Fermat’s principle of least time
Now in the further development of science, we want more than just a                         formula. First we have an observation, then we have numbers that we                         measure, then we have a law which summarizes all the numbers. But the                         real glory of science is that we can find a way of                             thinking such that the law is evident. The first way of thinking that made the law about the behavior of                         light evident was discovered by                         Fermat                         in about 1650, and it is called the principle of least                             time, or Fermat’s principle. His idea                         is this: that out of all possible paths that it might take to get from                         one point to another, light takes the path which requires the                         shortest time. Let us first show that this is true for the case of the mirror, that                         this simple principle contains both the law                         of straight-line propagation and the law for the mirror. So, we are                         growing in our understanding! Let us try to find the solution to the                         following problem. In Fig. 26–3 are shown two points,                         $A$ and $B$, and a plane mirror, $MM'$. What is the way to get from                         $A$ to $B$ in the shortest time?  The answer is to go straight from                         $A$ to $B$! But if we add the extra rule that the light has                         to strike the mirror and come back in the shortest time, the                         answer is not so easy. One way would be to go as quickly as possible                         to the mirror and then go to $B$, on the path $ADB$. Of course, we                         then have a long path $DB$. If we move over a little to the right,                         to $E$, we slightly increase the first distance, but we greatly                         decrease the second one, and so the total path length, and                         therefore the travel time, is less. How can we find the point $C$ for                         which the time is the shortest? We can find it very nicely by a                         geometrical trick. We construct on the other side of $MM'$ an artificial point $B'$,                         which is the same distance below the plane $MM'$ as the point $B$ is                         above the plane. Then we draw the line $EB'$. Now because $BFM$ is a                         right angle and $BF = FB'$, $EB$ is equal to $EB'$. Therefore the sum                         of the two distances, $AE + EB$, which is proportional to the time it                         will take if the light travels with constant velocity, is also the sum                         of the two lengths $AE + EB'$. Therefore the problem becomes, when is                         the sum of these two lengths the least? The answer is easy: when the                         line goes through point $C$ as a straight line from $A$                         to $B'$! In other words, we have to find the point where we go toward the                         artificial point, and that will be the correct one. Now if $ACB'$ is a                         straight line, then angle $BCF$ is equal to angle $B'CF$ and thence to                         angle $ACM$. Thus the statement that the angle of                         incidence equals                         the angle of reflection is equivalent to the                         statement that the light goes to the mirror in such a way that it comes                         back to the point $B$ in the least possible time. Originally, the                         statement was made by Hero of Alexandria that the light travels in such                         a way that it goes to the mirror and to the other point in the shortest                         possible distance, so it is not a modern theory. It was this that                         inspired Fermat to suggest to                         himself that perhaps refraction operated on a similar basis. But for                         refraction, light obviously does not use the path of shortest                         distance, so Fermat tried                         the idea that it takes the shortest time. Before we go on to analyze refraction, we should make one more remark                         about the mirror. If we have a source of light at the point $B$ and it                         sends light toward the mirror, then we see that the light which goes                         to $A$ from the point $B$ comes to $A$ in exactly the same manner as                         it would have come to $A$ if there were an object at $B'$, and                         no mirror. Now of course the eye detects only the light which                         enters it physically, so if we have an object at $B$ and a mirror                         which makes the light come into the eye in exactly the same manner as                         it would have come into the eye if the object were at $B'$, then the                         eye-brain system interprets that, assuming it does not know too much,                         as being an object at $B'$. So the illusion that there is an                         object behind the mirror is merely due to the fact that the light                         which is entering the eye is entering in exactly the same manner,                         physically, as it would have entered had there been an object                         back there (except for the dirt on the mirror, and our knowledge of                         the existence of the mirror, and so on, which is corrected in the                         brain). Now let us demonstrate that the principle of                         least time will give Snell’s law of                         refraction. We must, however, make an assumption about the speed of                         light in water. We shall assume that the speed of light in water is                         lower than the speed of light in air by a certain factor, $n$.                          In Fig. 26–4, our problem is again to go from $A$ to $B$                         in the shortest time. To illustrate that the best thing to do                         is not just to go in a straight line, let us imagine that a beautiful                         girl has fallen out of a boat, and she is screaming for help in the                         water at point $B$. The line marked $x$ is the shoreline. We are at                         point $A$ on land, and we see the accident, and we can run and can                         also swim. But we can run faster than we can swim. What do we do? Do                         we go in a straight line? (Yes, no doubt!) However, by using a little                         more intelligence we would realize that it would be advantageous to                         travel a little greater distance on land in order to decrease the                         distance in the water, because we go so much slower in the                         water. (Following this line of reasoning out, we would say the right                         thing to do is to compute very carefully what should be done!)                         At any rate, let us try to show that the final solution to the problem                         is the path $ACB$, and that this path takes the shortest time of all                         possible ones. If it is the shortest path, that means that if we take                         any other, it will be longer. So, if we were to plot the time it takes                         against the position of point $X$, we would get a curve something like                         that shown in Fig. 26–5, where point $C$ corresponds to                         the shortest of all possible times. This means that if we move the                         point $X$ to points near $C$, in the first approximation there                         is essentially no change in time because the slope is zero at                         the bottom of the curve. So our way of finding the law will be to                         consider that we move the place by a very small amount, and to demand                         that there be essentially no change in time. (Of course there is an                         infinitesimal change of a second order; we ought to have a                         positive increase for displacements in either direction from $C$.) So                         we consider a nearby point $X$ and we calculate how long it would take                         to go from $A$ to $B$ by the two paths, and compare the new path with                         the old path. It is very easy to do. We want the difference, of                         course, to be nearly zero if the distance $XC$ is short. First, look                         at the path on land. If we draw a perpendicular $XE$, we see that this                         path is shortened by the amount $EC$. Let us say we gain by not having                         to go that extra distance. On the other hand, in the water, by drawing                         a corresponding perpendicular, $CF$, we find that we have to go the                         extra distance $XF$, and that is what we lose. Or, in time, we                         gain the time it would have taken to go the distance $EC$, but we lose                         the time it would have taken to go the distance $XF$. Those times must                         be equal since, in the first approximation, there is to be no change                         in time. But supposing that in the water the speed is $1/n$ times as                         fast as in air, then we must have                         \begin{equation}                         \label{Eq:I:26:3}                         EC=n\cdot XF.                         \end{equation}                         Therefore we see that when we have the right point, $X\!C\sin E\!X\!C =                         n\cdot X\!C\sin X\!C\!F$ or, cancelling the common hypotenuse length $XC$                         and noting that                                                  \begin{equation*}                         EXC=ECN=\theta_i                         \,\text{ and }\,                         XCF\approx BCN'=\theta_r\;                         (\text{when $X$ is near $C$}),                         \end{equation*}                         
                         \begin{gather*}                         EXC=ECN=\theta_i                         \,\text{ and }\,                         XCF\approx BCN'=\theta_r\\                         (\text{when $X$ is near $C$}),                         \end{gather*}                                                 we have                         \begin{equation}                         \label{Eq:I:26:4}                         \sin\theta_i=n\sin\theta_r.                         \end{equation}                         So we see that to get from one point to another in the least time when                         the ratio of speeds is $n$, the light should enter at such an angle                         that the ratio of the sines of the angles $\theta_i$ and $\theta_r$ is                         the ratio of the speeds in the two media.