Identify the angle of incidence and the angle of refraction when the ray passes from the transparent substance to the air:
The refractive index of the transparent substance is 1.09 and the refractive index of air is close to 1. Then:
![\begin{gathered} 1\cdot\sin(x)=1.09\cdot\sin(65º) \\ \\ \Rightarrow\sin(x)=1.09\sin(65º) \\ \\ \Rightarrow x=\sin^(-1)\left(1.09\sin(65º)\right) \\ \\ \therefore x=81.0688...º \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/odt8aohkr3if6a4netfl0jcoep754uvqvn.png)
The angle α is complementary to the angle x. Then:
![\begin{gathered} \alpha+x=90º \\ \\ \Rightarrow\alpha=90º-x=90º-81.0688...º=8.93...º \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/8tjwxf7tjkyuz233x07esti1z3ph5o5yj3.png)
Therefore, the angle α is equal to 8.93º.