# Why does Snell's law not factor in wavelength?

If dispersion is caused due to different wavelengths bending with different angles, how is Snell's law right in generalizing it as $\sin(i) / \sin(r)$? Or am I missing something?

Snell's law is given by $n_1 \sin i =n_2\sin r$ where $n_1$ is the refractive index of the material the light is initially in and $n_2$ the refractive index of the material the light is going into. These are not constants for a given material and change (although very slightly) with the wavelength of light. Thus for different wavelengths of light the ratio of $\sin i$ to $\sin r$ will change and thus the angle of refraction.

The different refractive indexes for different wavelengths also explains why say red light travels faster in water than blue light.

The term dispersion refers to the speed of light in a material having a dependence on frequency (or equivalently wavelength). The refraction angle's dependence on frequency is caused by the material dispersion, not the other way around.

In all materials the refractive index will have dispersion but it's often the case that certain materials in certain frequency/wavelength ranges have a relatively small change in refractive index, and the dispersion is neglected. Nevertheless, if you wanted the "dispersive" form of Snell's Law you'd use,

$n_1(\lambda) \sin \theta_1 =n_2(\lambda) \sin \theta_2$.

Once you've assumed some functional form for $n_1(\lambda)$ and $n_2(\lambda)$ and solved for either $\theta_1$ or $\theta_2$, you'd see the frequecy/wavelength dependence in the angle.

The whole point of Snell's law is about taking into account wavelength! Remember that the fundamental property of the light is its frequency. Wavelength, on the other hand, is not a fundamental property of a beam of light since the wavelength changes all the time as it passes through various media.

The index of refraction of a medium is a dimensionless ratio of wavelengths: the wavelength in the medium in question, and the wavelength in another reference medium (vacuum).