378 Week 11: Light
Fermat, observing that light must speed up or slow down as it passesbetween distinct
physical media, hypothesized that the trajectory followed by lightbetween pointAin
medium 1 and pointBin medium 2 wouldnotbe a straight line; it would instead be the
path that takes the minimum time. This, as we shall see, isanotherway to get Snell’s
law, but this time in a ray description of the light that is altogether independent of the
wavelength or wave properties of the light.
Although Fermat was not the first person to propose a variational/minimum principle for
optics (that honor belongs to Ibn al-Haytham in 1021, over 600 years earlier) he was the
first to do so post Descartes, with an analytic geometry capable offully exploiting the
idea. Although Fermat’s principle puts the cart a bit in front of the horse by making it
thecauseof the trajectory followed by light instead of afeatureof the trajectory followed
by light (that can be derived from other principles) variational principles based on his
original statement proved to be essential to a formulation of classical mechanics that
would translate, with minimal changes, into a formulation of quantummechanics. It is
therefore worth looking at in a bit of detail, especially for physics majors or minors.H
D
x D−x
y 1
H 1
(^2) y
2
θi
θl
A
B
Figure 146: The path withθi=θlis the one with the minimal time when the entire trajectory is
otherwise in a single medium with a constant speed.
In figure 146 illustrate and prepare to prove the law of reflection from Fermat’s require-
ment that the time required to go between points A and B on a path that reflects off of
the mirror is a minimum. From the result above we can ignore all trajectories that are
not straight except where they strike the reflecting surface. The total distance between
the two pointsAandBis therefore the sum of the two hypotenuses:H = H 1 +H 2
={
y^21 +x^2}^12
+
{
y^22 + (D−x)^2}^12
(925)
We need to find a condition that produces the minimum of this function. We therefore
differentiate with respect tox, set the result to zero, and solve for (say)xorθ 1 .y 1 ,y 2
andDare all constant, so (using the chain rule, note well):
dH
dx=
1
22 x
{y^21 +x^2 }12 −
1
2 2(D−x)
{y^22 + (D−x)^2 }12 = 0 (926)
or
sin(θi) =
√ x
y 12 +x^2=
x
H 1=
D−x
H 2=
√ (D−x)
y^22 + (D−x)^2= sin(θl) (927)