1076_Physics ProblemsTechnical Physics

# 1076_Physics - Chapter 38 P38.26 sin = 0.350 d= sin = 417 632.8 nm = 1.81 10 3 nm 0.350 Line spacing = 1.81 m(a d= 1 = 2.732 10 4 cm = 2.732 10 6 m

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

Chapter 38 417 P38.26 sin . θ = 0350: d == = × λ sin . . 632 8 181 10 3 nm 0.350 nm Line spacing = 181 . m µ P38.27 (a) d × = × = −− 1 3660 2 732 10 2 732 10 2 732 46 lines cm cm m nm .. = d m sin : At 10 09 . = 478 7 . nm At 13 71 ., = 647 6 At 14 77 = 696 6 (b) d = sin 1 and 2 2 λθ = d sin so sin sin sin λλ 2 1 1 22 2 = d . Therefore, if 1 10 09 . then sin sin . 2 21 0 0 9 af gives 2 20 51 . . Similarly, for 1 13 71 2 28 30 . and for 1 14 77 2 30 66 . . P38.28 sin = m d Therefore, taking the ends of the visible spectrum to be v = 400 nm and r = 750 nm , the ends the different order spectra are: End of second order: sin 2 2 1500 r r dd nm . Start of third order: sin 3 3 1200 v v nm . Thus, it is seen that θθ 23 rv > and these orders must overlap regardless of the value of the grating spacing d. P38.29 (a) From Equation 38.12, RN
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online