Refer to the Fourier transform infrared spectrum in Figure 20-33.

(a) The interferogram was sampled at retardation intervals of$\mathbf{1}\mathbf{.}\mathbf{2260}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{cm}$. What is the theoretical wavenumber range (0 to ?) of the

spectrum?

(b) A total of 4 096 data points were collected from $\mathbf{\delta }\mathbf{=}\mathbf{-}\mathbf{∆}\mathbf{}\mathbf{to}\mathbf{}\mathbf{\delta }\mathbf{=}\mathbf{+}\mathbf{∆}$. Compute the value of$\mathbf{∆}$, the maximum retardation.

(c) Calculate the approximate resolution of the spectrum.

(d) The interferometer mirror velocity is given in the figure caption. How many microseconds elapse between each datum?

(e) How many seconds were required to record each interfero gram once?

(f) What kind of beam splitter is typically used for the region 400 to 4 000${\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$? Why is the region below 400${\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$not observed?

Given,

$$\begin{gathered} \mathrm{\lambda }=\frac{1}{{\displaystyle \tilde{\mathrm{v}}}} \\ \mathrm{\lambda }=\mathrm{wavelength}\mathrm{of}\mathrm{spectral}\mathrm{line} \\ \stackrel{\mathbf{~}}{\mathbf{v}}=\mathrm{wave}\mathrm{number} \\ ∆\tilde{\mathrm{v}}=\frac{1}{28} \\ =\frac{1}{\left(2.12660\times {10}^{-4}\mathrm{cm}\right)} \\ =3949{\mathrm{cm}}^{-1} \end{gathered}$$,

The solution for theoretical wavenumber range is $3949{\mathrm{cm}}^{-1}$.

Each interval is,

$$\begin{gathered} =0.5186\mathrm{cm} \\ \mathrm{The}\mathrm{range}\mathrm{of}\pm ∆,\mathrm{so}∆=0.2593\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{resolution}\approx \frac{1}{∆} \\ =\frac{1}{\left(0.2593\mathrm{cm}\right)} \\ =3.86{\mathrm{cm}}^{-1} \end{gathered}$$

The solution for approximate resolution of the spectrum is 3.86${\mathrm{cm}}^{-1}$ .

The mirror velocity = 0.693cm/s

$$\begin{gathered} \mathrm{interval}=\frac{1.2660\times {10}^{-4}\mathrm{cm}}{0.693\mathrm{cm}/\mathrm{s}} \\ =183\mathrm{\mu S} \end{gathered}$$

The Solution for microseconds elapse between each datum is 183$\mathrm{\mu s}$.

Given,

total data point is =4096 points

interval =183$\mathrm{\mu S}$

Both total data point and intervals are multiplied we get

$\left(4096\mathrm{points}\right)\left(183\mathrm{\mu s}/\mathrm{point}\right)=0.748\mathrm{s}$

The solution for seconds were required to record each interfero gram once is 0.748s.

The background adjustment is clearly visible because the beam splitter is germanium in KBr. The KBR absorbs less than 400${\mathrm{cm}}^{-1}$ light