ch36-p097 - m max = d sin 90 = 12 (8900)(5 10-4 ) 2 where...

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97. The problem specifies d = 12/8900 using the mm unit, and we note there are no refraction angles greater than 90 ° . We convert λ = 500 nm to 5 × 10 4 mm and solve Eq. 36-25 for " m max " (realizing it might not be an integer):
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Unformatted text preview: m max = d sin 90 = 12 (8900)(5 10-4 ) 2 where we have rounded down. There are no values of m (for light of wavelength ) greater than m = 2....
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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