Inside a substance such as glass or water, light travels more slowly than it does in a vacuum. If c denotes the speed of light in a vacuum and v denotes its speed through some other substance, then $v=\frac{c}{n}$

Where n is a constant called the index of refraction.

To almost exact approximation, a substance’s index of refraction does not depend the wavelength of light. For instance, when red and blue light waves enter water, they both slow down by about the same amount. More precise measurements, however, reveal that n varies with wavelength. Table 1 presents some indices of refraction of Cutson glass, for different wavelengths of visible light. A nanometer (nm) is $10^{–9}$ meters. In a vacuum, light travesl at $c = 3.0 × 10^8 m/s$. Indices of refraction of Cutson glass (Table 1)

For visible light, which graph best expresses the index of refraction of Cutson glass as a function of the frequency of the light?

As just discussed, index of refraction decreases linearly with wavelength Graph b show a linear relationship. But the problem wants the relationship between frequency and n, not between wavelength and n. Since higher wavelengths correspond to lower frequencies, and vica versa, n increasing with frequency.

To clarify this reasoning, we should review the relationship between wavelength and frequency. For all waves, not just light waves, v = λf, where v denotes velocity, λ denotes wavelength, and f denotes frequency. In a vacuum, all light waves travel at speed c. So, frequency and wavelength are inversely proportional. $f=\frac{c}{λ}$ Higher wavelengths correspond to lower frequencies, and vice versa. Therefore, since n decreases at higher wavelengths, it increases at higher frequencies.

Again, extending table 1 can help us spot this relationship. Using $f=\frac{c}{λ}$, we can find the frequency of different colors of light. For instance

$f_{yellow} =\frac{c}{λ_{yellow}}=\frac{3.0×10^8m/s}{580×10^{-9}m}=5.17×10^{14}Hz$ 

$f_{yellow–orange} = \frac{c}{λ_{yellow–orange}}=\frac{3.0×10^8m/s}{600×10^{-9}m}=5.0×10^{14}Hz$