Full set of Notes with Fill-Ins - San Jose State University

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Chem 155 Unit 12 228 of 316Resolution in InterferometryInterferometers are easily made into highresolutionspectrometers.Consider two spectral lines differing in frequency byonly 2%:How do these lines appear in the time domain?786cos( θ ) 6cos( θ . 1.02 ) 3cos( θ ) cos( θ . 1.02 )24200.99901320 10 20 30 40 50 60 70 80 90 1000 θ2 π100Page 228 of 316
Chem 155 Unit 12 229 of 316Let us define resolution as: Δν = ν 1 - ν 2 when ν 1 canjust be separated from ν 2 .I assert that to separate, or resolve these lines in thefrequency domain spectrum, we must scan theretardation, δ, from 0 (where ν 1 and ν 2 are in phase)to a new value, let’s just call it “δ” where the twosignals are again in phase.The frequency of oscillation of the signal in time is:f = 2v(cm/s) / λ(cm) = 2 v νwhere ν is the light frequency in wavenumbers (cm -1 ).The light power function will be sinusoidal:P(δ) = P i cos(2 π f t) = P i cos(2 π (2vν) t)If we substitute v, the mirror velocity, with δ/2t = v weget intensity versus retardation:P(δ) = B cos(2 π δ ν)Page 229 of 316
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Full set of Notes with Fill-Ins - San Jose State University

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