65. (a) If both mirrors are perfectly reﬂecting, there is a node at each end of the crystal. With one end partially silvered, there is a node very close to that end. We assume nodes at both ends, so there are an integer number of half-wavelengths in the length of the crystal. The wavelength in the crystal is λ c = λ/n ,where λ is the wavelength in a vacuum and n is the index of refraction of ruby. Thus N ( λ/ 2 n )= L ,where N is the number of standing wave nodes, so N = 2 nL λ = 2(1 . 75)(0 . 0600 m) 694 × 10 − 9 m =3 . 03 × 10 5 . (b) Since λ = c/f ,where f is the frequency, N =2 nLf/c and ∆ N =(2 nL/c )∆
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.