hw11_solutions

# hw11_solutions - Phys 214 Spring 2008 PS#11 Solutions...

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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 cytoplasm-guanine interface there is a phase shift of π and at the guanine-cytoplasm inter- face there is no phase shift. Therefore, with this pattern there will always be a half-cycle 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 ( m = 0 ) 2 t c = parenleftbigg m + 1 2 parenrightbigg λ n c

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• Spring '08
• THORNE
• Light, Wavelength, YF Problem

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