Chapter 22, Problem 62AP

To determine

The angular dispersion of visible light passing through an equilateral prism.

Answer to Problem 62AP

The angular dispersion of visible light passing through an equilateral is $4.5°$.

Explanation of Solution

The ray diagram for the beam through the prism is

Formula to calculate the refraction angle at the first surface of the prism is,

${\theta }_{1r}={\sin }^{-1}\left(\frac{n_a\sin {\theta }_{1i}}{n_g}\right)$

Here,

${\theta }_{1i}\text{and}{\theta }_{1r}$ are the incident and refraction angle at the first surface

$n_a\text{and}{n}_g$ are the refractive index of air and glass

Substitute $1$ for $n_a$, $1.66$ for $n_g$, $50°$ for ${\theta }_{1i}$ to find ${\theta }_{1r}$.

$\begin{array}{c}{\theta }_{1r}={\sin }^{-1}\left(\frac{\left(1\right)\sin \left(50\right)}{1.66}\right)\\ =27.5°\end{array}$

Calculating the value of the angle $\left(\alpha \right)$ is,

$\begin{array}{c}\alpha =\left(90-27.5°\right)\\ =62.5°\end{array}$

Calculating the value of the angle $\left(\beta \right)$ is,

$\begin{array}{c}\beta =\left(180-60-62.5°\right)\\ =57.5°\end{array}$

Calculating the incident angle at the second surface of the prism is,

$\begin{array}{c}{\theta }_{2i}=\left(90-57.5\right)\\ =32.5°\end{array}$

Formula for the final angle of refraction for the violet light is,

${\theta }_{2r}={\sin }^{-1}\left(\frac{n_g\sin {\theta }_{2i}}{n_a}\right)$

Substitute $1$ for $n_a$, $1.66$ for $n_g$, $32.5°$ for ${\theta }_{2i}$ to find ${\theta }_{2r}$.

$\begin{array}{c}{\theta }_{2r}={\sin }^{-1}\left(\frac{\left(1.66\right)\sin \left(32.5°\right)}{\left(1\right)}\right)\\ =63.1°\end{array}$

Thus, the final refraction angle for violet is $63.1°$.

The ray diagram for the beam through the prism is

Formula to calculate the refraction angle at the first surface of the prism is,

${\theta }_{1r}={\sin }^{-1}\left(\frac{n_a\sin {\theta }_{1i}}{n_g}\right)$

Here,

${\theta }_{1i}\text{and}{\theta }_{1r}$ are the incident and refraction angle at the first surface

$n_a\text{and}{n}_g$ are the refractive index of air and glass

Substitute $1$ for $n_a$, $1.62$ for $n_g$, $50°$ for ${\theta }_{1i}$ to find ${\theta }_{1r}$.

$\begin{array}{c}{\theta }_{1r}={\sin }^{-1}\left(\frac{\left(1\right)\sin \left(50\right)}{1.62}\right)\\ =28.2°\end{array}$

Calculating the value of the angle $\left(\alpha \right)$ is,

$\begin{array}{c}\alpha =\left(90-28.2°\right)\\ =61.8°\end{array}$

Calculating the value of the angle $\left(\beta \right)$ is,

$\begin{array}{c}\beta =\left(180-60-61.8°\right)\\ =58.2°\end{array}$

Calculating the incident angle at the second surface of the prism is,

$\begin{array}{c}{\theta }_{2i}=\left(90-58.2\right)\\ =31.8°\end{array}$

Formula for the final angle of refraction for the red light is,

${\theta }_{2r}={\sin }^{-1}\left(\frac{n_g\sin {\theta }_{2i}}{n_a}\right)$

Substitute $1$ for $n_a$, $1.62$ for $n_g$, $31.8°$ for ${\theta }_{2i}$ to find ${\theta }_{2r}$.

$\begin{array}{c}{\theta }_{2r}={\sin }^{-1}\left(\frac{\left(1.62\right)\sin \left(31.8°\right)}{\left(1\right)}\right)\\ =58.6°\end{array}$

Thus, the final refraction angle for red is $58.6°$.

Formula for the angular dispersion is,

$\text{Dispersion}={\left({\theta }_{2r}\right)}_{violet}-{\left({\theta }_{2r}\right)}_{red}$

Here,

${\left({\theta }_{2r}\right)}_{violet}\text{and }{\left({\theta }_{2r}\right)}_{red}$ are the final refraction angle for the violet and red.

Substitute $63.1°$ for ${\left({\theta }_{2r}\right)}_{violet}$, $58.6°$ for ${\left({\theta }_{2r}\right)}_{red}$, to find Dispersion.

$\begin{array}{c}\text{Dispersion}=\left[\left(63.1°\right)-\left(58.6°\right)\right]\\ =4.5°\end{array}$

Conclusion:

Therefore, the dispersion of the emerging light is $4.5°$.