Question

What length of mirror drive in an FTIR spectrometer would be required to provide a resolution of (a) 0.010 cm^-1, (b) 0.50 cm^-1, and (c) 2.0 cm^-1?

Short answer

a) Length of mirror is 50 cm.

b) Length of mirror is 1.0 cm

c) Length of mirror is 0.25 cm

Step-by-step solution

Part (a) Step 1: Given information

Given resolution is:0.010 cm^-1 .

Find the length of the mirror drive in an FTIR spectrometer.

The expression for the distance between the mirror derives in a spectrometer is written as

$$\begin{gathered} \Delta \overline{v}=\frac{1}{\delta } \\ \delta =\frac{1}{\Delta \overline{v}}......................................\left(1\right) \end{gathered}$$
Where,

$\delta$= distance between the mirror derives and

$\Delta \overline{v}$= resolution of the FTIR spectrometer

And the expression for the length of the mirror derives is written as
$L=\frac{\delta }{2}.................\left(2\right)$

Part(a)- Step2 - Explanation

Substitute $\Delta \overline{v}=0.010{\mathrm{cm}}^{-1}$in Equation (l).

$$\begin{gathered} \delta =\frac{1}{0.010{\mathrm{cm}}^{-1}} \\ =100\mathrm{cm} \end{gathered}$$

Substitute $\overline{\delta }=100\mathrm{cm}$ in Equation (II).

$$\begin{gathered} L=\frac{100\mathrm{cm}}{2} \\ =50\mathrm{cm} \end{gathered}$$

So, the length of the mirror derives at resolution of 0.010 cm is 50 cm.

Part(b) - Step 1- Given information

Given the resolution is 0.50cm^-1.

Find the length of the mirror drive in an FTIR spectrometer.

Part(b) - Step 2- Explanation

Substitute $\Delta \overline{v}=0.50{\mathrm{cm}}^{-1}$in Equation (I).

$$\begin{gathered} \delta =\frac{1}{0.50{\mathrm{cm}}^{-1}} \\ \delta =2\mathrm{cm} \end{gathered}$$

Substitute $\delta =2\mathrm{cm}$in Equation (II).

$$\begin{gathered} L=\frac{2\mathrm{cm}}{2} \\ L=1\mathrm{cm} \end{gathered}$$

So, the length of the mirror derives in a spectrometer at resolution of 0.50/cm is 1 cm.

Part(c)-Step1-Given information

Given the resolution is 2.0cm^-1.

Find the length of mirror drive in an FTIR spectrometer .

Part(c)-Step2-Explanation

Substitute $\Delta \overline{v}=2{\mathrm{cm}}^{-1}$in Equation (I).

$$\begin{gathered} \delta =\frac{1}{2{\mathrm{cm}}^{-1}} \\ \delta =0.50\mathrm{cm} \end{gathered}$$

Substitute $\delta =0.50\mathrm{cm}$in Equation (II).

$$\begin{gathered} L=\frac{0.50\mathrm{cm}}{2} \\ L=0.25\mathrm{cm} \end{gathered}$$

So, the length of the mirror derives in a spectrometer at resolution of 2/cm is 0.25 cm.