#$&* course phy 242 july 2 Question: `q**** query univ phy problem 33.44 11th edition 33.38 (34.28 10th edition) 3 mm plate, n = 1.5, in 3 cm separation between 450 nm source and screen. How many wavelengths are there between the source and the screen?
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Given Solution: `a** The separation consists of 1.5 cm = 1.5 * 10^7 nm of air, index of refraction very close to 1, and 1.5 mm = 1.5 * 10^6 nm of glass, index of refraction 1.5. The wavelength in the glass is 450 nm / 1.5 = 300 nm, approx.. So there are 1.5 * 10^7 nm / (450 nm/wavelength) = 3.3 * 10^4 wavelengths in the air and 1.5 * 10^6 nm / (300 nm/wavelength) = 5.0 * 10^3 wavelengths in the glass. ** STUDENT QUESTION After reading the solution, I am unsure about why the path difference is equal to twice the plate thickness. INSTRUCTOR RESPONSE One ray is reflected from the 'top' surface of the plate. The other passes through the plate to the other surface, is reflected there. It has already passed through the thickness of the plate. It has to pass back through the plate, to the other surface, before it 'rejoins' the original ray. It has therefore traveled an extra distance equal to double that of the plate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!