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

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

View Full Document Right Arrow Icon
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
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 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