91Ó°ÊÓ

• The angle of incidence at the boundary between two materials${\mathrm{θ}}_1={40.1}^0$
• The refractive indices $n_1=1.30,n_2=1.40,n_3=1.32,n_4=1.45$

We can apply Snell’s law to the material of1and air to find thevalue of${\mathrm{θ}}_5$in the air. Similarly, by applying Snell’s law to each boundary, we get 4 equations and solving them, we can find thevalue of${\mathrm{θ}}_4$in the bottom material.

Formula:

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

According to Snell’s law,

$n_1\sin {\mathrm{θ}}_1=n_5\sin {\mathrm{θ}}_5$

We have for the air$n_5=1$and${\mathrm{θ}}_1={40.1}^0$,

$\begin{array}{c}{\mathrm{θ}}_5={\sin }^{−1}\left(\frac{n_1\sin \text{ }{\mathrm{θ}}_1}{n_5}\right)\\ {\mathrm{θ}}_5={\sin }^{−1}\left(\frac{1.30\text{ }\sin \text{ }{40.1}^0}{1}\right)\\ {\mathrm{θ}}_5={56.86}^0\\ ~56.9°\end{array}$

Therefore, the value of ${\mathrm{θ}}_5$ in the air is ${56.9}^0$

We have,

$\begin{array}{c}{n}_1\sin \text{ }{\mathrm{θ}}_1=n_2\sin \text{ }{\mathrm{θ}}_2=n_3\sin \text{ }{\mathrm{θ}}_3=n_4\sin \text{ }{\mathrm{θ}}_4\\ n_1\sin \text{ }{\mathrm{θ}}_1=n_4\sin \text{ }{\mathrm{θ}}_4\\ {\mathrm{θ}}_4={\sin }^{−1}\left(\frac{n_1\sin \text{ }{\mathrm{θ}}_1}{n_4}\right)\\ {\mathrm{θ}}_4={\sin }^{−1}\left(\frac{1.30\sin \text{ }{40.1}^0}{1.45}\right)\\ {\mathrm{θ}}_4={35.3}^0\end{array}$

Therefore, thevalue of${\mathrm{θ}}_4$in the bottom material is${35.3}^0$