21.1. Dispersion http://www.ck12.org
FIGURE 21.3
The dispersion of light through a prism.- White light is composed of seven dominant colors of light: red, orange, yellow, green, blue, indigo, and violet.
You may have heard the mnemonic (Roy G. Biv) which is helpful in remembering the order of the colors in
the visible spectrum. - The degree of refraction depends upon the color of the light. Red light is refracted the least and violet light
refracted the most. - Since the medium is uniform, the index of refraction must depend upon the color of the light.
It is well known that the color of light depends upon its frequency. In the visible spectrum (and within a vacuum),
red light has the smallest frequency, of about 4. 0 × 1014 Hzand the longest wavelength, about 750 nm. Violet light
has the largest frequency, of about 7. 5 × 1014 Hzand the shortest wavelength, about 400nm.
Since the speed of light is constant in a vacuum, the product of frequency and wavelength remains constant. Recall
the Wave Equation:
c=fλor v=fλ
The frequency of light isf, the wavelength of light isλ, and the velocity of light in vacuum isc. If, however, light
passes through any other medium, its velocity is given asv.
We know that light travels slower thancin any medium other than a vacuum. When light travels from one medium
into another, the frequency of the light does not change. However, its velocity, as we learned from Snell’s Law, does.
Sincen=cv, we can see that red light, for example, having the greatest speed through the prism inFigure21.3, has
the smallest index of refraction through the prism.
Thus, asvincreases,nmust decrease.
http://demonstrations.wolfram.com/LightRayInAPrism/
http://demonstrations.wolfram.com/DispersionOfLightThroughAPrism/
Check Your Understanding
Which light has the greater index of refraction: green or yellow?
Answer: We can see fromFigure21.3 that green light is refracted more than yellow. It follows, then, that the index
of refraction for green light is greater than yellow light.
Illustrative Example 22.1.1
Show that for light of wavelengthλin air, wavelength in a medium of refractive indexnisλn=λn.
Solution:
Using the wave equation and the definition for the index of refraction, we have
Equation 1:v=fλand