91Ó°ÊÓ

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:

$n_1\text{ s¾±²Ô}{\mathrm{θ}}_1=n_2\text{ s¾±²Ô }{\mathrm{θ}}_2$

The angle of incidence${\mathrm{θ}}_{B,1}$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.${\mathrm{θ}}_c$

$\text{sin}\text{ }{\mathrm{θ}}_{B,1}=\text{³¦´Ç²õ }{\mathrm{θ}}_c$

Using Snell’s law,

$n_2\text{sin}{\mathrm{θ}}_c=n_3\text{²õ¾±²Ô }90°$

$sinsin{\mathrm{θ}}_c=\frac{n_3}{n_2}$

Therefore,

$\begin{array}{c}\text{sin}\text{ }{\mathrm{θ}}_{B,1}=\text{³¦´Ç²õ }{\mathrm{θ}}_c\\ =\sqrt{1−{\mathrm{θ}}_c}\\ =\sqrt{1−{(\frac{n_3}{n_2})}^2}\end{array}$

Since

$n_3\text{sin}\text{ }{\mathrm{θ}}_{B,2}=n_2\text{ s¾±²Ô }{\mathrm{θ}}_{B,1}$

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

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

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

Using Snell’s law,

$\begin{array}{l}{n}_1\text{ s¾±²Ô }\mathrm{θ}=n_2\text{²õ¾±²Ô }{\mathrm{θ}}_c\\ n_1\text{ s¾±²Ô }\mathrm{θ}=n_2\left(\frac{n_3}{n_2}\right)\end{array}$

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

Hence, the initial angle$\mathrm{θ}$ is$49.9°.$

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

Its sine is given by

$\begin{array}{c}\text{²õ¾±²Ô }{\mathrm{θ}}_{A,1}=\text{cos}\text{ }{\mathrm{θ}}_c\\ =\sqrt{1-{(\frac{n_3}{n_2})}^2}\end{array}$

$n_3\text{²õ¾±²Ô }{\mathrm{θ}}_{A,2}=n_2\text{²õ¾±²Ô }{\mathrm{θ}}_{A,1}$

Therefore,

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

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

Using Snell’s law, we get

$\begin{array}{c}{n}_1\text{ s¾±²Ô }\mathrm{θ}=n_2\text{ s¾±²Ô}\text{ }{\mathrm{θ}}_{A,1}\\ =n_2\sqrt{1−{(\frac{n_3}{n_2})}^2}\\ =\sqrt{n_2^2−n_3^2}\end{array}$

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

Hence, the initial angle$\mathrm{θ}$ is$26.1°.$

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

Its sine is given by

$\text{²õ¾±²Ô }{\mathrm{θ}}_{B,1}=\frac{n_2}{\sqrt{n_2^2+n_3^2}}$

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

$n_3\text{ s¾±²Ô }{\mathrm{θ}}_{B,2}=n_2\text{ s¾±²Ô }{\mathrm{θ}}_{B,1}$

$\begin{array}{c}\text{sin}\text{ }{\mathrm{θ}}_{B,2}=\frac{n_2}{n_3}\text{²õ¾±²Ô }{\mathrm{θ}}_{B,1}\\ =\frac{n_2^2}{n_3\sqrt{n_2^2+n_3^2}}\end{array}$

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

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

Using Snell’s law,

$n_1\text{ s¾±²Ô }\mathrm{θ}=n_2\text{ s¾±²Ô }{\mathrm{θ}}_{Brewster}$

$\text{²õ¾±²Ô }{\mathrm{θ}}_{Brewster}=\frac{n_3}{\sqrt{n_2^2+n_3^2}}$

$n_1\text{ s¾±²Ô }\mathrm{θ}=n_2\frac{n_3}{\sqrt{n_2^2+n_3^2}}$

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

Hence, the initial angleis$35.3°.$