problem36_73

problem36_73 - -= = = -= -= -= -=-- d) Since , 2 ) 2 sin( 2...

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36.73: a) From the segment , y d the fraction of the amplitude of 0 E that gets through is ). sin( 0 0 t kx a y d E dE a y d E ϖ - = b) The path difference between each little piece is ). ) sin ( sin( ) sin ( sin 0 t y D k a y d E dE y D k kx y θ - - = - = This can be rewritten as )). cos( ) sin sin( ) sin cos( ) (sin( 0 t kD y k y k t kD a y d E dE - + - = c) So the total amplitude is given by the integral over the slit of the above. - - × + - = = 2 2 0 2 2 ) sin sin( ) sin cos( ) (sin( a a a a y k y k t kD y d a E dE E )). cos( t kD - But the second term integrates to zero, so we have: [ ] [ ] ). sin( 1 . . . . . . sin , 0 At . λ ) (sin ) λ ) (sin sin( ) sin( 2 ) (sin 2 ) (sin sin( ) sin( 2 sin sin sin( ) ( sin )) sin (cos( ) sin( 0 0 0 2 2 0 2 2 0 t kD E E a a t kD E ka ka t kD E E ka y k t kD E y k y d t kD a E E a a a a
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Unformatted text preview: -= = = -= -= -= -=-- d) Since , 2 ) 2 sin( 2 / ) (sin ) 2 / ) (sin sin( 2 2 2 = = = I ka ka I I E I where we have used ). ( sin 2 2 t kx E I-=...
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This note was uploaded on 01/24/2010 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

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