Question

An empty cell exhibited 17.5 interference fringes in the region of 1050 to 1580 cm^-1. What was the path length of the cell?

Short answer

The path length is $1.65\mathrm{x}{10}^{-2}\mathrm{cm}$.

Step-by-step solution

Step 1. Given information 

An empty cell exhibited 17.5 interference fringes in the region of 1050 to 1580 cm^-1. Determine the path length of the cell.

Step 2. Explanation

The path length of the cell is given as
$b=\frac{\Delta N}{2\left({\overline{v}}_1-{\overline{v}}_2\right)}\dots .\left(\mathrm{l}\right)$

Where,

$\Delta N:$Number of the interference fringes between the two wavelengths,

${\overline{v}}_1:$Wave number of the lower absorption bond, and the

${\overline{v}}_2:$Wave number of the higher absorption bond.

Substitute the values

${\overline{v}}_2=$1580cm^-1, ${\overline{v}}_1=$1050 cm^-1 and $\Delta N=$17.5, we get

$$\begin{gathered} b=\frac{17.5}{2\left(1050{\mathrm{cm}}^{-1}-1580{\mathrm{cm}}^{-1}\right)} \\ =\frac{17.5}{-1060{\mathrm{cm}}^{-1}} \\ =-0.01650\mathrm{cm} \\ =-1.65\times {10}^{-2}\mathrm{cm} \end{gathered}$$

The path length of the cell is not negative hence path length is $1.65\times {10}^{-2}\mathrm{cm}$.