Chapter 7.5, Problem 58E

a.

To determine

To find:the value of $r$ .

Answer to Problem 58E

  $13.71°$

Explanation of Solution

Given information:

Snell’s law is as follows:

  $n_1\sin i=n_2\sin r$

Here the refractive index of the light existing and entering the medium is $n_1$ and $n_2$ respectively.

The angle of incidence and refraction is $i$ and $r$ respectively.

  $\begin{array}{l}{n}_1=1.00\\ n_2=2.42\\ i=35°\end{array}$

Calculation:

Substitute all the given values in the Snell’s law and evaluate $r$ as follows:

  $\begin{array}{c}\left(1.00\right)\sin \left(35°\right)=\left(2.42\right)\sin r\\ \sin 35°=2.42\sin r\\ \sin r=\frac{\sin 35°}{2.42}\\ r={\sin }^{-1}\left(\frac{\sin 35°}{2.42}\right)\\ ={\sin }^{-1}\left(\frac{0.57}{2.42}\right)\\ ={\sin }^{-1}0.24\\ =13.71°\end{array}$

Thus, the value of $r$ is $13.71°$ .

b.

To determine

To describe:the use of Snell’s law by gemologist to identify whether the diamond is genuine.

Answer to Problem 58E

By comparing the actual value of refractive index for diamond with the evaluated refractive index value of the substance.

Explanation of Solution

Given information:

Snell’s law is as follows:

  $n_1\sin i=n_2\sin r$

Here the refractive index of the light existing and entering the medium is $n_1$ and $n_2$ respectively.

The angle of incidence and refraction is $i$ and $r$ respectively.

  $\begin{array}{l}{n}_1=1.00\\ n_2=2.42\\ i=35°\end{array}$

Calculation:

The gemologist may use Snell’s law to the evaluate the refractive index of light entering the sample for the value 1.00 of refractive index of light existing. If the refractive index of light entering the sample is 2.42, then the sample is genuine diamond.