Fermat’s Principle

Section 43.9 Fermat’s Principle

The three laws of geometric optics can be understood from the perspectives of a deeper principle concerning the propagation of light due to Pierre Fermat (1601-1665), the French lawyer and mathematician. In its original form, Fermat’s principle states that,

“The actual path between two points taken by a beam of light is the one which is traversed in the least time.”

This principle can be stated in terms of a quantity called the optical path length(OPL) defined as the product of geometric length and the refractive index of the medium. Let a ray of light travel a geometric length \(l\) in a medium of refractive index \(n\text{.}\) then, its optical path lenfth will be

\begin{equation} \text{OPL} = n l.\tag{43.27} \end{equation}

Let \(v\) be the speed of light in that medium. Then time taken will be

\begin{equation*} t = \dfrac{l}{v}. \end{equation*}

Since \(v\) in a medium is related to \(c\) in vacuum by \(v = c/n\text{,}\) we can write this as

\begin{equation*} t = \dfrac{nl}{c} = \frac{\text{OPL}}{c}. \end{equation*}

Therefore, minimizing \(t\) will mean minimizing \(\text{OPL}\) also. The laws of reflection and refraction through transparent media can be easily derived from Fermat’s principle as I will show now.

Subsection 43.9.1 (Calculus) Law of Reflection from Fermat’s Principle

Consider two points A and B in the same medium as shown in Figure 43.39. A ray of light traveling to a mirror reflects in the direction of B. Where on the mirror the light has to hit so that total time between A and B will be minimum? I know that you know the answer that \(\theta^\prime = \theta\text{.}\) But, here we pretend that we do not know this and want to apply Fermat’s principle of least time, or equivalently least optical path length, to figure this fact out.

Strategy: if we can find the location of point P such that time for AP+PB corresponds to the least time path between A and B that contains a reflection from the mirror, we should be able to prove the two angles are equal. Note: the least time path without reflection will just be the direct path from A to B, but we are not interested in that path.

Figure 43.39. Deriving law of reflection using Fermat’s principle.

Let the fixed distances in the figure be AC = BD = \(L\) and CD = \(h\text{.}\) Since we need to find P, let CP = \(x\text{,}\) unknown. Then, according to Fermat’s least time principle, for fixed A and B, point P will be at a spot such that the time for travel for light of speed \(v\) will be smallest. Let time for AP be \(t_{AP}\) and time for PB be \(t_{PB}\text{.}\) Since the rays are in the same medium we will use the same speed for both rays. We can write time \(t\) as a function of \(x\text{.}\)

\begin{align*} t \amp = t_{AP} + t_{PB} = \frac{AP}{v} + \frac{PB}{v} \\ \amp = \frac{\sqrt{L^2 + x^2}}{v} + \frac{\sqrt{L^2 + (h-x)^2}}{v}. \end{align*}

To minimize \(t\text{,}\) we take a derivative of \(t\) with respect to the independent variable \(x\) and set it to zero.

\begin{equation*} \frac{x}{\sqrt{L^2 + x^2}} = \frac{h-x}{\sqrt{L^2 + (h-x)^2}}. \end{equation*}

This relation can be written in terms of the angles \(\theta_1\) and \(\theta_1^\prime\text{.}\) \

\begin{equation*} \sin\theta_1 = \sin\theta_1^{\prime} \end{equation*}

Since both angles are less than \(90^{\circ}\text{,}\) we can immediately write down their equality.

\begin{equation*} \theta_1 = \theta_1^{\prime} \end{equation*}

Therefore, point P has to be such that the angle of incidence will be equal to the angle of reflection.

Subsection 43.9.2 (Calculus) Law of Refraction from Fermat’s Principle

To deduce the law of refraction based on Fermat’s principle, we fix points A and B in the two media, and find point P at the interface where a ray from point A in medium 1 will refract in the direction of B in medium 2 as illustrated in Figure 43.40.

Figure 43.40. Deriving law of refraction using Fermat’s principle.

Let points A and B be such that AC = BD = \(L\) and CD = \(h\text{.}\) AC and BD are chosen equal for the convenience in calculation. Let point P be at a distance \(x\) from C. We need to find point P such that time from A to B is least. Note that light travels with different speeds in the two media.

\begin{align*} \amp v_1 = \frac{c}{n_1},\\ \amp v_2 = \frac{c}{n_2}, \end{align*}

where \(c\) is the speed of light in vacuum. On path APB we can write the time as a function of \(x\) and then minimize this function.

\begin{align*} t \amp = t_{AP} + t_{PB} = \frac{AP}{v_1} + \frac{PB}{v_2}\\ \amp = \frac{\sqrt{L^2 + x^2}}{v_1} + \frac{\sqrt{L^2 + (h-x)^2}}{v_2}. \end{align*}

To minimize \(t\text{,}\) we take the derivative of \(t\) with respect to the independent variable \(x\) and set it to zero.

\begin{equation*} \frac{1}{v_1}\frac{x}{\sqrt{L^2 + x^2}} = \frac{1}{v_2}\frac{h-x}{\sqrt{L^2 + (h-x)^2}}. \end{equation*}

Writing the speeds in terms of \(c\) and the refractive indices, and replacing the ratios of the distances in the right angled triangles by trigonometric functions we immediately arrive at the Snell’s law.

\begin{equation*} n_1\sin\theta_1 = n_2\sin\theta_2. \end{equation*}

Subsection 43.9.3 Fermal’s Principle is Fundamental

The derivations of the laws of reflection and refraction from requiring that the true path of the ray be the one where light takes least amount of time captures some fundamental nature of light. Although, from experiments, we already knew the results of these derivations, the derivations themselves do not rely on experiments. Rather, it proposes a mathematical principle from which results of these experiments could have been predicted and experiments just test these predictions.

A general lesson of Fermat’s principle is that, in order to understand physical reality at the deepest levels, we need to find some physical aspect of a system that nature is trying to optimize - in the case of light, it is the time of travel that nature optimizes, and in other systems, it may be something else. This way of thinking led to the discovery of principle of least action, upon which classical mechanics can be understood. As a matter of fact, not only classical mechanics but also the entirety of fundamental physics relies on extensions of this principle.

Exercises 43.9.4 Exercises

1. Practice Fermat Principle 1.

What should be the value of \(a\) so that the integration of the following functions from \(x = 0\) to \(x = 1\) will be minimum. (a) \(f(x) = a^2 x^3 - a x\text{,}\) (b) \(f(x) = a^4 x^3 - a^2 x\text{.}\)

Answer.

(a) \(a\) = 1; max.; (b) \(a\) = -1, min; \(a\) =0 max; \(a\) = 1 max.

Solution 1. a

Let us call the integral of \(f(x)\) by \(g(a)\text{.}\)

\begin{equation*} g(a) = \int_0^1 f(x)\: dx = \int_0^1 (a^2 x^3 - a x)\: dx = \dfrac{a^2}{4} - \dfrac{a}{2}. \end{equation*}

To find the extrema of \(g(a)\) we will set the first derivative of \(g(a)\) to zero.

\begin{equation*} \dfrac{dg}{da} = 0,\ \ \Longrightarrow\ \ \dfrac{a}{2} - \dfrac{1}{2} = 0,\ \ \Longrightarrow\ \ a = 1. \end{equation*}

Does \(a=1\) correspond to max or min for the function \(g(a)\text{?}\) We check with the second derivative.

\begin{equation*} \left.\dfrac{d^2g}{da^2}\right|_{a=1} = \dfrac{1}{2}>0,\ \ \Longrightarrow\ \ g(a)\ \textrm{minimum at } a = 1. \end{equation*}

Solution 2. b

\begin{equation*} g(a) = \int_0^1 f(x)\: dx = \dfrac{a^4}{4} - \dfrac{a^2}{2}. \end{equation*}

To find the extrema of \(g(a)\) we will set the first derivative of \(g(a)\) to zero.

\begin{equation*} \dfrac{dg}{da} = 0,\ \ \Longrightarrow\ \ a^3 - a = 0,\ \ \Longrightarrow\ \ a = -1,0,1. \end{equation*}

Let us find which one of \(a= -1, 0, 1\) correspond to max or min for the function \(g(a)\text{?}\) We check with the second derivative.

\begin{equation*} \dfrac{d^2g}{da^2} = 3a^2 -1, \end{equation*}

which gives the following for \(a= -1, 0, 1\)

\begin{align*} \amp \left.\dfrac{d^2g}{da^2}\right|_{a=0} = -1,\ \Longrightarrow\ \ a = 0 \ \textrm{is max}.\\ \amp \left.\dfrac{d^2g}{da^2}\right|_{a=\pm 1} = 2,\ \Longrightarrow\ \ a = -1\ \textrm{and}\ a = 1 \ \textrm{are min}. \end{align*}

2. Practice Fermat Principle 2.

Using Fermat’s principle, find the point where a ray from A in air must be incident on the interface Y so that the refracted ray passes through point B inside the glass as shown in Figure 43.41. (\(n_\text{glass}=1.5\))

Answer.

\(y = 1.37\, \text{cm}\text{.}\)

Solution.

We need to find the location of the point P in the figure such that the time A-P-B is a minimum. Let \(y\) be the coordinate of point P with respect to the origin at the level of A. Then, the time \(t_{\textrm{APB}}\) will be a function of \(y\) given by

\begin{equation*} t_{\textrm{APB}} = t_{\textrm{AP}} + t_{\textrm{PB}}. \end{equation*}

Let \(v\) be the speed of light in vacuum and \(n\) the refractive index of glass.

\begin{equation*} t_{\textrm{APB}} = \dfrac{\sqrt{y^2 + a^2}}{v} + \dfrac{\sqrt{b^2 + (c-y)^2}}{v/n}. \end{equation*}

Now, we set the derivative of this \(t\) with respect to \(y\) to zero. This gives

\begin{equation*} \dfrac{dt_{\textrm{APB}}}{dy} = \dfrac{y}{v\sqrt{y^2 + a^2}} -\dfrac{n(c-y)}{v\sqrt{b^2 + (c-y)^2}} = 0. \end{equation*}

Solving this for \(y\) is not easy. After some algebra you will get a polynomial equation in y to the power 4.

\begin{equation*} (n^2-1)y^4 - 2c(n^2-1)y^3 +[(n^2-1)c^2 + n^2 a^2 - b^2]y^2 - 2n^2ca^2 y + n^2c^2 a^2 = 0. \end{equation*}

Using \(a= 2\text{,}\) \(b = 4\text{,}\) \(c = 3\text{,}\) and \(n= 1.5\text{,}\) we can solve this equation numerically to find \(y = -0.83 \pm 2.6 i\text{,}\) \(1.37\text{,}\) \(6.29\text{,}\) where \(i = \sqrt{-1}\text{.}\) The complex solutions are not the solution of the physical problem here and from the geometry in the problem, the answer will be \(y = 1.37\, \text{cm}\text{.}\)

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