91影视

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

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

Formula:

$n_1\sin {胃}_1=n_2\sin {胃}_2$

According to Snell鈥檚 law,

$n_1\sin {胃}_1=n_5\sin {胃}_5$

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

$\begin{array}{c}{胃}_5={\sin }^{鈭�1}\left(\frac{n_1\sin \text{鈥�}{胃}_1}{n_5}\right)\\ {胃}_5={\sin }^{鈭�1}\left(\frac{1.30\text{鈥�}\sin \text{鈥�}{40.1}^0}{1}\right)\\ {胃}_5={56.86}^0\\ ~56.9掳\end{array}$

Therefore, the value of ${胃}_5$ in the air is ${56.9}^0$

We have,

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

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