# Doubts about Fourier transform of IR spectroscopy

• I
• Salmone
In summary, Michelson interferometer can be used to measure infrared light. When the movable mirror is placed at a certain point, the intensity of the light is zero.
Salmone
I was studying a Michelson interferometer for infrared absorption in Fourier transform and I've found these two images (taken from https://pages.mtu.edu/~scarn/teaching/GE4250/ftir_lecture_slides.pdf ) representing an infrared monochromatic beam of light going into the interferometer and the resulting interferogram which is of course a cosine function and two monochromatic beams of light going into the interferometer which has the resulting interferogram.

With focus to the second interferogram, when the "optical path difference" (OPD from now on) is equal to ##0## of course we have a maximum of the intensity since both beams interfere constructively and when the movable mirror is moved so that the OPD is equal to ##\frac{\lambda_1}{2}## where ##\lambda_1## is the wavelenght of one of the beams, this one interfere desctructively with his part that traveled on the other path eliminating each other and the beam of wavelenght ##\lambda_2## also interfere with his part so that we get an intensity ##\neq 0## but smaller then the first one obtained when ##OPD=0##.

My question is:

At some points in the interferogram we see that total intensity of light is zero, in my opinion we get zero intensity when the movable mirror is placed at a point such that the OPD is equal to ##(n+\frac{1}{2})\lambda## for both beams, but this condition is only possible if ##\lambda_1=n\lambda_2## with ##n## being an integer, is this the only condition in which we get total intensity equal to zero?

Are we always sure that there will be a certain position for the movable mirror such that the total intensity is equal to zero?

Salmone said:
At some points in the interferogram we see that total intensity of light is zero, in my opinion we get zero intensity when the movable mirror is placed at a point such that the OPD is equal to (n+1/2)λ for both beams, but this condition is only possible if λ1=nλ2 with n being an integer, is this the only condition in which we get total intensity equal to zero?
Be careful how you use n; inconsistency may lead to confusion. The condition is that OPD = (n+1/2)λ1 and (m+1/2)λ2 simultaneously, where m and n are integers. This implies
λ12 = (2m+1)/(2n+1), i.e. the ratio of two odd integers.
It is also necessary that the intensities of the two beams are equal.
Diagram b appears to show λ1 = 2λ2; in this case the interference pattern would not have zero intensity at any point.

Gleb1964
mjc123 said:
It is also necessary that the intensities of the two beams are equal.
Actually I don't think this is true. I was thinking about A interfering destructively with B; but if A is interfering destructively with itself, and B with itself, simultaneously, then their individual intensities don't need to be equal.

Gleb1964 and hutchphd
Salmone said:
representing an infrared monochromatic beam of light going into the interferometer and the resulting interferogram
You left out a very important word here. Coherent. This is why the Michelson device uses plane waves from asingle source and a beam splitter.
The source material you supplied was terrible IMHO!

## 1. What is the Fourier transform of IR spectroscopy?

The Fourier transform of IR spectroscopy is a mathematical technique used to convert a signal from the time domain to the frequency domain. It is commonly used in IR spectroscopy to analyze the absorption of infrared light by a sample, which can provide information about the chemical bonds present in the sample.

## 2. How does the Fourier transform work in IR spectroscopy?

In IR spectroscopy, the Fourier transform works by breaking down the signal into its individual frequency components. This is achieved by passing the signal through a mathematical algorithm known as the Fourier transform. The resulting spectrum shows the intensity of absorption at different frequencies, which can be used to identify the functional groups present in the sample.

## 3. What are the advantages of using Fourier transform in IR spectroscopy?

One of the main advantages of using Fourier transform in IR spectroscopy is that it allows for a faster and more accurate analysis of the sample. It also provides a higher resolution spectrum, making it easier to identify and distinguish between different functional groups. Additionally, the use of Fourier transform allows for the collection of data over a wider range of frequencies, providing a more comprehensive analysis of the sample.

## 4. Are there any limitations to using Fourier transform in IR spectroscopy?

While Fourier transform is a powerful tool in IR spectroscopy, it does have some limitations. One of the main limitations is that it requires a high-quality, continuous signal. This means that any noise or interference in the signal can affect the accuracy of the results. Additionally, Fourier transform may not be suitable for samples with very low concentrations or complex mixtures, as it may not be able to accurately distinguish between different components.

## 5. How can I interpret the Fourier transform spectrum in IR spectroscopy?

To interpret the Fourier transform spectrum in IR spectroscopy, you can compare the peaks in the spectrum to known absorption frequencies of different functional groups. Each functional group has a unique set of absorption frequencies, which can help identify the presence of specific chemical bonds in the sample. Additionally, the intensity of the peaks can provide information about the relative concentrations of the different functional groups in the sample.

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