This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 75. (a) A path length diﬀerence of λ/2 produces the ﬁrst dark band, of 3λ/2 produces the second dark
band, and so on. Therefore, the fourth dark band corresponds to a path length diﬀerence of
7λ/2 = 1750 nm.
(b) In the small angle approximation (which we assume holds here), the fringes are equally spaced,
so that if ∆y denotes the distance from one maximum to the next, then the distance from the
middle of the pattern to the fourth dark band must be 16.8 mm = 3.5∆y . Therefore, we obtain
∆y = 16.8/3.5 = 4.8 mm. ...
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