Light dispersion in water

The dispersion that leads to the rainbow effect generated by transparent media results from an intrinsic property of the medium being considered: the dependence of its refractive index $n$ on the wavelength of light $lambda$ passing through it. In this sense, water in a glass is just as dispersive as water droplets in a rainbow.

When different wavelengths pass from air into a sample of water, they refract by different amounts according to Snell's law, and the fact that the index of refraction of water varies as a function of wavelength.

So why don't you see a rainbow in some situations, but you do see a rainbow in others? Well, it's because being able to see the rainbow depends on the geometry of the sample of water, at what angle the light is incident, and where the dispersed light is being viewed from.

For example, suppose we have a sample of white light (containing all wavelengths) that is normally (perpendicular to the surface) incident on a rectangular prism of water, then by Snell's law, the the light does not refract at all when it passes through the sample, and we see no dispersion.

On the other hand, if there is a droplet on a table, then we can arrange for white light to be shone on the droplet in such a way that it is not normally incident on the droplet and exits the droplet in such a way that it hits the table allowing us to see all of the separated colors.

As the other answer states, water is equally "dispersive" in the sense of how much the index of refraction changes for different wavelengths of light. However, the ability to see/detect/observe the effects of dispersion requires a good geometrical setup, and that requirement can be made more clear with some examples.

Firstly, note that a normal glass prism is always a triangular prism, and shown with a thin beam of light passing through it and splitting apart. This is all deliberate:

1. If you have the light pass through parallel sides, then the exit angle will be the same as the entry angle, and so you'd have to be able to tell the difference between colored beams of light all traveling parallel to one another. Very hard to see. So instead of using a rectangular prism, use a triangular prism so that the output angle depends on the index of refraction, not just the exit position.

2. If you use a wide beam of light (or open sunlight) then the splitting of one section of light will overlap with the splitting of the section next to it, making it all seem more "white" for having the mixed colors.

You can still see the effect of dispersion in broad sunlight sometimes, particularly when hanging a prism in front of a window. However, note that you have still simplified the setup to do so: by hanging in front of a window, you're guaranteed that all the sunlight is coming in at the same angle, that the floor will be dark enough to see the colors clearly, that the prism will be parallel to the floor or table it casts light on, and that the prism will slowly twist until you happen to notice the effect when the entry angle is best.

This is why uhoh suggested restricting the light to a narrow beam and letting it pass through a glass of water halfway between the center and the edge. The narrow beam solves (2), and the positioning ensures an angled entry and exit to solve (1). (You may notice that if you took the sides of the glass at that position and "flattened" them out, you'd get a triangular prism-like shape!)