrelation of the coherent transferfunction. The coherent transfer function in this case is probably a triangle functionwith half the cut-off frequency.

H(u) = F{h(x)}

2

3.h(x) = sinc2

(x

b

)(a) Incoherent iPSF

h(x) = |h(x)|2 = sinc4(x

b

)(b) MTF =

∣∣H(u)∣∣

H(u) = F{h(x)} = F{

sinc2(x

b

)· sinc2

(x

b

)}= F

{sinc2

(x

b

)}⊗F

{sinc2

(x

b

)}= bΛ(bu)⊗ bΛ(bu)

bΛ(bu)

MTF

3

2.710 Optics Spring �09Solutions to Problem Set #8

Due Wednesday, May 13, 2009Problem 4:a) Consider the system shown in Figure 1. At the input we place a sinusoidal

amplitude grating of perfect contrast,

gt(x) =1

2

h1 + cos

�2�x

�

�i; (1)

where � = 0:5�m, f1 = f2 = f = 20cm, � = 10�m, and the two apertures at the pupilplane 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 = jgt(x)j2 =1

4

h1 + cos

�2�x

�

�i2(2)

=1

4

h1 + cos2

�2�x

�

�+ 2 cos

�2�x

�

�i=

1

4

�1 +

1

2+1

2cos�4�x

�

�+ 2 cos

�2�x

�

��=

3

8+1

8cos�4�x

�

�+1

2cos�2�x

�

�;

where we used the following trigonometric identity,

cos2 (�) =1

2+1

2cos (2�) : (3)

Now we compute the frequency spectra of the input signal,

Figure 1: Optical system for problem 4.

1

Figure 2: Input signal spectrum.

Figure 3: Graphical calculation of OTF.

Gin = zfIing (4)

=3

8�(u) +

1

4

��(u� 1

�) + �(u+

1

�)

�+1

16

��(u� 2

�) + �(u+

2

�)

�;

which can be represented graphically as shown in Figure 2.

The ATF of the system is,

H = rect�x00 � xod

�+ rect

�x00+ xod

�; (5)

where xo is the lateral shift (1cm) and d is the aperture diamter (also 1cm). We com-pute the OTF of the optical system graphically as shown in Figure 3. The OTF is thenormalized cross-correlation of the ATF of the system.

To compute the spectrum of the output signal we multiply the OTF by the input spec-tra,

2

Gout = HGin (6)

=3

8�(u) +

1

32

��(u� 2

�) + �(u+

2

�)

�u=x

00�f

:

The output intensity is,

Iout = zfGoutg (7)

=3

8+1

16

24e�i2� 2x00� + ei2�2x00

�

2

35=

3

8+1

16cos

�2�2x

00

�

�:

(b) The constrast or visibility is given by,

V =Imax � IminImax + 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 theinput �eld is,

Gin = Ffgt(x)g =1

2�(u) +

1

4

��(u� 1

�) + �(u+

1

�)

�: (9)

The ATF of the system acts as a pass-band �lter as shown in Figure 4 and only allowsthe �rst di¤raction order to pass,

Gout =1

4

��(u� 1

�) + �(u+

1

�)

�u=x

00�f

: (10)

The output �eld is,

gout = zfGoutg (11)

=1

2

"e�i2�

x0� + ei2�

x0�

2

#

=1

2cos

�2�x0

�

�;

3

Figure 4: Band-pass �lter - coherent system.

and the output intensity is,

Iout = jgoutj2 (12)

=1

4cos2

�2�x0

�

�=

1

4

�1

2+1

2cos

�2�2x0

�

��=

1

8+1

8cos

�2�2x0

�

�:

For the coherent case we note the the contrast is V = 1.

4

2.710 Optics Spring �09Solutions to Problem Set #8

Due Wednesday, May 13, 2009Problem 5:a) Consider the system shown in Figure 1. The input transparency is a binary ampli-

tude 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 shifted2cm from the optical axis.

We begin by expressing the binary amplitude grating in a Fourier Series,

gt(x) = �

1Xq=�1

sinc (�q) ei2�qx� : (1)

Since equation 1 has a binary amplitude dependence that goes from 0 to 1, the intensityis also binary, Iin = jg(t)j2. The spectrum of the input signal is given by,

Gin(u) =

"�

1Xq=�1

sinc (�q) ��u� q

�

�#u=x

00�f

(2)

! Gin(x00) = �

1Xq=�1

sinc (�q) ��x00 � �fq

�

�;

and is shown in Figure 2.

The OTF, H, of the optical system is computed in the same way as in problem 4 andit is shown in Figure 3.

The output �eld spectrum is given by,

Gout = HGin (3)

=1

3�(x

00)�

p3

�16

��

�x00 � �f4

�

�+ �

�x00+�f4

�

��:

The output intensity is,

1

Figure 1: Optical system for problem 5.

Figure 2: Input signal spectrum.

Figure 3: System�s OTF.

2

Iout = zfGoutg (4)

=1

3�p3

�8

24e�i2� x04� + ei2�x04�

2

35=

1

3�p3

�8cos

�2�x04

�

�:

(b) The resulting contrast is,

V =Imax � IminImax + Imin

= 0:2067: (5)

(c) Comparying with the results from , for the coherent case, the outputintensity is,

Iout =3

8�2+

3

8�2cos

�2�x04

�

�;

and the contrast V = 1.

3

Lecture 19, p. 24

2.710 Optics Spring �09Solutions to Problem Set #8

Due Wednesday, May 13, 2009Problem 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 wave-length is � = 0:5�m, and the focal lengths are f1 = f2 = f = 20cm. The pupil maskis given by,

gPM(x00) =

�rect

�x00

a

�+ (i� 1) rect

�x00

b

��x00=u�f

(1)

! H(u) = gPM(u)

= rect�u�

�+ (i� 1) rect

�u

�

�;

where � = a=�f , � = b=�f , a = 3cm and b = 1cm. The magnitude and phase of thepupil mask are shown in Figure 2.

We now compute the coherent PSF,

h(x) = F�1fH(u)g (2)

= �sinc (�x) + (i� 1) �sinc (�x)= �sinc (�x)� �sinc (�x) + i�sinc (�x) :

The incoherent PSF is given by,

h(x) = jh(x)j2 = h � h� (3)

= [�sinc (�x)� �sinc (�x)]2 + �2sinc2 (�x)�2sinc2 (�x) + 2�2sinc2 (�x)� 2��sinc (�x) sinc (�x) :

The OTF of the system is found by computing the Fourier transform of equation 3,

H(u) = �triag�u�

�+ 2�triag

�u

�

�� 2

�rect

�u�

�� rect

�u

�

��; (4)

and is shown in Figure 3.

1

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) = �1X

q=�1sinc (�q) �

�u� q

�

�: (5)

To compute the spectrum of the output signal we multiply equations 5 and 6,

Gout(u) =1

3� (u) +

p3

12�

��

�u� 2

�

�+ �

�u+

2

�

��: (6)

The output intensity is given by,

Iout = FfGoutg (7)

=1

3+

p3

6�

24e�i2� x02� + ei2�x02�

2

35=

1

3+

p3

6�cos

�2�x02

�

�:

(b) The resulting contrast is,

V =Imax � IminImax + Imin

= 0:2757: (8)

3

(c) Finally, we compare our results to those from the coherent case as discussed in, where the output intensity is,

Iout(x0) =

�1

3

�2+

3

2�2+

3

2�2cos

�2�x02

�

�; (9)

and the contrast is V = 0:2548:

4

Lecture 19, p. 27

2.710 Optics Spring �09Solutions to Problem Set #8

Due Wednesday, May 13, 2009Problem 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 , we can see that the optical�eld at the Fourier plane for the special case of z = f is,

gF (x00; y

00) =

ei2�2f�

i�fG

�x00

�f;y00

�f

�: (1)

The intensity at that plane is,

IF = gF � g�F _ G �G�; (2)

which is proportional to the autocorrelation of the input �eld as they are related by aFourier transform,

=fgin ging = G �G�: (3)

Figure 1: Fourier transforming lens.

1

18

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