 ### There is an angle of incidence beyond which no light escapes a medium.

In the section on Refraction of light we learned, through Snell's law,

$$n_1 \, sin(\theta_1) = n_2 \, sin(\theta_2)$$

that when light passes from medium of high index of refraction (low speed of light) to low index (higher speed of light), its path bends away from the surface normal, as shown below. We can calculate the angle of refraction, θr, from the two indices of refraction and the angle of incidence with respect to the surface normal. It's just a simple matter of rearranging Snell's law:

$$sin(\theta_r) = \frac{n_1}{n_2} sin(\theta_1)$$

Rearranged in that way, we can see that the angle of refraction (its sine) is larger than the angle of incidence if n1 > n2, and smaller if n1 < n2. That leads us to the familiar rules in the gray box below. Note that for angles between 0 and 90˚, the sine of an angle is proportional to the angle (in fact, to a very good approximation for small angles, sin(θ) ≈ θ)

#### The critical angle

Now we can do an interesting thought experiment: What if there was an angle – we'll call it the critical angle, θc at which the refracted beam formed a 90˚ angle with the surface normal. In other words, that beam skimmed along the interface between the two media, as shown below. Well, we can actually calculate that angle because we know that θr = 90˚, and we know n1 and n2. We start with Snell's law:

$$n_1 \, sin(\theta_1) = n_2 \, sin(\theta_2)$$

Dividing by n1 on both sides gives

$$sin(\theta_1) = \frac{n_2}{n_1} sin(\theta_r)$$

and taking the inverse sine (arcsine) of each side gives us θi, which is the critical angle.

$$\theta_i = sin^{-1} \left(\frac{n_2}{n_1} sin(\theta_r)\right)$$

Now we've seen that n2 must be less than n1 in order to have such an angle of refraction, and given that θ = 90˚ and sin(90˚) = 1, we find that the critical angle is:

$$\theta_c = sin^{-1} \left( \frac{n_2}{n_1} \right), \; \; n_1 \gt n_2$$

Now the critical angle becomes interesting when our angle of incidence goes beyond it, θi > θc. Here's a picture: When the angle of incidence is greater than the critical angle, no light will pass into the second medium, and all of it will be reflected (following the law of reflection – angle of reflection = angle of incidence) from the interface.

#### Total internal reflection

That phenomenon is called total internal reflection, and it leads to several important devices useful techniques in research.

We can use the formula

$$\theta_c = sin^{-1} \left( \frac{n_2}{n_1} \right)$$

to calculate the critical angle for several materials. Here are a few:

Material n θc
Plexiglas ™ 1.49 21.1˚
Water 1.33 24.3˚
Window glass 1.52 20.6˚
sapphire 1.77 17.2˚
Cubic zirconia 2.15 13.9˚
Diamond 2.42 12.2˚
X

### Interface

An interface is a line or surface that separates two types of material.

The surface between air and water is an interface. So is the surface between the oil and water when oil floats on top of water.

X

### normal

A normal is a line that is perpendicular to a surface or to a tangent plane to that surface at a given point. It is more than a perpendicular line. It is perpendicular to a surface or tangent plane when viewed at any angle.

When light moves through an interface into a medium with a higher index of refraction, its path is bent toward the surface normal.

When light moves through an interface into a medium with a lower index of refraction, its path is bent away from the surface normal.

#### The critical angle

When light passes from a medium of relatively high index of refraction (low speed of light) to a medium with a lower index (faster speed of light), There is an angle of incidence with respect to the interface normal beyond which no transmission occurs. The light only reflects from the surface. The critical angle is The critical angle depends on the ratio of the indices of refraction of the two media.

### TIR in fiber optics

The basis for transmitting signals through fiber optic cable is total internal reflection.

Optical fiber is a long, thin fiber made of glass. When thin, glass can be reasonably flexible, and can be drawn into long fibers of diameter less than 10 microns (1 μm = 1.0 × 10-6 m)

Because of the thinness of the fiber, it would be very difficult to aim a light source (typically a laser) so that it both enters the end of a fiber and hits the wall at an angle less than the critical angle. Once inside, all reflection angles with the sides of the fiber are bound to be greater than θc, so the light beam is "trapped" inside the fiber and cannot escape through the side wall. For this reason, optical fiber is known as a wave guide – a "pipe" for light.

Optical fiber has several advantages over copper wire when used to transmit communication signals like telephone calls:

• Glass fibers, which are made of the silica in sand, are cheap and easier to obtain than copper
• Glass fibers are lightweight
• A typical single fiber can transmit over a million telephone calls at once. A copper wire of the same diameter would be limited to about 750 calls.
• Fiber optic cable is not electrically conductive, so there is no chance of shorting or other problems that plague systems that use metal conductors.
• Signals traveling through optical fiber are difficult to "tap." Copper cables can be probed and signals listened to with little possibility of detection.

The main disadvantage of fiber optics transmission is the difficulty in aligning small fibers for transmission of signals from one fiber to another. That has largely been solved, however. Once a light source has been aimed so that it actually enters an optical fiber, the angle of incidence with the wall on the first reflection is greater than the critical angle. Subsequent collisions maintain that, and the signal is "trapped" inside the fiber.  xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.