Michelson Lab.pdf

Section 3: Calculation of Spectral ResolutionTo calculate the spectral resolution of the FTS we will use a clever trick. Basicallywhat we need is the spectrum (given of course by A 2 (k)) that we would obtain for anidealized monochromatic source. If Fourier Transform Spectroscopy were perfect,then a monochromatic source would yield a spectrum which is non-zero only at thefrequency k 0 (a Dirac spike). But because the total path length difference D isnecessarily finite, what we actually obtain is a power spectrum peaked at k 0 whichfalls off quickly as you move away from the central frequency.First let’s figure out what the measured spectrum A 2 (k) will look like given amonochromatic source. Recall that the power output corresponding to amonochromatic source can be expressed as:2A0 I 1 cosk0z4Plugging this in to:yields:kD2A4kDDIkcoskzDdz222A A0A0 coskzdz cosk0zcoskzdz4 44DDHere the first term oscillates and for large D can be ignored. The second term issolved using trig identities:Thus:AndD2 D2AA020k A kz kzdz k k z k k z 0cos0coscos0cosD2D22 A 0sin k k0zsinA k 2 k k0k kk k00zk k 0Dk k DDdz2 2sink k0D sinA k A0 k k00By this point you may have lost track of the definitions of k 0 and k. k 0 is a constantdenoting the actual propagation constant of the monochromatic source, and k is avariable denoting the measured frequency range. Since k 0 is approximately equal to k(otherwise our spectrometer wouldn’t be of much use) we notice that:1k k k k so the first term in brackets dominates and we can neglect the second term.If we let: k k D , then we have:X0A20k1sin XXThe Sin(X)/X term acts as an envelope function, since the function is greatest in therange:012 of 13