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Double-pass Fourier transform imaging spectroscopy

R. Heintzmann, K. A. Lidke, and T. M. Jovin

Open Access Open Access

Abstract

Fourier Transform Imaging Spectroscopy (FTIS) has recently emerged as a widely used tool for spectral imaging of biological fluorescent samples. Here we report on a novel double-pass FTIS system capable of obtaining an excitation as well as an emission spectrum of a fluorescent sample with only a single sweep of the interferometer. This result is achieved by a modification of an existing FTIS system, which now places the excitation source before the interferometer so as to spectrally modulate the excitation as well as the detection. An analysis of the acquired signal allows for the reconstruction of the excitation as well as the emission spectrum of each fluorophore, assuming an independence of the two spectra for each fluorophore. Due to the patterned excitation generated by the Sagnac interferometer, a substantial degree of optical sectioning is achieved at excitation wavelengths. Further analysis of the acquired data also enables the generation of optically sectioned emission images. A theoretical analysis and experimental data based on fluorescent beads are presented.

©2004 Optical Society of America

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Figures (8)
Fig. 1.
Fig. 1. Sketch of the Fourier transform imaging spectrometer. The emission light of the sample, excited via the excitation filter (Ex) selects the maximal excitation bandwidth. The emission passes the dichroic mirror (Di), emission filter (Em), and interferometer with two alternate paths (A,B) to form an image on the CCD. These two paths interfere at the CCD, such that the detected intensity in every pixel is dependent on the wavelength and the optical path difference between A and B. The Sagnac based interferometer contains a beam splitter (BS) which is slightly detuned in angle (β) from its symmetrical position. This leads to a dependence of the path difference on the input angle of the light to the interferometer. This input angle (and thus the OPD) depends on the orientation (α) of the whole interferometric block (BS and mirrors M) and on the spatial position of the sample point. An interferogram is acquired by taking CCD images through a series of angular positions of the interferometer with equal increments in α. Note that the OPDs for light from different sample points in the image are not necessarily identical.

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Fig. 2.
Fig. 2. Modified setup allowing for the acquisition of an excitation spectrum along with the emission spectrum. The filter cube has been moved to a position in the infinity beam-path between interferometer and tube lens forming the image on the CCD camera. See also Fig. 1.

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Fig. 3.
Fig. 3. Sketch of the absolute magnitude of the Fourier transformed signal originating from a spectral band of uniform brightness. (a) Original spectrum Sem (λ). (b) Corresponding measured interferogram. (c) Recovered spectrum from the Fourier transform of the detected interferogram. Note the extra peak at zero frequency and the mirror of the spectrum on the left hand side.

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Fig. 4.
Fig. 4. Sketch of the absolute magnitude of the Fourier transformed double-pass signal originating from a spectral band of uniform brightness for each, excitation and emission. (a) Original spectra Sex (λ) [green] and Sem (λ) [red]. (b) Recovered spectrum from the Fourier transform. Note the extra peak at zero frequency, the difference frequency spectrum (A), emission (B), excitation spectrum (C) and the sum frequency spectrum (D).

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Fig. 5.
Fig. 5. Spectra of a green and a yellow bead as obtained by single (a) and double pass (b) Fourier transform imaging spectroscopy. The insets are plots of the respective normalized spectra of a green and a yellow (displayed in red) cluster of beads each. Their abscissa shows the wavelength in nanometers.

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Fig. 6.
Fig. 6. Measured intensity of a fluorescent plane scanned through the plane of best focus. It is observed that only the excitation signal and the two cross-peaks show optical section behavior, whereas the emission signal does not.

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Fig. 7.
Fig. 7. Setup for reducing unsuppressed excitation light by the use of two polarizers.

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Fig. 8.
Fig. 8. Difference light traces for reflected excitation light and emission light from the sample. The blue lines denote the excitation (illumination) light and the (~50%) reflected light from the interferometer and the red light mark the emission light (+scattered, +reflected) light from the object to measure. For the sake of clarity, only one of the two reflective paths is shown.

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Equations (13)

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I det ( OPD ) = ∫ 0 ∞ ε em ( k em , OPD ) I em ( k em ) d k em
ε em ( k em , OPD ) = ( 1 + m em cos ( k em OPD ) ) ⁄ 2
I ex ( k ex , OPD ) = ε ex ( k ex , OPD ) I Ls ( k ex )
I em ( k ex , k em , OPD ) = S ex , em ( k ex , k em , OPD ) I ex ( k ex , OPD ) ρ
I em ( k em , OPD ) = S em ( k em , OPD ) ρ ∫ 0 ∞ S ex ( k ex , OPD ) I ex ( k ex , OPD ) d k ex
I det ( OPD ) = ∫ 0 ∞ ε em ( k em , OPD ) I em ( k em , OPD ) d k em
= ρ ∫ 0 ∞ ∫ 0 ∞ [ ε em ( k em , OPD ) S em ( k em , OPD )
× S ex ( k ex , OPD ) I ex ( k ex , OPD ) ] d k ex d k em
= ρ ∫ 0 ∞ ε em ( k em , OPD ) S em ( k em , OPD ) d k em
× ∫ 0 ∞ ε ex ( k ex , OPD ) S ex ( k ex , OPD ) I Ls ( k ex ) d k ex
FT ( I det ( OPD ) ) = ρ [ a δ ( k ) + ( m em ⁄ 2 ) S em ( k ) + ( m em ⁄ 2 ) S em ( − k ) ]
⊗ [ b δ ( k ) + ( m ex ⁄ 2 ) S ex ( k ) I Ls ( k ) + ( m ex ⁄ 2 ) S ex ( − k ) I Ls ( − k ) ] ,
a = ∫ 0 ∞ S em ( k em ) d k em , b = ∫ 0 ∞ S ex ( k ex ) I Ls ( k ex ) d k ex
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