Chapter 25, Problem 24P

To determine

The difference in angular dispersion.

Answer to Problem 24P

The difference in angular dispersion is $\underset{\_}{\Delta {\theta }_4={\sin }^{-1}\left\{n_{\text{v}}\sin \left[\Phi -{\sin }^{-1}\left(\frac{\sin \theta }{n_{\text{v}}}\right)\right]\right\}-{\sin }^{-1}\left\{n_{\text{R}}\sin \left[\Phi -{\sin }^{-1}\left(\frac{\sin \theta }{n_{\text{R}}}\right)\right]\right\}}$.

Explanation of Solution

Write the expression for Snell’s law of refraction at the air-silica flint glass interface.

    $n_1\sin {\theta }_1=n_2\sin {\theta }_2$        (I)

Here, $n_1$ is the refractive index of the air medium, ${\theta }_1$ is the angle of incidence at the air-silica flint glass interface, $n_2$ is the refractive index of the silica flint glass and ${\theta }_2$ is the angle of refraction at the air-silica flint glass interface.

Rearrange the above equation to find ${\theta }_2$.

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

Given that the refractive index of the air is $n_1=1.0$.

The above equation is written as

  ${\theta }_2={\sin }^{-1}\left(\frac{\left(1.00\right)\sin {\theta }_1}{n_2}\right)$        (III)

For the incoming violet ray the above equation is written as

    ${\left({\theta }_2\right)}_{\text{violet}}={\sin }^{-1}\left(\frac{\sin \theta }{n_{\text{v}}}\right)$        (IV)

Here, ${\left({\theta }_2\right)}_{\text{violet}}$ is the angle at violet ray enters the silica flint glass, $\theta$ is the angle of incidence, and $n_{\text{v}}$ is the refractive index of violet ray in the silica flint glass.

For the incoming red ray the equation (III) is written as

    ${\left({\theta }_2\right)}_{\text{red}}={\sin }^{-1}\left(\frac{\sin \theta }{n_{\text{R}}}\right)$        (V)

Here, ${\left({\theta }_2\right)}_{\text{red}}$ is the angle at red ray enters the silica flint glass, $\theta$ is the angle of incidence, and $n_{\text{R}}$ is the refractive index of red ray in the silica flint glass.

Write the expression for Snell’s law of refraction at the silica flint glass-air interface.

    $n_1\sin {\theta }_4=n_2\sin {\theta }_3$        (VI)

Here, $n_1$ is the refractive index of the air medium, ${\theta }_3$ is the angle of incidence at the silica flint glass interface, $n_2$ is the refractive index of the silica flint glass and ${\theta }_4$ is the angle of refraction at the silica flint glass-air interface.

Rearrange the above equation to find ${\theta }_2$.

  $\begin{array}{c}\sin {\theta }_4=\frac{n_2\sin {\theta }_3}{n_1}\\ {\theta }_4={\sin }^{-1}\left(\frac{n_2\sin {\theta }_3}{n_1}\right)\end{array}$        (VII)

For the outgoing violet ray the above equation is written as

    ${\left({\theta }_4\right)}_{\text{violet}}={\sin }^{-1}\left(n_{\text{v}}\sin {\theta }_3\right)$        (VIII)

Here, ${\left({\theta }_4\right)}_{\text{violet}}$ is the angle at violet ray leaves the silica flint glass, $\theta$ is the angle of incidence, and $n_{\text{v}}$ is the refractive index of violet ray in the silica flint glass.

For the incoming red ray the equation (III) is written as

    ${\left({\theta }_4\right)}_{\text{red}}={\sin }^{-1}\left(n_{\text{R}}\sin {\theta }_3\right)$        (IX)

Here, ${\left({\theta }_2\right)}_{\text{red}}$ is the angle at red ray enters the silica flint glass, $\theta$ is the angle of incidence, and $n_{\text{R}}$ is the refractive index of red ray in the silica flint glass

Write the expression for sum of all the angles in the Triangle for the outgoing rays in Figure P25.23.

    $\left(90.0°-{\theta }_2\right)+\left(90.0°-{\theta }_3\right)+\Phi =180°$

Here, $\Phi$ is the apex angle of the prism.

Rearrange the above equation to find ${\theta }_3$.

    ${\theta }_3=\Phi -{\theta }_2$        (X)

Use equation (X) in (VIII).

    ${\left({\theta }_4\right)}_{\text{violet}}={\sin }^{-1}\left(n_{\text{v}}\sin \left(\Phi -{\theta }_2\right)\right)$        (XI)

Use equation (IV) in (XI).

    ${\left({\theta }_4\right)}_{\text{violet}}={\sin }^{-1}\left\{n_{\text{v}}\sin \left[\Phi -{\sin }^{-1}\left(\frac{\sin \theta }{n_{\text{v}}}\right)\right]\right\}$        (XII)

Use equation (VIII) in (IX).

    ${\left({\theta }_4\right)}_{\text{red}}={\sin }^{-1}\left(n_{\text{R}}\sin \left(\Phi -{\theta }_2\right)\right)$        (XIII)

Use equation (V) in (XIII).

    ${\left({\theta }_4\right)}_{\text{red}}={\sin }^{-1}\left\{n_{\text{R}}\sin \left[\Phi -{\sin }^{-1}\left(\frac{\sin \theta }{n_{\text{R}}}\right)\right]\right\}$        (XIV)

Write the expression for difference in angular dispersion.

    $\Delta {\theta }_4={\left({\theta }_4\right)}_{\text{violet}}-{\left({\theta }_4\right)}_{\text{red}}$        (XV)

Use equation (XII) and (XIV) in equation (XV).

    $\Delta {\theta }_4={\sin }^{-1}\left\{n_{\text{v}}\sin \left[\Phi -{\sin }^{-1}\left(\frac{\sin \theta }{n_{\text{v}}}\right)\right]\right\}-{\sin }^{-1}\left\{n_{\text{R}}\sin \left[\Phi -{\sin }^{-1}\left(\frac{\sin \theta }{n_{\text{R}}}\right)\right]\right\}$

Conclusion:

Therefore, the difference in angular dispersion is $\underset{\_}{\Delta {\theta }_4={\sin }^{-1}\left\{n_{\text{v}}\sin \left[\Phi -{\sin }^{-1}\left(\frac{\sin \theta }{n_{\text{v}}}\right)\right]\right\}-{\sin }^{-1}\left\{n_{\text{R}}\sin \left[\Phi -{\sin }^{-1}\left(\frac{\sin \theta }{n_{\text{R}}}\right)\right]\right\}}$.