ch36-p059 - 59. (a) We note that d = (76 106 nm)/40000 =...

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

View Full Document Right Arrow Icon
59. (a) We note that d = (76 × 10 6 nm)/40000 = 1900 nm. For the first order maxima λ = d sin θ , which leads to = F H G I K J = F H G I K J −− sin . 11 589 1900 18 λ d nm Now, substituting m = d sin / λ into Eq. 36-30 leads to D = tan / λ = tan 18°/589 nm = 5.5 × 10 –4 rad/nm = 0.032°/nm. (b) For m = 1, the resolving power is R = Nm = 40000 m = 40000 = 4.0 × 10 4 . (c) For m = 2 we have = 38°, and the corresponding value of dispersion is 0.076°/nm.
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