Consider an electromagnetic wave impinging upon an interface:
u
̂
n
×
E
⃗
0
i
e
i
(
k
⃗
i
⋅
r
⃗
+
ω
i
t
)
+
u
̂
n
×
E
⃗
0
r
e
i
(
k
⃗
r
⋅
r
⃗
+
ω
r
t
+
δ
r
)
=
u
̂
n
×
E
⃗
0
t
e
i
(
k
⃗
t
⋅
r
⃗
+
ω
t
t
+
δ
t
)
u
̂
n
×
E
⃗
0
i
e
i
(
k
⃗
i
⋅
r
⃗
+
ω
i
t
)
+
u
̂
n
×
E
⃗
0
r
e
i
(
k
⃗
r
⋅
r
⃗
+
ω
r
t
+
δ
r
)
=
u
̂
n
×
E
⃗
0
t
e
i
(
k
⃗
t
⋅
r
⃗
+
ω
t
t
+
δ
t
)
Where
(
k
i
⃗
,
ω
i
)
(
k
i
⃗
,
ω
i
)
describes the incoming wave,
(
k
r
⃗
,
ω
r
,
δ
r
)
(
k
r
⃗
,
ω
r
,
δ
r
)
the reflected wave, and
(
k
t
⃗
,
ω
t
,
δ
t
)
(
k
t
⃗
,
ω
t
,
δ
t
)
the transmitted wave. At the interface (ie. at points where the vector
r
⃗
r
⃗
points to the plane of the interface), all the waves must be in phase with
each other. This means that the frequencies must all be equal and there can be
no arbitrary phase between the waves. The net result of this is that we must
have (for an interface passing through the origin):
k
⃗
i
⋅
r
⃗
=
k
⃗
r
⋅
r
⃗
=
k
⃗
t
⋅
r
⃗
k
⃗
i
⋅
r
⃗
=
k
⃗
r
⋅
r
⃗
=
k
⃗
t
⋅
r
⃗
from which we get
k
i
sin
θ
i
=
k
r
sin
θ
r
.
k
i
sin
θ
i
=
k
r
sin
θ
r
.
It is important to note now that we are doing this at the interface. We have
chosen a coordinate system so that the interface is at
y
=
0
y
=
0
and contains the origin. This implies that the vector
r
⃗
r
⃗
is lying in the plane of the interface at the point where we say that the
above is true.
Finally, since the incident and reflected waves are in the same medium we must
have
k
i
=
k
r
k
i
=
k
r
and thus
θ
i
=
θ
r
θ
i
=
θ
r
Also,
we get that
k
⃗
i
,
k
⃗
r
,
u
̂
k
⃗
i
,
k
⃗
r
,
u
̂
all line in a plane (because
(
k
⃗
i
−
k
⃗
r
)
⋅
r
⃗
=
0
(
k
⃗
i
−
k
⃗
r
)
⋅
r
⃗
=
0
defines a plane). We also have
u
̂
×
(
k
⃗
i
−
k
⃗
t
)
=
0
u
̂
×
(
k
⃗
i
−
k
⃗
t
)
=
0
and
following the same arguments find that
k
⃗
i
,
k
⃗
r
,
k
⃗
t
,
u
̂
k
⃗
i
,
k
⃗
r
,
k
⃗
t
,
u
̂
all line in a plane and that
k
i
sin
θ
i
=
k
t
sin
θ
t
.
k
i
sin
θ
i
=
k
t
sin
θ
t
.
Now
we know that
ω
i
=
ω
t
ω
i
=
ω
t
so we can multiply both sides by
c
/
ω
i
c
/
ω
i
and get
n
i
sin
θ
i
=
n
t
sin
θ
t
n
i
sin
θ
i
=
n
t
sin
θ
t
At the interface, which we will set to
y
=
0
y
=
0
for convenience (you can always switch back to any coordinate system
afterwards. It is good practice to choose the coordinate system that makes
your problem easy)
[
k
⃗
i
⋅
r
⃗
+
ω
i
t
]

y
=
0
=
[
k
⃗
r
⋅
r
⃗
+
ω
r
t
+
δ
r
]

y
=
0
=
[
k
⃗
t
⋅
r
⃗
+
ω
t
t
+
δ
t
]

y
=
0
[
k
⃗
i
⋅
r
⃗
+
ω
i
t
]

y
=
0
=
[
k
⃗
r
⋅
r
⃗
+
ω
r
t
+
δ
r
]

y
=
0
=
[
k
⃗
t
⋅
r
⃗
+
ω
t
t
+
δ
t
]

y
=
0
now this must be true for all
r
⃗
r
⃗
on the surface and for all
t
t
so we must have
ω
i
=
ω
r
=
ω
t
ω
i
=
ω
r
=
ω
t
So now we have
[
k
⃗
i
⋅
r
⃗
+
ω
i
t
]

y
=
0
=
[
k
⃗
r
⋅
r
⃗
+
ω
i
t
+
δ
r
]

y
=
0
=
[
k
⃗
t
⋅
r
⃗
+
ω
i
t
+
δ
t
]

y
=
0
[
k
⃗
i
⋅
r
⃗
+
ω
i
t
]

y
=
0
=
[
k
⃗
r
⋅
r
⃗
+
ω
i
t
+
δ
r
]

y
=
0
=
[
k
⃗
t
⋅
r
⃗
+
ω
i
t
+
δ
t
]

y
=
0
which can be written
[
k
⃗
i
⋅
r
⃗
]

y
=
0
=
[
k
⃗
r
⋅
r
⃗
+
δ
r
]

y
=
0
=
[
k
⃗
t
⋅
r
⃗
+
δ
t
]

y
=
0
[
k
⃗
i
⋅
r
⃗
]

y
=
0
=
[
k
⃗
r
⋅
r
⃗
+
δ
r
]

y
=
0
=
[
k
⃗
t
⋅
r
⃗
+
δ
t
]

y
=
0
So now we can write
[
(
k
⃗
i
−
k
⃗
r
)
⋅
r
⃗
]

y
=
0
=
δ
r
[
(
k
⃗
i
−
k
⃗
r
)
⋅
r
⃗
]

y
=
0
=
δ
r
Since the interface passes through the origin, one of the allowed values of
r
⃗
r
⃗
is 0. So this is only true if
δ
r
=
0
δ
r
=
0
.
(If the interface does not include the origin then you can not make this
simplification, but clearly we can always choose a coordinate system such that
this is true and thereby simplify our lives) likewise we could have
written
[
(
k
⃗
i
−
k
⃗
r
)
⋅
t
⃗
]

y
=
0
=
δ
t
[
(
k
⃗
i
−
k
⃗
r
)
⋅
t
⃗
]

y
=
0
=
δ
t
and applied the same argument to get
δ
t
=
0
δ
t
=
0
.
Lets just use this henceforth and thus write:
[
k
⃗
i
⋅
r
⃗
]

y
=
0
=
[
k
⃗
r
⋅
r
⃗
]

y
=
0
=
[
k
⃗
t
⋅
r
⃗
]

y
=
0
[
k
⃗
i
⋅
r
⃗
]

y
=
0
=
[
k
⃗
r
⋅
r
⃗
]

y
=
0
=
[
k
⃗
t
⋅
r
⃗
]

y
=
0
"This book covers second year Physics at Rice University."