4.10 Stationary points 117
Division by 2βand multiplication by(1 1 − 1 c
2)
122gives
Thenε 1 = 1 α 1 ± 1 β.
EXAMPLE 4.25Snell’s law of refraction in geometric optics.
4A ray of light travels between points P and Q across a phase boundary at O. In the
upper region, the speed of light isv
11 = 1 c 2 η
1, where cis the speed of light in vacuum
and η
1is the refractive index of the phase. In the lower region the speed of light is
v
21 = 1 c 2 η
2. Snell’s law of refraction is
The law can be derived from a ‘principle of least time’, that the path followed is that of
least time.
5The total time travelled from P to Q through point O is (distance 2 speed in
each phase),
The problem is to find point O such that tis a minimum. Choosing x
1as the
independent variable, we have
r xy r xy Xx y
121211222222121222=+ , = + =− +
() ( )()
12t
rr
=+
1122vv
sin
sin
θ
θ
η
η
121221==
v
v
()10
1
2
22−−=,cc or c=±
4Willebrord van Roijen Snell (1591–1626), Dutch mathematician and physicist, formulated the law of
refraction in 1621.
5This use of the principle of least time, proposed by Fermat, was one of the examples used by Leibniz in his
1684 paper to demonstrate his method of finding maxima and minima.
......................................................................................................................................................................................................................................................................................................................................................
...
...
..
....
..
...
...
...
..
...
...
...
...
..
...
...
...
...
...
..
...
...
...
..
...
...
...
...
...
..
...
...
...
...
..
...
...
...
..
....
..
...
...
...
..
...
...
...
...
..
...
...
...
...
...
..
...
...
...
..
...
...
...
...
...
..
...
...
...
...
..
......
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
.....
....
................
.......
................................................................................................................
...........
....
....
...
...
...
...
.......................................................................................................
...
.....
...
...
..
...
...
..
...
...
..
...
...
..
...
...
....................................................................................................................................................................o
θ
1θ
2y
1
r1x
1x
2y
2r
2Q
P
phaseboundary
Figure 4.11