Total Internal Reflection
13.4 Total Internal Reflection
Learning Objectives
By the end of this section, you will be able to:
- Explain the phenomenon of total internal reflection
- Describe the workings and uses of optical fibers
- Analyze the reason for the sparkle of diamonds
Total Internal Reflection
| Practice 13.4.1 |
|---|
| You are cutting a sapphire to make it as brilliant as possible. Find the critical angle for the sapphire in air. The index of refraction of a sapphire is 1.77. |
| Practice 13.4.2 |
|---|
In the figure, five light rays enter a glass prism from the left. How many of these rays undergo total internal reflection at the slanted surface of the prism? |
| Practice 13.4.3 |
|---|
| In the figure, five light rays enter a glass prism from the left. Suppose the prism can be rotated in the plane of the paper. For all five rays to experience total internal reflection from the slanted surface, the prism should be rotated: |
| Practice 13.4.4 |
|---|
| A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). At what angle relative to the normal to the interface of the two liquids is light totally reflected? |
| Practice 13.4.5 |
|---|
| The speed of light in a material is 0.50c. What is the critical angle of a light ray at the interface between the material and a vacuum? |
Fiber Optics
Another example of total internal reflection is in optical fibers, or fiber optics, which are used all over the place. But one example is as an endoscope (a medical tool to look inside of bodies).
You take a thin, flexible fiber of glass (or really a whole bundle of them), which can be threaded into a cable. Light enters at one end inside the cable, but the light rays are all bouncing at a shallow angle (the incident angle θ is greater than θc, do you see this?) against the glass/air boundary, and totally reflect. They bounce their way along the cable, never getting out until they reach the end, outside the cable.
Discuss!
Oftentimes, fiber optic cables are covered in a material called cladding, which is really there to just protect the glass fibers of the cable. But the cladding has a smaller index of refraction than the glass, so total internal reflection will still occur. In the following problem, you are looking at a fiber optic cable that is surrounded by cladding.
Light enters the cable from the right, where it is initially traveling in air and enters the glass core at an incident angle α. Then the light refracts at an angle β when it enters the core of the cable. Total internal reflection occurs when the light travels in the core and is incident on the core/cladding boundary at the bottom, where the incident angle is equal to the critical angle for this core/cladding interface.
What you need to do is work backwards to determine the angle α that will result in total internal reflection. Basically, at what angle do you need to shine light into the optic so that it works correctly and transmits the light all the way down the cable?
Physics Education Research at UNG 