Chapter 10 Geometric Optics
10.4 Total Internal Reflection
Summary
- Explain the phenomenon of total internal reflection.
- Describe the workings and uses of fiber optics.
- Analyze the reason for the sparkle of diamonds.
A good-quality mirror may reflect more than 90% of the light that falls on it, absorbing the rest. But it would be useful to have a mirror that reflects all of the light that falls on it. Interestingly, we can produce total reflection using an aspect of refraction.
Consider what happens when a ray of light strikes the surface between two materials, such as is shown in Figure 1(a). Part of the light crosses the boundary and is refracted; the rest is reflected. If, as shown in the figure, the index of refraction for the second medium is less than for the first, the ray bends away from the perpendicular. Since n1 > n2, the angle of refraction is greater than the angle of incidence—that is, θ1 > θ2. Now imagine what happens as the incident angle is increased. This causes θ2 to increase also. The largest the angle of refraction θ2 can be is 90o, as shown in Figure 1(b).The critical angle θc for a combination of materials is defined to be the incident angle θ1 that produces an angle of refraction of 90o. That is, θc is the incident angle for which θ2= 90o. If the incident angle θ1 is greater than the critical angle, as shown in Figure 1(c), then all of the light is reflected back into medium 1, a condition called total internal reflection.
Critical Angle
The incident angle θ1 that produces an angle of refraction of 90o is called the critical angle,θc.
Snell’s law states the relationship between angles and indices of refraction. It is given by
When the incident angle equals the critical angle (θ1 =θc), the angle of refraction is 90o. θ2 = 90o). Noting that sin 90o = 1, Snell’s law in this case becomes
The critical angle θc for a given combination of materials is thus
Total internal reflection occurs for any incident angle greater than the critical angle θc, and it can only occur when the second medium has an index of refraction less than the first. Note the above equation is written for a light ray that travels in medium 1 and reflects from medium 2, as shown in the figure.
Example 1: How Big is the Critical Angle Here?
What is the critical angle for light traveling in a polystyrene (a type of plastic) pipe surrounded by air?
Strategy
The index of refraction for polystyrene is found to be 1.49 in Figure 2, and the index of refraction of air can be taken to be 1.00, as before. Thus, the condition that the second medium (air) has an index of refraction less than the first (plastic) is satisfied, and the equation can be used to find the critical angle θc. Here, then, n2 = 1.00. and n1 = 1.49.
Solution
The critical angle is given by
Substituting the identified values gives
Discussion
This means that any ray of light inside the plastic that strikes the surface at an angle greater than 42.2o will be totally reflected. This will make the insi