Chapter 35, Problem 87CP

To determine

The intensity of transmitted light in terms of the fraction of the incident intensity.

Answer to Problem 87CP

The intensity of transmitted light is $70.6\%$ of the incident intensity.

Explanation of Solution

The light ray path which is perpendicular to the diamond surface is as shown in the figure below.

Figure-(1)

Write the expression to obtain the refracted light intensity.

    $S_1^{\text{'}}={\left(\frac{n_2-n_1}{n_2+n_1}\right)}^2{S}_1$

Here, $S_1^{\text{'}}$ is the refracted light intensity, $n_1$ is the refractive index of air and $n_2$ is the refractive index of diamond and $S_1$ is the average Poynting vector magnitude.

Re-write the above equation.

    $\frac{S_1^{\text{'}}}{S_1}={\left(\frac{n_2-n_1}{n_2+n_1}\right)}^2$

Substitute $r$ for $\frac{S_1^{\text{'}}}{S_1}$ in the above equation.

    $r={\left(\frac{n_2-n_1}{n_2+n_1}\right)}^2$                                                                                                          (I)

Here, $r$ is the reflected light.

Write the expression to obtain the magnitude of the refracted light.

    $S_2=S_1-S_1^{\text{'}}$

Here, $S_2$ is the magnitude of the refracted light, $S_1$ is the average Poynting vector magnitude and $S_1^{\text{'}}$ is the refracted light intensity.

Substitute ${\left(\frac{n_2-n_1}{n_2+n_1}\right)}^2{S}_1$ for $S_1^{\text{'}}$ in the above equation.

    $\begin{array}{c}{S}_2=S_1-{\left(\frac{n_2-n_1}{n_2+n_1}\right)}^2{S}_1\\ S_2=\left[1-{\left(\frac{n_2-n_1}{n_2+n_1}\right)}^2\right]S_1\\ \frac{S_2}{S_1}=\left[1-{\left(\frac{n_2-n_1}{n_2+n_1}\right)}^2\right]\end{array}$

Substitute $t$ for $\frac{S_2}{S_1}$ in the above equation.

    $t=\left[1-{\left(\frac{n_2-n_1}{n_2+n_1}\right)}^2\right]$                                                                                             (II)

Here, $t$ is the transmitted light.

Write the expression to obtain the resultant magnitude of the transmitted light through $2^{\text{nd}}$ surface.

    $S_2^{\text{'}}=S_1{t}^2+S_1{t}^2{r}^2+S_1{t}^2{r}^4+S_1{t}^2{r}^6+\mathrm{.......}$

Here, $S_2^{\text{'}}$ is the resultant magnitude of the transmitted light through $2^{\text{nd}}$ surface, $S_1$ is the average Poynting vector magnitude, $r$ is the reflected light and $t$ is the transmitted light.

Further solve the above equation.

    $S_2^{\text{'}}=S_1{t}^2\left(1+r^2+r^4+r^6+\mathrm{.......}\right)$

Here, $\left(1+r^2+r^4+r^6+\mathrm{.......}\right)$ is the arithmetic progression.

Substitute $\left(\frac{1}{1-r^2}\right)$ for $\left(1+r^2+r^4+r^6+\mathrm{.......}\right)$ in the above equation.

    $\begin{array}{l}{S}_2^{\text{'}}=S_1{t}^2\left(\frac{1}{1-r^2}\right)\\ \frac{S_2^{\text{'}}}{S_1}=\left(\frac{t^2}{1-r^2}\right)\end{array}$                                                                                                  (III)

Here, $\frac{S_2^{\text{'}}}{S_1}$ is the ratio of intensity of transmitted light to the incident intensity

Conclusion:

Substitute $1.000$ for $n_1$ and $2.419$ for $n_2$ in equation (I) to calculate $r$.

    $\begin{array}{c}r={\left(\frac{2.419-1.000}{2.419+1.000}\right)}^2\\ ={\left(\frac{1.419}{3.419}\right)}^2\\ =0.172\end{array}$

Substitute $1.000$ for $n_1$ and $2.419$ for $n_2$ in equation (II) to calculate $t$.

    $\begin{array}{c}t=\left[1-{\left(\frac{2.419-1.000}{2.419+1.000}\right)}^2\right]\\ =\left[1-{\left(\frac{1.419}{3.419}\right)}^2\right]\\ =\left[1-0.172\right]\\ =0.827\end{array}$

Substitute $0.172$ for $r$ and $0.827$ for $t$ in equation (III) to calculate $\frac{S_2^{\text{'}}}{S_1}$.

    $\begin{array}{c}\frac{S_2^{\text{'}}}{S_1}=\left(\frac{{\left(0.827\right)}^2}{1-{\left(0.172\right)}^2}\right)\\ =\left(\frac{0.6839}{0.9704}\right)\\ =0.706\\ =70.6\%\end{array}$

Therefore, the intensity of transmitted light is $70.6\%$ of the incident intensity.