Diffraction. Diffraction and Polarization. Diffraction. Diffraction. Diffraction I. Angular Spread: θ ~ λ/a

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Emery Francis
5 years ago
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1 1 nd Polriztion Chpter 38 Rleigh s Criterion Polriztion n geometricl optics, we modeled rs like this! n fct wht hppens is this... A sphericl wve propgtes out from the perture. All wves do this.. For double slit interference we did this without considering the size of the perture. considers the size of the perture,. - ffects due to the fct tht pertures re not perfect point sources, but slits with finite width. Wves emerging from different loctions in the perture will interfere with ech other. Slit/perture width : Angulr Spred: ~ / Angulr Spred proportionl to / (Actul intensit distribution)

2 Consider the perture to be lrge collection of point sources which dd together t the screen. The more sources there re the weker ech source must be. First source (t top) At the electric field is the sum of ech of mn (wek) sources ll with different pth lengths. st source (t bottom) net n rel perture, the number of point sources pproches infinit, while ech one s strength simultneousl goes to zero. This sum of infinitesiml sources forms n rc. The sum of which is the stright line connecting the end points, net. ngle between the first nd lst infinitesiml electric field. β is exctl φ for the double slit experiment with d equl to the perture size!!! The rc of electric field hs some rdius R. net / = R sin(β/), R =? This sum of infinitesiml sources forms n rc. The sum of which is the stright line connecting the end points, net. πd sin φ = π sin mx = R β Thus, net = mx sin( β / ) ( β / ) min π sin sin( β / ) net = mx ( β / ) min π sin sin( β / ) net = mx ( β / ), so, so net mx sin ( β / ) ( β / ) net mx sin ( β / ) ( β / )

3 3 min will be minimum (dim) when the net sum field is zero. Actull, we hve n infinite mout of sources. We get zero intensit on the screen when net = 0. sin = min This mens tht the first source (the one t the top) nd the lst source (the one t the bottom) hve prllel field (constructive)! This is exctl the cse for mximum for two source interference seprted now b, not d. net = 0 Y mx (double source) Y min (single) nd d min mx = m, = 0 m 0 Minimums where sin = m, m 0 Young s Double Slit xperiment Most of the intensit of the light is between + min / nd - min /. Thus min / sin the ngulr spred totl 1 α α=φ/, 1 = = ϕ π φ = kd sin = d sin φ d sin d = π totl = cos(α) = cos(φ/) = cos(π dsin()/) totl = 4 1 cos (π dsin()/) 4 1 cos (πd/) where 1 is the intensit t S 1 (or S, the re the sme). o cos (π dsin()/) o cos (π d/) Where o is the pek intensit t the screen Young s Double Slit xperiment Rleigh s Criterion totl 1 α α=φ/, 1 = = ϕ o cos (π dsin()/) o cos (π d/) Where o is the pek intensit t the screen π φ = kd sin = d sin φ d sin d = π Add diffrction - net mx sin ( β / ) cos ( φ / ) ( β / ) πd sin π sin φ = = Cmer perture, ee Screen, film, retin, etc. First object produce diffrcted object on the screen.

4 4 Object Rleigh s Criterion xmple: Photogrphs in newsppers re mde up of lrge number of closel spced dots. ) How fr prt should these dots be such tht the re not quite resolved (Rleigh Criterion) when the printed picture is held 5 cm from our ee? b) How does this compre with low end lser printer with 300 x 300 = Cmer perture, ee Screen, film, retin, etc. Screen, film, retin, etc. First object produce diffrcted object on the screen. Second object lso produces diffrcted object on the screen. Rleigh s Criteri requires the ngulr seprtion of the two be minimum of sin = /. f this is true, then the two objects re resolved. xmple: Photogrphs in newsppers re mde up of lrge number of closel spced dots. ) How fr prt should these dots be such tht the re not quite resolved (Rleigh Criterion) when the printed picture is held 5 cm from our ee? b) How does this compre with low end lser printer with 300 x 300 ) Rleigh s Criteri requires the ngulr seprtion of the two be sin = /. = visible light, so = 600 nm. = perture, our pupil. Thus, = 3 mm. sin = /5cm, where is the dot spcing! /5cm = / = 600nm/3mm = 0.5 mm b) 300 dpi = 300 dots/inch 300 dpi = 300 dots/inch x 1nch/5.4mm = 11.8 dots/mm or 11.8 dots/mm 0.1 mm/dot 0.1 mm/dot well within Rleigh s Criteri, 0.5 mm in this cse Object R. C. for Circulr Aperture = Cmer perture, ee Rleigh s Criteri in Two Dimensions sin = /. ut circulr perture hs contributions from -D perture Screen, film, moving retin, etc. to screen in the third dimension! Screen, film, retin, etc. Rleigh s Criteri for circulr perture tkes these effects into ccount sin = 1. /. xmple: Photogrphs in newsppers re mde up of lrge number of closel spced dots. ) How fr prt should these dots be such tht the re not quite resolved (Rleigh Criterion) when the printed picture is held 5 cm from our ee? b) How does this compre with low end lser printer with 300 x 300

5 5 xmple: Photogrphs in newsppers re mde up of lrge number of closel spced dots. ) How fr prt should these dots be such tht the re not quite resolved (Rleigh Criterion) when the printed picture is held 5 cm from our ee? b) How does this compre with low end lser printer with 300 x 300 ) Rleigh s Criteri requires the ngulr seprtion of the two be sin = 1. /. = visible light, so = 600 nm. = perture, our pupil. Thus, = 3 mm. sin = /5cm, were is the dot spcing! /5cm = 1. / = nm/3mm = 0.6 mm, not big chnge!!! Polriztion Previousl we described the trnsverse nture of &M wves s follows nd view (coming t ou) i.e. plne wves Polriztion The nd fields must lws be perpendiculr. Unpolrized light consist of light with fields in ll rndom plnes. Polriztion iner polrized light hs the field in one plne (nd thus the field in plne t 90 o. n unpolrized light, ll positions (or plnes) re equll probble. Superposition ields the net field t n instnt! nd view Polriztion - Polrizer ight cn be polrized b device clled polrizer. A polrizer lets through light of onl one polriztion. Polriztion - Polrizer Polrized light cn be ltered b polrizer. 0 i /, where i ws unpolrized light = ε o o cos = ε o cos w of Mlus o cos

β 1 = 2 π and the path length difference is δ 1 = λ. The small angle approximation gives us y 1 L = tanθ 1 θ 1 sin θ 1 = δ 1 y 1

β 1 = 2 π and the path length difference is δ 1 = λ. The small angle approximation gives us y 1 L = tanθ 1 θ 1 sin θ 1 = δ 1 y 1 rgsdle (zdr8) HW13 ditmire (58335) 1 This print-out should hve 1 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. 001 (prt 1 of ) 10.0 points