We saw that when the index of refraction of the incident material is greater
than the transmitting material we can get total internal reflection at the
critical angle. An interesting question is "what happens at larger angles of
incidence?" This actually is somewhat subtle. From simple trigonometry we know
that
cos
θ
t
=
1
−
sin
2
θ
t
.
cos
θ
t
=
1
−
sin
2
θ
t
.
We also know from Snell's law that
sin
θ
t
=
n
i
n
t
sin
θ
i
sin
θ
t
=
n
i
n
t
sin
θ
i
so we have
cos
θ
t
=
1
−
n
i
2
n
t
2
sin
2
θ
i
.
cos
θ
t
=
1
−
n
i
2
n
t
2
sin
2
θ
i
.
So we see that if
n
i
>
n
t
n
i
>
n
t
cos
θ
t
cos
θ
t
can become an imaginary number! For convenience we will write this as
cos
θ
t
=
i
n
i
2
n
t
2
sin
2
θ
i
−
1
cos
θ
t
=
i
n
i
2
n
t
2
sin
2
θ
i
−
1
Now lets write down the expression for the transmitted wave:
E
t
=
E
0
t
e
i
(
K
t
⃗
⋅
r
⃗
−
ω
t
)
E
t
=
E
0
t
e
i
(
K
t
⃗
⋅
r
⃗
−
ω
t
)
For simplicity we will assume that the interface lies in the
y
=
0
y
=
0
plane and thus the
y
y
direction is normal to the interface. Also, we assume the
z
=
0
z
=
0
plane defines then plane of incidence. Then we can write
K
t
⃗
=
(
K
t
x
,
K
t
y
,
0
)
K
t
⃗
=
(
K
t
x
,
K
t
y
,
0
)
or
K
t
⃗
=
(
K
t
sin
θ
t
,
K
t
cos
θ
t
,
0
)
.
K
t
⃗
=
(
K
t
sin
θ
t
,
K
t
cos
θ
t
,
0
)
.
Also
r
⃗
=
(
x
,
y
,
0
)
.
r
⃗
=
(
x
,
y
,
0
)
.
So now we can write that the wave as
E
t
=
E
0
t
e
i
(
K
t
⃗
⋅
r
⃗
−
ω
t
)
E
t
=
E
0
t
e
i
(
K
t
⃗
⋅
r
⃗
−
ω
t
)
E
t
=
E
0
t
e
i
K
t
sin
θ
t
x
e
i
K
t
cos
θ
t
y
e
−
i
ω
t
E
t
=
E
0
t
e
i
K
t
sin
θ
t
x
e
i
K
t
cos
θ
t
y
e
−
i
ω
t
or
E
t
=
E
0
t
e
i
K
t
sin
θ
t
x
e
−
n
i
2
n
t
2
sin
2
θ
i
−
1
y
e
−
i
ω
t
.
E
t
=
E
0
t
e
i
K
t
sin
θ
t
x
e
−
n
i
2
n
t
2
sin
2
θ
i
−
1
y
e
−
i
ω
t
.
It is interesting to note the effect of the term
e
−
n
i
2
n
t
2
sin
2
θ
i
−
1
y
e
−
n
i
2
n
t
2
sin
2
θ
i
−
1
y
in that expression. This is an exponential decay. The amplitude of the wave
drops rapidly to zero.
So there is a transmitted wave but its amplitude drops precipitously. This is
referred to as the evanescent wave.
