a soap film has an index of refraction n=1.5. the film is viewed inreflected lig

Question

a soap film has an index of refraction n=1.5. the film is viewed inreflected light. a) at a spot where the film thickness is 910 nm,which wavelengths are missing in the reflected light? b) whichwavelengths are strongest in the reflected light?

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Explanation / Answer

{Film Thickness} = d = 910 nm = 910.0e-9 m {Index Of Refraction} = n = 1.5 {Visible Wavelengths} = {400 - 700 nm} Soap Film is assumed to be in theform of a bubble which has air on both sides of the film. (This causes a Pi radian phase shift of the reflected incident ray, and no phase shift of the internally reflected ray.) ITEM#A: {DESTRUCTIVE Interference (for missingwavelengths): ---->    2*n*d = (m)*   .... for m = 1, 2,3, 4, ... ---->    = 2*n*d/(m)   .... for m = 1, 2, 3,4, ... ---->    = 2*(1.5)*(910.0e-9)/((1)) = 2.73e-6 m = 2730 nm = 2*(1.5)*(910.0e-9)/((2)) = 1.365e-6 m = 1365 nm = 2*(1.5)*(910.0e-9)/((3)) = 9.1e-7 m = 910nm = 2*(1.5)*(910.0e-9)/((4)) = 6.825e-7 m = 683nm = 2*(1.5)*(910.0e-9)/((5)) = 5.46e-7m = 546 nm = 2*(1.5)*(910.0e-9)/((6)) = 4.55e-7 m = 455nm = 2*(1.5)*(910.0e-9)/((7)) = 3.9e-7 m = 390 nm = 2*(1.5)*(910.0e-9)/((8)) = 3.412e-7 m = 341.2 nm ---->   {Missing VisibleWavelengths} = { 455, 546, 683 nm} ITEM#B: {CONSTRUCTIVE Interference(for strongest wavelengths): ---->    2*n*d = (m -0.5)*   .... for m = 1, 2,3, 4, ... ---->    = 2*n*d/(m -0.5)   .... for m = 1, 2, 3, 4,... ---->    = 2*(1.5)*(910.0e-9)/((1) - 0.5) = 5.46e-6 m = 5460 nm = 2*(1.5)*(910.0e-9)/((2) - 0.5) = 1.82e-6 m = 1820 nm = 2*(1.5)*(910.0e-9)/((3) - 0.5) = 1.092e-6 m = 1092 nm = 2*(1.5)*(910.0e-9)/((4) - 0.5) = 7.8e-7 m = 780 nm = 2*(1.5)*(910.0e-9)/((5) - 0.5) = 6.067e-7 m = 606.7nm = 2*(1.5)*(910.0e-9)/((6) - 0.5) = 4.964e-7 m = 496.4nm = 2*(1.5)*(910.0e-9)/((7) - 0.5) = 4.2e-7 m = 420nm = 2*(1.5)*(910.0e-9)/((8) - 0.5) = 3.64e-7 m = 364 nm ---->   {Strongest VisibleWavelengths} = { 420, 496, 607 nm} .

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