Fermat's principle

Of all possible paths of light, only real paths whose optical path is maximum or minimal.

The Fermat's principle, in optics, is an extremal principle that establishes:

The path followed by light by spreading from one point to another is such that the time used to travel it is a minimum.

This statement is not complete and does not cover all cases, so there is a modern form of Fermat's principle. This says that:

The path followed by the light by spreading from one point to another is such that the time used to travel it is stationary regarding possible variations of the trajectory.

This means that if you express the journey through the light between two points ${displaystyle O_{1}}$ and ${displaystyle O_{2}}$ by means of a functional called optical path defined as ${displaystyle {mathcal {L}}_{O_{1}O_{2}}}[n({vec {r}}})}}$ the real trajectory of light will follow an extreme path to this functional one:

${displaystyle delta {mathcal {L}}_{O_{1}O_{2}[}n({vec {r}})]=delta int _{O_{1}}}{O_{2}}}{n({vec {r}})ds}=0. !$

The important characteristic, as the statement says, is that the paths close to the true require approximately equal times. In this form, Fermat's principle is reminiscent of Hamilton's principle or the Euler-Lagrange equations.

The principle in its modern form was stated by Pierre de Fermat in a letter of 1662, hence it bears his name.

The following are now some examples of the application of the principle to deduce the laws of geometric optics.

Equation of the path of a light ray

The equation of the trajectory of a real light ray in an optical system is:

${displaystyle {vec {nabla}n({textbf {r}})-{d over dsleft[n({textbf {r}}}){{{textbf {dr}} over {ds}}{right}=0}$

and follows from Fermat's principle.

Deduction of the equation

We have a light beam that moves through a medium with a continuous refractive index from a point ${displaystyle A}$ to a point ${displaystyle B}$.

Sea ${displaystyle {textbf {r}}}$ the position vector ${displaystyle {textbf {u}}}$ the unit vector tangent to the trajectory.

We have to ${displaystyle L=int _{A}^{B}n({textbf {r})ds=int dL}$. From this it follows that ${displaystyle nds=dL}$. Besides, you have to ${displaystyle ds={textbf {u}}cdot {textbf {dr}}}}$ and that ${displaystyle dL={vec {nabla }}Lcdot {textbf {dr}}}}$.

So. ${displaystyle n{textbf {u}}cdot {textbf {dr}={vec {nabla } lcdot {textbf {dr}}}}}}}}$. Being ${displaystyle {textbf {dr}}}$ of an arbitrary vector ${displaystyle n{textbf {u}}={vec {nabla} l}$ and therefore that ${displaystyle mid {{vec {nabla}}L}mid =n}$.

From this you get that ${displaystyle {{d(n{textbf {u}}}} over {ds}}={d over ds}{vec {nabla }}L}$. We note that${displaystyle {{d(n{textbf {u}}})} over {ds}={d over dsleft[n({textbf {r}}}){{{textbf {dr}} over ds}right]}}$.

On the other hand we have that (Cartesian will be used but it works for the rest of the orthonormal bases):

${displaystyle {d over ds}left({{partial l} over {partial x}}}right)={vec {nabla }}{{{partial l}{partial l}{partial x}}}{cdot {{textbf {dr}}}{over ds}{b$because ${displaystyle {dleft({partial l} over {partial x}right)={vec {nabla }}left({partial L} over {partial x}}}}right)cdot {textbf {dr}}}}}}$.

${displaystyle {d over ds}left({{partial L} over {partial x}}}}right)={vec {nabla }}{{{partial L} over {partial x}}}right)cdot {1over n}{vec {nabla }L}}$because ${displaystyle n{textbf {u}}={vec {nabla} l}$.

$############ ######################################################################################################################################### L} over {partial x}}{partial L} over {partial x}}}+{{partial } over {partial y}{{{partial L} over {partial x}}{partial l} over {partial y}}}{+{partial} over {partial z}}{{partial L} over {partial x}}{partial L} over {partial z}right]}$.

${displaystyle {d over ds}left({{partial L} over {partial x}}}right)={1 over n}left[{{partial } over {partial x}}{{partial L} over {partial x}}{partial L} over {partial x}}}+{{partial } over {partial x}{{{partial L} over {partial y}{partial l} over {partial y}}}{+{partial} over {partial x}}{partial L} over {partial z}{partial ♪ I'll be right ♪$because ${displaystyle {{partial } over {partial beta }}}left({{partial} L over {partial alpha }}}{2}=2{partial } over {partial beta }{{{partial }}{{partial L} over {partial alpha }{{partial) L} over {partial alpha}}}$.

${displaystyle {d over ds}left({{partial} L} over {partial x}}right)={1 over {2n}}{{{partial } over {partial x}}n^{2}={1 over {2n}}{{partial n} over {partial x}}}{{{{{partial n} over {partial x}}}}}}$because${displaystyle leftleft({partial L} over {partial x}right)^{2}+left({{partial L} over {partial y}{right)}{2}{2}{left({{{{partial L}}{partial l}}{right}{2mid}{$.

Making the same components for ${displaystyle and}$ e ${displaystyle z}$ is obtained that:

${cHFFFFFF}{cHFFFFFF}{cHFFFFFF}{cHFFFFFF}{cHFFFFFF}{cHFFFFFFFF}{cHFFFFFFFFFFFFFF}{cHFFFFFFFFFFFF}{cHFFFFFFFF}{cHFFFFFFFF}{cHFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF}{cHFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF}{cHFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF}{cHFFFFFFFFFFFFFFFFFF}{cHFFFFFFFFFFFFFFFFFF}{cHFFFFFFFFFFFFFFFF}{cHFFFFFFFFFFFFFFFFFFFFFF}{cHFFFFFF}{cHFFFFFF}{cHFFFF$.

Substituting we obtain that:

${displaystyle {{d(n{textbf {u}}}} over {ds}={vec {napp}n}$ or ${displaystyle {d over ds}left[n({textbf {r}}){{{textbf {dr}}} over ds}right]={vec {nable}n}$.

Analogy with Lagrangian mechanics

The optical path can be equated to action in Lagrangian mechanics. The refractive index can be treated as a Lagrangian composed of a potential. In this way the problem can be solved with the Lagrange equations. The ray of light is directed towards the area of greatest refractive index, of which the equivalent potential would be the opposite of the refractive index.

We have: ${displaystyle int _{A}^{B}n(lbrace q_{i}rbrace)ds=int _{A}^{B}n(lbrace q_{i}rbrace)Rdlambda equiv int _{A}^{B}L(lbrace q_{i}rbrace)dlambda$.

With: ${displaystyle L(lbrace q_{i}rbrace)=n(lbrace q_{i}rbrace)R}$ also ${displaystyle R={sqrt {sum left({{{dq_{}{d} over {dlambda }}right)^{2}}}}{sqrt {{{sum {{dot {q}}_{i}}}{2}}}}}}}}$ and ${displaystyle dlambda ={ds over R}}$.

So if we define ${displaystyle q_{i}(lambda)}$ we get that: ${displaystyle {d over {dlambda }}left({{{partial (nR)} over {partial {dot {q}}_{i}}}}}}{{{{partial (nR)} over {partial q_{i}}}}=0}$.

You get: ${displaystyle R{d over {ds}}left({{{partial (n)} over {partial {dot {dot {q}_{i}}}}R+{partial (R)} over {partial {dot {q}}}{i}{i}{right)}{partial}{i}{partial$ with ${displaystyle {d over {dlambda}}={1 over {1/R}{d over {ds}}=R{d over ds}}}$

Besides, you have to ${displaystyle {{partial(n)} over {partial {dot {q}}_{i}}}}}=0}$ and that ${displaystyle {{partial} ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫ ♫$.

From where: ${displaystyle R{d over {ds}}left(n{dq_{i} over ds}right)-R{{partial (n)} over {partial q_{i}}}}}$ or ${displaystyle {d over {ds}}left(n{dq_{i} over ds}right)-{{partial (n)} over {partial q_{i}}}}}}=0}$.

Creating a system with the three coordinates you get: ${displaystyle {d over {ds}}left(n{{textbf {dr} over dsright)-{vec {blah }n=0}$.

Equation of the path of a light ray in a homogeneous medium

We assume that:

${displaystyle {d over ds}left[n({textbf {r}}){{{textbf {dr}}} over ds}right]={vec {nable}n}$.

With ${displaystyle n}$constant, of what:

${displaystyle {d over ds}left[n({textbf {r}}){{{textbf {dr}}} over ds}right]=n{d over ds}{{{{textbf {dr}} over ds}}}{right$and ${displaystyle {vec {nabla}n=0}$.

Therefore:

${displaystyle {d over ds}{{textbf {dr}} over ds}=0}$ of what ${displaystyle {{textbf {dr}} over ds}={textbf {l}}}}$ being ${displaystyle {textbf {l}}}$ a constant, of what ${displaystyle {textbf {r}}={textbf {l}}s+C}$ being ${displaystyle {textbf {c}}}$ a constant.

The result is a starting point straight ${displaystyle {textbf {c}}}$ and vector director ${displaystyle {textbf {l}}}$.

Malus-Dupin Theorem

If on each ray emitted by a focus we go through equal optical paths, then the points that delimit them form a normal surface to all the rays. We call this surface a wave front. It coincides with the wave front given by the oscillatory theory. When deduced from Fernat's principle, it is valid regardless of the number of reflections or refractions that the ray may undergo before reaching its destination.

Law of reflection

If it is assumed that a ray of light leaves point A in the direction of the flat surface, which we assume reflective, and travels to point B, what will be the trajectory followed by the light? In this case, light travels all the way through the same medium, with the same refractive index, and therefore at the same speed. Thus, the time necessary to travel the path between A and B (passing through the surface P) will be the distance APB divided by the speed of light in that medium. Since velocity is a constant, the actual trajectory, according to Fermat's principle, will be the shortest.

It is easy to see that the distance APB is the same as the distance A'PB, where A' is the image of A. A' is on the perpendicular straight to the mirror passing through A, the same distance from the mirror as A and the other side of it. The A'PB minimum distance is obviously the A'P2B straight line, so the actual trajectory is AP2B. The complete analysis of the situation shows that P2 is such that the angles of incidence and reflection at the point are equal, from which the formula of the law of reflection is deduced: ${displaystyle theta _{i}=theta _{t}}$

Law of Refraction

Be a medium of propagation with refractive index ${displaystyle n_{1} }$ and a second medium of propagation with refractive index ${displaystyle n_{2} }$ such that we place the surface that separates the two media so that it coincides with the axis of the abcisas.

Sean. ${displaystyle A=(x_{A},;y_{A})}$ and ${displaystyle B=(x_{B},;y_{B})}$ two fixed points located from the plane, so that A is located in the first half, and B in the second half.

Be a beam of light that spreads from A to B crossing the surface that separates the two media at the point ${displaystyle P=(x,;0)}$.

The next step is to deduce the time the lightning takes to travel ${displaystyle {overline {AP}}}$ and ${displaystyle {overline {PB}}}$.

Sean. ${displaystyle v_{1}}$ and ${displaystyle v_{2}}$ the speed of propagation of light in the first and second half respectively.

${displaystyle t_{1}={frac {overline {AP}}{v_{1}}}}{frac {sqrt {(x_{A} -x)^{2}{2} +{y_{A}}}}}{{v_{1}}}}}}}}}}$; ${displaystyle t_{2}={frac {overline {PB}}{v_{2}}}}}{frac {sqrt {(x-x_{B})}{2}{2}}{ +{y_{B}}}}}{v_{2}}}}}}}}}$

${displaystyle t={frac {sqrt {(x_{A} -x)^{2}{y_{A}}}{2}}}}{v_{1}}}}}}}}{frac {sqrt {(x-x_{B})}{2}{2}{2}}}{{x-x-x-x}}}{2}}}}}}{$

If you want the value of ${displaystyle x }$ When ${displaystyle t }$ is minimal, it is equivalent if we find the value of ${displaystyle x }$ for which the derivative function ${displaystyle t }$ take the value 0.

${displaystyle {frac {dt}{dx}}=-{frac {x}{x}{x}{x}{x}{x}{x}{x}{x}{x}}{b}{x}{x}{x}{x}{x}}{x1⁄2}{x}{x1⁄2}{x}{x}{x}{x}{x1⁄2}{x}}}}}{x1⁄2}}}{x1⁄2}$

${displaystyle {frac {x_{A}{x}{v_{1} {sqrt {(x_{A}{x)^{2} +{y_{A}}}{2}}{2}}{x}{B}-x }{v_{2}{sqrt {(x-x}{B}}}{$

${displaystyle {frac {x}{x}{x}{v_{1} {overline {AP}}}}}}}}{frac {x_{B}-x }{v_{2}{overline {PB}}}}}}}}}}}{$

${displaystyle {frac {1}{v_{1} }{sin {alpha _{1}}}}}{{frac {1}{v_{2} }}}}}{sin {alpha _{2}}}}}}}}}$

${displaystyle {frac {c}{v_{1}}sin {alpha _{1}}}}{frac {c}{v_{2} }}{sin {alpha _{2}}}}}}}}$

${displaystyle n_{1} sin {alpha _{1}}=n_{2}sin {alpha _{2}}}}}}$

History

Pierre de Fermat

Hero of Alexandria (Heron) (c. 60) described a principle of reflection, which stated that a ray of light going from point A to point B, undergoing any number of reflections in plane mirrors, in the same medium, has a path with shorter length than any nearby path.

Ibn al-Haytham (Alhazen), in his Book of Optics (1021), extended the principle to both reflection and refraction, and expressed an early version of the principle of least time. His experiments were based on earlier work on refraction by the Greek scientist Claudius Ptolemy.

The generalized principle of lesser tense in its modern form was stated by Pierre de Fermat in a letter dated 1 January 1662 sent to the Cureau de la Chambre. It was met with objections made in May 1662 by Claude Clerselier, an optics expert and leading spokesperson for the Cartesians at the time. Among his objections, Clerselier stated:

(...) Fermat's principle can not be the cause, for otherwise we would be attributing knowledge to nature: and here, by nature, we understand only that order and lawfulness in the world, such as it is, which acts without foreknowledge, without choice, but by a necessary determination.

The principle of Fermat cannot be the cause, because otherwise we would be ascribed knowledge to nature: and here, by nature, we understand only that order and legality in the world, as it is, that acts without prior knowledge, without choice, but by a necessary determination.

Claude Clerselier (1622)

The original, in French, by Mahoney, is as follows:

Le principe que vous prenez pour fondement de votre démonstration, à savoir que la nature agit toujours par les voies les plus courtes et les plus simples, n’est qu’un principe moral et non point physique, qui n’est point et qui ne peut être la cause d’aucun effet de la nature.

The principle that you take to substantiate your demonstration, namely that nature always acts through the shortest and most simple pathways, is only a moral principle and not a physical one, which is neither nor can be the cause of any effect of nature.

Mahoney

In fact, Fermat's principle does not stand on its own, and it is now known that it can be derived from previous principles, such as Huygens' principle. Historically, Fermat's principle has served as a guiding principle in the formulation of the laws of Physics with the use of variational calculus (see the principle of least action).