Basic angles of refraction and incidence help needed

[Couperin]

Basic angles of refraction and incidence help needed (help still needed)

I feel silly for even asking this, but here goes.

http://img501.imageshack.us/img501/905/refractionpuzzlerg2.gif

The triangle represents a glass prism. The refractive index between air and the prism is 1.55. The angle of incidence (angle 'p' on my diagram) is 45 degrees, and by using the equation sin r = sin i / 1μ2, I've worked out angle 'q' to be roughly 21 degrees.

Knowing this, how could I work out angle 'r' on my diagram? I've tried visualising the two normal lines as two equals sides of an isoceles triangle, but I don't know whether that's just an illusion caused by the way I've drawn the diagram. I've also just assumed the angle 'r' is 90 degrees - 21 degrees, but I can't prove whether this is true or not.

So, could anyone assist me with working out angle 'r' on my diagram? A few hints or pointers?

 

[quasar987]

is the glass prism equilateral? If so, then the angle on the bottom left corner (let's call it A) is 60 ° and you can use the fact that the sum of the angles of the triangle Apr is 180°.

 

[Couperin]

Thankyou, Quasar :).

So if I call that little triangle in the bottom left corner (ABC), and A = 60°, and B = 90° - q which = 69°, then angle C, which is next to r, must = 51°.

Therefore, the angle of incidence against the back surface of the prism is 90° - 51°, which is 39°.

So, the sin of the angle of refraction (the angle at which the light emerges from the prism) must be sin 39° / 1.55^-1, which = 0.98.

So the angle itself must be sin^-1 * 0.98, which = 77° (aprox).

So the initial incidence beam that hit the prism in the first place has, by entering the prism, diverged by...

...hmm, what's the proper way to calculate divergence? Do you add 77° to the first angle of refraction (21°)? If so, my answer is that it's diverged 98°, but I'm probably wrong.