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U. Fresnel Diffraction I Modern Optics, Prof. Ruiz, UNCA doctorphys.com Chapter U. Fresnel Diffraction I U1. The Fresnel-Kirchhoff Framework. Spherical waves travel outward from a source indicated by the light bulb. According to the Huygens- Fresnel principle, any spherical wave can be considered to consist of tiny secondary wavelets, the baby waves. See the blue secondary wavelets in the figure. The red crest is the result of the outgoing “baby” wavelets. Since shadows are fairly sharp, the wavelets must interfere destructively at the extreme angles. Otherwise, you would not get the proper shadows due to the two vertical walls that border the opening. Also, the wavelets cannot be uniform in strength in all directions. After all, you cannot have baby waves traveling back to the light bulb. Therefore, in this model, we can at least say that the backward waves must be suppressed. The obliquity factor (1 cos ) 2 for the baby waves is the simplest form to do the trick. In the forward direction 0 and (1 cos 0 ) 1 1 1 22. In the backward direction 180 and (1 cos180) 1 1 0 22. Michael J. Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License Let the source strength be Eo . The strength at distance r ' from the source, somewhere along the aperture is then eikr ' E o r ' . The strength beyond the aperture at a distance introduces a baby wave factor: eeikr' ikr E o rr' . We want to integrate over the aperture to get the total strength at P: eeikr' ikr E E dA po Aperture rr' Now throw in the obliquity factor: (1 cos ) eeikr' ikr E E dA po Aperture 2'rr This equation is a simplified form of the Fresnel-Kirchhoff diffraction equation. For Fraunhofer diffraction the point P is very far away and the angle spread for is very small. Therefore (1 cos ) 1. 2 eikr ' const We analyzed cases of uniform light passing through the aperture, i.e., r ' . Then, in the far field of the Fraunhofer realm, (1 cos ) eikr'' e ikr e ikr e ikr E E dA E dA p o o . Aperture2r ' r r ' Aperture r Michael J. Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License eikr ' E Absorbing the constant r ' into o , we therefore have the starting point of our Fraunhofer cases in the previous chapters eikr E E dA po . Aperture r But the more general diffraction formula is . ikr' ikr (1 cos ) ee E E dA po Aperture 2'rr Augustin-Jean Fresnel Gustave Robert Kirchhoff 1788 - 1827 1824 - 1887 Civil Engineer and Physicist Physicist Remember Kirchhoff’s laws in circuits? Kirchhoff’s Junction Law: The algebraic sum of currents at a junction is zero. Kirchhoff’s Voltage Law: The voltages around any closed loop add to zero. Michael J. Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License U2. Wavefront and Optical Path. 1/2 h 2 22 bb 1 ( ) b b h => b 1 hh2 h '2 bb 1 ( )2 Similarly ' . 22bb 2a hh'22 ' => 22ab From crests near the horizontal axis, hh ' . h2 11 () 2 ab Michael J. Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License For Fraunhofer diffraction , which means that the waves reaching the aperture are virtually plane waves. Therefore, has to be very, very small compared to the wavelength. Another way of looking at things is that the distance “a” has to be very large in order to get the plane waves. Also note that large r on the other side of the aperture indicates smaller values due the small 1/r = 1/b in its formula. h2 11 () 2 ab Consider a slit width to estimate h. A typical grating has 250 lines per mm. That means the distance between each slit is about 1 mm 1000 microns 4 microns 250 250 . Each slit width must be less that this distance. Let’s take 1 micron for a single slit. Let r’ = a = 20 cm be not so large for the source and r = b = 100 cm for the screen distance. h21 1 (10 6 ) 2 1 1 ()() 2ab 2 1/ 5 1 (1062 ) (5 1) 310 12 m 310 3 nm 550 nm 2 The middle of the visible spectrum has wavelength 550e ikrnm.' const Far-Field Diffraction --- Fraunhofer Diffraction r ' Near-Field Diffraction --- Fresnel Diffraction Extreme-Near-Field --- Blinding Light through the Aperture --- “In Your Face” Michael J. Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License U3. Fresnel Zones (The Near Field). h2 11 () hR Return to the extra path length formula 2 ab where from the figure (one of the radii), az ' , and bz . The figure shows concentric regions of the wavefront where the extra path length is given by additional half wavelengths: 2 Rm 11 () m ikr' ikr m 2zz '(1 2 cos, where ) mee = 1, 2, 3, and so on. E E dA po Aperture 2'rr Fro small R we can take rz'' and rz . 1 (1 cos ) E E eik(') r r dA po zz '2A Also for small , we can take ≈ 0 and neglect the obliquity factor. Then. E E o eik(') r r dA p . zz ' A Michael J. Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License Consider a general zone with r ' , r , and . Then 2 2 2 2 r'' r z R z R RR22 r' r z '(1 )1/2 z (1 ) 1/2 zz'22 RR22 r' r z '(1 ) z (1 ) 2zz '22 2 R 2 11 r''() r z z 2'zz R 2 11 () Substitute 2'zz from before to simplify notation. ik(')(')(') r r ik z z ik z z ik e e e e EE Eoo eik(')(') r r dA e ik z z e ik dA p zz''AA zz E E o eik(') z z e ik dA p R zz ' A R2 11 m () m m 2zz ' 2 1 1 1zz ' zz' () L Define L z'' z z z . Then zz ' Michael J. Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License RR221 1 1 mm() m R2 m L m 2z ' z 2 L 2 => m Rm m L The mth Fresnel zone has area 2 2 2 Rmm11 R ( m 1) L m L L R All the Fresnel zones have equal area in this approximation. Using , the number R2 m m of Fresnel zones for a circular aperture with radius R is L , which we call R2 F the Fresnel number L . What is the Fresnel number if R 1 cm , zz' 10 cm , and 500 nm . zz' 10 10 100 L 5 cm zz' 10 10 20 . R 2(10 2 m) 2 F L (500 1092 m)(5 10 m) 1044 10 1 100,000 F 105 4000 25001011 2510 9 25 25 What is the radius of the central Fresnel zone for this case? => RL1 Michael J. Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License 9 2 11 10 R1 (500 10 m)(5 10 m) 2500 10 250 10 10 5 3 To two sig figs: R1 256 10 1610 0.1610 0.16 mm . Consider by comparison a diffraction grating for Fraunhofer diffraction. A grating can easily have 1000 lines per mm. So take R 1 m and for far field take L 1 m . R 2(10 6 ) 2 1 10 12 10 5 F 1 L (500 1097 )(1) 5 10 5 R 2 1 Another way of saying this result of L is R2 550 nm L , consistent with our earlier result that . U4. The Unobstructed Wave. Amplitudes at point P will be the sum of the secondary wavelets coming from the Fresnel zones. Note that the half wavelength added at each zone alternates phases by . The amplitudes will alternate being in phase and out of phase. EEEEEEp 1 2 3 4 5 ... Regroup these so that we have EE EEE EEE12 ( 3 ) ( 3 5 ) ... p 2 224 2 2 2 The middle one will be approximately the average of the neighbors so that E E 1 p 2 , where we do not worry about the last one, which will be much weaker. Michael J. Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License The unobstructed wave along the axis will have an irradiance 2 1E 1 1 2 IE1 2 2 4 2 1 , i.e., 25% of the strength for the first zone. Vibration Curve: Fresnel Zone Electric Field Vectors Consider each phasor arrow as coming from a subsection of a Fresnel zone. EEEEEEp 1 2 3 4 5 ... From the figure you can see that the final sum vector is about half the length of the first zone vector, as we indicated earlier with our equation . The above figure gives a nice visualization of the combinations of phasors to get each of the electric fields for each Fresnel zone: . In vector form, the sum vector for all the Fresnel zones, as seen graphically in the figure called a vibration curve, is 2 E1 E 11E1 2 E p 1 EE IEo 1 p 2 2 and the unobstructed irradiance is 2 2 8 . Michael J. Ruiz, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License U5. Circular Aperture. If you place an aperture that blocks off all the zones except the first one, then at point P along the axis, 1 2 EE IE p 1 and 2 1 , 4 times as bright as the unobstructed wave. But if the aperture lets the first two zones through, we have destructive interference due to the extra /2 optical path length from the second zone.
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