How Huygens gets reflection and refraction (technical)


One of the great successes of the ray model of light is the fact that it explained theoretically the well known and documented principles of reflection and refraction:

  • Reflection: When a ray of light hits a mirror it reflects back so that the angle of incidence as measured from the perpendicular is equal to the angle of reflection.
  • Refraction: (Snell's law) When a light ray passes into a different medium, it changes its angle according to the rule 

$$n_1 \sin{\theta_1} = n_2 \sin{\theta_2} $$
  where $n_1$ and $n_2$ are measurable parameters of each medium.

If Huygens can't explain how these results also arise in his models, it's dead in the water. Fortunately, he can. However, because of all the wavelets in Huygens' model, following the result is a bit tricky. The way to get it is to tell the story of what happens and follow the mechanism a step at a time. But you'll soon see why people preferred Newton's ray method to this one!

1. The setup

We'll start with a incoming line of pulses — like the wave coming in on the beach as shown on the page Huygens' principle -- the math. Let's imagine that we are consider a sheet of electromagnetic wave pulse (light) coming through air and encountering a volume of water. A 2D sketch of this is shown in the figure at the right. The air is shown as yellow, the water as blue, and the direction of motion of the wave pulse is shown by the arrows.

Imagine the surface of the water is made up of rows of water atoms. (Actually — electrons within the water atoms.) 

  • When the pulse of electric field reaches each atom (shown as a purple dots in the boundary between the two regions), it will shake it. That shaken atom will then cause a spherical wave to propagate outward from it.
  • Because the wave is coming in at an angle, each atom in the surface is not shaken at the same time. The ones at the left get hit first, the ones at the right last.

From our experience with still water (and jigsaw puzzles of New England lakes), we know that the surface of the water acts like a mirror for at least some of the light. From our experience swimming under water and seeing people standing by the side of the pool, we know that at least some of the light gets into the water from the air. So we can consider what happens to the outgoing ripples created by the vibrating atoms by light coming in from the air on both sides of the boundary. 

The picture describing the situation is shown below. It's quite complex and has a lot of features! Let's consider it just a piece at a time.

First note that it shows the direction of the incoming wave by black arrows (corresponding to the "rays of light") and the incoming wavefront by the solid black line perpendicular to the arrows.


Let's first look at what the wiggling atoms produce above the boundary. This is light back in the original region (air), so it corresponds to reflection.

When each atom is struck by the incoming wave, it vibrates, producing an outgoing circle that grows with the speed of light in air. (Since we are only looking in the region above the boundary, these are really semi-circles.)

Since the atom at the farthest left was struck first, the circle it produced is the largest. Since the atom just to the left of the center was the most recent one struck, the circle around it is the smallest. A bunch of intermediate circles are shown.

Pulling out the incoming wavefront and the outgoing wavefront and drawing the perpendicular from the farthest end of each wavefront to the surface gives the diagram below.

The wavefronts are drawn with heavier lines: the incoming one in black , the reflected one in blue. 

First focus on the incoming wavefront, line AB (in black). The normal to the wavefront (black arrow) indicating the direction a ray would be coming, is at an angle $\theta_{inc}$ to the normal to the surface. We draw the perpendicular, CB, from the right end of the wavefront to the surface to form a triangle ABC.

Now look at the outgoing (reflected) wavefront, DB (in blue), drawn to have the same length as AB. The position of D is determined by how far the wavelet from the surface traveled from the time it was started to the time shown. Since that wavelet is a circle, and since it must be tangent to the new wavefront (by how the new wavefront is defined), the line from D to the initiating atom must be perpendicular to the outgoing wavefront. This determines the position E. 

But since the time for the wavelet to get as far as it has must be the same as the time from the far end of the incoming wavefront to make it to the surface, distances a and b must be the same. Since we drew the incoming and outgoing wavefronts to be the same length, triangles ABC and ADE are similar. Therefore all the angles are the same and $\theta_{inc}$ must be equal to $\theta_{ref}$ or the angle of incidence equals the angle of reflection.

Phew! That's the kind of tricky geometrical argument one needed to show the result. But once you know that Huygens can also get it (even if the argument is more complex), the fact that Newton gets the reflection rule no longer serves as an argument to prefer Newton over Huygens when deciding what's "really" happening.

Now let's consider refraction, which turns out to be the more interesting and important result.



Article 730
Last Modified: July 5, 2019