ch35-p111 - 111. (a) The path length difference is 0.5 m =...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
111. (a) The path length difference is 0.5 µ m = 500 nm, which is represents 500/400 = 1.25 wavelengths — that is, a meaningful difference of 0.25 wavelengths. In angular measure, this corresponds to a phase difference of (0.25)2 π = π /2 radians 1.6 rad. (b) When a difference of index of refraction is involved, the approach used in Eq. 35-9 is quite useful. In this approach, we count the wavelengths between S 1 and the origin N Ln
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online