Download Reflection and critical angle

S. Blair
September 27, 2010
9
2.4. Wave propagation at an interface - reflection and refraction
Now, if we take the plane wave solution to the wave equation, we can determine what happens
when a wave strikes a dielectric interface. We know from our discussion on ray optics that
when a ray hits a mirror, the angle of reflection (with respect to the normal) equals the angle
of incidence. Also, we know that for a transmitted ray through a dielectric interface, the
angle of refraction and angle of incidence are related by Snell’s Law
ni sin θi = nt sin θt ,
where ni is the refractive index of the incidence medium, nt is the refractive index of the
transmission medium, and θ are the respective angle with respect to the normal at the
interface.
What ray optics doesn’t tell us is that at a dielectric interface, we will in general have
both transmitted and reflected components. In order to determine the amplitudes of the
refracted and reflected components, we have to use the wave optics description with plane
waves. The incident, reflected, and transmitted plane wave components can be written
Ei = Eio ej(ωt−ki ·r)
Er = Ero ej(ωt−kr ·r)
Et = Eto ej(ωt−kt ·r)
The field amplitudes Eio , etc., can represent either the TE (polarization perpendicular
to plane of incidence E⊥ ) or TM (electric field polarization in the plane of incidence E|| )
polarizations. Similar equations can be written for the magnetic fields, where B⊥ = (n/c)E||
and B|| = (n/c)E⊥ .
E t,//
y
θt
z
x into paper
ki
E i,//
θi θ r
E i,⊥
Incident
wave
E r,//
Transmitted wave
kt
E t,⊥
E r,⊥
kr
Reflected
wave
The next thing we have to specify are the boundary conditions, which state that the
tangential components of the electric and magnetic fields are continuous across an interface.
This leads to the following relationships for the TE wave:
Ei,⊥ + Er,⊥ = Et,⊥
Bi,|| cos θi + Br,|| cos θr = Bt,|| cos θt ,
which states that the x component of the electric field and z component of the corresponding
magnetic field are continuous. This leads to the following equations:
Eio,⊥ ej(ωt−kiz z) + Ero,⊥ ej(ωt−krz z) = Eto,⊥ ej(ωt−ktz z)
Bio,|| cos θi ej(ωt−kiz z) − Bro,|| cos θr ej(ωt−krz z) = Bto,|| cos θt ej(ωt−ktz z) .
S. Blair
September 27, 2010
10
Since these equations must be valid for all values of t and z, we must have that
(ωt − kiz z) = (ωt − krz z) = (ωt − ktz z) ,
Since the frequency ω is the same across the interface, the z wavevector components must
satisfy
kiz = krz = ktz ⇒ n1 sin θi = n1 sin θr = n2 sin θt ,
which give the laws of reflection and refraction. Canceling the exponential terms, the amplitudes are related by
Eio,⊥ + Ero,⊥ = Eto,⊥
Bio,|| cos θi − Bro,|| cos θr = Bto,|| cos θt .
Now, using the relationship between the electric and magnetic field amplitudes,
Eio,⊥ + Ero,⊥ = Eto,⊥
n1 Eio,⊥ cos θi − n1 Ero,⊥ cos θr = n2 Eto,⊥ cos θt ,
we can derive the reflection and transmission coefficients for the TE polarization
Er0,⊥
cos θi − n2 − sin2 θi
r⊥ =
=
Ei0,⊥
cos θi + n2 − sin2 θi
Et0,⊥
2 cos θi
t⊥ =
=
Ei0,⊥
cos θi + n2 − sin2 θi
where n = n2 /n1 . We can show that the following additional relationship holds:
r⊥ + 1 = t ⊥ .
Following the same steps for the TM polarization, we get
Er0,||
n2 − sin2 θi − n2 cos θi
r|| =
=
Ei0,||
n2 − sin2 θi + n2 cos θi
Et0,||
2n cos θi
t|| =
=
Ei0,||
n2 − sin2 θi + n2 cos θi
1 = r|| + nt|| .
S. Blair
September 27, 2010
11
2.4.1. Reflection amplitude and phase - n1 > n2
The solid line is TM polarization, and dotted line is TE polarization. Brewster’s angle
occurs when
n2
tan θp = .
n1
At this angle, r|| = 0, so that the TM polarization is completely transmitted. Brewester’s
angle is used in may different lasers to force the output to be in one state of polarization.
Total internal reflection occurs when
sin θc =
n2
,
n1
and only happens when n1 > n2 . Under TIR, both the TE and TM waves are completely
reflected back into the incidence medium. This effect is used extensively in optical waveguides.
2.4.2. Reflection amplitude and phase - n1 < n2
S. Blair
September 27, 2010
12
2.4.3. Reflectance and transmittance
The quantity reflectance is the ratio of the reflected intensity to the incident intensity:
R⊥ =
likewise
|Er0,⊥ |2
= |r⊥ |2 ,
2
|Ei0,⊥ |
R|| = |r|| |2 .
At normal incidence (i.e. θi = 0), these expression reduce to
R⊥ = R|| =
n1 − n2
n1 + n2
2
.
For air to glass with n1 = 1 and n2 ∼ 1.5, R ∼ 4%.
The transmittance is the ratio of transmitted intensity to incident intensity:
T⊥ =
n2
|t⊥ |2
n1
T|| =
n2 2
|t|| | ,
n1
for normal incidence. The reason for the refractive index ratio is that intensity is defined as
I = |E|2 /2η = v0 r |E|2 /2, and r (and therefore n and η) are different for the incident and
transmitted waves. More generally, in the absence of absorption, the transmittance can be
calculated from T = 1 − R.