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Principles of Fourier Transform Optical Measurements Chapter 7 Time Domain Spectroscopy • Encode spectral information about a source in the form of a time dependent electrical signal called interferogram. • The interferogram is the time dependent function of the intensity of the source. – Time domain signal: Intensity versus Time/Distance • The spectrum- the interferogram, when appropriately analyzed, yields the spectrum of the source. – Frequency Domain Signal: Signal versus Frequency. 1 The Michelson interferometer • Converts a high frequency signal to a measurable low frequency signal – Signal/frequency modulation • Interferogram is analyzed to obtain the spectrum Components of a Michelson Interferometer • Two plane mirrors at right angle – – • • Fixed mirror Movable mirror Beam splitter at an angle of 45° to the mirrors: divides incoming light (ideally 50 % transmitted and 50 % reflected) Compensator: equalizes the optical path lengths in both arms 2 Simple Case of Idealized Monochromatic Radiation • Assume for simplicity that there is a monochromatic input to the interferometer adjusted so that the optical path lengths in both arms are identical. • The two beams will be in phase when they return to the beam splitter and, as such, they will constructively interfere. • Looking into the interferometer at the output, the field will appear bright. • Move one mirror back 1/4 of a wavelength. The two beams will be 180° out of phase and they will destructively interfere, the field will appear dark. • If the mirror is continuously moved, the field will oscillate from light to dark for each quarter-wavelength movement of the mirror. • One cycle of the interferogram occurs when the mirror moves a distance equal to 1/2 λ. Relationship between the Frequency of Interferogram and Optical Frequency Let τ be the time required for the mirror to move 1/2 wavelength, VM the constant mirror velocity, f the frequency of the inteferogram, and ν the optical frequency. λ 2 = VM × τ f = 1 τ = 2VM λ (1) = 2VM ν = 2VM ν c (2) (τ is the period of the interferogram / time required to complete a cycle) if the mirror velocity is 1 cm/s, the frequency of the interferogram is, f = 6.67 × 10 −11ν 3 The equation for the interferogram P (δ ) = 1 P (ν ) cos(2πft ) 2 It is a simple cosine wave, the amplitude of which depends on the intensity of the monochromatic source δ :retardation: difference in path travelled by the two beams P (δ ) : radiant power of the output as a function of retardation P(ν ) : radiant power of the source as a function of optical frequency (source spectrum) Modified interferogram equation P(δ ) = B (ν ) cos(2πft ) The modified equation accounts for unequal splitting of the source power and the frequency dependence of the detector response Substitute f by equation 2, then substitute VM by P(δ ) = B (ν ) cos(2πδ ν ) δ /2 t The frequency of the oscillation depends on two factors: 1) the frequency of the incoming electromagnetic radiation 2) the velocity of the mirror Thus, using an interferometer a very high optical frequency can be uniquely encoded in the form of a low-frequency oscillation. 4 Simple Interferograms and Spectra What is the nature of a broadband spectral input? • Each input can be treated independently and hence the output will be the summation of cosine functions. • At zero path difference, all the waves are in phase • As the mirror is moved away from zero position the waves rapidly sum out to a steady average value. • The resulting ac signal/ the interferogram can be expressed mathematically as: +∞ (1) P (δ ) = ∫ B(ν ) cos(2πδν )dν −∞ 5 Fourier Transformation • The Fourier transform of the integral (1) is integral (2) • Integrals (1) and (2) form a cosine Fourier transform pair. +∞ (1) P(δ ) = ∫ B(ν ) cos(2πδν )dν −∞ FT +∞ (2) B (ν ) = ∫ P(δ ) cos(2πδν )dδ −∞ Effects of finite mirror movement and digitization • To reconstruct the exact spectrum of the source, the interferogram must be measured from - ∞ to ∞+. However, the mirror movement is finite. • Digitization of the interferogram requires use of finite-sized sampling interval. • These two practical constraints limit the resolution and the frequency range. • The width of the signal is inversely related to the maximum extent of the interferogram, hence the mirror movement. 6 Resolution • Resolution depends on the width of the interferogram (mirror travel distance) • The minimum mirror travel distance required for two lines to be resolved is given by: ∆ν = ν 2 − ν1 = 1 δ Advantages of FT Methods • • • • • • Multiplex or Fellgets advantage Throughput advantage High accuracy and reproducibility of frequency measurements High resolution Computerization Controlled resolution function 7 Multiplex or Fellgets advantage • • In scanning each resolution element is "seen" during a fraction of the total scanning time (T). If M is the number of resolution elements, then Number of resolution elements =(λmax − λmin ) ∆λ T Signal ∝ Noise ∝ S ∝ N M T M T M Fellgets advantage • With the interferometer, each resolution element is "seen" all the time (all optical frequencies are incident on the detector at once). The spectral information is said to be multiplexed. Signal ∝ T Noise ∝ T S ∝ T N • Superior by a factor of M 8 Throughput advantage • • • • Throughput is the amount of light that one can get through the spectrometer. One of the factor that limits the throughput of a grating or prism instrument is the necessity for an entrance slit. The interferometer has a large circular entrance aperture, roughly the size of the mirrors and, as such, has greater throughput. The improvement is a factor of 100. 9