Lecture 17. Fresnel Diffraction
J. Scott Tyo OPTI 512R College of Optical Sciences University of Arizona Last Updated August 10, 2011
1. THE FRESNEL REGION In the last lecture we looked at Fraunhofer diffraction in the far-field of the aperture. Given that the aperture was limited by a circle of radius L1, we found that we are in the the Fraunhofer region when πL 2 z >> 1 . 12 λ (1)
There are a number of qualitative ways to describe this far field. They include
• The distance after which the spherical waves look locally planar
• The distance after which the electromagnetic wavefronts become spherical • The distance after which the ratio of the electric and magnetic fields is η = μ/.
• The distance after which the field strength falls off as 1/r.
• The distance after which the shape of the field distribution is a function of angle of observation only and not distance z.
Before we get to the Fraunhofer region, we have the Fresnel region where these particular conditions are not satisfied. Consider the illustration in Fig. 1. Very close to the aperture, the intensity is approximately identical to the aperture shape. Very far from the aperture, in the Fraunhofer zone, the intensity is described by the Fourier transform of the aperture opening, and remains constant in shape as we move further away. Between
Figure 1. Conceptual diagram of Rayleigh-Sommerfeld diffraction. Figure 10.4 from Gaskill
1 fields u2222 (x ,y ,z ) y1 y Aperture Plane 2
x1 x2
L2 L1
z=z z=z1 2
z12
Figure 2. Geometry for Fresnel diffraction. the near-field and the far-field is a region known as the Fresnel region where the fields continuously change from having the shape of the aperture to having the shape of the Fourier transform. This change must happen slowly, and Maxwell’s equations must be satisfied at all times. Consider once again the Kirchoff approximation to the Rayleigh-Sommerfeld diffraction integral
∞ jkr z12 e 12 u2(x2,y2)= u1(x1,y1) dx1dy1. (2) −∞ jλr12 r12
As we did last lecture, we write the distance r12 as 2 2 x2 − x1 y2 − y1 r12 = z12 1+ + (3) z12 z12
As we move back toward the aperture from the Fraunhofer zone (z12 decreases), we will need to consider additional terms in the Taylor series approximation for the square root. Writing the first three terms of the Taylor series yields 2 2 x2−x1 y2−y1 kz12 1+ z + z 12 12 (4) k 2 2 k 2 2 2 ≈ kz12 + (x2 − x1) +(y2 − y1) − 3 (x2 − x1) +(y2 − y1) + ... 2z12 8z12 In the Fraunhofer zone we completely ignored the final term in Eq. 4 and even discarded the terms that varied 2 2 2 like (x1 + y1 )/z12 .Asz12 decreases, we will now search for conditions under which we can ignore the last term on the right-hand side of Eq. 4, which will produce the Fresnel condition.
We have already limited our aperture to having a radius L1. If we now limit our observations to be contained inside a circle of radius L2 as shown in Fig. 2, we can say that
2 2 2 (x2 − x1) +(y2 − y1) ≤ (L1 + L2) . (5)
If the last term is much less than one radian, we can ignore its effects on the exponential. This condition is
4 π(L1 + L2) z3 >> , (6) 12 4λ
2 3 Spherical Wavefront Parabolic Approximation 2 Planar Approximation
1
y 0
−1
−2
−3 0 1 2 3 4 x
Figure 3. Parabolic (Fresnel) and planar (Fraunhofer) approximations to the Spherical wave. and we can write jkz12 ∞ e π 2 2 u2(x2,y2)= u1(x1,y1)exp j (x2 − x1) +(y2 − y1) dx1dy1. (7) jλz12 −∞ λz12
Equation 6 tells us when we are inside the Fresnel region. It is important to note that the Fresnel region extends all the way to ∞, which means that the Fraunhofer region is itself a part of the Fresnel region. It is also important to note that the Fresnel region is not necessarily in the far-field of the aperture. Furthermore, the beginning of the Fresnel region depends on the radius of the circle L2 that inscribes our desired observations. The Fresnel zone actually begins closer to the aperture for observation points near the axis than for observation points further away. Let’s take a moment to consider the Fresnel approximation to the Kirchoff diffraction integral. In the Kirchoff integral, we have the kernel exp{jkr12}. We know that this kernel gives us spherical wave fronts, and provides us with a spherical wave. In the Fresnel approximation, we replace
π jkr12 2 2 e ≈ exp j (x2 − x1) +(y2 − y1) . (8) λz12
We see then that the Fresnel approximation takes a spherical wave and approximates it with quadratic phase curvature. When x1 =0andy1 = 0, that is, the point source is at the origin, we have the fields
jkz e 12 j π (x 2+y 2) u0(x2,y2) ≈ e λz12 2 2 . (9) jλz12
The right hand side of Eq. 9 is then the quadratic phase curvature approximation (Fresnel approximation) to a spherical wave. Whereas the Fraunhofer zone was reached when the spherical wave fronts are approximately planar, the Fresnel zone begins when the spherical wave fronts are approximately parabolic, as shown in Fig. 3 From this point forward whenever we see the term
j π x2+y2 u(x, y) ∝ e λR ( ), (10) we should recognize it as the quadratic approximation to a spherical wave with radius of curvature R in the plane of observation.
3 2. THE FRESNEL DIFFRACTION INTEGRAL
Equation 7 is known as a Fresnel transform of the fields in the aperture given by u1(x1,y1). The Fresnel transform is like a Fourier transform, only more complicated because it involves terms of the form exp{j2πξx} and exp j2πx2. The transform can be computed using the Fresnel integrals, which are tabulated. However, even though they are tabulated, they are not included in some software packages (like Matlab, though they are included in other packages, like Mathematica). If we take another look at Eq. 7, we recognize the Fresnel transform as having the properties of a convolution integral u2(x, y)=u1(x, y) ∗ h12(x, y) (11) where we define the impulse response as
jkz12 e π 2 2 h12(x, y)= exp jk (x + y ) . (12) jλz12 λz12
This impulse response has a tabulated Fourier transform in table 7-3 of Gaskill, and we have the transfer function 2 2 jkz12 −jπλz12(ξ +η ) H12(ξ,η)=e e . (13) This is our first hint that we will be able to consider optical propagation in terms of a LSI system, and consider both impulse responses – what we refer to as point-spread functions – and transfer functions in how images propagate.
2.1. Using Fourier Transforms to Evaluate Fresnel Diffraction If we now take the definition of the Fresnel transform and expand its terms, we can write
jkz ∞ e 12 π 2 2 π 2 2 2π j (x +y ) j (x +y ) −j (x1x2+y1y2) u2(x2,y2)= e λz12 2 2 u1(x1,y1)e λz12 1 1 e λz12 dx1dy1 (14) jλz12 −∞ jkz e 12 j π (x 2+y 2) j π (x 2+y 2) e λz12 2 2 F u x ,y e λz12 1 1 . = 1( 1 1) x y (15) jλz12 ξ= 2 ,η= 2 λz12 λz12
Equation 15 tells us that the Fresnel diffraction pattern can be computed by taking the Fourier transform of the aperture fields modified by a phase curvature term in the aperture plane. Let’s consider the various pieces of this diffraction integral. We have a spherical curvature term in terms of the observation position (x2,y2) that precedes the integral. Inside the integrand, we have the linear Fourier transform kernel that we saw in Fraunhofer diffraction, but we also have a spherical phase curvature term in terms of the source position (x1,y1). When we get far enough away (z12 →∞), the spherical curvature term inside the Fourier transform can be ignored, and the Fresnel diffraction approximation reduces to the Fraunhofer approximation that we saw last time. Let’s consider the following example. Our aperture is taken to be circular with radius 1 cm. The wavelength is 1 μm. The Fresnel distance is computed to be 20 cm and the Fraunhofer distance is computed as 315 m. For numerical reasons, we compute the Fresnel diffraction pattern at three different distances that are well inside the Fresnel zone, and we see the fields evolve from a reasonably complicated distribution to the Fraunhofer pattern. Note that the distribution is approximately Fraunhofer at a distance of 50 m, even though this is well below the far-field limit. These results are presented in Fig. 4.
4 Figure 4. Fresnel diffraction patterns for a circular aperture with a =1[cm],λ =1[μm].
2.2. Fresnel integrals and the Cornu Spiral In the previous section we saw how we could use the tools of the Fourier transform (typically evaluated numer- ically for realistic problems) to evaluate Fresnel diffraction integrals. In this section, we will look at the direct evaluation of the Fresnel transform through the Fresnel integrals. As an example evaluation, let’s consider the Fresnel diffraction from a rectangular aperture. The transmission function of the aperture is x y t x ,y 1 , 1 . ( 1 1) = rect a b (16)
In this case, assuming normally-incident plane wave illumination, we can write the diffraction as
jkz ∞ e 12 π 2 2 j (x2−x1) +(y2−y1) u2(x2,y2)= t(x1,y1)e λz12 dx1dy1 (17) jλz12 −∞ jkz ∞ ∞ e 12 x π 2 y π 2 1 j (x2−x1) 1 j (y2−y1) = rect e λz12 dx1 rect e λz12 dy1 (18) jλz12 −∞ a −∞ b ejkz12 = IxIy. (19) jλz12
The integrals Ix and Iy are closely related to a tabulated set of functions known as the Fresnel Integrals, which are presented in many mathematical handbooks (Abromowitz & Stegun, CRC Handbook, etc.). They are also included in some (but not all) mathematical packages. (Mathematica has them while Matlab does not, though there are workarounds to evaluate them in Matlab, try using helpdesk to find out). The Fresnel integral
5 The Cornu Spiral 0.8 0.8
0.6 0.6
I(2) − I(1) 0.4 0.4 x=2 I(1) − I(0) 0.2 0.2 x=1
0
S(x) 0 S(x)
−0.2 x=0 −0.2 I(.5)−I(−.5)
−0.4 −0.4
−0.6 −0.6
−0.8 −0.8 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 C(x) C(x)
Figure 5. The Cornu spiral that provides a graphical representation for the Fresnel sine and cosine integrals as defined in Eq. 22 and Eq. 23. is defined as ∞ j π α2 I(x)= e 2 dα (20) 0 C η jS η = ( )+ ( ) (21) η π C(x)= cos α2 dα (22) 0 2 η π S(x)= sin α2 dα. (23) 0 2
An interesting exercise is to plot S(x) as a function of C(x) parameterized by x. This plot creates a spiral known as the Cornu spiral that is plotted in Fig. 5. The spiral seems to imply that the integrals asymptotically evaluate to a finite number, but we should expect as much. We can evalutate ∞ j π α2 I(∞)= e 2 dα (24) 0 ∞ 1 j π α2 = e 2 dα. (25) 2 −∞
However, we can evaluate Eq. 25 using the moment theorem as √ ∞ 1 j π α2 1 j π α2 2 j π 1 j e 2 dα = F e 2 = e 4 = + . (26) 2 −∞ 2 ξ=0 2 2 2
1 1 1 1 Looking at Fig. 5, we see that the spiral asymptotically approaches the points 2 + j 2 and − 2 − j 2 . Our integrals Ix and Iy are not quite of the form of the Fresnel integrals. To remedy this, we have to massage
6 them a bit. Let’s start with Ix. ∞ x π 2 1 j (x2−x1) Ix = rect e λz12 dx1 (27) −∞ a x =a/2 1 π 2 j (x2−x1) = e λz12 dx1 (28) x1=a/2 u=x2+a/2 j π u2 = e λz12 du (29) u=x2−a/2 ηb λz12 j π η2 = e 2 dη (30) 2 ηa λz12 = (I(ηb) − I(ηa)). (31) 2 In Eq. 30 we have ηa = 2/z12λ(x2 − a/2) and ηb = 2/z12λ(x2 + a/2). We see then that the lower and upper limits of the integration in Eq. 30 depend on the observation position as well as the shape of the aperture. For example, when the observation postion is on-axis (x2 = 0), the integration range is symmetric. However, when the observation position moves off axis, the integration range loses its symmetry. This can be explained with an example. As the observer moves in the +x2 direction, the top portion of the aperture is closer to the observer, hence breaking the symmetry. To understand how to use the Cornu spiral, let’s consider the case where we have an aperture such that a = λz12/2 (this would correspond to a 1 mm aperture being illuminated by 1 micron light and observed at 1 1 1 1 2 m distance). When the observer is on-axis, we have ηa = − 2 and ηb =+2 . We can plot I( 2 )andI(− 2 )on the Cornu spiral (which is in the complex plane), and the total integral is equal to the red vector that connects the two points as shown in Fig. 5. If we now move a distance x2 = a/2offaxis,wehaveηz =0andηb =1. These points and the corresponding vector are plotted in green. Now if we move an additional a off axis so that x2 =3a/2, we have ηa =1andηb = 2, and these are plotted in pink. As we continue moving further and further off axis, the two points will chase each other around the spiral until ηa,ηb →∞and I(ηb) − I(ηa) → 0. Physically this means that the diffracted intensity will remain relatively constant until one end of the vector starts rotating around the asymptotic point. Once this happens, the intensity will oscillate somewhat, then drop towards zero. Far from the axis, ηa,ηb →∞and Ix → 0.
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