4500HWF02 - Fourier Transform Infrared Spectrometer 3 A...

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Unformatted text preview: Fourier Transform Infrared Spectrometer 3 A Fourier Transform Infrared (FTIR) spectrometer is based on a Michelson in— terferometer. An FTIR spectrometer is capable of determining the spectral content of a light source. This is accomplished by scanning one of the mirrors of the Michelson in- terferometer and recording the output interferogram as a function of the optical path difference between the two interfering beams. In an FTIR, the wavelength of a wave is typically represented by its wavenumber v where ’U‘—"X, and A is the freespace wavelength. A particular argon ion laser emits with a Gaussian lineshape given by Ba» = mr— ($302], ' where k is the wavevector magnitude (k = 27r/A 2 27m), k0 is the center wavevector magnitude, and Ak is the wavevector bandwidth. For this spectrum, calculate, showing all work, the interferogram, I (6), produced by the FTIR for this spectrum, where 6 is the optical path difference between the two arms of the Michelson interferometer. Sim- plify your answer as much as possible. Put your final answer in the space provided. 1(5) = Fourier Transform Infrared Spectrometer 3 The spectrum is k — kg 2 BO“) " empl_( Ak )l The corresponding interferogram is 1+°° 1(5):? _ B(k)cos(k6)dk, OI‘ I(6) = 2—17r—/+oo exp[—— cos(k6) dk 1(a) = 3’3 +°°ezp[ — (k ‘ '“°)2]cos[Ak6( k )1 emf-,5) 27r _°° Ak E ‘ = £1: (am m cos[Ak6(k;:o) + {Ska} = cos[Ak6( k gkk" )1 cos(6ko)'— sin[Ak6( k L?" )1 sin(6k0) 1(6) = $7]: [:0 exp[ — (k;:0)2]cos[Ak6(k)] cos(6ko) (at?) _ % +00 emp[- (k;:0)2] sinlAkngkkofl Sin(6ko) d(k;kk°) The second integral is an odd function sin[Ak6(k—KT’:9- ] multiplied by an even function emp[ — (kw—'99)? and integrated from —00 to +00. This integral is thus zero. The first integral is of the form +00 / emp(—a2:r2)cos(ba:) d3: E g exp(—b2/4a2), 0 except that it is over the range from —00 to +00 and so it is twice as large, +00 / emp(—a2w2)cos(b:v) dw E 3/;‘7: exp (—b2/4a2), —00 In the present case a = 1 and b = Ak 6 and thus, the inteferogram is I(6) = cos(ko 6)] [fiestfl — (AZ5)2)], II(6) = % égcosaco 6) emp[ — (A—Séf], which has the appearance of a (c0)sinusoid in 6 with a Gaussian envelope. ...
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This note was uploaded on 04/29/2008 for the course ECE 4500 taught by Professor Gaylord during the Spring '08 term at Georgia Tech.

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4500HWF02 - Fourier Transform Infrared Spectrometer 3 A...

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