Chapter 13, Problem 1Q

Interpretation Introduction

Interpretation:

The refractive index for a compound at $25°\text{C}$ is to be determined when the refractive index of the compound at $20.1°\text{C}$ is 1.3191.

Concept introduction:

Refractive index is an important physical property that determines the composition of the substance. It determines the amount of refraction the light undergoes due to the change in speed because of the difference in densities.

Refractive index $\left(n\right)$ is defined as the ratio of the velocity of light in vacuum or air to the velocity of light in another medium of the substance under study that is given as follows:

  $n=\frac{V_{\text{air/vacuum}}}{V_{\text{another}\text{medium}\text{under}\text{study}}}$

Where,

• $V_{\text{air/vacuum}}$ is the velocity of light in vacuum or air.
• $V_{\text{another}\text{medium}\text{under}\text{study}}$ is the velocity of light in another medium of the substance under study.

The density of organic compounds varies with temperature so does velocity of light passing through it. Density is inversely proportional to the temperature of the substance.

To determine the refractive index an instrument called refractometer is used.

The formula to calculate the variation of refractive index with temperature is given as follows:

  $\Delta n=4.5\times {10}^{-4}\times \left(T_1-T_2\right)$

Where,

• $\Delta n$is the change in the refractive index.
• $T_{\text{1}}$ and $T_2$ is the initial and final temperature, respectively.

Answer to Problem 1Q

The refractive index for the compound at $25°\text{C}$ is 1.3168.

Explanation of Solution

Given Information:

The refractive index for the compound at $20°\text{C}$ is 1.3191.

The formula to calculate the variation of refractive index with temperature is given as follows:

  $\Delta n=4.5\times {10}^{-4}\times \left(T_1-T_2\right)$

The value of $T_{\text{1}}$ is $20.1°\text{C}$ .

The value of $T_2$ is $25°\text{C}$ .

Substitute the values in the above formula to calculate the variation of refractive index with temperature.

  $\begin{array}{c}\Delta n=4.5\times {10}^{-4}\times \left(T_{\text{1}}-T_{\text{2}}\right)\\ =4.5\times {10}^{-4}\left(20.1°\text{C}-25°\text{C}\right)\\ =-0.002205\end{array}$

The formula to calculate the refractive index at $25°\text{C}$ is given as follows:

  $\Delta n=n_2-n_1$ ....... (1)

Where,

• $n_2$ is the refractive index at $25°\text{C}$ .
• $n_1$ is the refractive index at $20.1°\text{C}$ .

Rearrange the equation (1) to calculate the value of the refractive index $\left(n_2\right)$ at $25°\text{C}$. We get,

  $n_2=\Delta n+n_{\text{1}}$

The value of $\Delta n$ is 0.002205.

The value of $n_{\text{1}}$ is 1.3191.

Substitute the values in the above formula to calculate the refractive index ( ${\text{n}}_2$ ) at $25°\text{C}$ .

  $\begin{array}{c}{n}_2=n_1+\Delta n\\ =1.3191+\left(-0.002205\right)\\ =1.3168\end{array}$

Conclusion

The refractive index for the compound at $25°\text{C}$ is 1.3168.