W9_parallel_resonance.eps

Week 11: Light 377

θ 2 and thesame hypotenuse(the distance between wavefronts on the interface between

media), we note that:

D= λ^1

sin(θ 1 )

= λ^2

sin(θ 2 )

(922)

or (substituting from above and cancellingc/f):

1

n 1 sin(θ 1 )=

1

n 2 sin(θ 2 ) (923)

Inverting, we obtainSnell’s Law:

n 1 sin(θ 1 ) =n 2 sin(θ 2 ) (924)

Since the geometry is exactly the same going fromn 2 ton 1 , we conclude that it doesn’t

matter which medium has the greater or the lesser index of refraction.

Fermat’s Principle

S

S 1

0

Figure 145: For constant speed, the straight line path betweenAandBtakes the least time.

In figure 145, we note that any curved path such asS 1 is longer than the pathS 0 (some-

thing that can be proven using the calculus of variations, which we willnot introduce

here). The time required to traverseS 1 ist 1 =S 1 /vwhilet 0 =S 0 /v. The minimal time

path is therefore clearly the minimal distance path, the straight line. Fermat’s principle

thus correctly describes this case.

Fermat noted that a straight line is the path along which it takes theleast timeto travel

between two pointsAandBat constant speed in ordinary space. Any other path is

longer in distance than the straight line path, and hence takes longer to traverse at the

same speed. This is illustrated in figure 145 – the curved path is longer, so it takes more

time to traverse it if you have to move at exactly the speed of light (or the same speed

along both trajectories).

Thus when we say that light travels a constant speed (the speed oflight) in a straight

line betweenAandB, it isalsotrue that the path that it follows is the one that takes

the least time.

Now consider the Law of Reflection above. It is equally easy to see that any reflective

path betweenAandBthat doesn’t haveθi=θlis longer, and hence takes more time.

We will examine and prove this below using calculus.

What happens when the speed isnotconstant? In that case, one has to solve anopti-

mizationproblem, a problem ineconomy. It seems that one might be able to obtain some

benefit fromgoing furtherwhere the speed is greater and thereby reduce the amount of

distance one has to travel at the slower speed, and actually go betweenAandBinless

time than the straight line trajectory.