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Unformatted text preview: Phys 214, Spring 2008: PS #11 Solutions Problem 1. YF Problem 35.28 The bottom plate is l = 9 cm long, the metal strip prop is h = 0.0800 mm thick, and λ = 656 nm. Taking x = 0 to be at the left edge of the bottom plate and x = l to be the right edge with the prop, the height of the top plate is y = h l x. Interference maxima are observed when k air (2 y ) = 2 πm , for m ∈ Z . The factor of 2 is since the light travels once down, reflects off the bottom plate, then travels up again. Since k = 2 π / λ , we have 2 π λ parenleftbigg 2 h l x parenrightbigg = 2 πm 2 hx λl = m. When m =1 , the distance between maxima is thus Δ x = λl 2 h ≈ 0.0369cm . Hence the number of fringes per centimeter is N = 1 Δ x ≈ 27.1 fringes / cm . Problem 2. YF Exercise 35.54 a) At the cytoplasmguanine interface there is a phase shift of π and at the guaninecytoplasm inter face there is no phase shift. Therefore, with this pattern there will always be a halfcycle phase dif ference for two neighboring reflected beams. So for guanine layers we have ( m =0 ) 2 t g = parenleftbigg m + 1 2 parenrightbigg λ n g λ = 533nm , and for cytoplasm layers we have (...
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This note was uploaded on 03/31/2009 for the course PHYS 214 taught by Professor Thorne during the Spring '08 term at Cornell.
 Spring '08
 THORNE

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