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Unformatted text preview: PHYS 2214 Homework #8 Solutions 1. Deriving Snells Law. (a) From the figure in the book, t = t 1 + t 2 = d 1 v 1 + d 2 v 2 = q h 2 1 + x 2 v 1 + q h 2 2 + ( l x ) 2 v 2 . (b) The independent variable here is x , since the others are just pa rameters. So, we find the minimum time by setting the derivative of t with respect to x equal to zero: dt/dx = 0. This yields x v 1 q h 2 1 + x 2 = ( l x ) v 2 q h 2 2 + ( l x ) 2 . Using the trig relations sin( 1 ) = x/d 1 and sin( 2 ) = ( l x ) /d 2 gives sin( 1 ) v 1 = sin( 2 ) v 2 which, using n = c/v , becomes Snells Law: n 1 sin( 1 ) = n 2 sin( 2 ) . 2. Two source interference. (a) The phase difference between the two sources at the observer due to their path difference is = k x = 2 x/ . Constructive interference will take place when = 2 m , or = x/m , where m is an integer. x = 2 . 04 10 6 m , so the wavelengths within the visible range, 400 nm to 700 nm, that will give constructive interference are for m = 3 , 4 , and 5, or = { 680 , 510 , 408 } nm. 1 (b) The results would be the same as in part (a), since all that mat ters is the path length difference of the two light sources to the observer. (c) Destructive interference will take place when = 2 x/ = (2 m + 1) , or = 2 x/ (2 m + 1), where m is an integer. The wavelengths within the visible range that will give destructive in terference are for m =3 and 4, or = { 583 , 453 } nm....
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This note was uploaded on 01/02/2012 for the course PHYSICS 2214 at Cornell University (Engineering School).
 '11
 DAVIS,J.C.
 Work, Waves And Optics

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