This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 29. (a) In a manner similar to that discussed in Sample Problem 374, we ﬁnd the ratio should be d/a = 4. Our reasoning is, brieﬂy, as follows: we let the location of the fourth bright fringe coincide with the ﬁrst minimum of diﬀraction pattern, and then set sin θ = 4λ/d = λ/a (so d = 4a). (b) Any bright fringe which happens to be at the same location with a diﬀraction minimum will vanish. Thus, if we let sin θ = m1 λ/d = m2 λ/a = m1 λ/4a = m2 λ/a, or m1 = 4m2 where m2 = 1, 2, 3, · · · . The fringes missing are the 4th, 8th, 12th, and so on. Hence, every fourth fringe is missing. ...
View
Full
Document
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

Click to edit the document details