In Fig. 33-78, where ${\mathit{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.70}$, ${\mathit{n}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.50}$,and${\mathit{n}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.30}$,light refracts from material 1 into material 2. If it is incident at point A at the critical angle for the interface between materials 2 and 3, what are (a) the angle of refraction at pointBand (b) the initialangle$\mathit{\theta }$?If, instead, light is incident atBat the critical angle for the interface between materials 2 and 3, what are (c) the angle of refraction at pointAand (d) the initial angle$\mathit{\theta }$? If, instead of all that, light is incident at point Aat Brewster’s angle for the interface between materials 2 and 3, what are (e) the angle of refraction at point B and (f) the initialangle$\mathit{\theta }$?

Refractive index,$n_1=1.7$

Refractive index,$n_2=1.5$

Refractive index,$n_3=1.3$

We can use Snell’s law to find the required angles of refraction and the initial angles.

Formula:

$n_1\text{\hspace{0.17em}sin}{\theta }_1=n_2\text{\hspace{0.17em}sin\hspace{0.17em}}{\theta }_2$

The angle of incidence${\theta }_{B,1}$at$B$is the component of the critical angle at$A.$

From the figure, its sine angle is equal to the cosine of the critical angle.${\theta }_c$

$\text{sin}{\theta }_{B,1}=\text{cos\hspace{0.17em}}{\theta }_c$

Using Snell’s law,

$n_2\text{sin}{\theta }_c=n_3\text{sin\hspace{0.17em}}90°$

$sinsin{\theta }_c=\frac{n_3}{n_2}$

Therefore,

$\begin{array}{c}\text{sin}{\theta }_{B,1}=\text{cos\hspace{0.17em}}{\theta }_c\\ =\sqrt{1-{\theta }_c}\\ =\sqrt{1-{(\frac{n_3}{n_2})}^2}\end{array}$

Since

$n_3\text{sin}{\theta }_{B,2}=n_2\text{\hspace{0.17em}sin\hspace{0.17em}}{\theta }_{B,1}$

Thus, the angle of refraction ${\theta }_{B,2}$ at $B$becomes

$\begin{array}{c}{\theta }_{B,2}={\sin }^{-1}\left(\frac{n_2}{n_3}\sqrt{1-{(\frac{n_3}{n_2})}^2}\right)\\ ={\sin }^{-1}\sqrt{{\left(\frac{n_2}{n_3}\right)}^2-1}\\ ={\sin }^{-1}(\sqrt{{\left(\frac{1.50}{1.30}\right)}^2-1}\\ =35.1°\end{array}$

Hence, the angle of refraction at point $B$is$35.1°.$

Using Snell’s law,

$\begin{array}{l}{n}_1\text{\hspace{0.17em}sin\hspace{0.17em}}\theta =n_2\text{sin\hspace{0.17em}}{\theta }_c\\ n_1\text{\hspace{0.17em}sin\hspace{0.17em}}\theta =n_2\left(\frac{n_3}{n_2}\right)\end{array}$

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{n_3}{n_1}\right)\\ ={\sin }^{-1}\left(\frac{1.3}{1.7}\right)\\ =49.9°\end{array}$

Hence, the initial angle$\theta$ is$49.9°.$

The angle of incidence${\theta }_{A,1}$at$A$is the complement of the critical angle at.$B$

Its sine is given by

$\begin{array}{c}\text{sin\hspace{0.17em}}{\theta }_{A,1}=\text{cos}{\theta }_c\\ =\sqrt{1-{(\frac{n_3}{n_2})}^2}\end{array}$

$n_3\text{sin\hspace{0.17em}}{\theta }_{A,2}=n_2\text{sin\hspace{0.17em}}{\theta }_{A,1}$

Therefore,

$\begin{array}{c}{\theta }_{A,2}={\sin }^{-1}\left(\left(\frac{n_2}{n_3}\right)\sqrt{1-{(\frac{n_3}{n_2})}^2}\right)\\ ={\sin }^{-1}\left(\sqrt{{\left(\frac{n_2}{n_3}\right)}^2-1}\right)\\ ={\sin }^{-1}\sqrt{{\left(\frac{1.50}{1.30}\right)}^2-1}\\ =35.1°\end{array}$

Hence, the angle of refraction at point $A$is.$35.1°$

Using Snell’s law, we get

$\begin{array}{c}{n}_1\text{\hspace{0.17em}sin\hspace{0.17em}}\theta =n_2\text{\hspace{0.17em}sin}{\theta }_{A,1}\\ =n_2\sqrt{1-{(\frac{n_3}{n_2})}^2}\\ =\sqrt{n_2^2-n_3^2}\end{array}$

$\begin{array}{c}\theta ={\sin }^{-1}\frac{\sqrt{n_2^2-n_3^2}}{n_1}\\ ={\sin }^{-1}\frac{\sqrt{{\left(1.5\right)}^2-{\left(1.3\right)}^2}}{1.7}\\ =26.1°\end{array}$

Hence, the initial angle$\theta$ is$26.1°.$

The angle of incidence${\theta }_{B,1}$at$B$is the complement of Brewster angle at$A.$

Its sine is given by

$\text{sin\hspace{0.17em}}{\theta }_{B,1}=\frac{n_2}{\sqrt{n_2^2+n_3^2}}$

So, the angle of refraction${\theta }_{B,2}$at $B$will be

$n_3\text{\hspace{0.17em}sin\hspace{0.17em}}{\theta }_{B,2}=n_2\text{\hspace{0.17em}sin\hspace{0.17em}}{\theta }_{B,1}$

$\begin{array}{c}\text{sin}{\theta }_{B,2}=\frac{n_2}{n_3}\text{sin\hspace{0.17em}}{\theta }_{B,1}\\ =\frac{n_2^2}{n_3\sqrt{n_2^2+n_3^2}}\end{array}$

$\begin{array}{c}{\theta }_{B,2}={\sin }^{-1}\left(\frac{{\left(1.5\right)}^2}{\left(1.3\right)\sqrt{{1.3}^2+{1.5}^2}}\right)\\ =60.7°\end{array}$

Hence, the angle of refraction at point$B$is$60.7°.$

Using Snell’s law,

$n_1\text{\hspace{0.17em}sin\hspace{0.17em}}\theta =n_2\text{\hspace{0.17em}sin\hspace{0.17em}}{\theta }_{Brewster}$

$\text{sin\hspace{0.17em}}{\theta }_{Brewster}=\frac{n_3}{\sqrt{n_2^2+n_3^2}}$

$n_1\text{\hspace{0.17em}sin\hspace{0.17em}}\theta =n_2\frac{n_3}{\sqrt{n_2^2+n_3^2}}$

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{n_2{n}_3}{n_1\sqrt{n_2^2+n_3^2}}\right)\\ ={\sin }^{-1}\left(\frac{1.5\times 1.3}{1.7\sqrt{{1.5}^2+{1.3}^2}}\right)\\ =35.3°\end{array}$

Hence, the initial angleis$35.3°.$