Chapter 38, Problem 31PQ

Light is incident on a prism as shown in Figure P38.31. The prism, an equilateral triangle, is made of plastic with an index of refraction of 1.46 for red light and 1.49 for blue light. Assume the apex angle of the prism is 60.00°.

1. a. Sketch the approximate paths of the rays for red and blue light as they travel through and then exit the prism.
2. b. Determine the measure of dispersion, the angle between the red and blue rays that exit the prism.

Figure P38.31

To determine

The sketch of the appropriate path of the red and blue light rays through the prism.

Answer to Problem 31PQ

The sketch of the appropriate path of the red and blue light rays through the prism is as shown below.

Explanation of Solution

The wavelength of the violet light is $410\text{nm}$ and the wavelength of the red light is $680\text{nm}$.

The wavelength of light is inversely proportion to the refractive index of the material through which it is passing.

The wavelength of the violet light is lesser than the wavelength of the red light. Hence, the refractive index of the violet light is more as compared to the refractive index for the red light.

Therefore, violet light refracts more as compared to the red light as shown in the figure below.

Figure-(1)

To determine

The dispersion angle between the red and blue rays when exit the prism.

Answer to Problem 31PQ

The dispersion angle between the red and blue rays when exit the prism is $2.64°$.

Explanation of Solution

Write the expression for Snell’s law.

    $n_1\sin {\theta }_{\text{i}}=n_2\sin {\theta }_{\text{t}}$

Here, incidence angle is ${\theta }_{\text{i}}$, refractive index of the first medium is $n_1$, refraction angle is ${\theta }_{\text{t}}$ and the refractive index of the second medium is $n_2$.

Further solve the above equation for ${\theta }_{\text{t}}$.

    $\begin{array}{c}\sin {\theta }_{\text{t}}=\frac{n_1}{n_2}\sin {\theta }_{\text{i}}\\ {\theta }_{\text{t}}={\sin }^{-1}\left(\frac{n_1}{n_2}\sin {\theta }_{\text{i}}\right)\end{array}$                                                                                       (I)

The prism diagram on which light is incident at an angle $45°$ is as shown below.

Figure-(1)

Write the expression for dispersion angle between the red and blue rays when exit the prism.

    $\varphi ={\left({\theta }_{\text{t}}^{\text{'}}\right)}_{\text{b}}-{\left({\theta }_{\text{t}}^{\text{'}}\right)}_{\text{r}}$                                                                                                    (II)

Here, dispersion angle between the red and blue rays when exit the prism is $\varphi$, refracted angle at the exit of the prism for blue light is ${\left({\theta }_{\text{t}}^{\text{'}}\right)}_{\text{b}}$ and refracted angle at the exit of the prism for red light is ${\left({\theta }_{\text{t}}^{\text{'}}\right)}_{\text{r}}$.

Conclusion:

Case (i): For blue light.

Substitute $1$ for $n_1$, $1.49$ for $n_2$ and $45.0°$ for ${\theta }_{\text{i}}$ in equation (I) to find ${\left({\theta }_{\text{t}}\right)}_{\text{b}}$

    $\begin{array}{c}{\left({\theta }_{\text{t}}\right)}_{\text{b}}={\sin }^{-1}\left(\frac{1}{1.49}\sin 45°\right)\\ ={\sin }^{-1}\left(0.474\right)\\ =28.33°\end{array}$

Solve for $\alpha$.

    $\begin{array}{c}\alpha =90°-{\left({\theta }_{\text{t}}\right)}_{\text{b}}\\ =90°-28.33°\\ =61.67°\end{array}$

Solve for $\beta$.

    $\begin{array}{c}\beta =180°-60°-\alpha \\ =180°-60°-61.67°\\ =58.33°\end{array}$

Solve for ${\theta }_{\text{i}}^{\text{'}}$.

    $\begin{array}{c}{\theta }_{\text{i}}^{\text{'}}=90°-\beta \\ =90°-58.33°\\ =31.67°\end{array}$

Substitute $31.67°$ for ${\theta }_{\text{i}}^{\text{'}}$, $1.49$ for $n_1$ and $1$ for $n_2$ in equation (I) to find ${\left({\theta }_{\text{j}}^{\text{'}}\right)}_{\text{b}}$.

    $\begin{array}{c}{\left({\theta }_{\text{t}}^{\text{'}}\right)}_{\text{b}}={\sin }^{-1}\left(\frac{1.49}{1}\sin 31.67°\right)\\ ={\sin }^{-1}\left(0.782\right)\\ =51.47°\end{array}$

Case (ii): For red light.

Substitute $1$ for $n_1$, $1.46$ for $n_2$ and $45.0°$ for ${\theta }_{\text{i}}$ in equation (I) to find ${\left({\theta }_{\text{t}}\right)}_{\text{b}}$

    $\begin{array}{c}{\left({\theta }_{\text{t}}\right)}_{\text{r}}={\sin }^{-1}\left(\frac{1}{1.46}\sin 45°\right)\\ ={\sin }^{-1}\left(0.484\right)\\ =28.96°\end{array}$

Solve for $\alpha$.

    $\begin{array}{c}\alpha =90°-{\left({\theta }_{\text{t}}\right)}_{\text{b}}\\ =90°-28.96°\\ =61.04°\end{array}$

Solve for $\beta$.

    $\begin{array}{c}\beta =180°-60°-\alpha \\ =180°-60°-61.04°\\ =58.96°\end{array}$

Solve for ${\theta }_{\text{i}}^{\text{'}}$.

    $\begin{array}{c}{\theta }_i^{\text{'}}=90°-\beta \\ =90°-58.69°\\ =31.04°\end{array}$

Substitute $31.04°$ for ${\theta }_i^{\text{'}}$, $1.49$ for $n_1$ and $1$ for $n_2$ in equation (I) to find ${\left({\theta }_{\text{j}}^{\text{'}}\right)}_{\text{r}}$.

    $\begin{array}{c}{\left({\theta }_{\text{j}}^{\text{'}}\right)}_{\text{r}}={\sin }^{-1}\left(\frac{1.46}{1}\sin 31.04°\right)\\ ={\sin }^{-1}\left(0.752\right)\\ =48.83°\end{array}$

Substitute $51.47°$ for ${\left({\theta }_{\text{t}}^{\text{'}}\right)}_{\text{b}}$ and $48.83°$ for ${\left({\theta }_{\text{j}}^{\text{'}}\right)}_{\text{r}}$ in equation (II) to find $\varphi$.

    $\begin{array}{c}\varphi =51.47°-48.83°\\ =2.64°\end{array}$

Therefore, the dispersion angle between the red and blue rays when exit the prism is $2.64°$.