Question

The interferometer mirror of a Fourier transform infrared spectrophotometer travels pm1 cm .

(a) How many centimeters is the maximum retardation,$\mathbf{∆}$ ?

(b) State what is meant by resolution.

(c) What is the approximate resolution $\left({\mathrm{cm}}^{‐1}\right)$of the instrument?

(d) At what retardation interval,$\mathbf{∆}$ , must the interferogram be sampled (converted into digital form) to cover a spectral range of 0 to 2000 cm^-1 ?

Short answer

1. The greatest Fourier transform retardation in centimetre is +2 cm .
2. The ability to distinguish between neighbouring peaks is referred to as resolution.
3. The instrument's approximate resolution is $\frac{1}{∆}=0.5{\mathrm{cm}}^{‐1}$
4. The interferogram and retardation interval is $\mathrm{\delta }=2.5\mathrm{\mu m}$

Step-by-step solution

Definition of interferogram and retardation

• The concept behind this interferometry is simple: interference fringes are created by splitting a monochromatic light beam into two beams, one of which strikes a fixed mirror and the other a moveable mirror. When the reflected beams are brought back together, an interference pattern appears.
• When a body's velocity drops, its acceleration reduces. Negative acceleration is referred to as 'deceleration' or'retardation.'

Determine the retardation in centimetre

(a)

The maximum Fourier transform retardation in centimetre must be computed.

The difference in route length followed by the two waves is referred to as retardation.

$\mathrm{\delta }\left(=2\left[\mathrm{OM}‐\mathrm{OS}\right]\right)$retardation is the term for this disparity.

A Fourier transform IR spectrophotometer's interferometer mirror moves is$\pm 1\mathrm{cm}$

The greatest Fourier transform retardation in centimetre is +2 cm

Determine the resolution

(b)

It is necessary to define the resolution.

The ability to distinguish between neighbouring peaks is referred to as resolution.

Determine the approcximate resolution

(c)

The instrument's approximate resolution must be established.

The instrument's approximate resolution is $\frac{1}{∆}=0.5{\mathrm{cm}}^{‐1}$

Determine the retardation interval

(d)

To span a spectral range, the retardation interval and interferogram must be determined.

The difference in route length followed by the two waves is referred to as retardation.

$\mathrm{\delta }\left(=2\left[\mathrm{OM}‐\mathrm{OS}\right]\right)$retardation is the term for this disparity.

Interferogram: Fourier analysis of an interferogram returns the spectrum from whence it was created. The spectrum is the interferogram's Fourier transform.

The interferogram and retardation interval should be sampled to cover a spectral range as calculated in

$$\begin{gathered} \mathrm{\delta }=\frac{1}{\left(2∆\mathrm{v}\right)} \\ =\frac{1}{\left(4000{\mathrm{cm}}^{-1}\right)} \\ =2.5\mathrm{\mu m} \end{gathered}$$

The interferogram and retardation interval is $\mathrm{\delta }=2.5\mathrm{\mu m}$