Question

A laser beam strikes one end of a slab of material, as in the figure below. The index of refraction of the slab is 1.64. (Assume θ = 46.0

Answer 1

This problem is an application of Snell's law, which states:

n1 sin x1 = n2 sin x

Where n1 and n2 are the indices of refraction for materials 1 and 2, and x1 and x2 are the angles of the beam's path with respect to the normal (perpendicular to the surface which the beam strikes).

So, the beam first hits the slab at an angle of 46 degrees to the normal. n1 is the index of refraction for air, which is approximately 1.00, and n2 is for the slab, which is 1.55. Therefore:

1.00 sin 46 = 1.64 sin x2

Solving for x2 gives you26.01 degrees. Now for the next part - the beam will hit the top of the slab at an angle of (90 - 26.01) = 63.98 degrees with respect to the normal. Draw it out.

This is total internal reflection because none of the beam escapes from the slab. How do we know this? Use Snell's law, where n1 = 1.55, x1 = 63.98 degrees, and n2 = 1.00:

1.64 sin 63.98 = 1.00 sin x2 = 1.47
There is no angle x2 that satisfies this equation since sin x is at max equal to 1.

So what will happen is that beam inside the slab will keep bouncing up and down, hitting the edges and exhibiting total internal reflection because the angle at which it strikes the air-slab interface is too great. Find how far HORIZONTALLY the beam travels in between reflections. The first time, it travels upward a distance of 1.64 mm ( The angle with respect to horizontal is 26.01 degrees, as determined earlier. Thus, the x distance is (1.64 mm)/(tan 26.01) = 3.36 mm.

Okay, so now after this first reflection, what angle is the beam traveling at? A reflected beam has the same angle with respect to the normal as the incident beam. This was 63.98 degrees with respect to the normal, or 26.01 degrees with respect to the horizontal. Now it travels 3.10 mm downward in the y direction before hitting the surface again. So the x distance traveled here is 3.10 mm/(tan 26.01) = 6.35 mm. Now it will reflect again at the same angle, so it travels another 6.35 mm before hitting the top. Therefore, the total distance it travels is:

3.33 mm + 6.35 mm + 6.35 mm + 6.35 mm + .....

Which goes on until the end is reached. The slab is 42.0 cm, which is 420 mm. The above formula for distance traveled in terms of the number of reflections "n" is:

x = 3.33 + 6.35n = 420
Solve for n, and you get 65.622. This means that there are 65 internal reflections before the beam exits the slab.