Which of the following phenomena of light are responsible for the formation of a rainbow?

1) reflection, refraction, dispersion
2) refraction, dispersion, total internal reflection
3) refraction, dispersion, internal reflection
4) dispersion, scattering, total internal reflection

I know the answer’s got to be 2) or 3). My book says 3) but does not give an explanation.

My confusion lies here: What is the difference between internal reflection and total internal reflection?


The word "total" makes the difference. Depending on the angle of incidence the reflection is either total or partial.

Quoted from Encyclopedia Britannica - Total internal reflection (emphasis added by me):

Total internal reflection, in physics, complete reflection of a ray of light within a medium such as water or glass from the surrounding surfaces back into the medium. The phenomenon occurs if the angle of incidence is greater than a certain limiting angle, called the critical angle.
At all angles less than the critical angle, both refraction and reflection occur in varying proportions.

Applying this to the rainbow:

According to Wikipedia - Rainbow - Mathematical derivation the angle of incidence inside the rain drop ($\beta$ in the image below) is $\beta_\text{max}\approx 40.2°$.
rain drop optics
According to Wikipedia - Total internal reflection the critical angle for light from water to air is $\theta_c=49°$.

So you have $\beta_\text{max} < \theta_c$, and therefore you have partial reflection. This makes 3) the correct answer.

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  • $\begingroup$ So internal reflection is at less than the critical angle and total internal reflection is at the critical angle? $\endgroup$ – Dora Jan 26 at 12:05
  • 2
    $\begingroup$ @Dora Yes. And total internal reflection is also at more than the critical angle. $\endgroup$ – Thomas Fritsch Jan 26 at 12:12
  • $\begingroup$ thanks a lot. It’s all cleared up now. :) $\endgroup$ – Dora Jan 26 at 13:19
  • $\begingroup$ But is the answer 2) or 3)? $\endgroup$ – Dora Jan 26 at 13:19
  • $\begingroup$ @Dora See my additions to the answer. $\endgroup$ – Thomas Fritsch Jan 26 at 13:51

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.

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