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Fresnel Diffraction.Nb Optics 505 - James C

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Fresnel Diffraction.Nb Optics 505 - James C Fresnel Diffraction.nb Optics 505 - James C. Wyant 1 13 Fresnel Diffraction In this section we will look at the Fresnel diffraction for both circular apertures and rectangular apertures. To help our physical understanding we will begin our discussion by describing Fresnel zones. 13.1 Fresnel Zones In the study of Fresnel diffraction it is convenient to divide the aperture into regions called Fresnel zones. Figure 1 shows a point source, S, illuminating an aperture a distance z1away. The observation point, P, is a distance to the right of the aperture. Let the line SP be normal to the plane containing the aperture. Then we can write S r1 Q ρ z1 r2 z2 P Fig. 1. Spherical wave illuminating aperture. !!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!! 2 2 2 2 SQP = r1 + r2 = z1 +r + z2 +r 1 2 1 1 = z1 + z2 + þþþþ r J þþþþþþþ + þþþþþþþN + 2 z1 z2 The aperture can be divided into regions bounded by concentric circles r = constant defined such that r1 + r2 differ by l 2 in going from one boundary to the next. These regions are called Fresnel zones or half-period zones. If z1and z2 are sufficiently large compared to the size of the aperture the higher order terms of the expan- sion can be neglected to yield the following result. l 1 2 1 1 n þþþþ = þþþþ rn J þþþþþþþ + þþþþþþþN 2 2 z1 z2 Fresnel Diffraction.nb Optics 505 - James C. Wyant 2 Solving for rn, the radius of the nth Fresnel zone, yields !!!!!!!!!!! !!!!!!!! !!!!!!!!!!! rn = n l Lorr1 = l L, r2 = 2 l L, , where (1) 1 L = þþþþþþþþþþþþþþþþþþþþ þþþþþ1 + þþþþþ1 z1 z2 Figure 2 shows a drawing of Fresnel zones where every other zone is made dark. Note that in the center the zones are widely spaced and the spacing decreases as the radius increases. As shown above, the radius of the zones increase as the square root of integers. 3 2 1 0 -1 -2 -3 -3 -2 -1 0 1 2 3 Fig. 2. Fresnel zones. If rn and rn+1are inner and outer radii of the nth zone then the area of the nth zone is given by 2 2 Area of nth Fresnel zone =prn+1 -prn =p Hn + 1L l L -p HnL l L (2) 2 =plL =pr1 , independent of n. That is, the area of all zones are equal. If the higher order terms in the expansion for SRQ are maintained the area of the zones would slightly increase with increasing r. Generally, it is assumed that z1and z2 are sufficiently large compared to r that the higher order terms can be neglected and the area of all zones are equal. Fresnel Diffraction.nb Optics 505 - James C. Wyant 3 13.2 Circular Aperture On-Axis Irradiance In discussing the Fresnel diffraction of a circular aperture it is convenient to first look at the on-axis irradiance. Consider a circular aperture of radius R, and let S and P lie on the normal through the center of the circular aperture as shown in Figure 3. S r1 z1 Q ρ R r2 z2 P Fig. 3. Circular aperture of radius R. If R is small compared with z1and z2 the usual approximations can be made in the Kirchhoff integral that the obliquity factor is equal to 1 and the 1 r1 r2 in the denominator can be replaced with 1 z1 z2. The cylindrical symmetry about the axis SP suggests using polar coordinates in the integration. Then R A É H + L @ D = þþþþþþþþþþþþþþþþþþþþþ È k r1 r2 prÇr (3) uo P Él Å 2 z1 z2 0 2 2 2 2 2 2 Since r1 = z1 +r and r2 = z2 +r , rÇr =r1 Ç r1 = r2 Çr2 1 1 ÇHr1 + r2L = J þþþþþþþ + þþþþþþþN rÇr r1 r2 Therefore, r1 r2 z1 z2 rÇr= þþþþþþþþþþþþþþþþþ ÇHr1 + r2L þþþþþþþþþþþþþþþþþ ÇHr1 + r2L r1 + r2 z1 + z2 Fresnel Diffraction.nb Optics 505 - James C. Wyant 4 l@RD 2 p A É H + L @ D = þþþþþþþþþþþþþþþþþþþþþþþþþþþþþþþþ È k r1 r2 ÇH + L uo P Él H + L Å r1 r2 z1 z2 l@0D where l@0D = z1 + z2 and l@RD is equal to r1 + r2 when r=R, that is when point Q is on the rim of the aperture. It is convenient to introduce a variable q defined by l r + r = z + z + q þþþþ 1 2 1 2 2 Thus q measures the difference between r1 + r2 and z1 + z2 in units of half-wavelengths and is equal to the number of Fresnel zones within the circular aperture. uo@pDcan now be written as p @ D z @ D = þþþþþþþþþþþþþþþþuoo Pþþþþþ ÈÉpq Ç uo P É Å q 0 where z=q[R], that is the value of q associated with a point on the rim of the aperture, and uoo@pD is the unob- structed value of the amplitude at point P. A É k Hz +z L uoo@PD = þþþþþþþþþþþþþþþþþ È 1 2 z1 + z2 Solving the integral yields Épz uo@PD = uoo@PD H1 -È L (4) The on-axis irradiance is given by zp 2 i@PD = 2 i @PD H1 - Cos@pzDL = 4 i @pD SinA þþþþþþþþE (5) oo oo 2 As the radius of the aperture increases, the irradiance at point P oscillates between zero and 4 times the unob- structed irradiance. Thus, the irradiance produced by the first Fresnel zone is equal to 4 times the irradiance obtained with a clear aperture. As the equation shows, for odd numbers of Fresnel zones present i@PD = 4ioo@PD (6) For even numbers of Fresnel zones present the Fresnel zones cancel one another and the on-axis irradiance is zero. The relationship between z and R is found as follows: At the edge of the aperture r = R and 2 2 2 12 R r1 = Hz1 + R L z1 + þþþþþþþþþþþ and 2 z1 2 2 2 12 R r2 = Hz2 + R L z2 + þþþþþþþþþþþ 2 z2 But l r + r = z + z +z þþþþ 1 2 1 2 2 Fresnel Diffraction.nb Optics 505 - James C. Wyant 5 Solving for z, the number of Fresnel zones present, yields R2 Hz + z L z= þþþþþþþþþþþþþþþþ1 þþþþþþþþþþþþþ2 (7) l z1 z2 Off-Axis Irradiance Distribution Now we want to calculate the irradiance of the Fresnel diffraction of a circular aperture at off-axis positions. Fig. 4 shows the schematic setup. Aperture Observation η Plane ξ y Source x z1 z2 Fig. 4. Fresnel diffraction of circular aperture. The general equation for the amplitude of the Fresnel diffraction pattern is given by u@x, yD = É kz È 2 É þþþþþþþþþp Hx2+y2L É þþþþþþþþþp Hx2+h2L -É þþþþþþþþþ2 p Hx x+y hL (8) þþþþþþþþþþþþþþþ È l z2 Å Å Iu@x, hD È l z2 M È l z2 Çx Çh Élz2 - - É þþþþþþþþþp Hx2+h2L One important fact to note about this equation is that we have the Fourier transform of u@x, hD È l z2 , where u@x, hD is the amplitude of the wavefront illuminating the aperture which we will take as the amplitude of a É spherical wave. For an on-axis spherical wave illuminating the aperture u@x, hD can be written as þþþþþþA È kz1 z1 É þþþþþþþþþþþk Hx2+h2L È 2 z1 . If Pupil[x, h] describes the aperture amplitude transmission function we can write É k Hz +z L È 1 2 É þþþþþþþþþp Hx2+y2L u@x, yD = A þþþþþþþþþþþþþþþþþþþþþþþþ È l z2 Élz1 z2 (9) É þþþþþþþþp Hzþþþþþþþþ1+z2þþþþþL Hx2+h2L -É þþþþþþþþþ2 p Hx x+y hL Å Å IPupil@x, hD È l z1 z2 M È l z2 Çx Çh - - !!!!!!!!!!!!!!!!!!!! 2 2 Let rmax be the maximum value of Hx +h L and nFZones be the number of Fresnel zones in the aperture. Then Hz1 + z2L 2 nFZones = þþþþþþþþþþþþþþþþþþþþþþ rmax and we can write l z1 z2 Fresnel Diffraction.nb Optics 505 - James C. Wyant 6 É k Hz +z L È 1 2 É þþþþþþþþþp Hx2+y2L u@x, yD = A þþþþþþþþþþþþþþþþþþþþþþþþ È l z2 Élz z 1 2 (10) Ix2+h2M i É HnFZones pL þþþþþþþþþþþþþþþþþ y -É þþþþþþþþþ2 p Hx x+y hL Å Å jPupil@x, hD È rmax 2 z È l z2 Çx Çh - -k { r 2 nFZones But þþþþþþþþþþþþþþþþþmax = þþþþþþþþþþþþþþþþþþþþþþ,so l z1 z2 Hz1 + z2L É k Hz +z L È 1 2 nFZones É þþþþþþþþþp Hx2+y2L u@x, yD = A þþþþþþþþþþþþþþþþþþþþþþþþþþ þþþþþþþþþþþþþþþþþþþþþ È l z2 É Hz + z L r 2 1 2 max (11) Ix2+h2M i É HnFZones pL þþþþþþþþþþþþþþþþþ y -É þþþþþþþþþ2 p Hx x+y hL Å Å jPupil@x, hD È rmax 2 z È l z2 Çx Çh - -k { nFZones É þþþþþþþþþp Hx2+y2L @ D = þþþþþþþþþþþþþþþþþþþþþ È l z2 u x, y Uoo 2 É rmax Ix2+h2M (12) i É HnFZones pL þþþþþþþþþþþþþþþþþ y -É þþþþþþþþþ2 p Hx x+y hL Å Å jPupil@x, hD È rmax 2 z È l z2 Çx Çh - -k { É H + L where U = A þþþþþþþþþþþþþþþþþþþÈ k z1 z2 is the unobstructed on-axis irradiance. oo Hz1+z2L é Fourier-Bessel transform (Hankel transform of zero order) Since the aperture is circular we have the Fourier transform of a rotationally symmetric function which can be written as a Fourier-Bessel transform (Hankel transform of zero order). If the irradiance is normalized so the result is in units of the unobstructed irradiance, the irradiance as a function of the number of Fresnel zones in the aperture can be written as 1 2 2 irradianceBessel@nFZones_, r_D := AbsAnFZones Å BesselJ@0, p r r D ÈÉ HnFZones pL r 2 prÇrE 0 The pupil radius, r, is normalized to one at the edge of the aperture and the units of r are wavelengths/(pupil diameter).
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