EBI831 said:

ok so i can't seem to wrap my head around this b/c it seems counterintuitive. (1) index of refraction is inversely proportional to velocity: ie increase velocity= lower n (2) index of refraction technically inversely proportional to sin theta or the angle that you bend(**so larger index of refractions bend less toward the normal and thus have larger angles of refraction**) (3) frequency stays the same but wavelength changes in refraction SO if red light has a greater wavelength than blue light then it has a greater velocity than blue light which makes it have a smaller index of refraction than blue light, then why oh why oh why DOESN'T IT HAVE A LARGER ANGLE OF REFRACTION THAN BLUE LIGHT ACCORDING TO THIS BOOK I'M WORKING FROM? what about the rule blue bends best doesn't that mean it bends more toward the normal b/c of its larger n and thus the angle is reduced?

Actually, the larger the index of refraction, the more the light bends toward the normal. Consider white light, light that is composed of all the colors of the visible spectrum. If it were incident on a glass surface you would witness chromatic dispersion because the index of refraction for light in any medium except a vacuum is dependent on the wavelength of light. Now, you must keep in mind snell's law:

n1 sin theta 1 = n2 sin theta2

The two sides are equal to one another. The value for n can be found by dividing the speed of light in vacuum to the speed of light in the respective medium. Now, if n is equal to:

c/v where c is the speed of light in vacuum and v is the speed of light in the medium

you can also denote it this way: c/lambda * frequency where lambda is the wavelength of the light.

A good mneumonic for the spectrum is (roygbiv) with the frequency increasing toward the indigo color side. So, red light has a greater wavelength than blue light. We also know wave theory---as light goes from one medium to another only the velocity and wavelength change and not the frequency. Let us assume that the white light is incident from air onto glass. The index of refraction of glass will always be greater than that for air so, n2>n1. However, both sides of snell's law must be equal. In order for that to happen, theta 2<theta 1, meaning the light bends toward the normal line. As n2 becomes more big, theta 2 becomes smaller so that the two sides can stay equal. Now, n is equal to c/lambda * frequency. The frequency of light for red and blue light is the same. So, if red light has a greater wavelength than blue light, it has a smaller index of refraction than blue light. We just determined that as n2 becomes bigger, theta 2 becomes smaller. So, blue light, with a bigger n2, will bend more toward the normal when refracted if light is incident from air. I hope this helps, and if you need further help, just pm me. I used the example of light incident from air, but any general case can be used. PM if you need further clarification. Good luck!