problem36_75

# problem36_75 - 36.75: a) I = I 0 sin 2 ( N ) 2 sin 2 2 0...

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Unformatted text preview: 36.75: a) I = I 0 sin 2 ( N ) 2 sin 2 2 0 lim I . 0 0 N 2 cos( N ) sin ( N ) 2 2 = lim . 0 1 2 cos( ) sin 2 2 =N ^ Use l' Hopital' s rule : lim 0 2 So lim I = N I 0 . 0 b)The location of the first minimum is when the numerator first goes to zero at N 2 min = or min = . The width of the central maximum goes like 2 min , so it is 2 N 1 proportional to . N N = n where n is an integer, the numerator goes to zero, giving a c)Whenever 2 2n . This is true minimum in intensity. That is, I is a minimum wherever = N assuming that the denominator doesn't go to zero as well, which occurs when = m , 2 where m is an integer. When both go to zero, using the result from part(a), there is a n maximum. That is, if is an integer, there will be a maximum. N n d)From part c), if is an integer we get a maximum. Thus, there will be N - 1 N n minima. (Places where is not an integer for fixed N and integer n .) For example, N n = 0 will be a maximum, but n = 1, 2. . ., N - 1 will be minima with another maximum at n = N. 3 , etc.) and if N is e)Between maxima is a half-integer multiple of i.e. , 2 2 2 2 sin ( N 2 ) 1, so I I 0 . odd then sin 2 2 ...
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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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