To determine whether the laser beam will pass out of the plastic into the air, we can use Snell's Law, which states:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
where:
- \(n_1\) is the index of refraction of the first medium (plastic, \(n_1 = 1.62\)),
- \(\theta_1\) is the angle of incidence in the first medium (plastic, \(\theta_1 = 54^\circ\)),
- \(n_2\) is the index of refraction of the second medium (air, \(n_2 = 1.00\)),
- \(\theta_2\) is the angle of refraction in the second medium (air).
We need to find \(\theta_2\) and check if it is possible for \(\theta_2\) under the boundary conditions of refraction. It is critical to find the critical angle, \( \theta_c \), for total internal reflection, which occurs when the light moves from a denser medium to a less dense medium (from plastic to air in this case).
The critical angle can be found using:
\[ \sin(\theta_c) = \frac{n_2}{n_1} \]
Substituting the values:
\[ \sin(\theta_c) = \frac{1.00}{1.62} \]
Calculating this gives:
\[ \sin(\theta_c) \approx 0.6173 \]
Now we find \(\theta_c\):
\[ \theta_c \approx \arcsin(0.6173) \approx 38.12^\circ \]
Now, we observe that the angle of incidence in the plastic, \(54^\circ\), is greater than the critical angle \(38.12^\circ\). Consequently, this means the laser beam will undergo total internal reflection and will not pass out into the air.
Thus, the correct answer is:
Since the angle of incidence exceeds 38.12°, the laser beam will not pass out of the plastic.