We will use Snell's law, which states:
[tex]n_1\sin \theta_1=n_2\sin \theta_2[/tex]Where n1 and n2 are the refraction indexes and their respective angles are "theta1" and "theta2".
For part A we replace the values:
[tex]1\sin 40=1.33\sin \theta_2[/tex]Now we solve for "theta2" first by dividing both sides by 1.33:
[tex]\frac{\sin40}{1.33}=\sin \theta_2[/tex]Now we use the inverse function for sine:
[tex]\arcsin (\frac{\sin 40}{1.33})=\theta_2[/tex]Solving the operation:
[tex]28.9=\theta_2[/tex]For part B, since "theta1" and "theta3" are angles of reflection, according to the reflection law, these angles are equal, therefore:
[tex]\theta_3=\theta_1=40[/tex]For part C. The index of refraction is defined as:
[tex]n=\frac{c}{v}[/tex]Where "c" is the speed of light in a vacuum and "v" is the speed of light in the medium. Replacing the values:
[tex]1.33=\frac{3\times10^8\text{ m/s}}{v}[/tex]Now we solve for "v":
[tex]v=\frac{3\times10^8\text{ m/s}}{1.33}[/tex]Solving the operation:
[tex]v=2.26\times10^8\text{ m/s}[/tex]For part d. We will use the following formula:
[tex]\lambda=\frac{v}{f}[/tex]Where "v" is the speed and "f" is the frequency. Replacing we get:
[tex]\lambda=\frac{2.26\times10^8\text{ m/s}}{7.85\times10^{16}s^{-1}}[/tex]Solving the operations:
[tex]\lambda=0.288\times10^{-8}m[/tex]For part e. The largest value of the angle of incidence that will result in refraction is 90 degrees.
