ch36-p077 - 77 Letting d sin = m we solve for = d sin(10 mm...

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

View Full Document Right Arrow Icon
λ= d mm m sin ( . θ = ° = 10 2500 mm / 200)(sin30 ) nm where 1, 2, 3 . m = In the visible light range m can assume the following values: m 1 = 4, m 2 = 5 and m 3 = 6. (a) The longest wavelength corresponds to m 1 = 4 with λ 1 = 2500 nm/4 = 625 nm. (b) The second longest wavelength corresponds to
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