course Phy 232
017. `Query 15*********************************************
Question: `q**** query univ phy problem 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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Your solution: The separation consists of 1.5*10^7nm and an index of refraction of 1.5*10^6nm,index of refraction 1.5. The wavelength in the glass would be 450nm/1.5=300 nm. Then you would take 1.5*10^7/450nm/wavelength = 3.33 *10^4 wavelengths in the air and 1.5*10^6/300nm/wavelength = 5.0 *10^3 wavelengths in the glass.
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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. **
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