r
⊥
≡
(
E
0
r
E
0
i
)
⊥
=
n
i
cos
θ
i
−
n
t
cos
θ
t
n
i
cos
θ
i
+
n
t
cos
θ
t
r
⊥
≡
(
E
0
r
E
0
i
)
⊥
=
n
i
cos
θ
i
−
n
t
cos
θ
t
n
i
cos
θ
i
+
n
t
cos
θ
t
t
⊥
≡
(
E
0
t
E
0
i
)
⊥
=
2
n
i
cos
θ
i
n
i
cos
θ
i
+
n
t
cos
θ
t
t
⊥
≡
(
E
0
t
E
0
i
)
⊥
=
2
n
i
cos
θ
i
n
i
cos
θ
i
+
n
t
cos
θ
t
r
∥
≡
(
E
0
r
E
0
i
)
∥
=
n
t
cos
θ
i
−
n
i
cos
θ
t
n
t
cos
θ
i
+
n
i
cos
θ
t
r
∥
≡
(
E
0
r
E
0
i
)
∥
=
n
t
cos
θ
i
−
n
i
cos
θ
t
n
t
cos
θ
i
+
n
i
cos
θ
t
t
∥
≡
(
E
0
t
E
0
i
)
∥
=
2
n
i
cos
θ
i
n
t
cos
θ
i
+
n
i
cos
θ
t
t
∥
≡
(
E
0
t
E
0
i
)
∥
=
2
n
i
cos
θ
i
n
t
cos
θ
i
+
n
i
cos
θ
t
We can rewrite these equations using Snell's Law to eliminate the
cos
θ
t
cos
θ
t
term. 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
.
We can substitute this into
r
⊥
=
n
i
cos
θ
i
−
n
t
1
−
n
i
2
n
t
2
sin
2
θ
i
n
i
cos
θ
i
+
n
t
1
−
n
i
2
n
t
2
sin
2
θ
i
=
n
i
cos
θ
i
−
n
t
2
−
n
i
2
sin
2
θ
i
n
i
cos
θ
i
+
n
t
2
−
n
i
2
sin
2
θ
i
=
cos
θ
i
−
n
t
2
n
i
2
−
sin
2
θ
i
cos
θ
i
+
n
t
2
n
i
2
−
sin
2
θ
i
r
⊥
=
n
i
cos
θ
i
−
n
t
1
−
n
i
2
n
t
2
sin
2
θ
i
n
i
cos
θ
i
+
n
t
1
−
n
i
2
n
t
2
sin
2
θ
i
=
n
i
cos
θ
i
−
n
t
2
−
n
i
2
sin
2
θ
i
n
i
cos
θ
i
+
n
t
2
−
n
i
2
sin
2
θ
i
=
cos
θ
i
−
n
t
2
n
i
2
−
sin
2
θ
i
cos
θ
i
+
n
t
2
n
i
2
−
sin
2
θ
i
Similarly we can derive that
r
∥
=
n
t
2
n
i
2
cos
θ
i
−
n
t
2
n
i
2
−
sin
2
θ
i
n
t
2
n
i
2
cos
θ
i
+
n
t
2
n
i
2
−
sin
2
θ
i
r
∥
=
n
t
2
n
i
2
cos
θ
i
−
n
t
2
n
i
2
−
sin
2
θ
i
n
t
2
n
i
2
cos
θ
i
+
n
t
2
n
i
2
−
sin
2
θ
i
t
⊥
=
2
cos
θ
i
cos
θ
i
+
n
t
2
n
i
2
−
sin
2
θ
i
t
⊥
=
2
cos
θ
i
cos
θ
i
+
n
t
2
n
i
2
−
sin
2
θ
i
t
∥
=
2
n
t
n
i
cos
θ
i
n
t
2
n
i
2
cos
θ
i
+
n
t
2
n
i
2
−
sin
2
θ
i
t
∥
=
2
n
t
n
i
cos
θ
i
n
t
2
n
i
2
cos
θ
i
+
n
t
2
n
i
2
−
sin
2
θ
i
This form allows us to easily plot the coefficients for different values of
θ
i
.
θ
i
.
For example,here are the coefficients for the case where
n
t
n
i
=
1.5
.
n
t
n
i
=
1.5
.
It
is interesting to note that the sign of the coefficient can change on
reflection.
E
r
=
−

r

E
=
e
i
π

r

E
0
e
i
(
K
⃗
⋅
r
⃗
−
ω
t
)
=

r

E
0
e
i
(
K
⃗
⋅
r
⃗
−
ω
t
+
π
)
E
r
=
−

r

E
=
e
i
π

r

E
0
e
i
(
K
⃗
⋅
r
⃗
−
ω
t
)
=

r

E
0
e
i
(
K
⃗
⋅
r
⃗
−
ω
t
+
π
)
This corresponds to a phase change by
π
π
upon reflection.
We can also look at what happens to the reflection coefficients when
n
i
n
t
=
1.5
.
n
i
n
t
=
1.5
.
That
is going from a high index of refraction material to a lesser. In this case we
see at the critical angle we get total internal reflection. What happens to
the phase here is complicated.
When we have
n
i
n
t
>
1
n
i
n
t
>
1
(or
n
t
n
i
<
1
)
n
t
n
i
<
1
)
it is convenient to write
r
⊥
=
cos
θ
i
−
i
sin
2
θ
i
−
n
t
2
n
i
2
cos
θ
i
+
i
sin
2
θ
i
−
n
t
2
n
i
2
r
⊥
=
cos
θ
i
−
i
sin
2
θ
i
−
n
t
2
n
i
2
cos
θ
i
+
i
sin
2
θ
i
−
n
t
2
n
i
2
Now to understand what this implies we need to digress a little. Recall that
e
i
α
=
cos
α
+
i
sin
α
.
e
i
α
=
cos
α
+
i
sin
α
.
We could have written this as
e
i
α
=
a
+
i
b
e
i
α
=
a
+
i
b
then we see that
α
=
tan
−
1
b
a
.
α
=
tan
−
1
b
a
.
Now consider
cos
α
−
i
sin
α
cos
α
+
i
sin
α
=
e
−
i
α
e
i
α
=
e
−
2
i
α
cos
α
−
i
sin
α
cos
α
+
i
sin
α
=
e
−
i
α
e
i
α
=
e
−
2
i
α
Now looking back at
r
⊥
r
⊥
we see that
r
⊥
=
e
−
i
φ
r
⊥
=
e
−
i
φ
where
tan
φ
2
=
sin
2
θ
i
−
n
t
2
n
i
2
cos
θ
i
.
tan
φ
2
=
sin
2
θ
i
−
n
t
2
n
i
2
cos
θ
i
.
We could go through a similar excersize for
r
∥
r
∥
and
get the same result with
tan
φ
2
=
sin
2
θ
i
−
n
t
2
n
i
2
n
t
2
n
i
2
cos
θ
i
.
tan
φ
2
=
sin
2
θ
i
−
n
t
2
n
i
2
n
t
2
n
i
2
cos
θ
i
.
Figure 208 in Pedrotti and Pedrotti summarizes all the possible phase
changes.
"This book covers second year Physics at Rice University."