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
A C B, 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
X C is short. First, look
at the path on land. If we draw a perpendicular
X E, we see that this
path is shortened by the amount
E C. 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,
C F, we find that we have to go the
extra distance
X F, and that is what we lose. Or, in
time, we
gain the time it would have taken to go the distance
E C, but we lose
the time it would have taken to go the distance
X F. 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
we have
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 θi and θr is
the ratio of the speeds in the two media.