4500HWF02

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

## 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.

### Page1 / 3

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online