The goal of the problem is to determine angles X, Y, and Z made by light traveling through the equilateral prism. In stating the problem, I've been tricky in two ways. First of all, the 31

^{o}angle is NOT the angle of incidence at the left-hand surface. In optics, all angles are measured to the normal, not to a surface; so the incident angle to plug into Snell's Law is 59

^{o}. A student who plugs in 31

^{o}has neither paid attention in class nor collaborated with his classmates, and thus earns a red and blue VOID stamp.

Most students, though, figure out the correct incident angle. They plug in to get the refracted angle, 35

^{o}(from the normal, of course).

What's crazy here is the disconnect between geometry and the angles my students use for the right-hand interface. The vast majority of my class at this point makes two incompatible assertions:

1. Angles X and Y are each 60

^{o}, because the prism is equilateral.

2. The angle of incidence at the right interface is 35

^{o}, giving a refracted angle in air of 59

^{o}. What's the justification? Since the light entered the glass at 35

^{o}, it must make the same angle when it leaves.

This second assumption likely comes from class, where I show light incident on a water-air surface. If the light hits the water from the air at 45

^{o}, we show using Snell that the light refracts at 33

^{o}. Then we show again that light incident the other way, from water to air, can follow the same path: incident at 33

^{o}gives refraction into air at 45

^{o}. Perhaps, also, some have seen the calculation that light that hits a rectangular block of glass must exit the glass parallel to the original incident beam.* My class has incorrectly made the blanket interpretation that the angle of refraction in the air must, in all circumstances, be equal to the original angle of incidence. Oops.

The first assumption, though, is either lazy, or born of an inability to understand what a "refracted angle measured from the normal" means. Some careful but straightforward Euclidian geometry can show that angle X must be (90-35)=55

^{o}-- after all, angle X and the 35

^{o}angle of refraction make up the normal to the surface, which by definition is a 90

^{o}angle. Then, 180

^{o}in a triangle, so angle Y must be (180-55-60)=65

^{o}. The angle of incidence at the right-hand surface, therefore, is (90-65)=25

^{o}.

Snell's law shows that a 25

^{o}incident angle from glass to air give a refracted angle of 39

^{o}. Angle Z is the complement of 39

^{o}, and so is

**51**

^{o}. Done.

I assigned this problem in my general physics class after I had assigned a plug-and-chug-level problem set on Snell's Law. Sure enough, no one at all thought to trace through the geometry. That's partially my fault for giving them the plug-and-chug problems. I made them believe that Snell's Law didn't require any thinking. At the general physics level, I think it useful to do some plug-and-chug just to make the point that "Hey, this math is NOT too hard for you, so don't even think about using that as an excuse." Of course, then it is incumbant on me to guide the class a step beyond the plugging and the chugging. This problem is a good way to do that.

So, what did I do? I RE-assigned this problem, gave a brief hint on the order of "Don't assume that the light makes an equilateral triangle, actually do some geometry," and gave the quiz below. I won't grade the re-submitted problem, only the quiz.

GCJ

P.S. This problem is good for AP physics, too... I probably would not draw in the normal lines, though.

* Check out Giancoli chapter 23 problem 38, which asks students to prove this fact geometrically.

1. What is the angle of incidence at the left-hand edge? Why?

2. Explain how to figure out angle X.

3. Explain how to figure out angle Y.

4. Explain how to figure out the angle of incidence at the right hand edge.

5. Calculate angle Z.