A Geometric Proof Demonstrating That the Path Described by Snell’s Law Follows Fermat’s Principle of Least Time

Refraction artifacts in ultrasound imaging can produce the appearance of ghost side-by-side structures and, in some cases, spurious color Doppler jets. Similarly, the refraction of light causes a fish to appear farther away than its actual position. These phenomena result from a bent path taken by the sound waves or light rays. This bent path can also be observed when a lifeguard approaches a drowning swimmer: running a longer distance on sand before entering the water at an angle to minimize total rescue time. In all cases—ultrasound waves, light rays, and the lifeguard—the path bends as it crosses media with different propagation speeds, a behavior governed by Snell’s law and Fermat’s principle of least time. This fastest-path problem can be solved analytically by setting the derivative of the total travel time with respect to the point of path bending to zero, yielding a local minimum [Serway et al., 2000, p.1128]. Notably, in the lifeguard-swimmer scenario, minimizing total time does not mean minimizing the swimming distance. The fastest route requires more running and less swimming, but not the minimum swim distance along the perpendicular. In this paper, we present an intuitive geometric proof that shows this path is indeed the shortest combination in terms of distances from the interface to the “wave fronts”, and thus the fastest overall.

Main

Suppose light travels from point A to point B according to Snell’s law, following the path AOB. We can construct a circle as shown below, with diameter GF, and two right triangles GEF and GHF. Using this construction, we will demonstrate—purely through geometry—that AOB represents the fastest path. In the diagram, the angles of incidence ($\alpha$) and refraction ($\beta$) are determined by Snell’s law: where $v_1\mathrm{a}\mathrm{n}\mathrm{d}{v}_2$ are the propagation speeds in the respective media [Serway et al., 2000, p.1117].

Firstly, we will establish EF and GH are equivalent in terms of the travel time.

Suppose the propagation speeds in medium 1 and medium 2 are $v_1\mathrm{a}\mathrm{n}\mathrm{d}{v}_2$ respectively.

In Figure 1, the following angles are observed:

According to Snell’s Law,

And by trigonometry, we know

Therefore, after substituting the sine terms and cancelling $GF$, we obtain

This implies that the travel time from point E to point F is equal to the travel time from point H to point G.

Let Q be any point on the boundary other than O, and suppose a light ray travels along the path AQB. In this scenario, AQ is longer than AO, while BQ is shorter than BO. As a result, it is not immediately evident whether the total path AQB is longer or shorter than AOB. The objective is to demonstrate that the travel time along AQB is greater than that of AOB—in other words, that AOB represents the fastest path.

From point Q, we construct the path CQD such that CQ is perpendicular to GE and QD is perpendicular to FH.

Fermat’s principle implies the path obeying Snell’s law is the fattest path. Any other paths will result in longer time. Below is the proof by geometry.

Question to prove: From point A to point B, the path AOB follows Snell’s Law. Q is any other point on the boundary. Prove that the path AQB takes longer to travel than AOB.

From point Q, we can find point C on GE such that CQ is perpendicular to GE. Similarly, we can find point D on HF such that QD is perpendicular to HF. Because the shortest distance from a point to a line is along the perpendicular, therefore AQ > CQ and QB > QD. It follows that AQ + QB > CQ + QD. Thus, AQB is a longer path than CQD.

By applying the properties of similar triangles, we obtain the following.

Calculating travel time:

As shown earlier, by the definition of Snell’s law, ${\displaystyle \frac{\frac{EF}{GF}}{v_1}}={\displaystyle \frac{\frac{GH}{GF}}{v_2}}$. Therefore, ${\displaystyle \frac{EF}{v_1}}={\displaystyle \frac{GH}{v_2}}$Thus,

Similarly, we can derive ${\displaystyle \frac{CQ}{v_1}}+{\displaystyle \frac{QD}{v_2}}={\displaystyle \frac{EF}{v_1}}$:

Calculating travel time:

As we know, ${\displaystyle \frac{EF}{v_1}}={\displaystyle \frac{GH}{v_2}}$Thus,

From (1) and (2), we derive

That is, AOB and CQD have equal travel time.

As established in (1), the path AQB is longer than the path CQD. Since AOB and CQD have equal travel time, it follows that AQB requires more time to travel than AOB.

Q.E.D.

The intuition. The above proof can be further simplified by considering the behavior of light. While AO and OB represent the directions of propagation, GE and FH are wavefronts perpendicular to those paths. Since all points on a wavefront advance simultaneously, when point E reaches F, point C reaches D, A reaches B, and G reaches H. This implies that paths AOB and CQD are equivalent in terms of travel time. In contrast, AQB is visibly longer than CQD, and therefore must also be “longer” than AOB.

Lifeguard and swimmer. Given the lifeguard’s running speed on land and swimming speed in water, and any fixed positions A and B for the lifeguard and the drowning swimmer, a corresponding circle can be constructed. In every case, it can be shown in the same fashion that the path AOB is indeed the fastest route.

Point of refraction by geometry. Given points A and B, we initially do not know the exact point at which the path should refract. To begin, we randomly select a point Q on the boundary between A and B. For some different propagation speed ratio ${\displaystyle \frac{u_1}{u_2}}$, this path AQB will be the fastest one. As established during the proving process, we know ${\displaystyle \frac{u_1}{u_2}}={\displaystyle \frac{E\text{'}F\text{'}}{G\text{'}H\text{'}}}$. If ${\displaystyle \frac{E\text{'}F\text{'}}{G\text{'}H\text{'}}}>{\displaystyle \frac{EF}{GH}}$, like this case, then Q must lie to the right of O, suggesting that a slower swimmer will aim to minimize swimming distance. This insight guides us to test another point Q located further to the left. By iteratively adjusting the position of point Q and recalculating the ratio ${\displaystyle \frac{E\text{'}F\text{'}}{G\text{'}H\text{'}}}$, we will eventually converge on point O, which corresponds to the path of least time.

Ants and food. Although ants do not understand physics or calculus, studies have shown that they can approximate Snell’s law when crossing different surfaces and follow the faster path to a food source [Oettler et al., 2013]. Such geometric intuition is likely the result of millions of years of evolution

Refraction artifacts in ultrasound imaging. Refraction artifacts can produce misleading side-by-side structures or color doppler jets in ultrasound imaging [Hadzik et al., 2017; Naganuma et al. 2021, Ochi et al., 2020]. A clear understanding of the underlying physics and geometry of these artifacts can aid in making an accurate diagnosis.