1
PH 316 Critical Angle and Beyond
(similar to Griffiths sec 9.3.3, p. 386 ff. )
10/26/06
A plane wave reflected from a boundary between two dielectric media of indices n1 and and n2 can show
a ‘critical angle’ for reflection. This discussion will examine that phenomenon from the point of view of
electromagnetic radiation.
n1
Eref
r
Etr

n2
All B fields out of page
tr
y
x
Ein
The incident wave is given by Ein exp(i kinr - it) . The propagation vector kin has x and y
components, and can be written in terms of unit vectors i and j as
kin = i kin x + j kin y = i kin cos  + j kin sin .
kin is the magnitude of the propagation vector ( kin = 2/in ). The phase velocity of any wave, including
light is v = f. In vacuum we have c = f vac, and in index n we have vn = f n . The frequency of the
light in all media is the same. This is because the light makes atoms and electrons oscillate as it passes
through. The incident light ‘drives’ the medium at its frequency, and in a steady-state driven system, the
driven element responds at the driving frequency. The oscillating atoms and electrons then re-radiate at
the same frequency as the incoming light. Because n = c/vn, we can write
n = c/vn = (f vac)/(f n),
or n = vac/n .
This says the wavelength in a medium of index of refraction n is smaller than in vacuum by a factor of n.
n = vac/n
We will refer to kvac =2/vac as k. Then we may note that kn = 2/n = n kvac . The propagation constant
k is larger in a medium of index n by a factor of n.
Then
kin = n1 k, and
ktr = n2 k,
since the transmitted wave travels in the medium of index n2. (k is the propagation constant in vacuum.)
Now we write
kin = i kin x + j kin y = i n1 k cos  + j n1 k sin .
2
For the reflected k we have
kref = i kref x + j kref y = - i n1 k cos r + j n1 k sin r.
The transmitted k travels at angle ’, so
ktr = i ktr x + j ktr y = i n2 k cos ’ + j n2 k sin ’.
We will now rewrite the incident electric field
Ein exp(i kinr - it)
The phase of this wave is kinr - t (it’s the argument of the sine or the cosine wave). Since
r = i x + j y,
the phase of the incoming wave is
kinr - t = n1 k (x cos  + y sin ) -t .
The reflected electric field is
Eref exp(i krefr - it).
The phase of the reflected wave is
krefr - t = n1 k ( -x cos r + y sin r) -t .
The transmitted wave is Etr exp(i ktrr - it). The phase of this wave is
ktrr - t = n2 k ( x cos tr + y sin tr ) -t
Remember that the parallel component of E must be continuous at the boundary, at every instant of time
and at every point on the boundary. In order for this to happen, all waves must have the same
frequency. They must oscillate back and forth at exactly the same rate. If any of the three waves failed to
do this, we could not have the parallel components matching on both sides of the boundary. (Earlier we
talked about how this is true because the incident wave ‘drives’ the other two waves.)
But something else must happen. Suppose the boundary is in the plane x=0. The parallel components of
E must still match for every value of y on the boundary. If the equality broke down someplace, then this
boundary condition could not be satisfied. For the waves to match, there is a condition on their
amplitudes, but in addition, the phases of all three waves must be the same for all times t, and for all
values y on the boundary. So we must have
phase of incident wave = phase of reflected wave = phase of transmitted wave
3
When we set all three phases equal, and plug in x=0 for the boundary we have
n1 k (0 cos  + y sin ) -t = n1 k ( -0 cos r + y sin r) -t = n2 k ( 0 cos tr + y sin tr ) -t
This must be true for all values of y, so we conclude that
n1 sin  = n1 sin r = n2 sin tr
The lefthand equality says that the angle of incidence  must equal the angle of reflection r. The
righthand equality says that n1 sin  = n2 sin tr . This is Snell’s law, which follows out of the boundary
condition on the waves at the interface
When n1 > n2, the angle tr is greater than the incident angle . (From n1 sin  = n2 sin tr .) As we
increase , the transmitted angle tr can reach 90o. When tr = 90o,  is at the ‘critical angle’. After this,
the sine of tr is greater than 1, according to Snell’s law. The cosine of tr is then pure imaginary. We
will write
cos tr = i (sin tr/sin )2-1)
when   crit .
The transmitted wave is then written
Etr exp(i ktrr - it) = Etr exp(i k n2 [cos tr x + sin tr y ] - it).
Because of cos tr being imaginary, we find
Etr exp(i ktrr - it) = Etr exp(i k n2 sin tr y - it) exp(-(sin tr/sin )2-1) k x) .
The transmitted wave oscillates in the y-direction, parallel to the boundary, but it decreases
exponentially in the direction perpendicular to the boundary: it ‘dies’ going away from the boundary.