OPTICAL DIFFRACTION
3.4 DIFFRACTION-LIMITED IMAGING
A circular lens captures light falling on the lens, so it not only performs a lens function but also acts as a circular aperture of diameter equal to that of the lens. In an imaging system, there will be a limiting aperture that limits the ability of the system to focus an input point source to an output point source. If the output is distorted only by the diffraction of the limiting aperture, the system is called a diffraction-limited system.
A diffraction-limited system has no other aberrations and is therefore the best possible imaging system [49].
3.4.1 Intuitive Effect of Aperture in Imaging System
Figure 3.10 shows that the aperture of a convex lens loses high spatial frequencies in the focused output image of an input point source. Light from the input point source is expanded as a set of plane waves at different angles, the angular spectrum, as shown on the left-hand side of Figure 3.10. On striking a vertical plane at the lens, the source plane wave components at the widest angle from the axis correspond to those with the highest frequency at the plane of the lens. These fall outside the lens aperture so that they are lost in forming the output image. The loss of high spatial frequencies in passing through the aperture results in rings around the point source image and smoothing of the image. Smoothing arises because the high spatial frequencies provide the fine detail in an image. A larger diameter lens will capture higher spatial frequencies and the output image will appear closer to a point source, making the point sharper. This is the reason that a photographic camera will provide
DIFFRACTION-LIMITED IMAGING 57
Low spatial frequency High spatial frequency
P(x,y) Aperture
Image of point source Point
source
k y
ky = k sin θ
kz = k cos θ
y
θ
FIGURE 3.10 Diffraction through aperture loses high spatial frequencies, causing smoothing of an image.
less sharp pictures with small apertures: highf-numbers. The large apertures used in dimmer light will also be less sharp but for a different reason: the edges of the lens are now used and these often have more aberrations than the center.
3.4.2 Computing the Diffraction Effect of a Lens Aperture on Imaging
We now show how to compute the effects of the aperture on an output image. The aperture is called a pupil function in the following. Figure 3.11 shows a simple imaging system with input objectUowith coordinatesxoandyoat distanced0from a lens. The lens with coordinatesxandyhas an optical field labeledUl in front of the lens and Ulafter passing through the lens. An imageUiwith coordinatesxiandyiis formed
U0
Ui
x0
y0
x y
xi
yi
d0 di
Ul Ul′
FIGURE 3.11 Imaging a point source (at arrowhead) with a limited aperture lens.
58 OPTICAL DIFFRACTION
at distancedifrom the lens. Owing to linearity of wave propagation, Ui(xi, yi)=
∞
∞
∞
∞ h(xi, yi, xo, yo)Uo(xo, yo)dxodyo (3.55) The effects of diffraction on imaging can be computed in three steps:
1. Compute the geometric image by ignoring the effects of diffraction (Section 3.4.2.1).
2. Compute the impulse response describing diffraction through the system (Section 3.4.2.2).
3. For output image, compute the convolution between the impulse response from step 2 and the geometric image from step 1 (Section 3.4.2.3).
We proceed with equations for the three steps to prove that this procedure computes the correct image with diffraction effects.
3.4.2.1 Step 1: Compute the Geometric Image We would like the image to be an identical copy of the input, except for scaling byMto allow for magnification.
This requires the impulse response from input to output to be a delta function of the form (|M|2is added to preserve total power)
h(xi, yi, xo, yo)= 1
|M|2δ(xi−Mxo, yi−Myo) (3.56) The negatives account for image inversion throughM= −di/do= −xi/xo. Substi- tuting equation (3.56) into equation (3.55) shows that a scaled copy of the input is obtained:
Ui(xi, yi)= ∞
∞
∞
∞
1
|M|2δ(xi−Mxo, yi−Myo)Uo(xo, yo)dxodyo (3.57) From the sampling theorem for delta functions, theδmeans sampleUoatxi−Mx0= 0 orx0=xi/Mto give a geometric image that is a scaled version of the input:
Ug(xi, yi)= 1
|M|2Uo xi
M,yi
M (3.58)
3.4.2.2 Step 2: Compute the Impulse Response Describing Diffraction Through the System The propagation impulse response from the output of the lens in Figure 3.11 to the output screen is adapted from Fresnel diffraction equa- tion (3.45) (Section 3.3.4),
h(xi, yi, x, y)= 1 jλdi
∞
−∞Ul(x, y) exp jk
2di
(xi−x)2+(yi−y)2
dxdy (3.59) where we neglected the leading phase term that is independent ofxandy.
DIFFRACTION-LIMITED IMAGING 59
Propagation through the lens with pupilP(x, y) is represented by Ul(x, y)=Ul(x, y)P(x, y) exp
−jk 2f
x2+y2 (3.60)
Propagation from point source input object to the lens is represented by Ul(x, y)= 1
jλdoexp jk
2do
(x−xo)2+(y−yo)2
(3.61) Substituting equation (3.61) into equation (3.60) and the result into equation (3.59) gives the overall impulse response from the input point source to the output point source:
h(xi, yi, xo, yo)
= 1 λ2dido
(a) exp
jk 2do
x2o+y2o
(b) exp
jk 2do
x2i +yi2
∞
−∞P(x, y) exp
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩ jk
2
(c) 1
do + 1 di − 1
f x2+y2
⎫⎪
⎪⎪
⎬
⎪⎪
⎪⎭ exp
−jk xo
do+xi di
x+
yo do +yi
di
y
dxdy (3.62)
For imaging, our interest is in magnitude; therefore, we ignore the phase terms (a) and (b). If we assume perfect imaging according to the lens law, equation (1.24), 1/do+1/di−1/f =0, term (c) is zero and as exp{0} =1, the exponential contain- ing term (c) vanishes. We eliminatedousing equation (1.24) ordo= −di/M. Then equation (3.62) becomes
h(xi, yi, xo, yo)= 1 λ2dido
∞
−∞P(x, y) exp jk
di(xi−Mxo)x+(yi−Myo)y
dxdy (3.63) Equation (3.63) is not space invariant because of the magnificationM. We force it to be space invariant, that is, in convolution form, by redefining variables ˆxo= |M|xo
and ˆyo= |M|yo. We also incorporate the scaling in new variables, ˆx=x/(λdi) and ˆ
y=y/(λdi) from which dx=λdid ˆxand dy=λdid ˆy. In addition, usingk=2π/λ, equation (3.63) becomes (cancel|M|from 1/(λ2d1do) andP)
h(xi−xˆo, yi−yˆo)
= ∞
−∞P(λdix, λdˆ iy) expˆ (
−j2π
(xi−xˆo)ˆx+(yi−yˆo)ˆy )
d ˆxd ˆy (3.64)
60 OPTICAL DIFFRACTION
Equation (3.64) is the Fraunhofer diffraction of the aperture; note the scaling byλdi in the pupil function for Fraunhofer.
3.4.2.3 Step 3. Show That Output Image is Convolution of Geometric Image and Fraunhofer Diffraction of Aperture Function: The output image may be written (convolution version of equation (3.55)) as
Ui(xi, yi)= ∞
∞
∞
∞ h(xi−xˆo, yi−yˆo)Uo(xo, yo)dxodyo (3.65) From text preceding equation (3.64), ˆxo= |M|xoand ˆyo= |M|yo, so dxo=d ˆxo/|M| and dyo=d ˆyo/|M|in equation (3.65) give
Ui(xi, yi)= ∞
∞
∞
∞ h(xi−xˆo, yi−yˆo)Uo
xˆo M,yˆo
M 1
|M| 1
|M|d ˆxod ˆyo (3.66) From equation (3.58),
Uo xˆo
M,yˆo
M
= |M|2Ug(ˆxo,yˆo) (3.67)
SubstitutingUo(ˆxo/M,yˆo/M) from equation (3.67) into equation (3.66) gives Ui(xi, yi)=
∞
∞
∞
∞ h(xi−xˆo, yi−yˆo)Ug(ˆxo,yˆo) d ˆxod ˆyo (3.68) The final result for imaging limited by diffraction of an aperture, equation (3.68), is the convolution of the geometric image,Ug(x, y), with the Fraunhofer diffraction of the pupil (or aperture) function,h(x, y) (scaled byλdi), equation (3.64),
Ui(x, y)=h(x, y)∗Ug(x, y) (3.69) For a square aperture, the pupil function has rectangular transmission function inxor y. The aperture is scaled byλdiand then Fourier transformed to find the Fraunhofer diffraction of the aperture. It will have the shape shown in Figure 3.2b. Convolving with the geometric image will smooth the image out, that is, remove high spatial frequencies. The Fraunhofer diffraction function is wider for smaller apertures, by the Fourier transform scaling property, so that smaller apertures will provide more smoothing.