p36_080 - m will give a value of which falls into the...

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

View Full Document Right Arrow Icon
80. (a) Since n 2 >n 3 , this case has no π -phase shift, and the condition for constructive interference is =2 Ln 2 .W eso lv efo r L : L = 2 n 2 = m (525 nm) 2(1 . 55) =(169nm) m. For the minimum value of L ,let m = 1 to obtain L min =169nm . (b) The light of wavelength λ (other than 525 nm) that would also be preferentially transmitted satis±es m 0 λ =2 n 2 L ,or λ = 2 n 2 L m 0 = 2(1 . 55)(169 nm) m 0 = 525 nm m 0 . Here m 0 =2 , 3 , 4 ,... (note that m 0 = 1 corresponds to the λ = 525 nm light, so it should not be
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: m will give a value of which falls into the visible light range. So no other parts of the visible spectrum will be preferentially transmitted. They are, in fact, reected. (c) For a sharp reduction of transmission let = 2 n 2 L m + 1 / 2 = 525 nm m + 1 / 2 , where m = 0 , 1 , 2 , 3 , . In the visible light range m = 1 and = 350 nm. This corresponds to the blue-violet light....
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