Robert shines a laser beam through a slab of plastic and onto the interface between the slab of plastic and the air on the other side. The index of refraction for the plastic is 1.62 and the index of refraction for the air is 1.00. If the angle of incidence in the plastic is 54°, will the laser beam pass out of the plastic into the air?

Group of answer choices

Since the angle of incidence exceeds 38.12°, the laser beam will pass out of the plastic.

Since the angle of incidence exceeds 38.12°, the laser beam will not pass out of the plastic.

Since the index of refraction exceeds 38.12°, the laser beam will not pass out of the plastic.

Since the index of refraction exceeds 38.12°, the laser beam will pass out of the plastic.

1 answer

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.