ch35-p088 - S 2 is 1360 nm ⇔ 1360/400 = 3.4 from P 2...

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

View Full Document Right Arrow Icon
88. (a) Since P 1 is equidistant from S 1 and S 2 we conclude the sources are not in phase with each other. Their phase difference is φ source = 0.60 π rad, which may be expressed in terms of “wavelengths” (thinking of the λ 2 π correspondence in discussing a full cycle) as source = (0.60 π / 2π ) λ = 0.3 λ (with S 2 “leading” as the problem states). Now S 1 is closer to P 2 than S 2 is. Source S 1 is 80 nm ( 80/400 λ = 0.2 λ ) from P 2 while source
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: S 2 is 1360 nm ( ⇔ 1360/400 λ = 3.4 λ ) from P 2 . Here we find a difference of ∆ path = 3.2 λ (with S 1 “leading” since it is closer). Thus, the net difference is ∆ net = ∆ path – ∆ source = 2.90 λ , or 2.90 wavelengths. (b) A whole number (like 3 wavelengths) would mean fully constructive, so our result is of the following nature: intermediate, but close to fully constructive....
View Full 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