I Problem with light-ray construction

[psychics_xxx]

Hey, this is my first post so I am looking forward to see your answers! The problem is rather simply, it's about origin of rainbow. As you can see in image below, my calculations implies that always alpha = 2*betha which is of course wrong. Can someone prove why that is? It's all included in my notes but if some things need claryfication, just let me know.

 

[A.T.]

View attachment 362586

Why do you assume that B is on the dashed horizontal line through O?

 

[psychics_xxx]

Why do you assume that B is on the dashed horizontal line through O?

Well, maybe I am wrong, but if point B was a little bit higher or lower, can't we just "rotate" our refference and the image would be the same?
Ps. Yeah it does not make sens but the equation: theta = 4 beta - 2 alpha is correct and gives right angle 42 for caustic

 

[A.T.]

can't we just "rotate" our refference and the image would be the same?

In your image, the interior rays are symmetrical to a different axis than the exterior rays. Rotating the reference frame cannot fix that.

 

[kuruman]

You are showing the ray internally reflected at B. That's fine. However, it means that angle β is equal or greater than the angle of internal reflection. So why do you show the ray reflected at A but not also at C where the internal angle of incidence is also β?

Also note that, for the water-air interface, the critical angle for internal reflection is about 41° not 30° as implied in your diagram. This means that your drawing should show the ray exiting at A. The drawing on the right is to scale.
The incoming horizontal ray is tangent to the circle at point A which is at the 12 o'clock position.
The refracted ray is at 41° withe respect to the vertical radius and intersects the circle at point B.
This internal ray is reflected (at what angle?) and intersects the circle at point C.
What is the internal angle of incidence at C and how should you draw the ray past that point?