Here is an extremely sensitive method for measuring nitrite () down to in natural waters. The water sample is treated with sulfanilamide and N-(1-naphthylethylenediamine) in acid solution to produce a colored product with a molar absorptivity of $\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}{\mathbf{M}}^{\mathbf{-}\mathbf{1}}\mathbf{}{\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$at 540 nm. The colored solution is pumped into a 4.5-meter-long, coiled Teflon tube whose fluorocarbon wall has a refractive index of 1.29. The aqueous solution inside the tube has a refractive index near 1.33. The colored solution is pumped through the coiled tube. An optical fiber delivers white light into one end of the tube, and an optical fiber at the other end leads to a polychromator and detector.

(a) Explain the purpose of the coiled Teflon tube and explain how it functions.

(b) What is the critical angle of incidence for total internal reflection at the Teflon/water interface?

(c) What is the predicted absorbance of a$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{nM}$ solution of colored reagent?

(a)

The purpose and function of the coiled Teflon tube must be described.

Because of the optical structure of the Teflon tube, the inside solution has a much greater refractive index than the walls. Internal solution has a higher refractive index (1.33) than walls (1.29). The sample cell is a 4.5 m long tube. The sample was coiled properly to fit into a reasonable volume. The incident radiation travelled the entire length of the tube. If the path length is long, the observable absorbance indicates a very low analyte concentration.

(b)

The crucial angle must then be determined.

Snell's law:

${\mathrm{n}}_1{\mathrm{sin\theta }}_1={\mathrm{n}}_2{\mathrm{sin\theta }}_2$

${\mathrm{n}}_1$and ${\mathrm{n}}_2$are the two media's refractive indices

${\theta }_1$and ${\theta }_2$

The angle of incidence is the same as the angle of reflection of the light.

0.95499 is the crucial angle.

Snell's law states that

The smallest possible angle ${\theta }_i$whole refraction is a term used to describe the total refraction of light.

Given

$$\begin{gathered} {\mathrm{n}}_{\mathrm{cladding}})=1.29 \\ {\mathrm{n}}_1{\mathrm{sin\theta }}_1={\mathrm{n}}_2{\mathrm{sin\theta }}_2 \\ {\mathrm{n}}_{\mathrm{core}}{\mathrm{sin\theta }}_{\mathrm{i}}={\mathrm{n}}_{\mathrm{cladding}}{\mathrm{sin\theta }}_{\mathrm{r}} \end{gathered}$$

The overall impression

$$\begin{gathered} {\mathrm{sin\theta }}_2\ge 1 \\ {\mathrm{sin\theta }}_{\mathrm{i}}\ge \frac{{\mathrm{n}}_{\mathrm{ctaliane}}}{{\mathrm{n}}_{\mathrm{mene}}} \end{gathered}$$

Then there's the crucial aspect.

$$\begin{gathered} {\mathrm{n}}_{\mathrm{cladding}}=1.2900 \\ {\mathrm{n}}_{\mathrm{core}}=1.33,\mathrm{sin\theta i}³\frac{1.29}{1.33} \\ {\sin }^{-1}=0.9699 \\ \mathrm{\theta }=1.3248 \end{gathered}$$

The sin of inves valesis was transformed into$\theta$

It will get

$\theta i\ge 75.{9054}^{°}$

It get the ratio of the cladding's index of refraction to the core's index of refraction $\theta$i.

(c)

It is necessary to anticipate the absorbance of a particular solution.

According to Beer-Law, Lambert's an absorbing species' absorbance A is equal to $A=\epsilon bc$

The molar absorption coefficient is $\epsilon$, and the optical path length is b.

The concentration is given by c, meaning that absorbance is proportional to path length.

Given,

$$\begin{gathered} \mathrm{A}=\mathrm{\epsilon bc} \\ \mathrm{\epsilon }=4.5\times {10}^4{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1} \\ \mathrm{b}=450\mathrm{cm} \\ \mathrm{c}=1.0\times 10-9\mathrm{M} \end{gathered}$$

The absorbance of a solution can be estimated using the formula:

$$\begin{gathered} \mathrm{A}=\left(4.5\times {10}^4{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}\right)\left(450\mathrm{cm}\right)\left(1.0\times 10-9\mathrm{M}\right) \\ =0.02025 \end{gathered}$$

0.02025 is the absorbance of the given solution.