The answer is 3), although the question misses the actual cause of rainbows.
Consider a ray of a single color of light that hits the drop with angle of incidence A. Some of this light enters the drop with an angle of refraction B=arcsin(sin(A)/n)), where n is the index of refraction. Note that B is less than the critical angle C.
The surface normal of a spherical raindrop contains a radius of the sphere. The path that this light takes inside the drop forms an isosceles triangle with the radii at either end of the path. This means that the angle of incidence at the back of the drop is also B, which is less than the critical angle. Total Internal Reflection is impossible.
So what causes a rainbow? The ray deflects through the angle A-B when entering the drop, 180°-2B when it reflects internally off of the back, and another A-B when it exits. The total deflection is 180°+2A-4B, so it makes an angle D=4B-2A with the original ray.
If you plot D as a function of A, for 0°<=A<90° and n~=1.33, you will find that D(A) has a maximum somewhere near A=60° and D=40°. The intensity of the light will be inversely proportional to D'(A), so it is infinite at this maximum.
This isn't an energy-conservation paradox, since the band of angles over which it is infinitely bright, is infinitesimally small. What it does mean, is that red light reflects at all angles from 0° to about 42°, and is much brighter at 42°. Dispersion means that the maximum is at a different angle for violet light, about 40°. These bright bands become the rainbow bands, and the sky inside the rainbow is a little brighter than outside.