λ=dmmmsin( .θ=°=102500mm / 200)(sin30 )nmwhere 1, 2, 3.m=…In the visible light range mcan assume the following values: m1= 4, m2= 5 and m3= 6. (a) The longest wavelength corresponds to m1= 4 with λ1= 2500 nm/4 = 625 nm. (b) The second longest wavelength corresponds to
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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.