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Unformatted text preview: This week Please hand in or upload to elms Lab3 lab report HW 5 due next class (after spring break) Today we start lab 4 (continued after spring break) FabryPerot Since n depends on lambda, difference in path length depends on lambda Thin Film δ =( k 2 i s 2 − k 1 • s 1 ) + φ 2 − φ 1 k 2 = nk 1 also φ 2 = φ 1 for normal incidence δ = k ( n 2 t ) (n2t is effective difference in path length) δ = 2 π : constructive interference 2 π λ n 2 t = 2 π λ = 2 nt : constructive interference if phase change on reflection is same for both reflections. if opposite, need to add π : destructive interference For normally incident rays: I total = 4 I cos 2 δ 2 Thin films δ = n 2 2( L + d ) − n 1 D sin θ = D 2 w and sin α = d w n 1 sin θ = n 2 sin α n 1 D 2 w = n 2 d w d = n 1 2 n 2 D δ = 2 n 2 L cos α = L t (see bottom α in pic) δ =2n 2 t cos α if no change of phase on reflection, constructive interference for 2n 2 t cos α = m λ If rays are not normally incident… Previous works if the source is far away. If instead we have a point source near by: Looks like 2 narrow slits e with source spacing along the line as shown… Don’t need lens to see fringes...
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This note was uploaded on 12/29/2011 for the course PHYSICS 375 taught by Professor Eno during the Spring '11 term at Maryland.
 Spring '11
 ENO
 Physics

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