Chapter 34, Problem 25P

To determine

The angular spread of visible light passing through a prism.

Answer to Problem 25P

The angular spread of visible light passing through a prism is ${\sin }^{-1}\left[n_V\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin \theta }{n_V}\right)\right)\right]-{\sin }^{-1}\left[n_R\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin \theta }{n_R}\right)\right)\right]$.

Explanation of Solution

Given Info:

Explanation:

Write the expression for snell’s law for prism to calculate angle of refraction for red light.

$\begin{array}{c}{n}_1\sin {\theta }_1=n_2\sin \left(\varphi -{\theta }_2\right)\\ \sin \left(\varphi -{\theta }_2\right)=\left(\frac{n_1\sin {\theta }_1}{n_2}\right)\\ {\theta }_2=\varphi -{\sin }^{-1}\left(\frac{n_1\sin {\theta }_1}{n_2}\right)\end{array}$ (1)

Here,

$n_1$ is refractive index for air.

${\theta }_1$ is angle of incidence for red light.

$n_2$ is refractive index for red light.

${\theta }_{2^{\prime }}$ is the angle of refraction for red light.

$\varphi$ is the apex angle.

Substitute $1$ for $n_1$, $\theta$ for ${\theta }_1$ and $n_R$ for $n_2$ in above equation

$\begin{array}{c}{\theta }_2=\varphi -{\sin }^{-1}\left(\frac{1\cdot \sin {\theta }_{}}{n_R}\right)\\ =\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_R}\right)\end{array}$

Write the expression for snell’s law,

$\begin{array}{c}{n}_1\sin {\theta }_{1^{\prime }}=n_2\sin \left({\theta }_2\right)\\ \sin \left({\theta }_{1^{\prime }}\right)=\left(\frac{n_2\sin {\theta }_2}{n_1}\right)\\ {\theta }_{1^{\prime }}={\sin }^{-1}\left(\frac{n_2\sin {\theta }_2}{n_1}\right)\end{array}$

Here,

${\theta }_{1^{\prime }}$ is deviation of red light from incidence light.

${\theta }_2$ is angle of refraction for red light

$n_1$ is the refractive index for air.

$n_2$ is refractive index for red light

Substitute ${\theta }_2=\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_R}\right)$ for ${\theta }_2$, $1$ for $n_1$, ${\theta }_R$ for ${\theta }_{1^{\prime }}$ and for $n_R$ for $n_2$ in above equation.

$\begin{array}{c}{\theta }_R={\sin }^{-1}\left(\frac{n_R\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_R}\right)\right)}{1}\right)\\ ={\sin }^{-1}\left(n_R\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_R}\right)\right)\right)\end{array}$

Write the expression for snell’s law for prism to calculate angle of refraction for violet light.

$\begin{array}{c}{n}_1\sin {\theta }_1=n_2\sin \left(\varphi -{\theta }_2\right)\\ \sin \left(\varphi -{\theta }_2\right)=\left(\frac{n_1\sin {\theta }_1}{n_2}\right)\\ {\theta }_2=\varphi -{\sin }^{-1}\left(\frac{n_1\sin {\theta }_1}{n_2}\right)\end{array}$ (2)

Here,

$n_1$ is refractive index for air.

${\theta }_1$ is angle of incidence for violet light.

$n_2$ is refractive index for violet light.

${\theta }_{2^{\prime }}$ is the angle of refraction for violet light.

$\varphi$ is the apex angle.

Substitute $1$ for $n_1$, $\theta$ for ${\theta }_1$ and $n_V$ for $n_2$ in equation (2).

$\begin{array}{c}{\theta }_2=\varphi -{\sin }^{-1}\left(\frac{1\cdot \sin {\theta }_{}}{n_V}\right)\\ =\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_V}\right)\end{array}$

Write the expression for Snell's law.

$\begin{array}{c}{n}_1\sin {\theta }_{{1^{\prime }}^{\prime }}=n_2\sin \left({\theta }_2\right)\\ \sin \left({\theta }_{{1^{\prime }}^{\prime }}\right)=\left(\frac{n_2\sin {\theta }_2}{n_1}\right)\\ {\theta }_{{1^{\prime }}^{\prime }}={\sin }^{-1}\left(\frac{n_2\sin {\theta }_2}{n_1}\right)\end{array}$

Here,

${\theta }_{{1^{\prime }}^{\prime }}$ is deviation of violet light from incidence light.

${\theta }_2$ is angle of refraction for violet light

$n_1$ is the refractive index for air.

$n_2$ is refractive index for violet light.

Substitute ${\theta }_2=\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_V}\right)$ for ${\theta }_2$, $1$ for $n_1$, ${\theta }_V$ for ${\theta }_{{1^{\prime }}^{\prime }}$ and for $n_V$ for $n_2$ in above equation.

$\begin{array}{c}{\theta }_V={\sin }^{-1}\left(\frac{n_V\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_V}\right)\right)}{1}\right)\\ ={\sin }^{-1}\left(n_V\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_V}\right)\right)\right)\end{array}$

Write the expression for angular spread of visible light,

$\omega ={\theta }_V-{\theta }_R$

Here,

${\theta }_V$ is deviation of violet light from visible light.

${\theta }_R$ is the deviation of red light from visible light.

Substitute ${\sin }^{-1}\left(n_V\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_V}\right)\right)\right)$ for ${\theta }_V$ and ${\sin }^{-1}\left(n_R\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_R}\right)\right)\right)$ for ${\theta }_R$ in above equation.

$\omega ={\sin }^{-1}\left(n_V\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_V}\right)\right)\right)-{\sin }^{-1}\left(n_R\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin {\theta }_{}}{n_R}\right)\right)\right)$

Conclusion:

Therefore, the angular spread of visible light passing through a prism is ${\sin }^{-1}\left[n_V\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin \theta }{n_V}\right)\right)\right]-{\sin }^{-1}\left[n_R\sin \left(\varphi -{\sin }^{-1}\left(\frac{\sin \theta }{n_R}\right)\right)\right]$.