Chapter 38, Problem 124PQ

(a)

To determine

The proof that total deviation of a ray from its original path is ${\theta }_1-\beta +{\sin }^{-1}\left(\sin \beta \sqrt{n^2-\sin {\theta }_1{}^2}-\cos \beta \sin {\theta }_1\right)$.

Answer to Problem 124PQ

The total deviation of a ray from its original path is

${\theta }_1-\beta +{\sin }^{-1}\left(\sin \beta \sqrt{n^2-\sin {\theta }_1{}^2}-\cos \beta \sin {\theta }_1\right)$ and is proved below.

Explanation of Solution

Write the expression for refraction law at left side of the given prism.

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

Here, ${\theta }_1$ is incident angle of ray coming from left at left surface, ${\theta }_2$ is the refracted angle at left face of the prism and $n$ is refractive index of prism.

Write the expression for refraction law at right side of the given prism.

    $n\sin {\theta }_3=\sin {\theta }_4$                                                                                                    (II)

Here, ${\theta }_3$ is incident angle of ray coming from left at right surface, ${\theta }_4$ is the refracted angle at left face of the prism

Write the expression angle of deviation of the given prism.

    $\alpha ={\theta }_1+{\theta }_4-\beta$                                                                                                  (III)

Here, $\alpha$ is angle of deviation and $\beta$ is apex angle of the prism.

Write the relation among various angle from the geometry of the prism.

    $\begin{array}{l}{\theta }_2+{\theta }_3=\beta \\ {\theta }_3=\beta -{\theta }_2\end{array}$

Substitute $\beta -{\theta }_2$ for ${\theta }_3$ in equation (II) to find ${\theta }_4$.

    $\begin{array}{c}n\sin \left(\beta -{\theta }_2\right)=\sin {\theta }_4\\ {\theta }_4={\sin }^{-1}\left(n\sin \left(\beta -{\theta }_2\right)\right)\end{array}$

Further solve this for ${\theta }_4$.

    ${\theta }_4={\sin }^{-1}\left(n\sin \beta \cos {\theta }_2-n\cos \beta \sin {\theta }_2\right)$                                                           (IV)

Use general trigonometric relation for conversion.

    ${\sin }^2\theta +{\cos }^2\theta =1$

Substitute $\sqrt{1-{\sin }^2{\theta }_2}$ for $\cos {\theta }_2$ in equation (IV) to find ${\theta }_4$.

    ${\theta }_4={\sin }^{-1}\left(n\sin \beta \sqrt{1-{\sin }^2{\theta }_2}-n\cos \beta \sin {\theta }_2\right)$

Substitute $\frac{\sin {\theta }_1}{n}$ for $\sin {\theta }_2$ in above equation to find ${\theta }_4$.

    $\begin{array}{c}{\theta }_4={\sin }^{-1}\left(n\sin \beta \sqrt{1-{\left(\frac{\sin {\theta }_1}{n}\right)}^2}-n\cos \beta \left(\frac{\sin {\theta }_1}{n}\right)\right)\\ ={\sin }^{-1}\left(\sin \beta \sqrt{n^2-\sin {\theta }_1{}^2}-\cos \beta \sin {\theta }_1\right)\end{array}$

Substitute ${\sin }^{-1}\left(\sin \beta \sqrt{n^2-\sin {\theta }_1{}^2}-\cos \beta \sin {\theta }_1\right)$ for ${\theta }_4$ in equation (III) to find angle of deviation.

    $\begin{array}{c}\alpha ={\theta }_1+{\sin }^{-1}\left(\sin \beta \sqrt{n^2-\sin {\theta }_1{}^2}-\cos \beta \sin {\theta }_1\right)-\beta \\ ={\theta }_1-\beta +{\sin }^{-1}\left(\sin \beta \sqrt{n^2-\sin {\theta }_1{}^2}-\cos \beta \sin {\theta }_1\right)\end{array}$

Conclusion:

Therefore, it is proved that the angle of deviation is ${\theta }_1-\beta +{\sin }^{-1}\left(\sin \beta \sqrt{n^2-\sin {\theta }_1{}^2}-\cos \beta \sin {\theta }_1\right)$.

(b)

To determine

The plot of the angle of deviation versus the angle of incidence for $n=1.5$ and apex angle $60.0°$

Answer to Problem 124PQ

The graph is following.

Explanation of Solution

Write the expression for the angle of deviation of the given prism.

    $\alpha ={\theta }_1-\beta +{\sin }^{-1}\left(\sin \beta \sqrt{n^2-\sin {\theta }_1{}^2}-\cos \beta \sin {\theta }_1\right)$                                             (V)

Conclusion:

Substitute $1.5$ for $n$, $60.0°$ for $\beta$ in above equation to find $\alpha$

    $\begin{array}{c}\alpha ={\theta }_1-60+{\sin }^{-1}\left(\sin \left(60.0°\right)\sqrt{{\left(1.5\right)}^2-\sin {\theta }_1{}^2}-\cos \left(60.0°\right)\sin {\theta }_1\right)\\ ={\theta }_1-60+{\sin }^{-1}\left(0.87\sqrt{2.25-\sin {\theta }_1{}^2}-\left(0.5\right)\sin {\theta }_1\right)\end{array}$

Deviation varies linearly, decreases and then increases approximately with incidence angle ${\theta }_1$.

Therefore, the plot of the angle of deviation versus the angle of incidence is as follows.

Figure-(1)

(c)

To determine

The incidence angle for which the derivation angle is minimum.

Answer to Problem 124PQ

The incidence angle for which deviation is minimum is $48°$ if the light deviated by this angle passes symmetrically through the prism.

Explanation of Solution

The minimum deviation occurs when the incident and the refracted ray are identical and make equal angles to the normal of the prism.

The angle exhibits a minimum at $48°$, if the light deviated by this angle passes symmetrically through the prism. So the angle of incidence to the prism and angle of departure are equal.

Conclusion:

The minimum angle of deviation depends on the refractive index for different wavelength, the refractive index is different.