p36_039 - 36-7 bears this out. We disregard the m = 0 value...

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

View Full Document Right Arrow Icon
39. The situation is analogous to that treated in Sample Problem 36-5, in the sense that the incident light is in a low index medium, the thin flm has somewhat higher n = n 2 , and the last layer has the highest reFractive index. To see very little or no reflection, according to the Sample Problem, the condition 2 L = µ m + 1 2 λ n 2 where m =0 , 1 , 2 ,... must hold. The value oF L which corresponds to no reflection corresponds, reasonably enough, to the value which gives maximum transmission oF light (into the highest index medium – which in this problem is the water). (a) IF 2 L = ( m + 1 2 ) λ n 2 (Eq. 36-34) gives zero reflection in this type oF system, then we might reasonably expect that its counterpart, Eq. 36-35, gives maximum reflection here. A more careFul analysis such as that given in §
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 36-7 bears this out. We disregard the m = 0 value (corresponding to L = 0) since there is some oil on the water. Thus, For m = 1 , 2 , . . . maximum reection occurs For wavelengths = 2 n 2 L m = 2(1 . 20)(460 nm) m = 1104 nm , 552 nm , 368 nm . . . We note that only the 552 nm wavelength Falls within the visible light range. (b) As remarked above, maximum transmission into the water occurs For wavelengths given by 2 L = m + 1 2 n 2 = = 4 n 2 L 2 m + 1 which yields = 2208 nm , 736 nm , 442 nm . . . For the dierent values oF m . We note that only the 442 nm wavelength (blue) is in the visible range, though we might expect some red contribution since the 736 nm is very close to the visible range....
View Full 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.

Ask a homework question - tutors are online