Understanding Light Dispersion in Equilateral Prisms for Beginners

[iamnotbatman]

I feel rather stupid for not understanding this, but it's really thrown me for a loop, so any help would be greatly appreciated!

A ray of light is incident at 45 degrees on an equilateral glass prism (n=1.5).

So far, I have determined the angle of refraction to be 28.135 degrees, using Snell's Law of course, and the measures of several arbitrary internal angles using basic geometry, including what I believe to be the second angle of incidence, at 31.865 degrees. What has me stumped is the angle of refraction back out of the prism. Snell's Law fails me here, and I do believe it is because I'm overlooking something about the dispersion of the light into its component colors... but I have no idea what to do, because other than the fact that red bends the least and violet bends the most, I don't understand dispersion... a nudge in the right direction would be awesome, please

Thank you!
-Dani

 

[SpaceTiger]

What has me stumped is the angle of reflection back out of the prism. Snell's Law fails me here, and I do believe it is because I'm overlooking something about the dispersion of the light into its component colors...

First of all, I'm going to assume that you mean refraction, not reflection (the angle of reflection is the same as the angle of incidence). If you assume all the light is refracted at the same angle (that is, constant n), then chromatic dispersion is being neglected. Using Snell's law:

$$sin(\theta_i)=nsin(\theta_r)$$

where the angle on the left is the incident and angle and the one on the right is the refracted angle, while n is the index of refraction in water. You say this formula fails you for the outgoing ray, but consider that there is symmetry here. That is, the path that a light ray travels from one medium to another is the same path as would be followed going in the other direction. This means that you can reverse the problem and imagine a light ray hitting the prism with an unknown angle and then bending to an angle of 31.865 degrees (assuming you did the geometry right). Solving from the above equation, this means:

$$sin(\theta_r)=\frac{sin(\theta_i)}{n}$$

This should give you the same angle as for a light ray coming from the prism at 31.865 degrees and bending in the air.

 

[iamnotbatman]

Yes, I did mean refraction. Thanks for pointing out the horrid typo lol. *slaps self on wrist*

Thank you very much for your help!