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Optics Learning by Computing with Examples using MATLAB_Dieter Moller_2007

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392	10. IMAGING USING WAVE THEORY

Unnormalized

Application 10.6.

1.Change the distance between the two object bars and study the resolution. If the distance is too small, the image is just one peak.

2.Change the calculation and apply the spread function to the total image over a large range of integration. The image is now closer to the object.

10.4.5 Transfer Function

10.4.5.1 Transfer Function and Fourier Transformation

The spread function is used in the coordinate space of the image to calculate how the object appears on the observation screen (image coordinates). If the lens did not alter the object we would observe a perfect image of the object. The image was obtained by convolution of the object function Iob(Y ) with the spread function S(Y − Y ), both using image space coordinates

Iim(Y ) Iob(Y )S(Y − Y )dY .

(10.32)

From the convolution theorem we know that the integral in Eq. (10.32) expresses the Fourier transform of the product of the Fourier transform of Iob(Y ) and the Fourier transform of S(Y ). If we call these Fourier transforms

ω(µ)	Iob(Y )e−i2πµY dY	(10.33)
and
τ (µ)	S(Y )e−i2πµY dY ,	(10.34)

one may write Eq. (10.32) as
Iim(Y ) C ω(µ)τ (µ)ei2πµY dµ,		(10.35)

10.4. IMAGE FORMATION USING INCOHERENT LIGHT

393

where C contains all constants.

The Fourier transform of S(Y ), which is τ (µ), is called the (unnormalized) transfer function. Equation (10.35) tells us that the Fourier transform of Iim(Y ), which we call φim(µ), is equal to the product of the Fourier transforms ω(µ)τ (µ),

which is
φim ω(µ)τ (µ).	(10.36)

For the coordinates we have 1/Y µ, where µ may be interpreted as the spatial frequency in the frequency domain. The transfer function operates in the spatial frequency domain.

In FileFig 10.7 we calculate the Fourier transform of an object, which is a rectangular pulse, and the Fourier transform of a spread function for a cylindrical lens (transfer function). Then we multiply the Fourier transform of the object with the transfer function and perform the Fourier transformation of the product. The resulting image should look like a rectangular pulse.

FileFig 10.7 (W7PUTRAS)

Demonstration of the use of a transfer function. Object is a pulse and calculation of its Fourier transformation. Spread function and calculation of its Fourier transformation. Product of both Fourier transformations and their Fourier transformation (inverse). The resulting image of these operations looks like the object.

W7PUTRAS is only on the CD.

10.4.5.2 Examples

Fourier Transform of (sin x/x)2 as Transfer Function

We take as an object a grid presented as a series of square pulses (FileFig 10.8) and calculate and perform the Fourier transformation ωob(µ). The spread function for a cylindrical lens is (Eq. (10.27)),

S(X) 4a2{sin(πX/λf #)/(πX/λf #)}2,

(10.37)

where we use f # f/2a, for X (the space coordinate) i/255, and i is the running number of the Fourier transformation from 1 to 256 (0-255). We have to use here i/255 in the frequency domain since we used i in the space domain. The Fourier transform of S(X) is τ (µ). The product of φ(µ) ωob(µ)τ (µ) is shown in FileFig 10.8, and the Fourier transform (inverse) of φ(µ) is the ﬁnal image. In FileFig 10.8 we have used the complex Fourier transform of 256 points for the object and its mirror image. We should remember that the longest wavelength of the spatial waves is 128, and the corresponding smallest frequency 1/128. We

394	10. IMAGING USING WAVE THEORY

also use the scale 1/256 in the transfer function and have the smallest frequency at 1/128. The wavelength is 0.0005 mm and the f # f/2a which is 10. Changing the f # in FileFig 10.8 will change the triangle corresponding to τi and will admit more or fewer spatial frequencies for image formation.

FileFig 10.8 (W8TRASIS)

Demonstration of the use of a transfer function. Object is a grid and calculation of it’s Fourier transformation. Spread function (sin y/y)2 and calculation of its Fourier transformation. Product of both Fourier transformations and their Fourier transformation (inverse). The resulting image of these operations looks more or less like the object.

W8TRASIS is only on the CD.

Application 10.8. Change the f # and observe through the changing triangle that more or fewer frequencies are used for image formation.

Fourier Transformation of the Bessel Function as Transfer Function

We may use as the transfer function (J 1(y)/y)2 corresponding to a circular lens instead of (sin y/y)2, which corresponds to a cylindrical lens. Keeping the argument of the transfer function the same but taking R instead of X we use

J 1(πR/λfn)/(πR/λfn),

(10.38)

where f n is the f -number. The variable R is replaced by i/255 and i is the running number of the Fourier transformation, from 1 to 256 (0–255). We have to use here i/255 in the frequency domain, since we used i in the space domain. In FileFig 10.9 the object is a grid presented as a series of square pulses in the space domain, but the spread function is now (J 1(y)/y)2, and its Fourier transformation is the transfer function τ (µ). The product of ωob(µ)τ (µ) is calculated and shown in FileFig 10.9. The Fourier transform (inverse) of φ(µ) ω(ob)(µ)τ (µ) is the ﬁnal image.

Changing the f # in FileFig 10.9 will change τk as shown and will admit more or fewer spatial frequencies for image formation.

FileFig 10.9 (W9TRAJ1S)

Demonstration of the use of a transfer function. Object is a grid and calculation of its Fourier transformation. Spread function (J 1(y)/y)2 and calculation of its Fourier transformation. Product of both Fouriers transformations and their

10.4. IMAGE FORMATION USING INCOHERENT LIGHT

395

Fourier transformation (inverse). The resulting image of these operations looks more or less like the object.

W9TRAJ1S is only on the CD.

Application 10.9. Change the f # and observe through the changing transfer function that more or fewer frequencies are used for image formation.

10.4.6 Resolution

The Rayleigh criterion of resolution states that two image peaks are considered resolved if the maximum of the diffraction pattern of one is at the position of the ﬁrst minimum of the other. This is demonstrated in FileFig 10.10. We consider two round objects and a round lens, and apply the Rayleigh criterion. The object is made of two “rectangles" of width b1 - b2 and b3 - b4 (FileFig 10.11). The aperture of the lens of diameter 2a forms a diffraction pattern of each point on the image screen.

The argument of the Bessel function 2πaR/λf is again written as

πR/λ(f/2a) πR/λ(f #),

(10.39)

where f # f/2a, and we use Y for R in FileFig 10.11. The ﬁrst zero of the Bessel function J1(πR /λf #) is at πR /λf # 3.83. Therefore if the maximum of one peak is at the minimum of the other, the distance 2b is (3.83/π)λf # 1.22λf #. In our example in FileFig 10.11 we have used the diameter 0.0005 mm for the round objects, which is the same as the wavelength. The f # is equal to 10 and the product λf # has a value 0.005. The center-to-center distance of the two objects is 0.0061. In FileFig 10.11 we show the image of the two objects at the minimum distance corresponding to the resolution according to the Rayleigh criterion.

FileFig 10.10 (W10BES3DS)

Demonstration of Resolution. Two Bessel functions in 3-D presentation of radius a 0.05, center-to-center distance of d 24.5, wavelength λ 0.0005, and distance to screen X 4000.

W10BES3DS

Rayleigh Distance and 3-D Graph for Two Round Apertures at Distance D

Radius of apertures is a, coordinate on the observation screen R, wavelength λ, and distance from aperture to screen is X.

396	10. IMAGING USING WAVE THEORY

1. Determination of Rayleigh distance for two round apertures

a ≡ .05

X ≡ 4000

R : 0, 1 . . . 50

R−d

g1(R) :

gg1(R) :

J 1

X·λ

J 1

X·λ

R−d

X·λ

2. 3-D Graph of pattern of two round apertures at distance d

i : 0 . . . N

j : 0 . . . N

xi : (−40) + 2.000 · i yj : (−40) + 20001 · j

λ ≡ .0005

N ≡ 60

X : 4000

RR(x, y) : (x)2 + (y)2

RR(x,y)

RR(x,y−d)

g2(x, y) : J 1

X·λ

gg2(x, y) :

J 1

X·λ

RR(x,y)

RR(x,y−d)

X·λ

10.4. IMAGE FORMATION USING INCOHERENT LIGHT		397
Mi,j : g2(xi , yj ) + gg2(xi , yj )	d ≡ 24.5.

FileFig 10.11 (W11TWOROJ1S)

Resolution. Two round objects at diameter 0.0005 and center-to-center distance of 0.0061. The convolution with the spread function (J 1(y)/y)2 results in a resolved image.

W11TWOROJ1S

Imaging: Two Round Apertures as Objects

At Rayleigh distance, round lens, and Y used for R.

Object

Y : −.01, −.0099 . . . 02		λ : .0005			for choice of f# f/2a
T ol :	.1 k :		2 · π	f :	500 a :	25
			λ

I o1(Y ) : ( (b2 − Y ) − (b1 − Y ))				I o2(Y ) : ( (b4 − Y ) − (b3 − Y ))

I o(Y ) : I o1(Y )I o2(Y ).

398	10. IMAGING USING WAVE THEORY

Image

J 1 k·a·(Y −Y Y )

lim(Y ) :

4 · a2 ·

dY Y

(Y −Y Y )

J 1

k·a·(Y −Y Y )

4 · a2 ·

dY.

(Y −Y Y )

Limits of integration

b1 ≡ −.00025 b2 ≡ +.00025

b3 ≡ +.00585

b4 ≡ +.00635.

Application 10.11.

1.Read the distance of the peaks of the image from the graph.

2.Change the distance between the objects so that the two are barely not resolved.

3.Change the distance between the objects so that the two are barely separated.

4.Increase the size of the objects until they are no longer resolved.

10.5IMAGE FORMATION WITH COHERENT LIGHT

10.5.1 Spread Function

We assumed in Section 10.4 that incoherent light is used for image formation. We now discuss the image formation process using coherent light and show that the spread function must be changed. The general process is the same as in Section 10.4. Coherent radiation is now assumed to emerge from an object. Using Huygens’ principle in the ﬁrst process, the wavelets emerging from the object are summed and travel to the lens. In the second process, the wavelets emerging from the lens are summed. Taking the action of the lens into account

	10.5. IMAGE FORMATION WITH COHERENT LIGHT	399
we found from Eq. (10.8) for the ﬁnal image,
g(Y ) C	h(y) exp{−ikηy/x0}dy α(η) exp{−ikηY/xi }dη,	(10.40)

where the constant C before the integral takes care of all constant factors. Similarly to the way we proceeded for the incoherent case, we use a δ function for a point in the object plane. Substituting for h(y) a delta function into Eq. (10.40) one obtains for the ﬁrst Fourier transformation

δ(y) exp{−ikηy/x0}dy 1	(10.41)
and then has for the remaining part of the integral of Eq. (10.8),
s(Y ) C α(η) exp{−ikηY/xi }dη.	(10.42)

Equation (10.42) is the spread function s(Y ) for the case of coherent light. For the image formation we again use a convolution integral, similar to the incoherent case (see Eq. (10.21)) where h(Y ) describes the object. But unlike Eq. (10.21), we use the spread function s(Y ), containing phase information. The convolution integral

h(Y )s(Y	−	Y )dY	(10.43)
	−
has to be squared for the description of the image
Iim(Y ) \|C		h(Y )s(Y − Y )dY \|2.	(10.44)

In contrast to the incoherent case, we have to use s(Y ) for the spread function instead of S(Y ), which is s(Y )2. The spread function is ﬁrst convolved with all points of the object and then we have to square the integral to get the image.

10.5.2 Resolution

As an example we discuss the problem of resolution. We consider two round objects of diameter 0.0005 mm, as we did for the incoherent case, and a wavelength of 0.0005 mm and a f # 10. The distance to be determined should be such that the ﬁnal image looks similar to the image of the two objects as we found for the incoherent case, when applying the Rayleigh criterion. The calculation is presented in FileFig 10.12.

The image is calculated from the convolution integrals, where we call h(Y ) here iob(Y ) and omit the constants before the integral,

Y b2		Y b4	2
	iob(Y )s(Y − Y )dX +		iob(Y )s(Y − Y )dY .
Iim(Y )		Y b3
Y b1

(10.45)

400	10. IMAGING USING WAVE THEORY

We use for s(Y ) the Bessel function similar to Eq. (10.38). The intensity of the image is obtained by squaring the result of the integration (note that s(Y − Y ) contains phase information).

The distance of the two peaks in the ﬁnal image of the two objects for the coherent case is at a center-to-center distance of 0.0082. This is larger than for the incoherent case. The ﬁnal image is shown for the same distance as used for the incoherent case, and one observes that the two peaks are not resolved. Since the distance used in the coherent case is larger than the one for the incoherent case, for a comparable set of parameters, we have the result: Better resolution is obtained by using incoherent light.

FileFig 10.12 (W12TWOROCOHS)

Two round objects of diameter 0.0005 and center-to-center distance of 0.0061, as used for the incoherent case. The convolution with the spread function, (J 1(y)/y), results in an image of one peak, not resolved. For larger center- to-center distance, 0.0082, the peaks are resolved.

W12TWOROCOHS

Imaging with Coherent Light

Two round apertures at Rayleigh distance, round lens, and Y used for R .

Y : −.01, −.0099 . . . 02

λ : .0005

f/10 f/2a

T o1 :

k :

2 · π

f :

500 a :

25.

Object Amplitudes

iob(1) : ( (b2 − Y ) − (b1 − Y ))

iob2(Y ) : ( (b4 − Y ) − (b3 − Y ))

iob(Y ) : iob1(Y ) + iob2(Y ).

Image

J 1

k·a·(Y −Y Y )

lim(Y ) :

4 · a2 ·

dY Y

(Y −Y Y )

J 1

k·a·(Y −Y Y )

4 · a2

dY Y .

(Y −Y Y )

Integration limits

b1 ≡ −.00025

b2 ≡ +.00025

b3 ≡ +.00585

b4 ≡ +.00635.

10.5. IMAGE FORMATION WITH COHERENT LIGHT

401

b4 − b3 + b3 6.1 · 10−3. 2

Resolution is obtained for b3 .00795, b4 .00845.

Application 10.12. Introduce the center-to-center distance of 0.0082 between the two objects, b3 0.00795 and b4 0.00845 and show that the two objects appear resolved.

10.5.3 Transfer Function

We proceed much as we did in the incoherent case and call

ω(µ)		iob(Y )e−i2πµY dY	(10.46)

and
τ (µ)	s(Y )e−i2πµY dY	,	(10.47)
˜

where τ˜(µ) is the transfer function in the coherent case. The difference in Eq. (10.47) with respect to Eq. (10.34) is that s(Y ) is an amplitude function that carries phase information. The image is obtained by a Fourier transformation, similar to Eq. (10.35) for the incoherent case. To obtain the description of the intensity of the image, the result has to be squared. We have for the coherent case

Iim(Y ) \|C ω(µ)τ˜(µ)ei2πµY dµ\|2.	(10.48)
We call φim(µ) the Fourier transformation of Iim(Y ) and get
Iim(Y ) {FT of the product of the FT of Iob and FT ofs}2.	(10.49)
In the frequency domain we have, similar to Eq. (10.36),
φim(µ) {ω(µ)τ˜(µ)}.	(10.50)

Comparing (10.50) with (10.36) one sees that a symmetric blocking function that does not carry phase information has the same effect of eliminating certain spatial frequencies for the incoherent and coherent cases.

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