Interference fringes (peaks or troughs) can be used to determine the pathlength

Question

Interference fringes (peaks or troughs) can be used to determine the pathlength of an empty (containing air n = 1.00) sample cell used for infrared spectroscopy. These interference fringes are observed as waves (or ripples) in the spectra, and are the result of light reflecting many times between the internal walls of the sample cell while undergoing constructive and destructive inference, which depends on the ratio of the pathlength and the wavelength of the light. When this effect is observed, the pathlength of the cell can be determined using the following equality for a empty cell, b = N/2(v_1 - v_2) where b is the pathlength, N is the number of peaks OR troughs, and v_1 and v_2 are the beginning and end points of the measured spectrum in cm^-1. An FTIR spectrum of an empty sample cell was obtained and 29 peaks were observed between 1806 cm^-1 and 544 cm^-1. For this sample cell, determine its pathlength in millimeters. Pathlength = mm

Explanation / Answer

Here N= 29 ,V1 = 1806 cm-1 and V2 = 544 cm-1

Now b= N/2(V1 - V2)

=29 / 2 x (1806 - 544)

= 29 / 2 x 1262

= 29 / 2524

= 0.0114 cm

= 0.0114 x 10 mm

= 0.114 mm

Thererofe the required pathlength in mm =0.114

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