Index of Refraction for AP Physics B & C
Practice problems for these concepts can be found at:
Light also undergoes interference when it reflects off of thin films of transparent material. Before studying this effect quantitatively, though, we have to examine how light behaves when it passes through different materials.
Light—or any electromagnetic wave—travels at the speed c, or 3 × 108 m/s. But it only travels at this speed through a vacuum, when there aren't any pesky molecules to get in the way. When it travels through anything other than a vacuum, light slows down. The amount by which light slows down in a material is called the material's index of refraction.
The index of refraction can be calculated using this equation.
This says that the index of refraction of a certain material, n, equals the speed of light in a vacuum, c, divided by the speed of light through that material, v.
For example, the index of refraction of glass is about 1.5. This means that light travels 1.5 times faster through a vacuum than it does through glass. The index of refraction of air is approximately 1. Light travels through air at just about the same speed as it travels through a vacuum.
Another thing that happens to light as it passes through a material is that its wavelength changes. When light waves go from a medium with a low index of refraction to one with a high index of refraction, they get squished together. So, if light waves with a wavelength of 500 nm travel through air (nair = 1), enter water (nwater = 1.33), and then emerge back into air again, it would look like Figure 23.14.
The equation that goes along with this situation is the following.
In this equation, λn is the wavelength of the light traveling through the transparent medium (like water, in Figure 23.14), λ is the wavelength in a vacuum, and n is the index of refraction of the transparent medium.
It is important to note that, even though the wavelength of light changes as it goes from one material to another, its frequency remains constant. The frequency of light is a property of the photons that comprise it (more about that in Chapter 25), and the frequency doesn't change when light slows down or speeds up.
When light hits a thin film of some sort, the interference properties of the light waves are readily apparent. You have likely seen this effect if you've ever noticed a puddle in a parking lot. If a bit of oil happens to drop on the puddle, the oil forms a very thin film on top of the water. White light (say, from the sun) reflecting off of the oil undergoes interference, and you see some of the component colors of the light.
Consider a situation where monochromatic light (meaning "light that is all of the same wavelength") hits a thin film, as shown in Figure 23.15. At the top surface, some light will be reflected, and some will penetrate the film. The same thing will happen at the bottom surface: some light will be reflected back up through the film, and some will keep on traveling out through the bottom surface. Notice that the two reflected light waves overlap; the wave that reflected off the top surface and the wave that traveled through the film and reflected off the bottom surface will interfere.
The important thing to know here is whether the interference is constructive or destructive. The wave that goes through the film travels a distance of 2t before interfering, where t is the thickness of the film. If this extra distance is precisely equal to a wavelength, then the interference is constructive. You also get constructive interference if this extra distance is precisely equal to two, or three, or any whole number of wavelengths.
But be careful what wavelength you use… because this extra distance occurs inside the film, we're talking about the wavelength in the film, which is the wavelength in a vacuum divided by the index of refraction of the film.
The equation for constructive interference turns out to be where m is any whole number, representing how many extra wavelengths the light inside the film went.
So, when does destructive interference occur? When the extra distance in the film precisely equals a half wavelength… or one and a half wavelengths, or two and a half wavelengths… so for destructive interferences, plug in a half-integer for m.
There's one more complication. If light reflects off of a surface while going from low to high index of refraction, the light "changes phase." For example, if light in air reflects off oil (n – 1.2), the light changes phase. If light in water reflects off oil, though, the light does not change phase. For our purposes, a phase change simply means that the conditions for constructive and destructive interference are reversed.
Summary: For thin film problems, go through these steps.
- Count the phase changes. A phase change occurs for every reflection from low to high index of refraction.
- The extra distance traveled by the wave in the film is twice the thickness of the film.
- The wavelength in the film is λn = λ/n, where n is the index of refraction of the film's material.
- Use the equation 2t = mλn. If the light undergoes zero or two phase changes, then plugging in whole numbers for m gives conditions for constructive interference. (Plugging in half-integers for m gives destructive interference.) If the light undergoes one phase change, conditions are reversed—whole numbers give destructive interference, half-integers, constructive.
Finally, why do you see a rainbow when there's oil on top of a puddle? White light from the sun, consisting of all wavelengths, hits the oil. The thickness of the oil at each point only allows one wavelength to interfere constructively. So, at one point on the puddle, you just see a certain shade of red. At another point, you just see a certain shade of orange, and so on. The result, over the area of the entire puddle, is a brilliant, swirling rainbow.
Practice problems for these concepts can be found at:
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